Tristan Needham: Visual Complex Analysis Clarendon · PDF fileTristan Needham: Visual Complex Analysis Clarendon Oxford University Press z.B. umv.science.upjs.sk/hutnik/NeedhamVCA.pdf

Tristan Needham: Visual Complex AnalysisClarendon Oxford University Press

z.B. umv.science.upjs.sk/hutnik/NeedhamVCA.pdf (9.4 MB)

Prof. Dr. Thomas RisseFaculty Electrical&Electronic Engineering and Computer Science

Hochschule Bremen, City University of Applied Sciences

February 7, 2018

umv.science.upjs.sk/hutnik/NeedhamVCA.pdf

2

Contents

1 Geometry and Complex Arithmetic 71.3 Some Applications . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3.3 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Transformations and Euclidean Geometry* . . . . . . . . . . . 7

1.4.4 Similarities and Complex Arithmetic . . . . . . . . . . 71.5 Exercises pp 45 . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Complex Functions as Transformations 312.2 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.2.3 Cassinian Curves* . . . . . . . . . . . . . . . . . . . . 312.3 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.3.2 The Disc of Convergence . . . . . . . . . . . . . . . . . 312.3.5 Manipulating Power Series . . . . . . . . . . . . . . . . 322.3.6 Finding the Radius of Convergence . . . . . . . . . . . 322.3.7 Fourier Series* . . . . . . . . . . . . . . . . . . . . . . 32

2.5 Cosine and Sine . . . . . . . . . . . . . . . . . . . . . . . . . . 322.5.1 Definitions and Identities . . . . . . . . . . . . . . . . . 322.5.3 The Geometry of the Mapping . . . . . . . . . . . . . . 33

2.7 The Logarithm Function . . . . . . . . . . . . . . . . . . . . . 332.7.1 Inverse of the Exponential Function . . . . . . . . . . . 332.7.3 General Powers . . . . . . . . . . . . . . . . . . . . . . 34

2.8 Averaging over Circles* . . . . . . . . . . . . . . . . . . . . . . 342.8.3 Averaging over Circles . . . . . . . . . . . . . . . . . . 34

2.9 Exercises pp 111 . . . . . . . . . . . . . . . . . . . . . . . . . 34

3

4 CONTENTS

3 Möbius Transformations and Inversion 59

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.1.3 Decomposition into Simple Transformations . . . . . . 59

3.2 Inversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.2.2 Preservation of Circles . . . . . . . . . . . . . . . . . . 59

3.2.6 Inversion in a Sphere . . . . . . . . . . . . . . . . . . . 60

3.4 The Riemann Sphere . . . . . . . . . . . . . . . . . . . . . . . 61

3.4.3 Transferring Complex Functions to the Sphere . . . . . 61

3.4.5 Stereographic Formulae* . . . . . . . . . . . . . . . . . 61

3.5 Mobius Transformations: Basic Results . . . . . . . . . . . . . 62

3.5.3 The Group Property . . . . . . . . . . . . . . . . . . . 62

3.6 Mobius Transformations as Matrices* . . . . . . . . . . . . . . 62

3.6.3 Eigenvectors and Eigenvalues* . . . . . . . . . . . . . . 62

3.6.4 Rotations of the Sphere as Mobius Transformations* . 63

3.7 Visualization and Classification* . . . . . . . . . . . . . . . . . 63

3.7.4 Parabolic Transformations . . . . . . . . . . . . . . . . 63

3.7.5 Computing the Multiplier* . . . . . . . . . . . . . . . . 64

3.9 Automorphisms of the Unit Disc* . . . . . . . . . . . . . . . . 65

3.9.3 Interpreting the Simplest Formula Geometrically* . . . 65

3.10 Exercises pp 181 . . . . . . . . . . . . . . . . . . . . . . . . . 65

4 Differentiation: The Amplitwist Concept 83

4.10 Exercises pp 211 . . . . . . . . . . . . . . . . . . . . . . . . . 83

CONTENTS 5

5 Further Geometry of Differentiation 935.6 Visual Differentiation of the Power Function . . . . . . . . . . 935.9 An Application of Higher Derivatives:

Curvature* . . . . . . . . . . . . . . . . . . . . . . . . . . . . 935.9.3 Complex Curvature . . . . . . . . . . . . . . . . . . . . 93

5.10 Celestial Mechanics* . . . . . . . . . . . . . . . . . . . . . . . 945.10.1 Central Force Fields . . . . . . . . . . . . . . . . . . . 945.10.2 Two Kinds of Elliptical Orbit . . . . . . . . . . . . . . 945.10.5 An Explanation . . . . . . . . . . . . . . . . . . . . . . 94

5.11 Analytic Continuation* . . . . . . . . . . . . . . . . . . . . . . 955.11.5 Analytic Continuation via Reflections . . . . . . . . . . 95

5.12 Exercises pp 258 . . . . . . . . . . . . . . . . . . . . . . . . . 95

6 Non-Euclidean Geometry* 1116.9 Exercises pp 328 . . . . . . . . . . . . . . . . . . . . . . . . . 111

7 Winding Numbers and Topology 1217.9 Exercises pp 369 . . . . . . . . . . . . . . . . . . . . . . . . . 121

8 Complex Integration: Cauchy’s Theorem 1338.13 Exercises pp 420 . . . . . . . . . . . . . . . . . . . . . . . . . 133

9 Cauchy’s Formula and Its Applications 1459.5 Exercises pp 446 . . . . . . . . . . . . . . . . . . . . . . . . . 145

6 CONTENTS

Chapter 1

Geometry and ComplexArithmetic

1.3 Some Applications

1.3.3 Geometry

On p.18 we have RθO(z) = eiθz + k with k = a(1 − eiθ) implying Rθ

a =Tk Rθ

O or Tv RαO = Rα

c with c = v/(1 − eiα). Then RθO Tv = Rθ

p withp = eiθv/(1− eiθ).The effect of two rotationsRθ

a andRφb is given byRφ

b Rθa(z) = Rφ

b (eiθz+a(1−

eiθ)) = eiφ(eiθz+a(1−eiθ))+b(1−eiφ) = ei(φ+θ)z+aeiφ(1−eiθ)+b(1−eiφ) =ei(φ+θ)z+ v with v = aeiφ(1− eiθ) + b(1− eiφ), hence for φ+ θ 6∈ 2πZ we haveRφb Rθ

a = Rφ+θc with c = v

1−ei(φ+θ) = aeiφ(1−eiθ)+b(1−eiφ)1−ei(φ+θ) .

On p.24 let Pn ∈ R[z], i.e. Pn(z) = ∑nν=0 dνz

ν with all dν ∈ R and Pn(c) = 0for some c ∈ C. Then 0 = Pn(z) = ∑n

ν=0 dνzν = ∑n

ν=0 dνzν = ∑n

ν=0 dν zν =

Pn(z) so that complex zeroes come in conjugate pairs.

1.4 Transformations and Euclidean Geome-try*

1.4.4 Similarities and Complex Arithmetic

on p.40 a given similarity S expands every distance by the same (non-zero)factor r, namely fixing cd in |ab||cd| = |SaSb|

|ScSd| we get |SaSb| = |ScSd||cd| |ab| =: r|ab|.

7

8 CHAPTER 1. GEOMETRY AND COMPLEX ARITHMETIC

1.5 Exercises pp 45

1. Given the general cubic Y = X3 + AX2 +BX + C.

(i) Show that the point of inflection occurs at −A3 .

Y ′(X) = 3X2 + 2AX + B and Y ′′(X) = 6X + 2A = 0 ⇐⇒X = −A

3 .(ii) Deduce geometrically that the substitution X = x− A

3 will reducethe cubic to the form Y = x3 + bx+ c.

The substitution moves the point of inflection to the Y -axis.Such a cubic cannot have a quadratic term because the cubicand linear terms all have points of inflection on the Y -axis, henceY = x3 + bx+ c.

(iii) Verify this by calculation.Y = (x − A

3 )3 + A(x − A3 )2 + B(x − A

3 ) + C = (x3 − Ax2 +A2

3 x−A3

27 )+A(x2−2A3 x+ A2

9 )+Bx−BA3 +C = x3 +(B− A2

3 )x+(2A3

27 −BA3 + C) = x3 + bx+ c.

2. In order to solve the cubic equation x3 = 3px+ 2q, do the following:

(i) Make the inspired substitution x = s+t, and deduce that x solvesthe cubic if st = p and s3 + t3 = 2q.

x3 = (s+ t)3 = s3 + 3s2t+ 3st2 + t3 = 3st(s+ t) + (s3 + t3) =3p(s+ t) + 2q = 3px+ 2q if st = p and s3 + t3 = 2q.

(ii) Eliminate t between these two equations, thereby obtaining aquadratic equation in s3.

t = psimplies s3 + p3

s3= 2q and thus (s3)2 − 2qs3 + p3 = 0.

(iii) Solve this quadratic to obtain the two possible values of s3. Bysymmetry, what are the possible values of t3 ?

Let v = s3, then v2 − 2qv + p3 = 0 with solutions v1,2 =q ±√q2 − p3 = s3 and t3 = 2q − s3 = q ∓

√q2 − p3.

(iv) Given that we know that s3 + t3 = 2q, deduce the formula x =3√q +√q2 − p3 + 3

√q −√q2 − p3 (4).

x = s+ t = 3√q +√q2 − p3 + 3

√q −√q2 − p3.

3. In 1591 François Viète published another method of solving cubics.The method is based on the identity (see p15) cos(3θ) = 4C3 − 3Cwith C = cos θ.

(i) Substitute x = 2√pC into the (reduced) general cubic x3 = 3px+2q to obtain 4C3 − 3C = q

p√p.

1.5. EXERCISES PP 45 9

x3 = 8p√pC3 = 6p√pC + 2q = 3px + 2q implies 4p√pC3 −3p√pC = q and hence 4C3 − 3C = q

p√p.

(ii) Provided that q2 ≤ p3, deduce that the solutions of the originalequation are x = 2√p cos (1

3(φ + 2mπ)), where m is an integerand φ = cos−1 q

p√p

= arccos qp√p.

x = 2√pC with C = cos θ and cos(3θ) = 4C3 − 3C = qp√p,

hence θ =arccos q

p√p

+2mπ3 and x = 2√p cos

arccos qp√p

+2mπ3 for m ∈ Z.

(iii) Check that this formula gives the correct solutions of x3 = 3x,namely, x = 0,±

√3.

Here, p = 1 and q = 0, hence x = 2 cos (13(φ + 2mπ)) with

φ = cos−1(0) = π2 which for m = 0, 1, 2 in turn gives x = 2 cos π

6 =√3, x = 2 cos π/2+2π

3 = 2 cos 5π6 = 2 cos(π − π

6 ) = −√

3 and x =2 cos π/2−4π

3 = 2 cos 3π2 = 0.

4. Here is a basic fact about integers that has many uses in number theory:If two integers can be expressed as the sum of two squares, then socan their product. With the understanding that each symbol denotesan integer, this says that if M = a2 + b2 and N = c2 + d2, thenMN = p2 + q2. Prove this result by considering |(a+ ib)(c+ id)|2.

MN = (a2 + b2)(c2 + d2) = |a+ ib|2 |c+ id|2 = |(a+ ib)(c+ id)|2 =|ac− bd+ i(ad+ bc)|2 = (ac− bd)2 + (ad+ bc)2.

5. The figure below shows how two similar triangles may be used to con-struct the product of two complex numbers.

O1

ab

ab

Explain this.The construction guarantees that ab has the correct angle. The two

triangles are similar because they have two angles in common, hence|a|1 = |ab|

|b| implies |ab| = |a| · |b|.

6. (i) If c is a fixed complex number, and R is a fixed real number,explain with a picture why |z− c| = R is the equation of a circle.


All z ∈ C with |z − c| = R have the distance R to c: theyconstruct the circle with radius R around c.

(ii) Given that z satisfies the equation |z+ 3− 4i| = 2, find the mini-mum and maximum values of |z|, and the corresponding positionsof z.

All z lie on the circle with radius 2 about c = −3 + 4i with|c| =

√32 + 42 = 5, hence min|z| : |z + 3− 4i| = 2 = 5− 2 = 3

which is attained at zmin = 35(−3+4i) and max|z| : |z+3−4i| =

2 = 5 + 2 = 7 which is attained at zmax = 75(−3 + 4i).

7. Use a picture to show that if a and b are fixed complex numbers then|z − a| = |z − b| is the equation of a line.

L = z : |z − a| = |z − b| is the perpendicular bisector consistingof all z with identical distance to a and b.

8. Let L be a straight line in C making an angle φ with the real axis, andlet d be its distance from the origin. Show geometrically that if z isany point on L then d = |=(e−iφz)|. [Hint: Interpret e−iφ using (9).]

If in the sketch the angles in ∆(0ZA) are φ, 90o + 90o − ψ and δ,then δ = ψ − φ.

x = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φ

δ

Z ′

x = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φ

δ

Z ′ψ

x = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φ

δ

Z ′

x = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φx = <z

y = =z

OA

Z

F

φ

δ

Z ′ψφ

F ′

As ∠(F ′OZ) = ψ−φ = δ the two triangles ∆(OZF ) and ∆(OF ′Z ′) arecongruent due to identical angles and hypotenuses so that d = |OF | =|F ′Z ′| = |=(Z ′)| = |=(e−iφz)|.

9. Let A, B, C, D be four points on the unit circle. If A+B+C+D = 0,show that the points must form a rectangle.

We have A + B + C + D = 0 and |A| = |B| = |C| = |d| = 1.Considering the four points1 as elements of R2, by (A+B) · (A−B) =A2−A·B+A·B−B2 = 1−1 = 0 we see that (A+B) ⊥ (A−B) and by(A+B)·(C−D) = −(C+D)·(C−D) = 0 we also see (A+B) ⊥ (C−D)hence (A−B)‖(C −D). In the same way we get (A−D)‖(B −C) so

1 https://math.stackexchange.com/questions/348758

https://math.stackexchange.com/questions/348758


that (A,B,C,D) is a parallelogram and thus A−B = C−D implyingA + D = B + C which by 0 = A + B + C + D = 2(A + D) vanishes.Thus working from A either A+B = 0, A+C = 0, or A+D = 0. Bysymmetry it suffices to consider when A+B = 0, then C +D = 0.(A−C) · (B−C) = (A−C) · (−A−C) = −A2−A ·C+A ·C+C2 = 0so AC ⊥ BC. Similarly the other three angles are also right angles, sowe get a rectangle.

10. Show geometrically that if |z| = 1 then = z(z+1)2 = 0. Apart from the

unit circle, what other points satisfy this equation?In general 2=w = w − w, hence for 1 = |z| =

√zz ⇒ zz = 1 and

for N = (z+ 1)2(z+ 1)2 we have 2= z(z+1)2 = 1

N (z(z+ 1)2− z(z+ 1)2) =1N ((zz)z + 2zz + z − (zz)z − 2zz − z) = 0.Alternatively, let z = eiϕ. Then arg z

(z+1)2 = ϕ − 2 arg(z + 1) = ϕ −2 arctan sinϕ

1+cosϕ = ϕ− 2 arctan tan ϕ2 = 0.

Whereas geometrically arg(z+1) = 12 arg z implies arg ((z+1)2) = arg z

so that arg z(z+1)2 = 0 which is equivalent with = z

(z+1)2 = 0.The origin 0 ∈ C satisfies the equation also.

11. Explain geometrically why the locus of z such that arg z−az−b = const is

an arc of a certain circle passing through the fixed points a and b.Without loss of generality let a = 0 (by shifting) and <b = 0 (by

rotating) and let the constant be ∆. Let z(ϕ) ∈ C with arg (z(ϕ)−a) =ϕ for some ϕ lie on the circle about M with radius r = |aM | = |bM | =|z(0)M | = |z(∆)M | = |z(ϕ)M |. Let α = ϕ − ∆. Now, we showarg z(ϕ)−a

z(ϕ)−b = ∆.

δ

2α

α

∆

2∆

∆

M

a <z

=z

b

z(0)

z(∆)

z = z(ϕ)

First, ∆(aMb) is equilateral so that ∠(aMb) = π − 2(π2 − ∆) =2∆. Second, ∆(aMz) is equilateral so that ∠(aMz) = π − 2α and∠(zMz(∆)) = 2α and ∠(bMz) = ∠(aMz) − 2∆ = π − 2(α + ∆) =π − 2ϕ. Third, ∆(bMz) is equilateral so that π = ∠(bMz) + 2∆ +


2δ = π − 2ϕ + 2∆ + 2δ ⇐⇒ α = ϕ − ∆ = δ which impliesarg(z − a)− arg(z − b) = ϕ− δ = ϕ− α = ∆.

12. By using pictures, find the locus of z for each of the following equations:< z−1−iz+1+i = 0 and = z−1−i

z+1+i = 0. [Hints: What does <(w) = 0 imply aboutthe angle of w? Now use the previous exercise.]

Let a = 1 + i and b = −a = −1 − i. Then < z−1−iz+1+i = 0 = < z−a

z−b isequivalent to arg z−a

z−b = ±π2 , i.e. (z − a) ⊥ (z − b) so that by Thales’

theorem the locus of points z with arg z−az−b = ±π

2 is the circle about theorigin with radius

√2.

= z−1−iz+1+i = 0 = = z−a

z−b is equivalent to arg z−az−b ∈ 0, π which means that

z − a and z − b are collinear. The locus is the line through a and b.

13. Find the geometric configuration of the points a, b, and c if b−ac−a = a−c

b−c .[Hint: Separately equate the lengths and angles of the two sides.]

If arg(b−a)−arg(c−a) = ∠(cab) = ∠(bca) = arg(a−c)+arg(b−c)then the triangle ∆(abc) is isosceles. Now, with |b−a||a−c| = |a−c|

|c−b| there arethree cases: firstly |c − b| = |a − c| ⇒ |b−a|

|a−c| = 1 ⇒ |b − a| = |a − c|,secondly |c− b| = |b−a| ⇒ |c−b|

|a−c| = |a−c||c−b| ⇒ |c− b| = |a− c|, and thirdly

|a − c| = |b − a| ⇒ |a−c||c−b| = 1 ⇒ |a − c| = |c − b|. Summarizing, the

triangle ∆(abc) is equilateral.

14. By considering the product (2 + i)(3 + i), show that π4 = arctan 1

2 +arctan 1

3 .zo = z1z2 = (2+ i)(3+ i) = 5+5i with arg zo = π

4 , arg z1 = arctan 12

and arg z2 = arctan 13 , so that arg zo = π

4 = arg z1 + arg z2 = arctan 12 +

arctan 13 .

15. Draw eiπ/4, eiπ/2 and their sum. By expressing each of these numbersin the form x+ iy, deduce that tan 3π

8 = 1 +√

2.z1 = eiπ/4 =

√2

2 + i√

22 , z2 = eiπ/2 = i and z = z1 + z2 =

√2

2 +i(1 +

√2

2 ), so that on one hand arg z = arctan 1+√

2/2√2/2 = arctan(1 +

√2)

and together with |z1| = 1 = |z2| and in ∆(Oz1z) on the other hand|z| =

√2 +√

2 and ∠(zOz1) = arctan h1√2+√

2/2= arctan

√2−√

22+√

2 =

arctan(√

2− 1) = π8 as tan π

8 = sin π4

1+cos π4=

√2/2

1+√

2/2 =√

2− 1.

16. Starting from the origin, go one unit east, then the same length north,then (1/2) of the previous length west, then (1/3) of the previous lengthsouth, then (1/4) of the previous length east, and so on. What pointdoes this "spiral" converge to?

S = 1 + i − 12 − i1

213 + 1

213

14 + . . . = ∑∞

ν=o(−1)ν(2ν)! + i

∑∞µ=o

(−1)µ(2µ+1)! =

cos 1 + i sin 1 = ei.


17. If z = eiθ 6= −1, then (z − 1) = i(z + 1) tan θ2 . Prove this (i) by

calculation, (ii) with a picture.(i) As |z− 1| =

√(c− 1)2 + s2 with c = cosϑ and s = sinϑ, and as

|i(z + 1) tan ϑ2 | =

√(c+ 1)2 + s2 |s|

|1+c| we have |lhs|2 = (c − 1)2 + s2 =((c + 1)2 + s2) s2

(c+1)2 = |rhs|2 ⇐⇒ (c − 1)2(c + 1)2 + s2(c + 1)2 =(c + 1)2s2 + s4 ⇐⇒ (c2 − 1)2 = s4. To show arg(lhs) = arg(rhs) wehave arg(lhs) = arctan s

c−1 = − arctan cot ϑ2 = − arctan tan(π2 −

ϑ2 ) =

ϑ2 −

π2 = ϑ

2 + π2 and arg(rhs) = π

2 + arg(z+ 1) = π2 + arctan s

c+1 = π2 + ϑ

2 .(ii) Geometrically, when rotating ∆(O1(z+1) by π

2 around O then z−1lies on the long edge because (z − 1) ⊥ (z + 1) as |z|2 = x2 + y2 = 1and (x− 1, y)(x+ 1, y)> = (x− 1)(x+ 1) + y2 = x2 + y2 − 1 = 0.

1<z

=z

O

zz − 1z + 1

iz

i(z + 1)

∠((z + 1)Oz) = θ2 implies ∠(Oz(z − 1)) = θ and that ∆(Oz(z − 1))

is an isosceles triangle with12 |z−1|

1 = sin θ2 , hence |z − 1| = 2 sin θ

2 =|z + 1| tan θ

2 as |z + 1| = 2 cos θ2 .

18. Prove that eiϑ+eiϕ = 2 cos ϑ−ϕ2 ei(ϑ+ϕ)/2 and eiϑ−eiϕ = 2i sin ϑ−ϕ

2 ei(ϑ+ϕ)/2

(i) by calculation, and (ii) with a picture.eiϑ + eiϕ = cosϑ + cosϕ + i sinϑ + i sinϕ = 2 cos ϑ+ϕ

2 cos ϑ−ϕ2 +

i2 sin ϑ+ϕ2 cos ϑ−ϕ

2 = 2 cos ϑ−ϕ2 ei(ϑ+ϕ)/2; eiϑ−eiϕ = cosϑ−cosϕ+i sinϑ−

i sinϕ = −2 sin ϑ+ϕ2 sin ϑ−ϕ

2 + i2 sin ϑ−ϕ2 cos ϑ+ϕ

2 = 2i sin ϑ−ϕ2 ei(ϑ+ϕ)/2.

(ii) Let α = 12(ϕ − δ) and z1 = eiϕ, z2 = eiδ and z = z1 + z2, then

arg z = δ + α = 12(ϕ + δ) and |z| = 2 · 1 · cosα = 2 cos ϕ−δ

2 , hencez = 2 cos ϑ−ϕ

2 ei(ϑ+ϕ)/2.


<z

=z

eiδ

eiϕ

eiϕ+eiδ

<z

=z

eiδ

eiϕ

eiϕ+eiδ

αα

similar for eiϑ − eiϕ.

19. The ’centroid’ G of a triangle T is the intersection of its medians. Ifthe vertices are the complex numbers a, b, and c, then you may assumethat G = 1

3(a + b + c). On the sides of T we have constructed threesimilar triangles [dotted] of arbitrary shape, so producing a new triangle[dashed] with vertices p, q, r.

a

b

c

G

q

r

p

Using complex algebra, show that the centroid of the new triangle is inexactly the same place as the centroid of the old triangle!

Let the vertices of the triangle be O, a and a + b with centre ofgravity s = 1

3(0 + a + a + b). For constants α, β ∈ R we then havep = αa ± iβa = (α ± iβ)a, q = a + αb ± iβb = a + (α ± iβ)b andr = a+ b+ (α± iβ)(−a− b) so that 1

3(p+ q+ r) = 13((α+ iβ)a+ (a+

(α± iβ)b) + (a+ b− (α± iβ)(a+ b))) = 13(a+ a+ b) = s.

20. Gaussian integers are complex numbers of the form m + in, where mand n are integers – they are the grid points in [1]. Show that it isimpossible to draw an equilateral triangle such that all three verticesare Gaussian integers. [Hints: You may assume that one of the verticesis at the origin; try a proof by contradiction; if a triangle is equilateral,you can rotate one side into another; remember that

√3 is irrational.]


Without loss of generality we assume that A = O and B = m forsome m ∈ Z, then C = m

2 (1+ i√

3). As√

3 ∈ R\Q the point C cannotbe a Gaussian integer no matter what m was used.

21. Make a copy of [12a], draw in the diagonal of the quadrilateral shownin [12b], and mark its midpoint m. As in [12b], draw the line-segmentsconnecting m to p, q, r, and s. According to the result in [12b], whathappens to p and to r under a rotation of π2 about m? So what happensto the line-segment pr ?

Deduce the result shown in [12a].In order to show [12b] assume without loss of generality a = 0,

b = 1, then p = 12(1 − i) and q = 1

2c(1 + i) and m = 12(c + 1). We

show m − p = i(m − q) in order to prove (m − p) ⊥ (m − q). Now,2(m−p) = c+1−(1−i) = i+c = i(1−ic) = i(c+1−c(1+i)) = 2i(m−q).so that ∠(pmq) = π

2 and |m− p| = |m− q|. ???

22. Will the result in [12a] survive if the squares are instead constructedon the inside of the quadrilateral?

. ???

23. Draw an arbitrary triangle, and on each side draw an equilateral tri-angle lying outside the given triangle. What do you suspect is specialabout the new triangle formed by joining the centroids (cf. Ex. 19) ofthe equilateral triangles? Use complex algebra to prove that you areright. What happens if the equilateral triangles are instead drawn onthe inside of the given triangle?

As in the example on p.16/17 let c = −a − b for a, b ∈ C so thata + b + c = 0. Construct triangle ∆(O, a, a + b). On each side of∆(O, a, a + b) construct the equilateral triangle and mark its centerpoint. We show that the emerging triangle is equilateral.


a

b

c

a+ b+ c = 0

<z

=z

Without loss of generality let the first vertex be the origin, the secondbe 1 and the third be 1 + b. Then c = −1− b. The emerging new redtriangle is ∆(1

2−i√

36 , 1+ b

2−i√

36 b, 1+b+ c

2−i√

36 c) = ∆(k, 1+kb, 1+b+kc)

for c = −1− b and k = 3−i√

36 . We have kk = 1

3 and cc = 1 + b+ b+ bb.To show ∆ to be equilateral we show e.g. |1+kb−k|2 = |1+b+kc−k|2⇐⇒ (1 + kb− k)(1 + kb− k) = (1 + b+ kc− k)(1 + b+ kc− k) ⇐⇒1 + kb− k+ kb+ kkbb− kkb− k− kkb+ kk = 1 + b+ kc− k+ b+ bb+kbc− kb+ kc+ kbc+ kkcc− kkc− k − kb− kkc+ kk ⇐⇒ kb+ kb =b− k(1+ b)+b+bb− kb(1+ b)− kb−k(1+b)−kb(1+b)+(1+b+ b)−kb⇐⇒ 2(k+ k)b+ 2(k+ k)b = b− k+ b+ bb− kbb− k− kbb+ (1 + b+ b)⇐⇒ 0 = bb− kbb−kbb and also |1+kb−k|2 = |1+b+kc−(1+kb)|2 =|b− k(1 + b)− kb|2 = |b− k − 2kb|2 ⇐⇒ (1 + kb− k)(1 + kb− k) =(b − k − 2kb)(b − k − 2kb) ⇐⇒ 1 + kb − k + kb + kkbb − kkb − k −kkb+kk = bb− kb−2kbb−kb+kk+ 2kkb−2kbb+ 2kkb+ 4kkbb ⇐⇒23(b+ b) = 2bb− 2kbb+ 2kkb− 2kbb+ 2kkb = 2bb(1− k− k) + 2

3(b+ b).Obviously, due to 1

3(k + (1 + kb) + (1 + b+ kc)) = 13(k + 1 + kb+ 1 +

b−k−kb)) = 13(0 + 1 + (1 + b)) the two triangles have the same center

of gravity. Geometrically??????

24. From (15), we know that 1 + z + z2 + . . .+ zn−1 = zn−1z−1 .

(i) In what region of C must z lie in order that the infinite series1 + z + . . . converges?

The series converges for |z| < 1 as the zn → 0.(ii) If z lies in this region, to which point in the plane does the infinite

series converge?


The series converges to 11−z for |z| < 1.

(iii) In the spirit of figure [9], draw a large, accurate picture of theinfinite series in the case z = 1

2(1 + i), and check that it doesindeed converge to the point predicted by part (ii).

Let sn = ∑nν=o z

ν .

<z

=z

1

z 1+zz2

1+z+z2

z3

1+z+z2+z3

z4

1+z+z2+z3+z4

z5

s5

z6

s61

1−z

25. Let Sn = cos θ+cos 3θ+cos 5θ+. . .+cos(2n−1)θ. Show that S = sin 2nθ2 sin θ

or equivalently S = sin(nθ) cos(nθ)sin θ . [Hint: Use Ex. 24, then Ex. 18 to

simplify the result.]Let Tn = sin θ+sin 3θ+sin 5θ+ . . .+sin(2n−1)θ. Then Sn+ iTn =

eiθ+ei3θ+ei5θ+ . . .+ei(2n−1)θ = eiθ(1+ei2θ+(ei2θ)2 + . . .+(ei2θ)n−1) =eiθ (ei2θ)n−1

ei2θ−1 = ei2nθ−1eiθ−e−iθ = cos(2nθ)+i sin(2nθ)−1

2i sin θ = i−i cos(2nθ)+sin(2nθ)2 sin θ so that

Sn = <(Sn + iTn) = sin(2nθ)2 sin θ .

26. (i) By considering (a+ ib)(cos θ+ i sin θ), show that b cos θ+a sin θ =√a2 + b2 sin(θ + arctan b

a).

On one hand z = (a + ib)(cos θ + i sin θ) = a cos θ − b sin θ +i(a sin θ+b cos θ) and on the other hand z = |a+ ib|ei arg(a+ib)eiθ =√a2 + b2ei(θ+arctan(b/a)), then taking the imaginary part we get=z = a sin θ + b cos θ =

√a2 + b2 sin(θ + arctan b

a).

(ii) Use this result to prove by the method of inductiondn

d tn (eat sin(bt)) =

√a2 + b2neat sin(bt+ n arctan b

a). (14)

The result is obviously true for n = 0. Let c = arctan ba.

Suppose the result holds for some n, then it holds for n+ 1 also:

dn+1

d tn+1 (eat sin(bt)) = d

d t(√a2 + b2

neat sin(bt+ nc))

=√a2+b2

n(aeat sin(bt+nc) + beat cos(bt+nc))

=√a2 + b2

n+1eat sin(bt+ nc+ c)

=√a2 + b2

n+1eat sin (bt+ (n+ 1) arctan b

a).


27. Show that the polar equation of the spiral Z(t) = eateibt in [15b] isr = e(a/b)θ.

r(θ) = |Z(t)| = eat where argZ(t) = bt = θ, hence t = θband

r(θ) = e(a/b)θ.

28. Reconsider the spiral Z(t) = eateibt in [15b], where a and b are fixedreal numbers. Let τ be a variable real number. According to (9), z 7→Fτ (z) = eaτeibτz is an expansion of the plane by factor eaτ , combinedwith a rotation of the plane through angle bτ .

(i) Show that Fτ [Z(t)] = Z(t + τ), and deduce that the spiral is aninvariant curve (cf. p 38) of the transformations Fτ .Fτ [Z(t)] = e(a+ib)τZ(t) = e(a+ib)τe(a+ib)t = ea(t+τ)eib(t+τ) =

Z(t + τ) for any t, τ ∈ R, hence the spiral is invariant under Fτfor any τ ∈ R.

(ii) Use this to give a calculus-free demonstration that all rays fromthe origin cut the spiral at the same angle.

z(t) = teiφ is ray from the origin with angle φ between rayand positive abscissa and intersection zo = τoe

iφ = e(a+ib)to =eatoeibto , that is for to = φ

b∈ R and 0 ≤ τo = eato = eaφ/b ∈ R

with the spiral. ddtZ(to) = (a + ib)e(a+ib)to = (a + ib)eaφ/beiφ is

the direction of the spiral in zo which differs from direction φ byconstant arg(a+ ib) independent of φ. calculus free????

(iii) Show that if the spiral is rotated about the origin through anarbitrary angle, the new spiral is again an invariant curve of eachFτ .

z 7→ ei%z is the rotation by % so that W (t) = ei%Z(t) =eatei(bt+%) is the rotated spiral with FτW (t) = e(a+ib)τe(a+ib)tei% =W (t+ τ).

(iv) Argue that the spirals in the previous part are the only invariantcurves of Fτ .

When restricting Fτ (Z(t)) = Z(t + κ) for all t and τ whereκ = κ(τ) to Fτ (Z(t)) = Z(t + τ) = e(a+ib)τZ(t) then we getddτZ(t+τ) = Z(t+τ) = (a+ib)e(a+ib)τZ(t) and for τ = 0 especially

Z(t) = (a+ ib)Z(t) which implies Z(t) = c(a+ ib)e(a+ib)t for someintegration constant c ∈ C. ??????

29. (i) If V (t) is the complex velocity of a particle whose orbit is Z(t), anddt is an infinitesimal moment of time, then V (t) dt is a complexnumber along the orbit. Thinking of the integral as the (vector)sum of these movements, what is the geometric interpretation ofs =

∫ thitloV (t) dt?


s =∫ thitloV (t) dt ≈ ∑n

ν=1 V (t∗ν)∆tν ≈∑nν=1 (s(tν) − s(tν−1)) =

s(thi)− s(tlo) is (an approximation of) the travelled distance s.(ii) Referring to [15b], sketch the curve Z(t) = 1

a+ibeateibt = eat

a2+b2 (a−ib)eibt.

<z2

=z

2

a+ ib

1a+ib

(iii) Given the result (13), what is the velocity of the particle in theprevious part.

V (t) = Z(t) = 1a+ib

ddte(a+ib)t = eateibt.

(iv) Combine the previous parts to deduce that∫ 1o e

ateibt dt = eateibt

a+ib

∣∣∣1o

and draw in this complex number in your sketch for part (ii).∫ 1o e

ateibt dt = eateibt

a+ib

∣∣∣1o

= ea+ib−1a+ib = (ea cos b−1+iea sin b)(a−ib)

a2+b2 =a(ea cos b−1)+bea sin b

a2+b2 + iaea sin b−b(ea cos b−1)

a2+b2 .

(v) Use this to deduce that∫ 1o e

at cos(bt) dt = a(ea cos b−1)+bea sin ba2+b2 and∫ 1

o eat sin(bt) dt = b(1−ea cos b)+aea sin b

a2+b2 .∫ 1o e

at cos(bt) dt = < ea+ib−1a+ib = a(ea cos b−1)+bea sin b

a2+b2 and∫ 1o e

at sin(bt) dt = = ea+ib−1a+ib = b(1−ea cos b)+aea sin b

a2+b2

30. Given two starting numbers S1, S2, let us build up an infinite sequenceS1, S2, S3, S4, . . . with this rule: each new number is twice the differenceof the previous two. For example, if S1 = 1 and S2 = 4, we obtain 1, 4,6, 4, -4, -16, -24,. . . Our aim is to find a formula for the nth number Sn.

(i) Our generating rule can be written succinctly as Sn+2 = 2(Sn+1−Sn). Show that Sn = zn will solve this recurrence relation ifz2 − 2z + 2 = 0.

Dividing zn+2 = 2(zn+1 − zn) by zn implies z2 = 2z − 2.(ii) Use the quadratic formula to obtain z = 1± i, and show that if A

and B are arbitrary complex numbers, Sn = A(1+ i)n+B(1− i)nis a solution of the recurrence relation.

z2 − 2z + 2 = 0 is solved by z = 1±√

1− 2 = 1± i. Further,the set of solutions is a vector space because with solutions Sn


and Tn also Sn + Tn as well as C Sn for any C ∈ C are solutions.Hence, Sn = A(1 + i)n +B(1− i)n is a solution of the recurrencerelation.

(iii) If we want only real solutions of the recurrence relation, show thatB = A and deduce that Sn = 2<[A(1 + i)n].

Any solution is of the type A(1+i)n+B(1−i)n. Such a solutionis real if S = S, i.e. A(1 + i)n +B(1− i)n = A(1− i)n + B(1 + i)n⇐⇒ A + B = A + B for n = 0 and A + B + i(A − B) =A(1 + i) +B(1− i) = A(1− i) + B(1 + i) = A+ B+ i(B− A) forn = 1 or A−B = B − A which added together implies 2A = 2Band then Sn = A(1 + i)n + A(1 + i)n = 2<(A(1 + i)n).

(iv) Show that for the above example A = −12 − i, and by writing this

in polar form deduce that Sn = 2n/2√

5 cos (n−44 π + arctan 2).

With A =√

52 e

i(−π+arctan 2) we have Sn = 2<(A(1 + i)n) =√5√

2n<(ei(nπ/4−π+arctan 2)) = 2n/2√

5 cos (n−44 π + arctan 2).

(v) Check that this formula predicts S34 = 262144, and use a com-puter to verify this.

S34 = 217√5 cos (304 π+arctan 2) = 217√5 cos (3

2π+arctan 2) =217√5( cos 3π

2 cos arctan 2−sin 3π2 sin arctan 2) = 217√5 sin arctan 2

= 217√5 tan arctan 2√1+tan2 arctan 2

= 218 = 262144.

[Note that this method can be applied to any recurrence relation of theform Sn+2 = pSn+1 + qSn.]

31. With the same recurrence relation as in the previous exercise, use acomputer to generate the first 30 members of the sequence given byS1 = 2 and S2 = 4. Note the repeating pattern of zeros. One gets 2, 4,4, 0, -8, -16, -16, 0, 32, 64, 64, 0, -128, -256, -256, 0, 512, 1024, 1024,0, -2048, -4096, -4096, 0, 8192, 16384, 16384, 0, -32768, -65536, . . . .

(i) With the same notation as before, show that this sequence corre-sponds to A = −i so that Sn = 2<[−i(1 + i)n].

We have S1 = 2<(A(1 + i)) = 2<(1 − i) = 2 and S2 =2<(A(1 + i)2) = 2<(−i2i) = 4, hence Sn = 2<[−i(1 + i)n].

(ii) Draw a sketch showing the locations of−i(1+i)n for n = 1, 2, . . . , 8,and hence explain the pattern of zeros.

n 1 2 3 4 5 6 7 8−i(1 + i)n 1− i 2 2 + 2i 4i −4 + 4i −8 −8− 8i −16i

2<(−i(1 + i)n) 2 4 4 0 −8 −16 −16 0


<z

=z

Because (1 + i)4 = 1 + 4i+ 6i2 + 4i3 + 1 = 1 + 4i−6−4i+ 1 = −4we have −i(1 + i)4m = −i(−4)m is purely imaginary for m ∈ Nso that S4m = 2<(−i(1 + i)n) = 0 for m ∈ N.

(iii) Writing A = a + ib, our example corresponds to a = 0. Moregenerally, explain geometrically why such a repeating pattern ofzeros will occur if and only if a

b∈ −1, 0, 1 or b = 0.

Sn = 2<(A(1 + i)n) = 0 ⇐⇒ A(1 + i)n is purely imaginary⇐⇒ arg ((A(1+i)n) = arg(A)+n arg(1+i) = arg(A)+nπ4 = ±π

2⇐⇒ arg(A) = arctan b

a∈ π

4Z ⇐⇒ab∈ −1, 0, 1 or b = 0.

(iv) Show that S1S2

= 12(1 − a

b) and deduce that a repeating pattern

of zeros will occur if and only if S2 = 2S1 (as in our example),S1 = S2, S1 = 0, or S2 = 0.

S1S2

= <((a+ib)(1+i))<((a+ib)2i) = 1

2a−b−b = 1

2(1 − ab). With part (iii) a

repeating pattern of zeros occurs ⇐⇒ ab∈ −1, 0, 1 or b = 0

⇐⇒ S1 = S2 or S1 = 2S2 or S1 = 0 or S2 = 0.(v) Use a computer to verify these predictions.

see https://sage.informatik.hs-bremen.de/home/risse/296/

32. The Binomial Theorem says that if n is a positive integer, (a + b)n =∑nν=0

(nν

)aνbn−ν where

(nν

)= n!

ν!(n−ν)! are the binomial coefficients. Thealgebraic reasoning leading to this result is equally valid if a and b arecomplex numbers. Use this fact to show that if n = 2m is even then(

2m1

)−(

2m3

)+(

2m5

)± . . . (−1)m+1

(2m

2m−1

)= 2m sin mπ

2 .As S = ∑2m

ν=o

(2mν

)iν = (1 + i)2m = (

√2eiπ/4)2m = 2meimπ/2 =

2m cos mπ2 + i2m sin mπ

2 implies =S = ∑mµ=o

(2m

2µ+1

)(−1)µ = 2m sin mπ

2 .

33. Consider the equation (z − 1)10 = z10.

(i) Without attempting to solve the equation, show geometricallythat all 9 solutions [why not 10?] must lie on the vertical line,<(z) = 1

2 . [Hint: Ex. 7.]First, the equation (z − 1)10 = z10 is polynomial of degree 9

because the term z10 cancels out. Second it implies |z − 1| = |z|which with Ex. 7 establishes that all z lie on the perpendicularbisector of 0 ∈ C and 1 ∈ C, namely the line <z = 1

2 .

https://sage.informatik.hs-bremen.de/home/risse/296/


(ii) Dividing both sides by z10, the equation takes the form w10 = 1,where w = z−1

z. Hence solve the original equation.

As the tenth roots of unity wk = eikπ/5 for k = 0, 1, 2, . . . , 9,then z wk = z − 1 or zk = −1

wk−1 = 1−wk|1−wk|2

for 0 6= k = 1, 2, . . . , 9.(iii) Express these solutions in the form z = x+ iy, and thereby verify

the result in (i). [Hint: To do this neatly, use Ex. 18.]With Ex. 18 we have zk = −1

wk−1 = −12i sin kπ

10 eikπ/10 = i e−ikπ/10

2 sin(kπ/10) =cos(π/2−kπ/10)+i sin(π/2−kπ/10)

2 sin(kπ/10) = 12 + i cos(kπ/10)

sin(kπ/10) = 12 + i cot kπ

10 .

34. Let S denote the set of 12th roots of unity shown in [19], one of whichis ξ = eiπ/6. Note that ξ is a primitive 12th root of unity, meaning thatits powers yield all the 12th roots of unity: S = ξ1, ξ2, ξ3, . . . , ξ12.

(i) Find all the primitive 12th roots of unity, and mark them on acopy of [19].

S = ξ, eiπ/3, i, ei2π/3, ξ5,−1, ξ7, ei4π/3,−i, ei5π/3, ξ11, 1 gives

<z

=z

ξ

ξ2ξ3

ξ4

ξ5

ξ6

ξ7

ξ8

ξ9ξ10

ξ11

ξ0

where the primitive roots ξ, ξ5, ξ7, ξ11 are marked in red.(ii) Write down, in the form of (16), the factorization of the polyno-

mial Φ12(z) whose roots are the primitive 12th roots of unity. [Ingeneral, Φn(z) is the monic polynomial whose roots are the prim-itive nth roots of unity; it is called the nth cyclotomic polynomial.]

Φ12(z) = (z − ξ)(z − ξ5)(z − ξ7)(z − ξ11) is monic and the setof its zeros is the set of the primitive 12th roots of unity.

(iii) By first multiplying out pairs of factors corresponding to conju-gate roots, show that Φ12(z) = z4 − z2 + 1.

Φ12(z) = ((z−ξ)(z−ξ11))((z−ξ5)(z−ξ7)) = (z2−z(ξ+ξ11)+ξ12))(z2−z(ξ5 +ξ7)+ξ12)) = (z2−2z<ξ+1)(z2−2z<(ξ5)+1) =(z2 − 2z<ξ + 1)(z2 + 2z<ξ + 1) = (z4 − 2z3<ξ + z2) + (2z3<ξ −4z2<2ξ+2z<ξ)+(z2−2z<ξ+1) = z4+z2(2−4<2ξ)+1 = z4−z2+1as 2− 4<2ξ = 2− 4 cos2 π

6 = 2− 4(√

32 )2 = 2− 43

4 = −1.


(iv) By repeating the above steps, show that Φ8(z) = z4 + 1.For ξ = eiπ/4 is S = ξ, i, ξ3,−1, ξ5,−i, ξ7, 1 the set of the 8th

roots of unity with primitive elements ξ, ξ3, ξ5, ξ7. Hence Φ8(z) =((z−ξ)(z−ξ7))((z−ξ3)(z−ξ5)) = (z2−z(ξ+ξ7)+ξ8)(z2−z(ξ3+ξ5) + ξ8) = (z2− 2z<ξ+ 1)(z2 + 2z<ξ+ 1) = (z4− 2z3<ξ+ z2) +(2z3<ξ−4z2<2ξ+2z<ξ)+(z2−2z<ξ+1) = z4+z2(2−4<2ξ)+1 =z4 + 1 as <ξ = cos π

4 =√

22 so that 4<2ξ = 2.

(v) For a general value of n, explain the fact that if ζ is a primitiventh root of unity, then so is ζ. Deduce that Φn(z) always has evendegree and real coefficients.

ζn = 1 implies ζn = 1 so ζ = 1z

= zn−1 is also a nth rootof unity. which is also primitive as ζk : k = 0, 2, . . . , n − 1 =zkn−k : k = 1, 2, . . . , n = zn−k : k = 1, 2, . . . , n = zk : k =1, 2, . . . , n.Thus, the primitive nth roots of unity lie symmetric to the abscissaso that the number of primitive roots must be even. In the repre-sentation Φn(z) = ∏

ζ is primitive(z− ζ) all products (z− ζ)(z− ζ) =z2− z(ζ+ ζ) + ζζ = z2−2z<ζ+ |ζ|2 are real polynomials, so thatΦn(z) also is a real polynomial.

(vi) Show that if p is a prime number then Φp(z) = 1+z+z2+. . .+zp−1.[Hint: Ex. 24.]

If p is prime then all roots 1 6= ξk = ξk with ξ = ei2π/p areprimitive so that Φp(z) = ∏p−1

k=1(z − ξk). Now, Φp(z) is monic ofdegree p − 1 with zeroes ξ, ξ2, . . . , ξp−1. Due to Ex. 24 P (z) :=∑p−1k=o z

k = zp−1z−1 is monic of degree p − 1 with the same zeroes,

hence Φp(z) = P (z) = ∑p−1k=o z

k.

[In these examples it is striking that Φn(z) has integer coefficients. Infact it can be shown that this is true for every Φn(z) ! For more onthese fascinating polynomials, see Stillwell [1994].]

35. Show algebraically that the formula A = 12=(ab+ bc+ cd+ da) (21)

is invariant under a translation by k, i.e., its value does not change ifa becomes a + k, b becomes b + k, etc. Deduce from [22a] that theformula always gives the area of the quadrilateral.[Hint: Remember, (z + z) is always real.]A = 1

2=((a+k)(b+k)+(b+k)(c+k)+(c+k)(d+k)+(d+k)(a+k)) =12=(ab+ ak + kb+ kk + bc+ bk + kc+ kk + cd+ ck + kd+ kk + da+dk + ka+ kk) = 1

2=(ab+ bc+ cd+ da).Using the translation invariance we can make sure that the origin al-ways lies within the quadrangle so that A = 1

2=(ab+ bc+ cd+ da) holdsin any case – not only in [22a].


36. According to the calculation on p.18, Rφb Rθ

a = Rθ+φc , where c =

aeiφ(1−eiθ)+b(1−eiφ)1−ei(θ+φ) . Let us check that this c is the same as the one given

by the geometric construction in [30b].

(i) Explain why the geometric construction is equivalent to sayingthat c satisfies the two conditions arg c−b

a−b = φ2 and arg c−a

b−a = − θ2 .

Choose L2 = L′1 with Rθa = RL1 RL2 and Rφ

b = RL′1 RL′2

.Now, c is the intersection of L1 and L′2. In ∆(a, b, c) we haveφ2 = ∠(cba) = arg(c − b) − arg(a − b) and θ

2 = ∠(bac) = arg(c −a)− arg(b− a) so that arg c−b

a−b = φ2 and arg c−a

b−a = θ2 respectively.

(ii) Verify that the calculated value of c (given above) satisfies thefirst of these conditions by showing that

c−ba−b = sin(θ/2)

sin((θ+φ)/2)eiφ/2. (33)

[Hint: Use 1− eiα = −2i sin α2 e

iα/2.]

c−ba−b =

aeiφ(1−eiθ)+b(ei(θ+φ)−eiφ)1−ei(θ+φ)

a−b =aeiφ(1−eiθ)−beiφ(1−eiθ)

1−ei(θ+φ)

a−b = eiφ(1−eiθ)1−ei(θ+φ) =

eiφ sin θ2 eiθ/2

sin((θ+φ)/2)ei(θ+φ)/2) = sin(θ/2)sin((θ+φ)/2)e

iφ/2 so that arg c−ba−b = φ

2 .(iii) In the same way, verify that the second condition is also satisfied.

c−ab−a =

aeiφ(1−eiθ)−a(1−ei(θ+φ))+b(1−eiφ)1−ei(θ+φ)

b−a =−a(1−eiφ)+b(1−eiφ)

1−ei(θ+φ)

b−a = 1−eiφ1−ei(θ+φ)

= sin(φ/2)eiφ/2sin((θ+φ)/2)ei(θ+φ)/2 = sin(φ/2)

sin((θ+φ)/2)e−iθ/2 so that arg c−a

b−a = − θ2 .

37. Deduce c−ba−b = sin(θ/2)

sin((θ+φ)/2)eiφ/2 (33) directly from [30b].

[Hint: Draw in the altitude through b of the triangle ∆(abc), and ex-press its length first in terms of sin θ

2 , then in terms of sin θ+φ2 .]

|hb| = |a − b| sin θ2 and |hb| = |c − b| sin(π − θ+φ

2 ) = |c − b| sin θ+φ2

so that |c−b||a−b| = |hb|/ sin((θ+φ)/2)

|hb|/ sin(θ/2) = sin(θ/2)sin((θ+φ)/2) . Because of arg c−b

a−b =arg(c− b)− arg(a− b) = φ

2 we have c−ba−b = sin(θ/2)

sin((θ+φ)/2)eiφ/2.

38. On page 18 we calculated that for any non-zero α, Tv RαO is a rotation:

Tv RαO = Rα

c , where c = v1−eiα . However, if α = 0 then Tv Rα

O = Tv isa translation. Try to reconcile these facts by considering the behaviourof Rα

c in the limit that α tends to zero.For the left hand side limα→0 TvRα

O = Tvlimα→0RαO = TvE = Tv

whereas for the right hand side and with p.18 we have limα→0Rαc (z) =

limα→0 (eiαz + c(1− eiα)) = limα→0 (eiαz + v) = z + v = Tv(z).

39. A glide reflection is the composition Tv RL = RL Tv of reflectionin a line L and a translation v in the direction of L. For example, ifyou walk at a steady pace in the snow, your tracks can be obtainedby repeatedly applying the same glide reflection to a single footprint.Clearly, a glide reflection is an opposite motion.


(i) Draw a line L, a line-segment AB, the image AB of the segmentunder RL, and the image A′B′ of AB under the glide reflectionTv RL.

<z

<z

A

B

A

B

A′

B′

(ii) Suppose you erased L from your picture; by considering the line-segments AA′ and BB′, show that you can reconstruct L.

The two dots in the figure above mark the points 12(A + A′)

and 12(B +B′), respectively.

(iii) Given any two segments AB and A′B′ of equal length, use theprevious part to construct the glide reflection that maps the for-mer to the latter.

L is specified by passing through 12(A + A′) and 1

2(B + B′),then A := RL(A) and v = A′ − A.

(iv) Deduce that every opposite motion is a glide reflection.The result Every opposite motion M can be represented as

a motionM mapping AB to A′B′ of equal length followed by thereflection in the line A′B′. (25)on p.36 together with part (iii) shows that every opposite motionM is a glide reflection. ???

(v) Express a glide reflection as the composition of three reflections.By definition a glide reflection as an opposite motion is the

combination of a direct motion and a reflection where by (28) onp.37 the direct motion can be expressed by two reflections.

40. Let L be a line making angle φ (or φ + π) with the real axis, and letp be the point on L that is closest to the origin, so that |p| is thedistance to the line. Consider the glide reflection [cf. previous exercise]G = Tv RL, where the translation is through distance r parallel to L.Let us fix the value of φ by writing v = +reiφ.


(i) Use a picture to show that p = ±i|p|eiφ, and explain the geometricsignificance of the ±.±i|p|eiφ = ±|p|ei(φ+π/2)

<z

<z

ϕ

<z

<z

ϕ

<z

<z

ϕ

<z

<z

p

−p

ϕ+ π2

<z

<z

ϕ

<z

<z

ϕ

<z

<z

ϕ

<z

<z

p

−p

The sign depends on on which side of the line the origin lies.(ii) What transformation is represented by the complex functionH(z)

= z + r?H is the reflection in the abscissa followed by the translation

by rex so that H is a glide reflection.(iii) Use pictures to explain why G = Tp Rφ

O H R−φO T−p.

p

−p

<z

<z

−pp

−p

<z

<z

−p

−ϕ

p

−p

<z

<z

−pp

−p

<z

<z

−p

−ϕϕ

p

RφO H R

−φO transforms the glide reflection H on the abscissa

followed by a translation by rex into the glide reflection G on lineL and followed by a translation by r units into the direction of L.???

(iv) Deduce that G(z) = ei2φz + eiφ(r ± 2i|p|).G(z) = eiφ(e−iφ(z − p) + r) + p = eiφ(eiφ(z − p) + r) + p =

ei2φz − ei2φ(±i|p|e−iφ) + reiφ ± i|p|eiφ = ei2φz + eiφ(r ± 2i|p|).(v) Hence describe (in geometric terms) the glide reflection repre-

sented by G(z) = iz + 4i. Check your answer by looking at theimages of -2, 2i, and 0.


G(z) = iz + 4i = ei2π/4z + eiπ/44eiπ/4 = ei2π/4z + eiπ/42√

2(1 +i) = ei2φz+eiφ(r+2i|p|) for φ = π

4 , r = 2√

2 and |p| =√

2. HenceG is the glide reflection in line L =graph(f) with y = f(x) = x+2combined with the translation by r units in the direction of L.G(−2) = −2i+ 4i = 2i, G(2i) = i(−2i) + 4i = 2 + 4i and G(0) =

4i are verified by −2 ≡(−20

)refl−→

(−20

)trans−→

(−20

)+(

22

)=(

02

)≡ 2i, 2i ≡

(02

)refl−→

(02

)trans−→

(02

)+(

22

)=(

24

)≡ 2 + 4i

and 0 ≡(

00

)refl−→

(−22

)trans−→

(−22

)+(

22

)=(

04

)≡ 4i because

reL = 2√

2 1√2

(11

)=(

22

).

41. Let M(z) be the representation of a general opposite motion as a com-plex function.

(i) Explain why M(z) is a direct motion, and deduce from (27) thatM(z) = eiαz + w, for some α and w.

A reflection of some opposite motion is a direct motion forwhich by (27), M(z) = eiθz+v holds. Hence M(z) = e−iθz+ v =:eiαz + w.

(ii) Using the previous exercise, deduce that every opposite motion isa glide reflection.

Since any opposite motion M(z) =M(z) is the complex con-jugate of some direct motionM(z) = eiθz+v it can be representedas a glide reflection by iv) of Ex. 40.

42. On p.19 we calculated that if θ + φ = 2π then Rφb Rθ

a = Tv, wherev = (1− eiφ)(b− a).

(i) Let Q = b−a be the complex number from the first centre of rota-tion to the second. Show algebraically that v has length 2|Q| sin θ

2 ,and that its direction makes an angle of π−θ

2 with Q.|v| = |1 − eiφ| · |b − a| =

√(1− cosφ)2 + sin2 φ|b − a| =√

2− 2 cosφ|b− a| = 2|Q| sin φ2 as cosφ = cos(φ2 + φ

2 ) = cos2 φ2 −

sin2 φ2 = 1 − 2 sin2 φ

2 ⇐⇒ 2 sin2 φ2 = 1 − cosφ ⇐⇒ 2 sin φ

2 =√2− 2 cosφ. With 1− eiα = −2i sin α

2 eiα/2 we have ∠(v, b− a) =

∠(1− eiφ) = π + π+φ2 = 3π+2π−θ

2 = π−θ2 .

(ii) Give direct geometric proofs of these results by redrawing figure[30b] on p.39 in the case θ + φ = 2π.

The two triangles ∆(z, a+eiφ(b−a), a) and ∆(Rφb Rθ

a(z), b, b−


eiφ(b − a)) are congruent because Rφb Rθ

a transforms the firstone into the second one: obviously z is mapped to Rφ

b Rθa(z),

a + eiφ(b − a) is mapped to Rφb Rθ

a(a + eiφ(b − a)) = Rφb (a +

ei2π(b − a)) = b and a is mapped to Rφb Rθ

a(a) = Rφb (a) =

Rφb (b+ eiπ(b− a)) = b+ ei(π+π)(b− a) = b− eiπ(b− a).

z

a

b

b−a

Rθa(z)

Rφb Rθa(z)eiφ (b− a)

−eiφ (b− a)

v

43. On p.18 we calculated that Tv RαO = Rα

c , where c = v1−eiα .

(i) Show algebraically that the complex number from the old centreof rotation (the origin) to the new centre of rotation (c) has length|v|

2 sin(α/2) and that its direction makes an angle of π−α2 with v.

As 1− eiα = −2i sin α2 e

iα/2 we have |c| = |v||1−eiα| = |v|

2 sin(α/2) asalso ∠(c, v) = arg(c)−arg(v) = − arg(1−eiα) = − arg(−ieiα/2) =− arg(−ei(π+α)/2) = −(π + π+α

2 ) = π − π+α2 = π−α

2 .(ii) Representing both Rα

O and Tv as the composition of two reflec-tions, use the idea in [30b] on p.39 to give direct, geometric proofsof these results.

.???

44. Just as in [13b], a dilative rotation Dθp centred at an arbitrary pointp may be performed by translating p to the origin, doing Dr,θO , thentranslating O back to p. Representing these transformations as complexfunctions, show that Dr,θp (z) = reiθz + v, where v = p(1− reiθ).Conversely, if v is given, deduce that Tv Dr,θO = Dr,θp where p = v

1−reiθ .Dr,θO (z) = Tp Dr,θO T−p(z) = reiθ(z− p) + p = reiθz+ p(1− reiθ) =

reiθz + v. Conversely, Tv Dr,θO (z) = reiθz + v = Dr,θp (z).

45. In the previous exercise you showed that an arbitrary dilative rotationor translation can be written as a complex function of the form f(z) =az + b, and, conversely, that every such function represents a uniquedilative rotation or translation.

(i) Given two pairs of distinct points A,B and A′, B′, show [byfinding them explicitly] that a and b exist such that f(A) = A′

and f(B) = B′.


With f(z) := A′ + B′−A′B−A (z − A) = az + b for a = B′−A′

B−A andb = A′− B′−A′

B−A A we have f(A) = A′ and f(B) = B′−A′+A′ = B′.(ii) Deduce the result (31) which states that every direct similarity

(preserving ratios of lengths and preserving angles) is a dilativerotation, i.e. Dr,θO := Rθ

O DrO = DrO RθO with Dr,θO (z) = reiθz or

(exceptionally) a translation.A direct similarity f is specified by any pair of points A and B

and their images A′ = f(A) and B′ = f(B): |A′B′||AB| specifies howlengths are preserved and ∠(AB,A′B′) how angles are preserved.By Ex.44 is any such function f(z) = az+ b a dilative rotation ortranslation.


Chapter 2

Complex Functions asTransformations

2.2 Polynomials

2.2.3 Cassinian Curves*

On p.62 for Q(z) = (z − a1)(z − a2) we have Q(z) ≈ z2 for z with |z| →∞ due to lim|z|→∞ Q(z)

z2 = lim|z|→∞ (z−a1)(z−a2)z2 = lim|z|→∞ z2−(a1+a2)z+a1a2

z2 =lim|z|→∞ (1− a1+a2

z+ a1a2

z2 ) = 1.Further we have limz→a1 |Q(z)| = limz→a1 |z−a1| |z−a2| ≈ r1|a1−a2| =: r1Dwith r1 = |z − a1|.

2.3 Power Series

2.3.2 The Disc of Convergence

On p.68 P (z) = ∑∞ν=0 cνz

ν is convergent in z = a ⇐⇒ Pn(z) = ∑nν=0 cνz

ν

for z = a is a Cauchy sequence.On p.68 P (z) = ∑∞

ν=0 zν/ν is convergent in z = −1 by the Leibniz criterion

but not absolutely convergent (harmonic series).

On p.72 h(z) = 1/(1+z2) = ∑∞ν=0(−1)νz2ν is the only complex function that

(i) agrees with H(x) = 1/(1 + x2) on the real axis, and (ii) can be expressedas a power series in z. because h fulfills both requirements. Its uniquenessfollows from the fact that the complex power series here is determined by thereal one.

31

32 CHAPTER 2. COMPLEX FUNCTIONS AS TRANSFORMATIONS

2.3.5 Manipulating Power Series

On p.74 comparison of coefficients leads to the infinite system of linearequations 1 = c0, 0 = c0 + c1, 0 = c0/2! + c1/1! + c2/0!, 0 = c0/3! + c1/2! +c2/1! + c3/0! etc. which leads to the guess cn = (−1)n/n! easily verified byconsidering the binomial expansion of 0 = (1 − 1)m = ∑m

k=0

(mk

)(−1)k =

m!∑mk=0

(−1)kk!(m−k)! = m!∑m

k=0 ck/(m− k)! where m is a positive integer.

2.3.6 Finding the Radius of Convergence

On p.75 applying the root test to the series P (z) = ∑∞j=1 (

j−3j )

j2zj yields

R = limj→∞1

j√|cj |

= limj→∞1

( j−3j )

j = limj→∞1

(1−3/j)j = 1e−3 = e3.

On p.77 given in 11+z2 = 1

2i∑∞j=0 (

1(−i−k)j+1 − 1

(i−k)j+1 )(z − k)j (17)on p.76 assume k to be real, then |i− k| =

√1 + k2 and i− k =:

√1 + k2 eiφ

where φ = arg(i− k) is the appropriate value of arctan(− 1k) so that 1

1+x2 =∑∞j=0

12i(

1(√

1+k2e−iφ)j+1 − 1(√

1+k2eiφ)j+1 )Xj = ∑∞j=0

12i

ei(j+1)φ−e−i(j+1)φ

(√

1+k2)j+1 Xj and so1

1+x2 = ∑∞j=0

sin((j+1)φ)(√

1+k2)j+1 (x− k)j. (18)

Generalization ??????

2.3.7 Fourier Series*

On p.78 let f(z) = 11−z with z = r eiθ so that f(z) = 1−z

(1−z)(1−z) = 1−r cos θ+ir sin θ1−2r cos θ+r2

with <f(z) = 1−r cos θ1−2r cos θ+r2 and =f(z) = r sin θ

1−2r cos θ+r2 .

2.5 Cosine and Sine

2.5.1 Definitions and Identities

On p.85 the definitions imply eiz = cos z + i sin z and e−iz = cos z − i sin zfor z ∈ C. So for a, b ∈ C we havecos(a + b) + i sin(a + b) = ei(a+b) = eia eib = (cos a + i sin a)(cos b + i sin b) =(cos a cos b− sin a sin b) + i(sin a cos b+ cos a sin b)cos(a+b)−i sin(a+b) = e−i(a+b) = e−ia e−ib = (cos a−i sin a)(cos b−i sin b) =(cos a cos b− sin a sin b)− i(sin a cos b+ cos a sin b)adding the two equations gives cos(a+ b) = cos a cos b− sin a sin b (23)and subtracting sin(a+ b) = sin a cos b+ cos a sin b (24)

2.7. THE LOGARITHM FUNCTION 33

On p.86 the definitions cosh x = 12(ex + e−x) and sinh x = 1

2(ex − e−x)imply 4(cosh r cosh s+ sinh r sinh s) = (er + e−r)(es + e−s) + (er − e−r)(es −e−s) = er+s + er−s + es−r + e−r−s + er+s − er−s − es−r + e−r−s = 2er+s +2e−r−s = 4 cosh(r + s) and correspondingly 4(sinh r cosh s + cosh r sinh s) =(er−e−r)(es+e−s)+(er +e−r)(es−e−s) = er+s+er−s−es−r−e−r−s+er+s−er−s + es−r − e−r−s = 2er+s − 2e−r−s = 4 sinh(r + s) so that

cosh(r + s) = cosh r cosh s+ sinh r sinh s (25)sinh(r + s) = sinh r cosh s+ cosh r sinh s (26)

On p.87 | sin z| = 0 ⇐⇒ sin z = 0 ⇐⇒ eiz − e−iz = 0 ⇐⇒ e−yeix −eye−ix = 0 ⇐⇒ (e−y − ey) cosx = 0 ∧ (e−y + ey) sin x = 0 ⇐⇒ x ∈πZ ∧ e−y = ey ⇐⇒ z ∈ πZ ⊂ R

2.5.3 The Geometry of the Mapping

On p.89, the image orbit under z 7→ w = cos z = 12(eiz + e−iz) of the

original particle travelling on the line y = −c is w = w(t) = 12(ei(t−ic) +

e−i(t−ic)) = 12(ec eit + e−ce−it) = 1

2(ec(cos t + i sin t) + e−c(cos t − i sin t)) =

cosh c cos t+i sinh c sin t, the parameter representation of an ellipsis with halfaxes a = cosh c and b = sinh c.

On p.90 confocal? ???

2.7 The Logarithm Function

2.7.1 Inverse of the Exponential Function

On p.98 the logarithm log z is defined to fullfil e< log z+i= log z = elog z = z =:|z| ei arg z implies e< log z = |z|, hence < log z = log |z| and = log z = arg z,hence log z = log |z|+ i arg z.

On p.99/100 as multivalued functions log(ab) = log |ab| + i arg(ab) =log |a| + log |b| + i(arg a + arg b) = log a + log b and log(a/b) = log |a/b| +i arg(a/b) = log |a| − log |b|+ i(arg a− arg b) = log a− log b.


2.7.3 General Powers

On p.102 taking zk = ek log z (30)on p.101 as the definition of a complex power we find that the principalbranch of za+ib = (r eiθ)

a+ib is given byza+ib = (eLogz)

a+ib = e(a+ib)Logz = e(a+ib)(ln r+iθ) = ra ebθ ei(aθ+b ln r).On p.102???

2.8 Averaging over Circles*

2.8.3 Averaging over Circles

On p.110 〈ez〉C = 12π∫ 2π0 ee

iθdθ = e0 = 1 implies 2π =

∫ 2π0 ee

iθdθ =∫ 2π

0 ecos θ+i sin θ dθ =∫ 2π0 ecos θ(cos sin θ + i sin sin θ) dθ, hence∫ 2π

0 ecos θ cos sin θ dθ = 2π and∫ 2π

0 ecos θ sin sin θ dθ = 0.

2.9 Exercises pp 111

1. Sketch the circle |z−1| = 1. Find (geometrically) the polar equation ofthe image of this circle under the mapping z 7→ z2. Sketch this imagecurve, which is called a cardioid.z = x+ iy ∈ C : |z − 1| = 1 is the circle around 1 with radius 1,

hence (x − 1)2 + y2 = 1 ⇐⇒ x2 + y2 − 2x = 0 = r2 − 2r cosφ ⇐⇒r = 0 or r = 2 cosφ. With z = reiφ 7→ w = r2ei2φ the picture of thecircle under the mapping z 7→ z2 is w ∈ C : w = 4 cos2(φ)ei2φ, φ ∈ R.

<z

=z

w-plane

<z

=z

z-plane

2. Consider the complex mapping z 7→ w = z−az−b . Show geometrically

that if we apply this mapping to the perpendicular bisector of the line-segment joining a and b, then the image is the unit circle. In greaterdetail, describe the motion of w round this circle as z travels along the


line at constant speed.The perpendicular bisector of the line-segment joining a and b is

z(t) = 12(a+ b)+ ti(b−a) for t ∈ R. Its image is w(t) =

12 (b−a)+ti(b−a)12 (a−b)+ti(b−a) =

1/2+it−1/2+it with |w(t)| = 1 and arg (w(t)) = 2 arctan(2t) so that w(t) :t ∈ R is the whole unit circle.

<z

=z

a

b

z(t)

z(t)− a

z(t)− b

3. Consider the family of complex mappings z 7→ Ma(z) = z−aaz−1 for a

constant. [These mappings will turn out to be fundamental to non-Euclidean geometry.] Do the following problems algebraically; in thenext chapter we will provide geometric explanations.

(i) Show thatMa[Ma(z)] = z. In other words,Ma is self-inverse.Ma[Ma(z)] =

z−aaz−1−aa z−aaz−1−1 = z−a−a(az−1)

a(z−a)−(az−1) = z(1−aa)1−aa = z.

(ii) Show thatMa(z) maps the unit circle to itself.Ma(z)Ma(z) = z−a

az−1z−aaz−1 = zz−az−az+aa

aazz−az−az+1 = 1−az−az+aaaa−az−az+1 = 1

for |z|2 = zz = 1.(iii) Show that if a lies inside the unit disc thenMa(z) maps the unit

disc to itself. Hint: Use |q|2 = qq to verify that |az−1|2−|z−a|2 =(1− |a|2)(1− |z|2).

With |az− 1|2− |z− a|2 = (az− 1)(az− 1)− (z− a)(z− a) =aazz− az− az + 1− (zz− az− az + aa) = aazz− aa− zz + 1 =(1−|a|2)(1−|z|2) we have |az−1|2−|z−a|2 = (1−|a|2)(1−|z|2) > 0for |a|, |z| < 1, i.e. for a and z lying within the unit disc, hence|az−1|2 > |z−a|2 ⇐⇒ |az−1| > |z−a| implying |Ma(z)| < 1.

4. In figure [7] on p.60 we saw that if q2 ≤ p3 then the solutions of x3 =3px+2q are all real. Draw the corresponding picture in the case q2 > p3,and deduce that one solution is real, while the other two form a complexconjugate pair.

Let q2 > p3, d =√q2 − p3 and xo = s+t for s3 = q+d and t = q−d.


Then xo ∈ R solves x3o = 3px+ 2q as

x3o = ( 3

√q + d+ 3

√q − d)3

= (q + d) + 3 3√q + d

23√q − d+ 3 3

√q + d 3

√q − d

2+ (q − d)

= 2q + 3 3√

(q + d)(q − d) 3√q + d+ 3 3

√(q + d)(q − d) 3

√q − d

= 2q + 3 3√q2 − d2 3

√q + d+ 3 3

√q2 − d2 3

√q − d

= 2q + 3p( 3√q + d+ 3

√q − d) = 3pxo + 2q.

<

=

to so xo = so + to

t1

s1

x1 = s1 + t1

t2s2

x2 = s2 + t2

<0

=

q−d

t3o

q q+d

s3oz 7→z3−→

???The other two solutions are just x1 = xoe

i2π/3 and x2 = xoei4π/3.

5. Show that the mapping z 7→ z2 doubles the angle between two rayscoming out of the origin. Use this to deduce that the lemniscate (see[9] on p.62) must self-intersect at right angles.

z = reiφ 7→ z2 = r2ei2φ so that this mapping doubles the anglebetween two rays coming out of the origin. Der Rest kommt irgendwiedaher, daß für die Foki ±1 die Lemniskate die Ortskurve der Punktemit Q(z) = (z− 1)(z+ 1) = z2− 1 = 1 ist, d.h. z2 = 2, d.h. z = ±

√2???

6. This question refers to the Cassinian curves in [9] on p.62.

(i) On a copy of this figure, sketch the curves that intersect eachCassinian curve at right angles; these are called the orthogonaltrajectories of the original family of curves.

..???(ii) Give an argument to show that each orthogonal trajectory hits

one of the foci at ±1...???

(iii) If the Cassinian curves are thought of as a geographical contourmap of the modular surface (cf. [10]) of (z2 − 1), then what isthe interpretation of the orthogonal trajectories in terms of thesurface?

..???


(iv) In Chapter 4 we will show that if two curves intersect at somepoint p 6= 0, and if the angle between them at p is φ, then theimage curves under z 7 → w = z2 will also intersect at angle φ, atthe point w = p2. Use this to deduce that as z travels out fromone of the foci along an orthogonal trajectory, w = z2 travelsalong a ray out of w = 1.

.. ???(v) Check the result of the previous part by using a computer to draw

the images under w 7→√w of (A) circles centred at w = 1; (B)

the radii of such circles... ???

(vi) Writing z = x + iy and w = u + iv, find u and v as functionsof x and y. By writing down the equation of a line in the w-plane through w = 1, show that the orthogonal trajectories of theCassinian curves are actually segments of hyperbolas.

.. ???

7. Sketch the modular surface of C(z) = (z + 1)(z − 1)(z + 1 + i). Hencesketch the Cassinian curves |C(z)| = const, then check your answerusing a computer. To answer the following questions, recall that if R(z)is a real function of position in the plane, then R(p) is a local minimumof R if R(p) < R(z) for all z 6= p in the immediate neighbourhood ofp. A local maximum is defined similarly.

(i) Referring to the previous exercise, what is the significance ofthe orthogonal trajectories of the Cassinian curves you have justdrawn?

In four figures |C(z)| is shown around the origin including ±1and around the zeroes 1, −1, and −1−i. In any case the x-y-planeis shown in black.

???.

(ii) Does |C(z)| have any local maxima?h(x, y) := |C(z)|2 = x6 + 64y6 + 2x5 + 16(3x2 + 2x + 4)y4 +

64y5 + 32(x2 + 1)y3− 4x3 + 4(3x4 + 4x3 + 4x2 + 4x+ 5)y2− 3x2 +4(x4 − 2x2 + 1)y + 2x+ 2 with ∂h

∂x= 6x5 + 32(3x+ 1)y4 + 10x4 +

64xy3 +16(3x3 +3x2 +2x+1)y2−12x2 +16(x3−x)y−6x+2 and


∂h∂y

= 384y5 + 4x4 + 64(3x2 + 2x+ 4)y3 + 320y4 + 96(x2 + 1)y2 −8x2 + 8(3x4 + 4x3 + 4x2 + 4x+ 5)y + 4 As also the graph, SAGEdoes not indicate any local maxima.???

(iii) Does |C(z)| have any non-zero local minima?As also the graph, SAGE indicates minima in (1, 0), (−1, 0)

as well as in (−1,−12)..???

(iv) If D is a disc (or indeed a more arbitrary shape), can the max-imum of |C(z)| on D occur at a point inside D, or must themaximum occur at a boundary point of D? What about the min-imum of |C(z)| on D?

Without any local maxima a maximum of |C(z)| on D occursat a boundary point of D. Any of the local minima occur exactlywhere they lie: either inside of D or on the boundary of D.

(v) Do you get the same answers to these questions if C(z) is replacedby an arbitrary polynomial? What about a complex function thatis merely known to be expressible as a power series?

..???

8. On page 62 we saw that the polar equation of the lemniscate with foci at±1 is r2 = 2 cos(2φ). In fact James Bernoulli and his successors workedwith a slightly different lemniscate having equation r2 = cos(2φ). Letus call this the standard lemniscate.

(i) Where are the foci of the standard lemniscate?For example zo = z(π6 ) = r(π6 )( cos π

6 , sinπ6) =

√cos π

3(√

32 ,

12) =√

12(√

32 ,

12) = (

√6

4 ,√

24 ) is a point of the standard lemniscate. It

satisfies |zo − k| · |zo + k| = k2 ⇐⇒ ((√

64 − k)2 + (

√2

4 )2)((√

64 +

k)2 + (√

24 )2) = k4 ⇐⇒ ((1

2 −√

62 k + k2)(1

2 +√

62 k + k2) = k4

⇐⇒ 12k

2 = 14 ⇐⇒ k = ±

√2

2 .(ii) What is the value of the product of the distances from the foci to

a point on the standard lemniscate?k2 = 1

2 .(iii) Show that the Cartesian equation of the standard lemniscate is

(x2 + y2)2 = x2 − y2.r2 = cos(2θ) = cos2 θ − sin2 θ ⇐⇒ r4 = (x2 + y2)2 =

cos2(2θ) = cos(2θ) cos2 θ − cos(2θ) sin2 θ = r2 cos2 θ − r2 sin2 θ =x2 − y2.

9. Here is an attempt [ultimately doomed] at using real methods to expandH(x) = 1

1+x2 into a power series centred at x = k, i.e., into a series of


the form H(x) = ∑∞j=0 cjX

j, where X = x− k. According to Taylor’sTheorem, cj = 1

j!H(j)(k), where H(j)(k) is the jth derivative of H at k.

(i) Show that c0 = 11+k2 and c1 = −2k

(1+k2)2 , and find c2. Note how itbecomes increasingly difficult to calculate the successive deriva-tives.

c0 = H(k) = 11+k2 andH ′(x) = − 2k

(1+x2)2 so that c1 = −2 k(1+k2)2

and H ′′(x) = −21·(1+x2)2−4x2(1+x2)(1+x2)4 = (7x2−1)(1+x2)

(1+x2)4 so that c2 =(7k2−1)(1+k2)

2(1+k2)4 .

(ii) Recall (or prove) that the nth derivative of a product AB of twofunctions A(x) and B(x) is given by Leibniz’s rule (AB)(n) =∑nj=0

(nj

)A(j)B(n−j).

By applying this result to the product (1 + x2)H(x), deduce that(1 + k2)H(n)(k) + 2nkH(n−1)(k) + n(n− 1)H(n−2)(k) = 0.

Leibniz’ rule can be proved by induction in n.Applying it to 1 ≡ (1 + x2)H(x) yields 0 = ((1 + x2)H(x))(n) =∑nj=0

(nj

)(1+x2)(j)H(n−j)(x) = (1+x2)H(n)(x)+n 2xH(n−1)(x)+

2n(n− 1)H(n−2)(x).Because the coefficients in this recurrence relation depend on n,we cannot solve it using the technique of Ex.30 on p.50.

(iii) Deduce from the previous part that the recurrence relation for thecj’s is (1 + k2)cn + 2kcn−1 + cn−2 = 0 which does have constantcoefficients.

From part (ii) we have 0 = (1 + k2)H(n)(k)n! + 2nkH(n−1)(k)

n! +n(n− 1)H(n−2)(k)

n! = (1 + k2)cn + 2kcn−1 + cn−2.(iv) Solve this recurrence relation, and hence recover the result 1

1+z2 =12i∑∞j=0 (

1(−i−k)j+1 − 1

(i−k)j+1 )(z − k)j (17) on p.76.(1 + k2)cn + 2kcn−1 + cn−2 = 0 ⇒ cn = − 1

1+k2 (2kcn−1 + cn−2)or cn + pcn−1 + qcn−2 = 0 with p = 2k

1+k2 and q = 11+k2 . Now,

analogously to Ex.30 on p.50 we have that cn = zn will solvethis recurrence relation if z2 + pz + q = 0, hence z1,2 = −p

2 ±12√p2 − 4q = − k

1+k2 ±√

k2

(1+k2)2 − 1+k2

(1+k2)2 = − k1+k2 ±

√−1

(1+k2)2 =−k±i1+k2 . Then cn = Azn1 +Bzn2 solves the recurrence relation becausez2ν +pzν +q = 0 ⇐⇒ znν +pzn−1

ν +qzn−2ν = 0 ⇐⇒ (1+k2)Cznν +

2kCzn−1ν +Czn−2

ν = 0 for ν = 1, 2 and all C ∈ R ⇐⇒ (1+k2)cn+2kcn−1 + cn−2 = 0. Now, co = 1

1+k2 and c1 = −2k(1+k2)2 result in the

system of linear equations A + B = 11+k2 and A−k+i

1+k2 + B−k−i1+k2 =−2k

(1+k2)2 which by setting A′ := (1 + k2)A and B′ := (1 + k2)B canbe simplified to A′+B′ = 1 and (k− i)A′+ (k+ i)B′ = 2k giving


A′ = 12(1 + ik) and B′ = 1

2(1 − ki) and hence A = 12

1+ki1+k2 and

B = 12

1−ki1+k2 . Combined we get cj = 1

21+ki1+k2 (−k+i

1+k2 )j + 1

21−ki1+k2 (−k−i1+k2 )

j

which after some manipulation turns out to be identical to cj =12i(

1(−i−k)j+1 − 1

(i−k)j+1 ).

10. Reconsider the series 11+x2 = ∑∞

j=0sin((j+1)φ)√

1+k2j+1 Xj (18) on p.77 withX = x− k, φ = arg(i− k) and k ∈ R.

(i) Show that we recover the correct series (missing the odd powersof x) when the centre k of the series is at the origin.

If k = 0 then φ = arg(i) = π2 , hence sin ((j + 1)π2) = 1 for

j ∈ 4Z, sin ((j + 1)π2) = 0 for j ∈ (1 + 4Z) ∪ (3 + 4Z), andsin ((j + 1)π2) = −1 for j ∈ 2 + 4Z so that ∑∞j=0

sin((j+1)π/2)√1+k2j+1 xj =∑∞

n=0(x4)n − x2∑∞n=0(x4)n = 1

1−x4 − x2

1−x4 = 1−x2

(1−x2)(1+x2) = 11+x2 .

(ii) Find a value of k such that the series is missing all the powers Xn,where n = 2, 5, 8, 11, 14, . . .. Check your answer using a computer.

sin((n + 1)φ) vanishes for all n ∈ 2 + 3No, i.e. for n + 1 ∈3N only if (n + 1)φ ∈ 3φN = πN ⇐⇒ φ = arg(i − k) = π

3implying k = −

√3

3 . SAGE’s taylor(f,x,k,14) produces f(x) ≈1

23887872

√3(3x+

√3)13+ 1

3981312 (3x+√

3)12− 1331776 (3x+

√3)10

− 1165888

√3(3x+

√3)9 + 1

13824

√3(3x+

√3)7 + 1

2304 (3x+√

3)6−1

192 (3x+√

3)4 − 196

√3(3x+

√3)3 + 1

8

√3(3x+

√3) + 3

4 with3x +

√3 = 3(x − k) and vanishing terms of (x − k)2, (x − k)8,

(x− k)11 and (x− k)14.

11. Show that each of the following series has the unit circle as its circle ofconvergence, then investigate the convergence on the unit circle. Youcan guess the correct answers by ”drawing the series” in the manner of[18] on p.80.

(i) ∑∞n=0 zn

|∑∞n=o zn| ≤ ∑∞

n=o |z|n = limm→∞∑mn=o r

n = limm→∞1−rm+1

1−r= 1

1−r which exists if and only if r = |z| < 1. Obviously, the serieswith partial sums Sn does not converge for z = 1 where Sn →∞,for z = −1 or z = ±i where Sn oscillates. What is more, Snoscillates for any z 6= 1 with |z| = 1, so that the series divergeson the unit circle.

(ii) ∑∞n=11nzn

|∑∞n=11nzn| ≤ ∑∞

n=11nrn = ∑∞

n=1∫ ro x

n−1 dx =∫ ro

∑∞n=1 x

n−1 dx=∫ ro

∑∞n=0 x

n dx =∫ ro

11−x dx = − ln(1− r) which exists as long as


r = |z| < 1. Due to Abel’s criterion the series converges e.g. forz = −1 whereas it does not converge for z = 1 (harmonic series).

(iii) ∑∞n=11n2 z

n

|∑∞n=11n2 z

n| ≤ ∑∞n=1|z|nn2 ≤

∑∞n=1

1n2 = π2

6 for any z with |z| ≤ 1so the series converges also on the unit circle.Because of 1

n2 <1

n(n−1) a majorante 1 + 12·1 + . . . + 1

n(n−1) = 1 +(1− 1

2) + (12 −

13) + . . .+ ( 1

n−1 −1n) = 2− 1

n→ 2 converges.

[By virtue of (29), note that the second series is − log(1− z).]

12. Consider the geometric series P (z) = ∑∞j=0 z

j, which converges to 11−z

inside the unit disc. The approximating polynomials in this case arePm(z) = ∑m

j=0 zj.

(i) Show that the error Em(z) = |P (z)−Pm(z)| is given by Em(z) =|z|m+1

|1−z| .Em(z) = |P (z) − Pm(z)| =

∣∣∣∑∞j=m+1 zj∣∣∣ = |zm+1|

∣∣∣∑∞j=0 zj∣∣∣ =

|z|m+1

|1−z| .

(ii) If z is any fixed point in the disc of convergence, what happensto the error as m tends to infinity?

limm→∞Em(z) = 0 for fixed z with |z| < 1.(iii) If we fix m, what happens to the error as z approaches the bound-

ary point z = 1?limz→1Em(z) =∞ for fixed m.

(iv) Suppose we want to approximate this series in the disc z ∈ C :|z| ≤ 0.9, and further suppose that the maximum error we willtolerate is ε = 0.01. Find the lowest degree polynomial Pm(z) thatapproximates P (z) with the desired accuracy throughout the disc.

As |1−z| ≥ 0.1 we have Em(z) ≤ 10|z|m+1 ≤ 10 ·0.9m+1 ≤ 1100

⇐⇒ 0.9m+1 = e(m+1) ln 0.9 ≤ e−3 ln 10 = 11000 ⇐⇒ (m+ 1) ln 0.9 ≤

−3 ln 10 ⇐⇒ m+ 1 ≥ −3 ln 10ln 0.9 ≈ 65.563.

13. We have seen that if we set Pn(z) = zn, then the representation of acomplex function f(z) as an infinite series ∑∞n=0 cnPn(z) (i.e., a powerseries) is unique. This is not true, however, if Pn(z) is just any odd setof polynomials. The following example is taken (and corrected) fromBoas [1987, p. 33]. Defining Po(z) = −1, and Pn(z) = zn−1

(n−1)! −zn

n! forn = 1, 2, 3, . . ., show that −2Po − P1 + P3 + 2P4 + 3P5 + . . . = ez =P1 + 2P2 + 3P3 + 4P4 + . . ..

P1 +2P2 +3P3 +4P4 +. . . = ( z0

0! −z1

1! )+2( z1

1! −z2

2! )+3( z2

2! −z3

3! )+. . . =(1−z)+(2z−z2)+(3

2z2− 1

2z3)+(2

3z3− 1

6z4)+. . . = 1+z+ 1

2z2+ 1

6z3+. . . =


ez as well as −2Po−P1 +P3 + 2P4 + 3P5 + . . . = −2(−1)− ( z0

0! −z1

1! )+( z

2

2!−z3

3! )+2( z3

3!−z4

4! )+. . . = 2−(1−z)+(12z

2− 16z

3)+(13z

3− 112z

4)+. . . =1 + z + 1

2z2 + 1

6z3 + . . . = ez.

14. Consider two power series, P (z) = ∑∞j=0 pjz

j and Q(z) = ∑∞j=0 qjz

j,which have approximating polynomials Pn(z) = ∑n

j=0 pjzj andQm(z) =∑m

j=0 qjzj. If the radii of convergence of P (z) and Q(z) are R1 and R2

then both series are uniformly convergent in the disc z ∈ C : |z| < r,where r < minR1, R2. Thus if ε is the maximum error we will tol-erate in this disc, we can find a sufficiently large n such that Pn(z) =P (z) + E1(z) and Qn(z) = Q(z) + E2(z), where the (complex) errorsE1(z) and E2(z) both have lengths less than ε. Use this to show that bytaking a sufficiently high value of n we can approximate P (z) + Q(z)and P (z) ·Q(z) with arbitrarily high precision using Pn(z)+Qn(z) andPn(z) ·Qn(z), respectively.

There are n1, n2 ∈ N such that |P (z) − Pn(z)| ≤ |E1(z)| < 12ε and

|Q(z)−Qn(z)| ≤ |E2(z)| < 12ε for any n ≥ maxn1, n2 and any |z| < r,

then |(P (z) + Q(z)) − (Pn(z) + Qn(z))| ≤ |P (z) − Pn(z)| + |Q(z) −Qn(z)| < ε for any such n and z.P and Q are continous, hence Pmax = max|z|≤r |P (z)| <∞ and Qmax =max|z|≤r |Q(z)| < ∞. Now, there exist n1, n2 ∈ N such that |Q(z) −Qn(z)| < 1

2ε1

Pmaxfor any n > maxn1, n2 and any |z| < r as well as

|P (z) − Pn(z)| · |Qn(z)| ≤ 12ε

1Qmax

again for any n > maxn1, n2 andany |z| < r. Together we have |(P (z) · Q(z)) − (Pn(z) · Qn(z))| =|P (z) · (Q(z) − Qn(z)) + (P (z) − Pn(z)) · Qn(z)| ≤ |P (z)| · |Q(z) −Qn(z)|+ |P (z)− Pn(z)| · |Qn(z)| < ε

2 + ε2 = ε.

15. Give an example of a pair of origin-centred power series, say P (z) andQ(z), such that the disc of convergence for the product P (z) · Q(z) islarger than either of the two discs of convergence for P (z) and Q(z).[Hint: think in terms of rational functions, such as z2/(5 − z)3, whichare known to be expressible as power series.]

Let P (z) = z2/(5 − z)3 and Q(z) = (5 − z)3/z2 both be expandedsay around 2, then R1 = 3 because 3 is the distance to the singularityin 5, and R2 = 2 because 2 is the distance to the singularity in 0, butthe product P (z) ·Q(z) ≡ 1 has convergence radius ∞.

16. Our aim is to give a combinatorial explanation of the Binomial Theorem(14) for all negative integer values of n. The simple yet crucial first stepis to write n = −m and to change z to −z. Check that the desiredresult (14) now takes the form (1 − z)−m = ∑∞

r=0 crzr, where cr is the

binomial coefficient cr =(m+r−1

r

)(42). [Note

that this says that the coefficients cr are obtained by reading Pascal’s


triangle diagonally, instead of horizontally.] To begin to understandthis, consider the special case m = 3. Using the geometric series for(1−z)−1, we may express (1−z)−3 as [1+z+z2 +z3 + ...] · [1+z+z2 +z3 + ...] · [1 + z + z2 + z3 + ...], where · simply denotes multiplication.Suppose we want the coefficient c9 of z9. One way to get z9 is to takez3 from the first bracket, z4 from the second, and z2 from the third.

(i) Write this way of obtaining z9 as the sequence zzz · zzzz · zz ofnine z’s and two ·’s, where the latter keep track of which powerof z came from which bracket. [I got this nice idea from myfriend Paul Zeitz.] Explain why c9 is the number of distinguishablerearrangements of this sequence of eleven symbols. Be sure toaddress the meaning of sequences in which a dot comes first, last,or is adjacent to the other dot.

Any product of three factors resulting in z9 can be thoughtof an arrangement of the nine z’s and the two dots. A dot inthe first or in the last place indicates that either the first or thelast factor is 1. Two adjacent dots means that at least the middlefactor is 1.

(ii) Deduce that c9 =(

119

), in agreement with (42).

There are c9 =(

112

)=(

119

)arrangements of two (or nine)

symbols in eleven places.(iii) Generalize this argument and thereby deduce cr =

(m+r−1

r

)(42).

In case of m factors there are m − 1 dots to be put down inany of r+m−1 places when computing cr, hence cr =

(m+r−1m−1

)=(

m+r−1r

).

17. Here is an inductive approach to the result of the previous exercise.

(i) Write down the first few rows of Pascal’s triangle and circle thenumbers

(53

),(

42

),(

31

),(

20

). Check that the sum of these numbers

is(

63

)= 20. Explain this.

The first six rows of the Pascal triangle are(00

)(10

) (11

)(20

) (21

) (22

)(30

) (31

) (32

) (33

)(40

) (41

) (42

) (43

) (44

)(50

) (51

) (52

) (53

) (54

) (55

)(63

)=

11 1

1 2 11 3 3 1

1 4 6 4 11 5 10 10 5 1

20

Summing the first n elements on the 1st Nebendiagonale gives∑n−1j=0

(jj

)= n =

(n1

).

Summing the first n elements on the 2nd Nebendiagonale gives


∑n−1j=0

(j+1j

)= ∑n

j=1 j = n(n+1)2 =

(n+1

2

).

Summing the first n elements on the 3rd Nebendiagonale gives∑n−1j=0

(j+2j

)= ∑n

j=1j(j+1)

2 = 12∑nj=1 j

2 + 12∑nj=1 j = n(n+1/2)(n+1)

6 +12n(n+ 1) = n(n+1)(2n+1)+3n(n+1)

12 = (n+2)(n+1)n6 =

(n+2

3

).

Alternatively,(

63

)= |S ⊂ e1, ..., e6 : |S| = 3| = |S ⊂

e1, ..., e6 : |S| = 3, e6 6∈ S| + |S ⊂ e1, ..., e6 : |S| = 3, e6 ∈S| =

(53

)+ |S ⊂ e1, ..., e5 : |S| = 2| with |S ⊂ e1, ..., e5 :

|S| = 2| = |S ⊂ e1, ..., e5 : |S| = 2, e5 6∈ S| + |S ⊂e1, ..., e5 : |S| = 2, e5 ∈ S| =

(42

)+ |S ⊂ e1, ..., e4 : |S| = 1|

with |S ⊂ e1, ..., e4 : |S| = 1| = |S ⊂ e1, ..., e4 : |S| =1, e4 6∈ S| + |S ⊂ e1, ..., e4 : |S| = 1, e4 ∈ S| =

(31

)+ 1 so

that(

63

)=(

53

)+(

42

)+(

31

)+(

20

).

(ii) Generalize your argument to show that(nr

)= ∑r

j=0

(n−1−jr−j

)=∑r

`=0

(n−1−r+`

`

)=(n−1r

)+(n−2r−1

)+ . . .+

(n−1−r

0

).

Use the alternative idea of part (i) and induction(nr

)=(n−1r

)+(

n−1r−1

)=(n−1r

)+(n−2r−1

)+(n−2r−2

)=(n−1r

)+(n−2r−1

)+ . . .+

(n−1−r

0

).

(iii) Assume that (1−z)−M = ∑∞r=0

(M+r−1

r

)zr holds for some positive

integer M . Now multiply this series by the geometric series for(1 − z)−1 to find (1 − z)−(M+1). Deduce that the binomial seriesis valid for all negative integer powers.

Assuming (1−z)−M = ∑∞r=0

(M+r−1

r

)zr we have (1−z)−(M+1) =

(1−z)−M(1−z)−1 =∞∑r=0

(M+r−1

r

)zr∑∞s=0 z

s =∞∑t=0zt∑r+s=t

(M+r−1

r

)= ∑∞

t=0 zt∑t

r=0

(M+r−1

r

)= ∑∞

r=0

(M+r−1

r

)zr.

18. The basic idea of the following argument is due to Euler. Initially,let n be any real (possibly irrational) number, and define B(z, n) =∑∞r=0

(nr

)zr where

(nr

):= n(n−1)···(n−r+1)

r! and(n0

):= 1. We know from

elementary algebra that if n is a positive integer then B(z, n) = (1+z)n.To establish the Binomial Theorem (14) for rational powers, we mustshow that if p and q are integers then B(z, p

q) is the principal branch

of (1 + z)q/q.

(i) With a fixed value of n, use the ratio test to show that B(z, n)converges in the unit disc, |z| < 1.

R = limn→∞|cn||cn+1| = lim

n→∞

n(n−1)···(n−r+1)r!

(n+1)n···(n−r+2)r!

= limn→∞n−r+1n+1 = 1.

(ii) By multiplying the two power series, deduce that B(z, n)·B(z,m)= ∑∞

r=0Cr(n,m)zr where Cr(n,m) = ∑rj=0

(nj

)(mr−j

).


B(z, n)·B(z,m) = ∑∞r=0

(nr

)zr·∑∞s=0

(ms

)zs =

∞∑t=0

zt∑r+s=t

(nr

)(ms

)= ∑∞

r=0Cr(n,m)zr where Cr(n,m) = ∑rj=0

(nj

)(mr−j

).

(iii) If m and n are positive integers, then show thatB(z, n) ·B(z,m) = B(z, n+m) (43),and deduce that Cr(n,m) =

(n+mr

). But Cr(n,m) and

(n+mr

)are

simply polynomials in n and m, and so the fact that they agree atinfinitely many values of m and n [positive integers] implies thatthey must be equal for all real values of m and n. Thus the keyformula (43) is valid for all real values of m and n.

If m and n are positive integers, then B(z, n) = ∑∞r=0

(nr

)zr =∑n

r=0

(nr

)zr = (1 + z)n so that B(z, n) · B(z,m) = (1 + z)n+m =∑n+m

r=0

(n+mr

)zr = ∑n+m

r=0 Cr(n,m)zr where Cr(n,m) =(n+mr

).

(iv) By substituting n = −m in (43), deduce the Binomial Theoremfor negative integer values of n.

By (ii) we have B(z,−m) · B(z,m) = B(z,−m)(1 + z)m =B(z, 0) = 1, hence B(z,−m) = ∑−m

r=0

(−mr

)zr = (1 + z)−m.

(v) Use (43) to show that if q is an integer then [B(z, 1/q)]q = (1+z).Deduce that B(z, p/q) is the principal branch of (1 + z)p/q.

Assuming that (43) holds for rational second arguments also ???we have [B(z, 1/q)]q = B(z, 1) = 1+z andB(z, p/q) = [B(z, 1/q)]p= (1 + z)p/q. principal branch? ???

19. Show that the ratio test cannot be used to find the radius of convergenceof the power series 1

1+x2 = ∑∞j=0

sin((j+1)φ)√1+k2j+1 Xj (18)

on p.77 with X = x − k, φ = arg(i − k) = arctan 1−k and k ∈ R. Use

the root test to confirm that R =√

1 + k2.For cj = we have on one hand side R = limj→∞

| sin(jφ)|√

1+k2j+1

| sin((j+1)φ)|√

1+k2j=

√1 + k2 limj→∞

| sin(jφ)|| sin((j+1)φ)| which alledgedly does not exist and on the

other hand R = limj→∞1

j√|cj |

=√

1 + k2 limj→∞1

j√| sin(jφ)|

which alled-gedly does. ???

20. Show that if m and n are integers, then∫ 2πo cos(mθ) cos(nθ) dθ vanishes

unless m = n, in which case it equals π. Likewise, establish a similarresult for

∫ 2πo sin(mθ) sin(nθ) dθ . Use these facts to verify (19), at least

formally.If n 6= m then

∫ 2πo cos(mθ) cos(nθ) dθ =

∫ 2πo

cos((n−m)θ)+cos((n+m)θ)2 dθ

= sin((n−m)θ)2(n−m)

∣∣∣2πo

+ sin((n+m)θ)2(n+m)

∣∣∣2πo

= 0. If n = m then∫ 2πo cos2(mθ) dθ =

12n(nθ + sin(nθ) · cos(nθ))

∣∣∣2πo

= π.


If n 6= m then∫ 2πo sin(mθ) sin(nθ) dθ =

∫ 2πo

cos((n−m)θ)−cos((n+m)θ)2 dθ =

sin((n−m)θ)2(n−m)

∣∣∣2πo− sin((n+m)θ)

2(n+m)

∣∣∣2πo

= 0. If n = m then∫ 2πo sin2(mθ) dθ =

12n(nθ − sin(nθ) · cos(nθ))

∣∣∣2πo

= π.

21. Do the following problems by first substituting z = reiφ into the powerseries for ez, then equating real and imaginary parts.

(i) Show that the Fourier series for [cos(sin θ)]ecos θ is ∑∞n=ocos(nθ)n! ,

and write down the Fourier series for [sin(sin θ)]ecos θ.ez = ere

iφ = er cosφ+ir sinφ = er cosφ( cos(r sinφ) + i sin(r sinφ)).Now, [cos(sin θ)]ecos θ is 2π-periodic and an even function in θso that all bn vanish. Setting r = 1 in ere

iθ = ∑∞n=0

rn

n! einθ =∑∞

n=0rn

n! ( cos(nθ) + i sin(nθ)) we get <eeiθ = [cos(sin θ)]ecos θ =∑∞n=0

1n! cos(nθ).

(ii) Deduce that∫ 2πo ecos θ[cos(sin θ)] cos(mθ) dθ = π

m! , where m is apositive integer.

1π

∫ 2πo ecos θ[cos(sin θ)] cos(mθ) dθ = am = 1

m! .(iii) By writing x = r√

2 , find the power series for f(x) = ex sin x.With x = r√

2 = r√

22 = r sin θ4 = r cos θ for θ = π

4 we have=ereiθ = =ereiθ = ex sin x = ∑∞

n=0rn

n! sin(nθ) = ∑∞n=0

(√

2x)nn! sin(nπ4 )

= ∑∞n=0 cnx

n for cn =√

2nn! sin(nπ4 ).

(iv) Check the first few terms of the series for f(x) by multiplying theseries for ex and sin x.

ex sin x = (1 + x+ 12x

2 + 16x

3 + . . . )(x− 16x

3 + 1120x

5 ∓ . . . ) =

(∑∞n=o

xn

n! )(∑∞m=o

(−1)m(2m+1)!x

2m+1) =∞∑`=ox`∑n+(2m+1)=`

1n!

(−1)m(2m+1)! =∑∞

`=o c`x`, hence co = 0, c1 = 1, c2 = 1, c3 = −1

6 + 12 = 1

3 ,c4 = −1

6 + 16 = 0, . . .

(v) Calculate the nth derivative f (n)(0) for f(x) = ex sin x usingdn

d tn (eat sin(bt)) =

√a2 + b2neat sin (bt+ n arctan b

a) (14)on p.22 of Chapter 1. By using these derivatives in Taylor’s The-orem, verify your answer to part (iii).

f (n)(0) =√

2net sin(t+ nπ/4)∣∣∣t=0

=√

2n sin(nπ/4), so thatco = f(0)

0! = 0, c1 = f ′(0)1! =

√2√

22 = 1, c2 = f ′′(0)

2! = 12

√22 = 1,

c3 = f ′′′(0)3! = 1

6

√23 sin(3π/4) =

√2

3

√2

2 = 13 , c4 = 0, . . .

22. Reconsider the formula ez = limn→∞ Pn(z) with Pn(z) = (1 + z/n)n.

(i) Check that Pn(z) is the composition of a translation by n, followedby a contraction by 1/n, followed by the power mapping z 7→ zn.


Pn(z) = (1 + z/n)n = ( 1n(n+ z))n.

(ii) Referring to figure [4] on p.58, use the previous part to sketch theimages under Pn(z) of circular arcs centred at −n, and of raysemanating from −n.

z(t) = −n + reit is a circular arc centred at −n which ismapped to Pn(z(t)) = ( 1

n(n + −n + reit))n = ( rne

it)n, an arc

centred at the origin of 1nof the original radius and, if so, with

several revolutions.z(t) = −n+ reiφ is a ray emanating from −n which is for fixed nmapped to Pn(z(t)) = ( 1

n(n+−n+ teiφ))n = ( tne

iφ)n = tn

nneinφ, a

ray emanting from the origin with n fold the angle of the originalray.

(iii) Let S be an origin-centred square (say of unit side) in the z-plane.With a large value of n, sketch just those portions of the arcs andrays (considered in the previous part) that lie within S.

<z

=z

(iv) Use the previous two parts to qualitatively explain figure [19] onp.81.

. ???

23. If you did not do so earlier, sketch the image of a vertical line x = kunder z 7→ w = cos z = 1

2(eiz + e−iz) by drawing the analogue of [26].Deduce that the asymptotes of this hyperbola are argw = ±k. Checkthis using the equation of the hyperbola.

z(t) = k + it 7→ cos (z(t)) = 12(e

iz(t) + e−iz(t)) = 12(e

i(k+it) +e−i(k+it)) = 1

2(e−t+ik+et−ik) = 1

2(e−t(cos k+i sin k)+et(cos k−i sin k)) =

et+e−t2 cos k − i e

t−e−t2 sin k = cosh t cos k − i sinh t sin k, hence x(t) =

(cos k) cosh t =: a cosh t and y(t) = −(sin k) sinh t =: b sinh t whichis the parameter representation of the hyperbola x2

a2 − y2

b2= 1, cp

https://en.wikipedia.org/wiki/Hyperbola.

https://en.wikipedia.org/wiki/Hyperbola


24. Consider the multifunction f(z) =√z − 1 3

√z − i.

(i) Where are the branch points and what are their orders?1 is a branch point of first order, i is a branch point of second

order, cp. p.92.(ii) Why is it not possible to define branches using a single branch

cut of the type shown in [35b]?.???

(iii) How many values does f(z) have at a typical point z? Find andthen plot all the values of f(0).

Since√z − 1 and 3

√z − i are multivalued function of order

one and two respectively f has typically six values.

<z

=z

f(0) = (±1)( 3√−i) = ±ei(−π/2+m2π)/3 for m ∈ N, hence f(0) ∈

±e−iπ/6,±ei(−π/2+2π)/3,±ei(−π/2+4π)/3.(iv) Choose one of the values of f(0) which you have just plotted, and

label it p. Sketch a loop L that starts and ends at the origin suchthat if f(0) is initially chosen to be −1, then as z travels along Land returns to the origin, f(z) travels along a path from −1 to p.Do the same for each of the other possible values of f(0).

.???

25. Describe the branch points of the function f(z) = 1√1−z4 . What is

the smallest number of branch cuts that may be used to obtain single-valued branches of f(z)? Sketch an example of such cuts. [Remark:This function is historically important, owing to the fact (Ex.20, p.214)that

∫f(x) dx represents the arc length of the lemniscate. This integral

(the lemniscatic integral) cannot be evaluated in terms of elementaryfunctions – it is an example of a new kind of function called an ellipticintegral. See Stillwell [1989, Chap. 11], for more background anddetail.]


f(z) = 1√1−z4 = 1√

1+iz√

1−iz√

1−z√

1+z = 1√−i√i−z√i√−i−z

√1−z√

1+z =1√

i−z√−i−z

√1−z√

1+z , hence f has singularities in ±i and ±1 which alsoare branch points??? check =z = ±<z as branch cuts ... ???

26. For each function f(z) below, find and then plot all the branch pointsand singularities. Assuming that these functions may be expressed aspower series centred at k [in fact they can be], use the resultIf a complex function or a branch of a multifunction can be expressedas a power series, the radius of convergence is the distance to thenearest singularity or branch point.

(27)

on p.96 to verify the stated value of the radius of convergence R.

(i) If f(z) = 1eπz−1 and k = (1 + 2i), then R = 1.

Any z with eπz = 1 is a singularity, i.e. any z = 0,±2i,±4i, . . .,or z ∈ 2iZ. For k = 1+2i then 1 = |k−2i| ≤ |k−2iZ| is minimalin 2i so that R = 1.

(ii) If f(z) is a branch of 5√z4 − 1 and k = 3i, then R = 2.

All four fourth roots of unity 1, i, −1 and −i are branch points.Hence R = minu∈±1,±i |u− k| = |i− 3i| = 2.

(iii) If f(z) is a branch of√z−iz−1 and k = −1, then R =

√2.

1 is a singularity and i is a branch point. Hence R = min|i−k|, | − 1− k| = min

√2, 2 =

√2.

27. Until Euler cleared up the whole mess, the complex logarithm was asource of tremendous confusion. For example, show that log(z) andlog(−z) have no common values, then consider the following argumentof John Bernoulli: log[(−z)2] = log[z2] ⇒ 2 log(−z) = 2 log(z) ⇒log(−z) = log(z). What is wrong with this argument?!

Assume that there exists z with log(−z) = log z ⇒ log(−1) =log −z

z= log(−z)− log z = 0, which is not possible.

From log[(−z)2] = log[z2] one can only derive log(−z) = log z mod i2π.

28. What value does zi take at z = −1 if we start with the principal valueat z = 1 (i.e., 1i = 1), and then let z travel one and a half revolutionsclockwise round the origin?

With f(z) = zi = ei log z = ei(log |z|+i arg z) = ei log |z|e− arg z and z =z(θ) = eiθ for θ = 0, ...,−2π, ...,−3π we have f(z(θ)) = e− arg(z(θ)) =e−θ with f(−1) = . . . , f(z(−3π)), f(z(−π)), . . . and thus after oneand a half clockwise revolution f(−1) = f(z(−3π)) = e3π.

29. In this exercise you will see that the "multifunction" kz is quite differentin character from all the other multifunctions we have discussed. Forinteger values of n, define `n = [Log(k) + 2mπi].


(i) Show that the ”branches” of kz are e`nz.e`nz = (eLog(k)+2mπi)

z = kz.(ii) Suppose that z travels along an arbitrary loop, beginning and

ending at z = p. If we initially choose the value e`2p for kz, thenwhat value of kz do we arrive at when z returns to p? Deducethat kz has no branch points.

.???

Since we cannot change one value of kz into another by travelling rounda loop, we should view its ”branches” . . . , e`−1z, e`oz, e`1z, . . . as an infi-nite set of completely unrelated single-valued functions.

30. Show that all the values of ii are real! Are there any other points zsuch that zi is real?

With zk = ek log z we have ii = ei(log |i|+i arg(i)) = e−(π/2+2nπ) ∈ R.For example, (−i)i = ei log(−i) = ei(log |−i|−iπ/2+2nπi) = eπ/2−2nπ ∈ R.

31. In the case of a real variable, the logarithmic power series was originallydiscovered [see next exercise] as follows. First check that ln(1 +X) canbe written as

∫Xo [1/(1 + x)] dx, and then expand 1

1+x as a power seriesin x. Finally, integrate your series term by term. [Later in the bookwe will be able to generalize this argument to the complex plane.]

ln(1 +X) =∫X

0 [ 11+x ] dx =

∫X0∑∞n=o(−x)n dx = ∑∞

n=o∫X

0 (−x)n dx =∑∞n=o

−1n+1(−x)n+1

∣∣∣X0

= ∑∞n=o

−1n+1(−X)n+1 = −∑∞n=1

(−1)nnXn.

32. Here is another approach to the logarithmic power series. As before,let L(z) = Log(1 + z). Since L(0) = 0, the power series for L(z) mustbe of the form L(z) = az + bz2 + cz3 + dz4 + . . .. Substitute this intothe equation 1 + z = eL = 1 +L+ 1

2!L2 + 1

3!L3 + 1

4!L4 + . . ., then find a,

b, c, and d by equating powers of z. [Historically the logarithmic seriescame first – both Mercator and Newton discovered it using the methodin the previous exercise – then Newton reversed the reasoning of thepresent exercise to obtain the series for ex. See Stillwell [1989, p.108].]

1 + z = 1 + (az + bz2 + cz3 + dz4) + 12(a2z2 + 2abz3 + 2acz4 +

b2z4) + 16(a3z3 + 2a2bz4 + a2bz4) + 1

24a4z4 ⇐⇒ a = 1, 0 = b+ 1

2a2 = 1

2⇒ b = −1

2 , 0 = c + ab + 16a

3 = c − 12 + 1

6 = c − 13 ⇒ c = 1

3 , and0 = d+ac+ 1

2b2 + 1

2a2b 1

24a4 = d+ 1

3 + 18 −

14 + 1

24 = d+ 8+3−6+124 = d+ 6

24⇒ d = −1

4 .

33. (i) Use [26] to discuss the branch points of the multifunction arccos(z)..???

(ii) Rewrite the equation w = cos z as a quadratic in eiz. By solvingthis equation, deduce that arccos(z) = −i log[z+

√z2 − 1]. [Why


do we not need to bother to write ± in front of the square root?]w = cos z = 1

2(eiz + e−iz) ⇐⇒ 2weiz = (eiz)

2 + 1 ⇐⇒v2 − 2wv + 1 = 0 for v = eiz ⇐⇒ v1,2 = w ±

√w2 − 1 where

the multivalued function square root takes care of the ± ⇐⇒iz = log[w +

√w2 − 1] ⇐⇒ z = −i log[w +

√w2 − 1].

(iii) Show that as z travels along a loop that goes once round either1 or −1 (but not both), the value of [z +

√z2 − 1] changes to

1/[z +√z2 − 1].

. ???(iv) Use the previous part to show that the formula in part (ii) is in

accord with the discussion in part (i).. ???

34. Write down the origin-centred power series for (1 − cos z). Use theBinomial Theorem to write down the power series (centred at Z = 0)for the principal branch of

√1− Z, then substitute Z = (1 − cos z).

Hence show that if we choose the branch of√

cos z that maps 0 to 1,then

√cos z = 1− 1

4z2− 1

96z4− 19

5760z6 . . .. Verify this using a computer.

Where does this series converge?Inserting 1 − cos z = 1 − (1 − 1

2z2 + 1

24z4 . . . ) = 1

2z2 − 1

24z4 . . . in

√1− Z = (1−Z)1/2 = ∑∞

r=0

(1/2r

)(−Z)r = ∑∞

r=0( 1

2 )( 12−1)·...·( 1

2−r+1)r! (−Z)r

(cp (1 + z)n = 1 + nz + n(n−1)2! z2 + n(n−1)(n−2)

3! z3 + . . . (14) on p.73)gives

√cos z = 1 − 1

2(1 − cos z) +12 ( 1

2−1)2! (1 − cos z)2 −

12 ( 1

2−1)( 12−2)

3! (1 −cos z)3 + . . . = 1 − 1

2(12z

2 − 124z

4 + 1720z

6 . . . ) − 18(

12z

2 − 124z

4 . . . )2 −

12 (− 1

2 )(− 32 )

6 (12z

2 . . . )3 + . . . = 1 − 1

4z2 + ( 1

48 −132)z

4 + ( − 12

16! + 2

812

124 −

12

12

32

618)z

6 + . . . = 1− 14z

2 − 196z

4 − 195760z

6 . . ..

with R = π2 when expanding around 0 since the next singularities or

branch points are ±π2 .


35. What value does f(z) = zsin z approach as z approaches the origin? Use

the series for sin z to find the first few terms of the origin-centred powerseries for z

sin z . Check your answer using a computer. Where does thisseries converge?

z = f(z) sin z = (co + c1z + c2z2 + . . . )(z − 1

6z3 + 1

120z5 ∓ . . . ) =

coz − co6 z

3 + co120z

5 + c1z2 − c1

6 z4 + c2z

3 − c26 z

5 + c3z4 + c4z

5 + . . . =coz+ c1z

2− (16co− c2)z3− (1

6c1− c3)z4 +( 1120co−

16c2 + c4)z5 + . . ., hence

f(z) = 1 + 0 · z+ 16z

2 + 0 · z3 + 7360z

4 . . . which is at least consistent witheveness of f . Alternatively we compute f(0) = 1, f ′(z) = sin z−z cos z

sin2 z

with f ′(0) = 0, f ′′(z) = z−sin(2z)+z cos2 zsin3 z

with f ′′(0) = 13 , f

′′′(z) =3(1+cos2 z) sin z−5z cos z−z cos3 z

sin4 zwith f ′′′(0) = 0, and – again due to SAGE –

f (4)(z) = z sin4 z+4 cos z sin3 z−20z sin2 z−12 sin(2z)+24zsin5 z

with f (4)(0) = 715 .

where the convergence radius of the expansion around 0 is R = π thedistance to the next singularities at ±π.

36. By considering Log(1 + ix), where x is a real number lying between ±1deduce that arctan(x) = x − 1

3x3 + 1

5x5 − 1

7x7 + 1

9x9 − 1

11x11 ± . . .. In

what range does this value of tan−1(x) lie? Give another derivation ofthe series using the idea in Ex.31.

ix− 12(ix)2+ 1

3(ix)3− 14(ix)4+ 1

5(ix)5∓. . . = Log(1+ix) = ln |1+ix|+iArg(1+ix) = ln

√1 + x2 +i arctan x, hence arctan x = =Log(1+ix) =

x− 13x

3 + 15x

5∓ . . . the values of which should lie in the range of Arg, i.e.in (−π, π). Alternatively, arctan x =

∫ x0

11+u2 du =

∫ x0∑∞n=0(−u2)n du =∑∞

n=0∫ x

0 (−u2)n du = ∑∞n=0(−1)n

∫ x0 u

2n du = ∑∞n=0(−1)n 1

2n+1u2n+1

∣∣∣x0

=∑∞n=0(−1)n x2n+1

2n+1 .

37. (i) Show geometrically that as z = eiθ goes round and round the unitcircle (with ever increasing θ), =[Log(1 + z)] = (Θ/2), where Θ isthe principal value of θ, i.e., −π < Θ ≤ π.

With Ch.1, Ex.18 on p.47 we have eiθ+eiφ = 2 cos θ−φ2 ei(θ+φ)/2


so that =[Log(1 + z)] = =( ln |1 + z| + iArg(1 + eiθ)) = Arg(1 +eiθ) = Arg(2 cos 0−θ

2 eiθ/2) = Θ2 .

(ii) Consider the periodic ”saw tooth” function F (θ) whose graph isshown below.

θ

F (θ)

−ππ

π/2

−π/2

By substituting z = eiθ in the logarithmic seriesLog(1 + z) = z − 1

2z2 + 1

3z3 − 1

4z4 + 1

5z5 − 1

6z6 +− . . . (29)

on p.101, use the previous part to deduce the following Fourierseries: F (θ) = sin θ − 1

2 sin(2θ) + 13 sin(3θ)− 1

4 sin(4θ) +− . . ..Log(1 + eiθ) = eiθ − 1

2ei2θ + 1

3ei3θ − 1

4ei4θ +− . . . implies Θ

2 ==[Log(1 + z)] = sin θ − 1

2 sin(2θ) + 13 sin(3θ)− 1

4 sin(4θ) +− . . ..

(iii) Check this Fourier series by directly evaluating the integralsan = 1

π

∫ 2π0 F (θ) cos(nθ) dθ and bn = 1

π

∫ 2π0 F (θ) sin(nθ) dθ (19)

on p.77 for n ∈ No.The ”saw tooth” function F (θ) is odd, hence all an vanish.

bn = 1π

∫ π−π

θ2 sin(nθ) dθ = 1

2π(−θn

cos(nθ)∣∣∣π−π

+∫ π−π

1n

cos(nθ) dθ) =1

2πn( − π cos(−nπ) − π cos(nπ)) = 1n

cos(nπ) = 1n

1 if n even−1if n odd

=1n(−1)n.

(iv) Use a computer to draw graphs of the partial sums of the Fourierseries. As you increase the number of terms, observe the magicalconvergence of this sum of smooth waves to the jagged graphabove. If only Fourier could have seen this on the screen, not justin his mind’s eye!


38. As in the previous exercise, let Θ = Arg(z).

(i) Use Log(1 + z) = z− 12z

2 + 13z

3− 14z

4 + 15z

5−+ . . . (29) on p.101to show that 1

2Log 1+z1−z = z + 1

3z3 + 1

5z5 + 1

7z7 + . . ..

12Log 1+z

1−z = 12(Log(1 + z) − Log(1 − z)) = 1

2(z −12z

2 + 13z

3 −14z

4 + 15z

5 −+ . . . )− 12(− z −

12z

2 − 13z

3 − 14z

4 − 15z

5 −+ . . . ) =z + 1

3z3 + 1

5z5 + . . ..

(ii) Show geometrically that as z = eiθ goes round and round the unitcircle, =(1

2Log 1+z1−z) = π

4 sign(θ).With Ch.1, Ex.18 on p.47 we have =(1

2Log 1+z1−z) = 1

2(=Log(1+z)−=Log(1−z)) = 1

2(Arg(1+z)−Arg(1−z)) = 12(Arg(ei(0+θ)/2)−

Arg(iei(0+θ)/2)) = 12(

θ2 −

θ2 −

π2) = −π

4 . Geometrically??????? (iii) Consider the periodic "square wave" function G(θ) whose graph

is shown below.

θ

G(θ)

−ππ

π/4

−π/4

Use the previous two parts to deduce that its Fourier series isG(θ) = sin θ + 1

3 sin(3θ) + 15 sin(5θ) + 1

7 sin(7θ) + . . ..G(θ) = =(1

2Log 1+eiθ1−eiθ ) = =(eiθ + 1

3ei3θ + 1

5ei5θ + 1

7ei7θ + . . . ) =

sin θ + 13 sin(3θ) + 1

5 sin(5θ) + 17 sin(7θ) + . . ..

(iv) Finally, repeat part (iii) of the previous exercise: Check thisFourier series by directly evaluating the integrals (19).


G(θ) is odd, hence all an vanish. bn = 1π

∫ π−π G(θ) sin(nθ) dθ =

− 1π

∫ 0−π

π4 sin(nθ) dθ+ 1

π

∫ π0π4 sin(nθ) dθ = 1

4cos(nθ)n

∣∣∣0−π− 1

4cos(nθ)n

∣∣∣π0

=14

1n(1− cos(nπ)− cos(nπ) + 1) = 1

n1−(−1)n

2 = 1n

1 if n odd0 if n even

.

(v) Finally, repeat part (iv) of the previous exercise: Use a computerto draw graphs of the partial sums of the Fourier series.

39. Show thatThe centroid Z must lie in the interior of the convex hull H. (32)on p.104 is still true even if the (positive) masses of the particles arenot all equal.

With massesmn located in zn for n = 1, 2, . . . , N the centroid is Z =1M

∑Nn=1mnzn with total mass M = ∑N

n=1mn. Hence Z = ∑Nn=1wnzn

where wn = mnM

is the convex sum of the (zn)n and thus lies in theconvex hull of the (zn)n.

40. Here is another simple way of derivingThe centre of a regular n-gon is the centroid of its vertices. (34)on p.104. If the vertices of the origin-centred regular n-gon are rotatedby φ, then their centroid Z rotates with them to eiφZ. By choosingφ = 2π

n, deduce that Z = O.

The n-gon and therefore also its centroid is fixed under the rotationaround 0 by φ = 2π

n. But the only fixpoint is O, hence Z = O.

41. To establish


Unless m is a multiple of n, the image under zm of an origin-centred regular n-gon is an origin-centred regular N-gon, whereN = (n divided by the highest common factor of m and n). Ifm is a multiple of n, then the image is a single point.

(36)

on p.106, let z0, z1, z2, . . . , zn−1 be the vertices (labelled counterclock-wise) of the regular n-gon, and let C be the circumscribing circle. Also,let wj = zmj be the image of vertex zj under the mapping z 7→ w = zm.Think of z as a particle that starts at z0 and orbits counterclockwiseround C, so that the image particle w = zm starts at w0 and orbitsround another circle with m times the angular speed of z.

(i) Show that each time z travels from one vertex to the next, wexecutes m

nof a revolution. Thus as z travels from z0 to zk, w

executes kmnrevolutions as it travels from w0 to wk.

Without loss of generality we may assume zk = reik2π/n, thenwk = rmeikm2π/n. Going from one vertex z on C to the next,means 1

nrevolution. Going from one w on Cm to the next, hence

means mnrevolutions.

(ii) Let wk be the first point in the sequence w1, w2, etc., such thatwk = w0. Deduce that if M

Nis m

nreduced to lowest terms, then

k = N . Note that N = (n divided by the highest common factorof m and n).

As in (i) assume wk = rmeikm2π/n = wo = rm implying kmn

=kMN∈ Z which in turn implies that k is a multiple of N , with

k = N due to the minimality of k.(iii) Explain why wN+1 = w1, wN+2 = w2, etc.

wN+1 = rmei(N+1)2πM/N = rmei2πM/N = rmeim2π/n = (rei2π/n)m= zm1 = w1 etc.

(iv) Show that w0, w1, . . . , wN−1 are distinct.Assume wν = rmeiν2πM/N = rmeiµ2πM/N = wµ for some 0 ≤

ν, µ ≤ N − 1 with ν 6= µ. Hence ei(ν−µ)2πM/N = 1 implying0 6= ν − µ to be a multiple of N , a contradiction.

(v) Show that w0, w1, . . . , wN−1 are the vertices of a regular N -gon..???

42. Consider the mapping z 7→ w = Pn(z), where Pn(z) is a general poly-nomial of degree n ≥ 2. Let Sq be the set of points in the z-planethat are mapped to a particular point q in the w-plane. Show thatthe centroid of Sq is independent of the choice of q, and is therefore aproperty of the polynomial itself. [Hint: This is another way of lookingat a familiar fact about the sum of the roots of a polynomial.]

z with Pn(z) = q are the zeroes of Qn(z) = Pn(z) − q. Now,


Qn(z) = c∏nj=1(z − zj) = c(zn + cn−1z

n−1 + . . . + co) for some c ∈ Cand zeroes z1, z2, . . . , zn taking multiplicity into account. Now, cn−1 =−∑n

j=1 zj and the centroid 1n

∑Pn(z)=q z of Sq of Pn is the centroid

1n

∑Qn(z)=0 z = 1

n

∑nj=1 zj = − 1

ncn−1 of So of Qn which is independent

of q.

43. Use Gauss’ Mean Value Theorem [p.110] to find the average of cos zover the circle |z| = r. Deduce (and check with a computer) that forall real values of r,

∫ 2π0 cos[r cos θ] cosh[r sin θ] dθ = 2π.

〈cos(z)〉C = 12π∫ 2π0 cos(0 + reiθ) dθ = cos 0 = 1 and 2π =

∫ 2π0 cos(0 +

reiθ) dθ =∫ 2π0 cos(r cos θ+ir sin θ) dθ = <

∫ 2π0 ( cos(r cos θ) cosh(r sin θ)−

i sin(r cos θ) sinh(r sin θ)) dθ =∫ 2π

0 cos(r cos θ) cosh(r sin θ) dθ.


Chapter 3

Möbius Transformations andInversion

3.1 Introduction

3.1.3 Decomposition into Simple Transformations

On p.123/124 the Möbius transform M(z) = az+bcz+d is decomposed into the

four transformations first z 7→ z + dc, second z 7→ 1

z, third z 7→ −ad−bc

c2z and

fourth z 7→ z + acbecause z 7→ −ad−bc

c21

z+d/c + ac

= −(ad−bc)+a(cz+d)c(cz+d) = acz+bc

c(cz+d) =az+bcz+d = M(z).

3.2 Inversion

3.2.2 Preservation of Circles

On p.127 let z1 = IK1(z) and z2 = IK2(z), then |[qz2]||[qz1]| = R2

2/|[qz]|R2

1/|[qz]|= R2

2R2

1=: k

independently of z. Now, because z1 and z2 are both in the same directionfrom q as z we have z2 − q = k(z1 − q) or IK2 = Dkq IK1 with dilatationDkq identical to the expansion (centred at q) of the plane by the factor k. Itfollows that if (7) holds for K1, i.e. if inversion in K1 maps L to a circle C1that passes through q then inversion in K2 maps L to Dkq (C1) which againis a circle that passes through q because scaling preserves circles as K givenby x = x(t) = xo + r cos t and y = y(t) = yo + r sin t with x′ = k x(t) =kxo + kr cos t and y′ = k y(t) = kyo + kr sin t show.

59

60 CHAPTER 3. MÖBIUS TRANSFORMATIONS AND INVERSION

On p.130 let L = R and q = 0, then the circle K of radius R centred atq = iR is tangent to L at p. Using IK(z) = q + R2

z−q = qz+(R2−|q|2)z−q (4)

on p.125 here gives IK(z) = iRzz+iR = z

1−iz/R .Further, if R is fixed and |z| < R, then IK(z) = z

1−iz/R = z∑∞j=0 (

izR)

j =z + iz2

R− z3

R2 − iz4

R3 + . . .

3.2.6 Inversion in a Sphere

On p.134 it is proven that Under inversion in a sphere, the image of asphere that does not contain the centre of inversion is another sphere thatdoes not contain the centre of inversion.Now, a circle C in space may be thought of as the intersection of two spheres.If the circle does not pass through the centre q of inversion then these twospheres can be chosen to not contain q.

R

R

r1

d1

r2

d2

Just choose d1 and r1 with R2 + d21 = r2

1 and d2 and r2 with R2 + d22 = r2

2appropriately. Now, the two spheres are mapped to other spheres that do notcontain the centre of inversion. Hence their intersecting circle does neither.If otherwise C does pass through q then the two spheres do also. Thesespheres are mapped to planes each of which is parallel to the tangent planeof that sphere at q. The intersection of these plane is a line which is theimage of C and which is parallel to the tangent of C at q. So summarizing

Under inversion in a sphere, the image of a circle C thatdoes not pass through the centre q of inversion is anothercircle that does not pass through q. If C does pass throughq then the image is a line parallel to the tangent of C at q.

(12)

On p.136 suppose that orthogonal spheres S1 and S2 (i.e. the tangent planesin each point of the intersecting circle are orthogonal) are subjected to inver-sion in a third sphere K. Choose Π to be the unique plane passing throughthe centres of S1, S2 and K. Let C be the great circle in which K intersects

3.4. THE RIEMANN SPHERE 61

Π. Then the intersections C1 and C2 of the two spheres with Π are orthog-onal circles. Since IC maps C1 and C2 to orthogonal circles C1 and C2, byrotational symmetry around the line in Π connecting the two midpoints ofS1 and S2 also S1 and S2 are orthogonal. Hence

Orthogonal spheres invert to orthogonal spheres. (15)

3.4 The Riemann Sphere

3.4.3 Transferring Complex Functions to the Sphere

On p.143 when transfering f(z) = z to Σ, then by symmetry Complexconjugation in C induces a reflection of the Riemann sphere in the verticalplane passing through the real axis.

3.4.5 Stereographic Formulae*

On p.146 let z = x + iy ∈ C being mapped to z = (X, Y, Z) ∈ Σ ⊂ R3.For given z = (X, Y, Z) ∈ Σ let z′ = X + iY be the foot of the perpen-dicular from z to C. Then z = |z|

|z′|z′. The right triangles in [23] on p.147

with hypotenuses Nz and Nz are similar, hence |z||z′| = long cathetus in ∆ with Nzlong cathetus in ∆ with Nz =

short cathetus in ∆ with Nzshort cathetus in ∆ with Nz = 1

1−Z , hence we obtain the first stereographic formulax+ iy = z = |z|

|z′|z′ = X+iY

1−Z . (19)Second, (|z′|, Z) is on the unit circle, hence |z′|2 + Z2 = 1 which with|z′| = |z|(1− Z) implies |z|2(1− Z)2 + Z2 = 1 or |z|2 = 1−Z2

(1−Z)2 = 1+Z1−Z .

Third, 2z1+|z|2 = 2z

1+(1+Z)/(1−Z) = 2z(1−Z)1−Z+1+Z = z(1 − Z) = X + iY and further

|z|2−1|z|2+1 = (1+Z)/(1−Z)−1

(1+Z)/(1−Z)+1 = (1+Z)−(1−Z)(1+Z)+(1−Z) = Z, hence

X + iY = z′ = 2z1+|z|2 and Z = |z|2−1

|z|2+1 (20)

On p.147: If z is the stereo graphic projection of the point z having sphericalcoordinates (φ, θ), then clearly z = reiθ, and so it only remains to find r as afunction of φ. (The Riemann sphere is a Thales circle so that ∠(SzN) = π

2 .)


N

S

z

z = reiθC

φ

φ2

From [23b] on p.147 it is clear that the right triangles ∆(NzS) and ∆(NOz)are similar because they have the additonal angle ∠(zNS) = ∠(zNO) incommon. Now, r

1 = long cathetusshort cathetus = |[Sz]|

|[Nz]| = cot φ2 implying

z = cot(φ/2) eiθ (21)

3.5 Mobius Transformations: Basic Results

3.5.3 The Group Property

On p.150 for M(z) = az+bcz+d we have M−1(z) = dz−b

−cz+a because M−1 M(z) =daz+bcz+d−b

−caz+bcz+d+a = d(az+b)−b(cz+d)

−c(az+b)+a(cz+d) = (ad−bc)zad−bc = z, so that the inverse of a Möbius

transformation is again a Möbius transformation.On p.151 for M1(z) = a1z+b1

c1z+d1and M2(z) = a2z+b2

c2z+d2we have M2 M1(z) =

a2a1z+b1c1z+d1

+b2c2a1z+b1c1z+d1

+d2= a2(a1z+b1)+b2(c1z+d1)

c2(a1z+b1)+d2(c1z+d1) = (a2a1+b2c1)z+(a2b1+b2d1)(c2a1+d2c1)z+(c2b1+d2d1) , so that the combi-

nation of two Möbius transformations is again a Möbius transformation.On p.152 the fixpoints of the normnalized Möbius transformation M(z) =az+bcz+d with ad − bc = 1 are solutions of M(ξ) = ξ ⇐⇒ aξ + b = ξ(cξ + d)⇐⇒ cξ2 + (d − a)ξ − b = 0 ⇐⇒ ξ2 + d−a

cξ − b

c= 0 ⇐⇒ ξ1,2 =

a−d2c ±

12c

√(a2 − 2ad+ d2) + 4bc = a−d±

√a2+2ad+d2+4bc−4ad

2c = a−d±√

(a+d)2−42c .

3.6 Mobius Transformations as Matrices*

3.6.3 Eigenvectors and Eigenvalues*

On p.160 the characteristic equation of the matrix [M ] of a normalizedMöbius transformation M(z) = az+b

cz+d with ad − bc = 1 is det[M − λE ] =

3.7. VISUALIZATION AND CLASSIFICATION* 63∣∣∣∣∣a− λ bc d− λ

∣∣∣∣∣ = (a − λ)(d − λ) − bc = λ2 − (a + d)λ + ad − bc = 0 ⇐⇒

λ2 − (a+ d)λ+ 1 = 0 ⇐⇒ λ+ 1λ

= a+ d.

On p.160 we show tr(NP ) = tr(PN) for any quadratic matrices N and P

for 2×2-matrices only. For N =(a bc d

)and P =

(e fg h

)we have tr(NP ) =

tr(ae+bg af+bhce+dg cf+dh

)= ae+ bg + cf + dh = tr

(ea+fc eb+fdga+hc gb+hd

)= tr(PN).

3.6.4 Rotations of the Sphere as Mobius Transforma-tions*

On p.162 let [R] =[a bc d

]with [R]∗ =

[a c

b d

], ~p =

[p1p2

]with ~p∗ = [p1, p2],

then [R]~p∗ =[ap1 + bp2cp1 + dp2

]∗= [ap1+bp2, cp1+dp2] = [p1, p2]

[a c

b d

]= ~p

∗ [R]∗

3.7 Visualization and Classification*

3.7.4 Parabolic Transformations

On p.168 the case of exactly one fix point ξ is considered, i.e. |a + d| = 2.The Möbius transformation G(z) = 1

z−ξ takes ξ to ∞ and it maps circlesthrough ξ to lines, especially two families of circles through ξ orthogonal toeach other to two families of lines orthogonal to each other because G mapscircles through ξ to straight lines and preserves angles.

7→

On p.169 T = 1a/c−ξ = 1

a/c−(a−d)/(2c) = 2c2a−(a−d) = c

(a+d)/2 = ±c because|a+ d| = 2.


3.7.5 Computing the Multiplier*

On p.170 to characterise normalized Möbius transformations M(z) = az+bcz+d

with normal form w−ξ+w−ξ− = m z−ξ+

z−ξ− (41)on p.166, namely elliptic iff m = eiα, hyperbolic iff m = ρ 6= 1, loxodromicotherwise, i.e. m = ρeiα with ρ 6= 1, 0 6= α 6= π as given on pp164. Let M bea normalized Möbius transformation, then

√m + 1√

m= a+ d (43)

on p.170 where we deduce from the graph of x 7→ x + 1xwith (x + 1

x)′ =

1− 1x2 = 0 ⇐⇒ x = ±1 and ±1 + 1

±1 = ±2

x

y

1

−1

2

−2

• if a+ d is real and |a+ d| < 2 then???

• if a + d is real and |a + d| = 2 then M is parabolic according to p.169and (27) on p.152

• if a+ d is real and |a+ d| > 2 then m is real and M is hyperbolic.

• otherwise M is loxodromic.

For F (z) := z−ξ+z−ξ− we derived M = F M F−1 (39)

on p.164 with M(z) = mz, hence [M ] =[m 00 1

]and normalized [M ] =[√

m 00 1/

√m

]because

√m√m− 0 = 1.

For the inversion M(z) = 1zwith ξ± = ±1 and m = −1 for both fix points

we have [M ] =[0 11 0

]and normalized [M ] =

[0 ii 0

].

3.9. AUTOMORPHISMS OF THE UNIT DISC* 65

3.9 Automorphisms of the Unit Disc*

3.9.3 Interpreting the Simplest Formula Geometrically*

On p.179 RL IJ = Ma = IJ RL because RL IJ(0) = RL(a) = a,RL IJ(a) = RL(0) = 0, and RL IJ(1/a) = RL(∞) = ∞, whereas onthe other hand IJ RL(0) = IJ(0) = a, IJ RL(a) = IJ(a) = 0, andIJ RL(1/a) = IJ(1/a) =∞. Hence, left and right hand side are identical.


1. In each of the figures below, show that p and p are symmetric withrespect to the circle K around q with radius R. The dashed lines arenot strictly part of the constructions, rather they are intended to behelpful or suggestive.

q

p

p q

p

p

q p

p

s

r

b

ψ

R

R

ρR− ρ

θ1

θ2

x yz

x2 = ρ2 +R2 − 2ρR cosψ

y2 = 2R2(1− cosψ)

(R− ρ)2 = x2 + y2 − 2xy cos θ1

z2 = R2 + (R2/ρ)2 − 2R(R2/ρ) cosψ

(R2/ρ−R)2 = y2 + z2 − 2yz cos θ2

(i) The half circle on qp intersects the circle K in b. Then ∆(qpb)is a right triangle (Thales). The right triangle altitude theoremstates h2 = ρ(R2/ρ − ρ) = R2 − ρ2 obviously equivalent withρ = |qp| and |qp| = R2/ρ.

(ii) Let b be one of the two intersection points of the circle K andthe circle around p with Radius R2/ρ. Then the two triangles∆(bpq) and ∆(bpq) are similar because they both are equilateraland they have the additional angle at q in common. Hence forx = |qp| we have R2/ρ

R= R

xwhich implies x = ρ.

(iii) Let s be the intersection of the circle K and the line qp andlet b be the point on the circle K such that ∠(pbs) = θ = ∠(sbp).By the low of cosines, we havex2 = ρ2 +R2 − 2ρR cosψ y2 = 2R2(1− cosψ)(R− ρ)2 = x2 + y2 − 2xy cos θ1 z2 = R2 + (R2

ρ)2 − 2RR2

ρcosψ

(R2/ρ−R)2 = y2 + z2 − 2yz cos θ2

Solving for cos θ1 and cos θ2 respectively we get

cos θ1 = x2+y2−(R−ρ)2

2xy = ρ2+R2−2ρR cosψ+2R2(1−cosψ)−(R−ρ)2

2√ρ2+R2−2ρR cosψ

√2R2(1−cosψ)

= −ρ cosψ+R(1−cosψ)+ρ√ρ2+R2−2ρR cosψ

√2(1−cosψ)

= (ρ+R)(1−cosψ)√ρ2+R2−2ρR cosψ

√2(1−cosψ)


and if and only if |qp| · |qp| = R2

cos θ2 = y2+z2−(RRρ−R)2

2yz = 2R2(1−cosψ)+R2+(RRρ

)2−2R2 Rρ

cosψ−(RRρ−R)2

2√

2R2(1−cosψ)√R2+(R2

ρ)2−2R2 R

ρcosψ

= (1−cosψ)−Rρ

cosψ+Rρ√

2(1−cosψ)√

1+R2ρ2−2R

ρcosψ

= (ρ+R)(1−cosψ)√ρ2+R2−2ρR cosψ

√2(1−cosψ)

So |qp| · |qp| = R2 ⇐⇒ cos θ1 = cos θ2. construction??????

2. In 1864 a French officer named Peaucellier caused a sensation by dis-covering a simple mechanism (Peaucellier’s linkage1) for transforminglinear motion (say of a piston) into circular motion (say of a wheel).The figure below shows six rods hinged at the white dots, and anchoredat o. Two of the rods have length `, and the other four have lengthr. With the assistance of the dashed circle, show that p = IK(p),where K is the circle of radius

√`2 − r2 centred at o. Construct this

mechanism – perhaps using strips of fairly stiff cardboard for rods, anddrawing pins for hinges – and use it to verify properties of inversion.In particular, try moving p along a line.

om

p

p

`

`

rrm′

p′

p′

p′′ p′′

Let m be the centre of the dashed half circle. If p = p then ∆(opm)is a right triangle and any point in the distance R =

√`2 − r2 to o is

mapped to itself. Let K be the red circle centered at o with radius R.Again for p → p we see that the dashed (half-) circle is orthogonal toK hence by (10) on p.129 it is mapped by IK to itself and p must bemapped to p because p has to be on the dashed (half-) circle and onthe line through o and p.p, p′, p′′ on an arc of some circle, then p, p′, p′′ on a common line??????

1 https://en.wikipedia.org/wiki/Peaucellier-Lipkin_linkage

https://en.wikipedia.org/wiki/Peaucellier-Lipkin_linkage


3. Let S be a sphere, and let p be a point not on S. Explain why IS(p)may be constructed as the second intersection point of any three spheresthat pass through p and are orthogonal to S. Explain the preservationof three-dimensional symmetry in terms of this construction.

The centers of the three spheres lie in exactly one plane E. Because??? p ∈ E. Without loss of generality we may assume E to be C.Then the intersection of the three spheres with E looks exactly like???. compare p.135!!! ???

4. Deduce ’If p and q are antipodal points of the Riemann sphere Σ andp and q their stereographic projections then q = −1/p (22)on p.148 directly from Inversion of C in the unit circle induces a re-flection of the Riemann sphere Σ in its equatorial plane, C. (17)on p.143.

Inversion of C in the unit circle is given by z 7→ 1/z. The reflec-tion of the Riemann sphere Σ in its equatorial plane, C, is given by(x, y, z) 7→ (x, y,−z).

<o

N

p

p

˜p

1p

q

q

Now, the triangle ∆(˜qNq) is equilateral and so is the triangle ∆(1pNq)

so that q = −1/p follows.

5. Consider the following two-stage mapping: first stereo graphically pro-ject C onto the Riemann sphere Σ in the usual way; now stereographi-cally project Σ back to C, but from the south pole instead of the northpole. The net effect of this is some complex mapping z 7→ f(z) of C toitself. What is f?

Considering figure [23] on p.147 some point z = (φ, θ) on the Rie-mann sphere is mapped stereographically to z = reiθ with r = cot φ

2 =1

tan φ2. The stereographical projection from S maps z to the intersec-

tion of zN with C with distance tan φ2 = 1

rfrom the origin, hence

f(z) = f(reiθ) = 1reiθ = 1

z.

6. Both figures below show vertical cross sections of the Riemann sphere.


O

N

S

ˆz

z

z

z

6 (N ˆzS) = π2

O

N

p

q

p q

6 (pNq) = π2

(i) In the left figure, show that the triangles ∆(pON) and ∆(NOq)are similar. Deduce:If p and q are antipodal points of Σ, then their stereographic pro-jections p and q are related by q = −1/p (22).

∆(pON), ∆(NOq) and ∆(qNp) (Thales) are three right tri-angles. ∆(pON) and ∆(NOq) due to ∠(OpN) = π

2 −∠(ONP ) =∠(ONq) have another angle in common and hence are similar.The right triangle altitude theorem gives 1 = pq and the similar-ity also gives 1

p= q

1 which for points p and q on the real line withp left from O and q right from O gives with q = −1/p.

(ii) The right figure is a modified copy of [21 b]. Show that the tri-angles ∆(zNO) and ∆(NOz) are similar. Deduce:Inversion of C in the unit circle induces a reflection of the Rie-mann sphere in its equatorial plane, C. (17).

Both triangles are right triangles. The triangle ∆(zNO) is sim-ilar to the right triangle ∆(ˆzNS) because ∠(zNO) = ∠(NS ˆz).Now, the triangle ∆(ˆzNS) is similar to the right triangle ∆(zNO)because they have the angle at N in common.Now ∆(zNO) ≈ ∆(zNO) and |ON | = 1 (radius of the Riemannsphere) implies short cathetus

long cathetus = |z|1 = 1

|z| , hence |z| · |z| = 1 which isequivalent to z = IC(z) with unit circle C. The images z and ˆzof z and z, respectively under the steroegraphic projection thenare reflections in the equatorial plane C of the Riemann sphere.

7. (i) Use a computer to draw the images in C of several origin-centredcircles under the exponential mapping, z 7→ ez. Explain the obvi-ous symmetry of these image curves with respect to the real axis.

Let z = reiφ for fixed r > 0 and φ ∈ [0, 2π) be some origin-centred circle. Then ez = exp(reiφ) = exp(r cosφ + ir sinφ) =u+ iv with u = er cosφ cos(r sinφ) and v = er cosφ sin(r sinφ).


<

=

<

=

The graph is symmetrical to the abszissa because v is an evenfunction.

(ii) Now use the computer to draw these same image curves on theRiemann sphere, instead of in C. Note the surprising new sym-metry!

symmetry? ???

<

=

symmetry ?(iii) Use The mapping z 7→ 1/z in C induces a rotation of the

Riemann sphere about the real axis through an angle of π. (18)on p.143 to explain this extra symmetry.

. ???

8. This exercise continues the discussion of (2), p.123. If a point p inspace emits a flash of light, we claimed that each of the light rays couldbe represented by a complex number. Here is one, indirect method ofestablishing this correspondence. Once again, we choose units of spaceand time so that the speed of light is 1. After one unit of time, theexpanding sphere of light emitted by p – made up of particles of lightcalled photons – forms a unit sphere. Thus each photon may be identi-fied with a point on the Riemann sphere, and hence, via stereographicprojection, with a complex number. Indeed, if the photon has sphericalpolar coordinates (φ, θ), then z = cot(φ/2)eiθ (21)on p.147 tells us that the corresponding complex number in deed is


z = cot(φ/2)eiθ.Sir Roger Penrose (see Penrose and Rindler [1984, p.13]) discovered thefollowing remarkable method of passing from a light ray to the associ-ated complex number directly, without the assistance of the Riemannsphere. Imagine that p is one unit vertically above the origin of the(horizontal) complex plane. At the instant that p emits its flash, let Cbegin to travel straight up (in the direction φ = 0) at the speed of light1) towards p. Decompose the velocity of the photon F emitted by p inthe direction (φ, θ) into components perpendicular and parallel to C.Hence find the time at which F hits C. Deduce that F hits C at thepoint z = cot(φ/2)eiθ. Amazingly, we see that Penrose’s constructionis equivalent to stereographic projection!

.???

9. In order to analyse astronomical data, Ptolemy required accurate tri-gonometric tables, which he constructed using the addition formulaefor sine and cosine. The figures below explain how he discovered thesekey addition formulae. Both the circles have unit radius. In the rightfigure the diagonals are in magenta and the diagonal AC is a diameter.

2φ2θ

<

=

M

A

B

C

DM

2θ<

=

a

b

(i) In the left figure, show that a = 2 sin θ and b = 2 cos θ.The triangle ∆(ACM) is equilateral. It consists of two con-

gruent halves. Each half triangle is a right triangle the longcathetus of which is sin θ, hence a = 2 sin θ and analogouslyb = 2 sin π−2θ

2 = 2 sin(π2 − θ) = 2 cos θ.(ii) In the right figure, apply Ptolemy’s Theorem (cp p.138-139) to the

illustrated quadrilateral, and deduce the sine addition theorem,i.e. deduce that sin(θ + φ) = sin θ cosφ+ sinφ cos θ.

Ptolemy’s Theorem states |AC| · |BD| = |AB| · |CD|+ |BC| ·|AD|. Hence, with part (i) we have |AC| · |BD| = 2 ·2 sin(φ+θ) =(2 cos θ)(2 sinφ) + (2 sin θ)(2 cosφ) = |AB| · |CD|+ |BC| · |AD|.


10. The aim of this question is to understand the following result: Any twonon-intersecting, non-concentric circles can be mapped to concentriccircles by means of a suitable Möbius transformation.

(i) If A and B are the two circles in question, show that there existsa pair of points ξ± that are symmetric with respect to both A andB.

If R is the radius of A centered at Q = O and r is the radiusof B centered at q = (d, 0) where d is the distance of the twomidpoints Q and q and if the two points are ξ+ = (P, 0) and ξ− =(p, 0) then by similarity w.r.t. A we have R

P= p

Rand pP = R2 and

by similarity w.r.t. B we have rd−p = d−P

rand (d − p)(d − P ) =

r2 = (d−p)(d−R2/p) which gives p2− 1d(d2 +R2− r2)p+R2 = 0

with solutions p1,2 = 12d(c±

√c2 − 4d2R2) with c = d2 +R2 − r2.

Q qA Bξ+ ξ−

The green circle C intersects A and B orthogonally. Hence C issymmetric to both A and B and C intersects the line connectingthe centers of A and B in exactly ξ+ and ξ−.Let ρ be the radius of C. As radius and tangent are orthogonal d =√R2 + ρ2+

√r2 + ρ2 and d2 = R2+ρ2+2

√R2 + ρ2

√r2 + ρ2+r2+

ρ2 ⇐⇒ (d2−R2−r2−2ρ2)2 = 4(R2 +ρ2)(r2 +ρ2) ⇐⇒ d4 +R4 +r4+4ρ4−2d2R2−2d2r2−4d2ρ2+2R2r2+4R2ρ2+4r2ρ2 = 4R2r2+4(R2+r2)ρ2+4ρ4 ⇐⇒ d4+R4+r4−2d2R2−2d2r2−2R2r2 = 4d2ρ2

⇐⇒ ρ = 12d

√d4 +R4 + r4 − 2d2R2 − 2d2r2 − 2R2r2.

(ii) Deduce that if F (z) = (z − ξ+)/(z − ξ−), then F (A) and F (B)are concentric circles, as was desired.

As a Möbius transformation F is preserving circles. Hence itsuffices to show that diameters of f(A) and f(B) have the samemidpoint.Without loss of generality we again can assume Q = O and q =(d, 0) on the real line. Then ξ+, ξ− ∈ R ??? and 1

2(f(−R)+f(R))and 1

2(f(d− r) + f(d+ r)) are identical. ???


11. This exercise yields a more intuitive proof of the result of the previousexercise. Using different colours for each, draw two non-intersecting,non-concentric circles, A and B, then draw the line L through theircentres. Label as p and q the intersection points of B with L.

(i) Using corresponding colours, draw a fresh picture showing theimages A, b, L, q of A, B, L, q under inversion in any circlecentred at p. To get you started, note that L = L.

???

p

A

mA

B

mB

L = Lm

A

A

mB

B

q

q

.(ii) Now add to your figure by drawing the circle K, centred at q,

which cuts A at right angles, and let g and h be the intersectionpoints of K and L.

First construct the circle with mAq as a diameter. The inter-section of this Thales circle with A are points on K.

p

A

mA

B

mB

L = Lm

A

A

mB

B

q

q

K

g h

(iii) Now draw a new picture showing the images K ′, L′, h′ of K, L,h under inversion in any circle centred at g.

.???


(iv) By appealing to the anticonformal nature of inversion, deducethat A, B are concentric circles centred at h′.

. ???

Since the composition of two inversions is a Möbius transformation,you have proved the result of the previous exercise.

12. Figure (i) below shows two non-intersecting, non-concentric circles Aand B, together with a chain of circles C1, C2, . . . that touch one an-other successively, and that all touch A and B. As you would expect,the chain fails to ’close up’: C8 overlaps C1 instead of touching it.Figure (ii) shows that this failure to close is not inevitable. Given adifferent pair A, B, it is possible to obtain a closed chain where Cntouches C1. Here n = 5, but by considering the case where A, B areconcentric, you can easily see that any value of n is possible, given theright A and B.

184 Mobius Transformations and Inversion

12 Figure (i) below shows two non-intersecting, non-concentric circles A and B, together with a chain of circles C l, C2, . . . that touch one another successively, and that all touch A and B. As you would expect, the chain fails to "close up": Cg overlaps Cl instead of touching it. Figure (ii) shows that this failure to close is not inevitable. Given a different pair A, B, it is possible to obtain a closed chain where Cn touches Cl. Here n = 5, but by considering the case where A, B are concentric, you can easily see that any value of n is possible, given the right A and B.

Steiner discovered, very surprisingly, that if the chain closes for one choice of C l, then it closes for every choice of C l, and the resulting chain always contains the same number of touching circles. Explain this using the result of Ex. 10.

13 (i) Let P be a sphere resting on the fiat surface Q of a table. Let Sl, S2, . .. be a string of spheres touching one another successively and all the same size as P. If each S-sphere touches both P and Q, show that S6 touches Sl, so that we have a closed "necklace" of six spheres around P.

(ii) Let A, B, C be three spheres (not necessarily of equal size) all touching one another. As in the previous part, let Sl, S2, ... be a string of spheres (now of unequal size) touching one another successively, and all touching A, B, C. Astonishingly (cf. previous exercise), S6 will always touch Sl, forming a closed "necklace" of six spheres interlocked with A, B, C. Prove this by first applying an inversion centred at the point of contact of A and B, then appealing to part (i).

The chain of six spheres in part (ii) is called Soddy's Hexlet, after the amateur mathematician Frederick Soddy who discovered it (without inversion!). For further information on Soddy's Hexlet, see Ogilvy [1969]. Soddy's full time job was chemistry-in 1921 he won the Nobel Prize for his discovery of isotopes!

14 The figure below shows four collinear points a, b, c, d, together with the (necessarily coplanar) light rays from those points to an observer. Imagine that the collinear points lie in the complex plane, and that the observer is above the plane looking down. Show that the cross-ratio [a, b, c, d] can be expressed purely in terms of the directions of these light rays; more precisely, show that

Steiner discovered, very surprisingly, that if the chain closes for onechoice of C1, then it closes for every choice of C1, and the resultingchain always contains the same number of touching circles. Explainthis using the result of Ex.10.

There is a Möbius transformation mapping the two circles A andB to two concentric circles A and B.

A

B

α

A

B

7→


It is possible to fill the annulus with a sequence of circles with radiusρ touching each other and touching A and B as well if 2π

2α ∈ N whereα = arcsin ρ

rA+ρ . In addition if the chain closes for one choice of C1,then it closes for every choice of C1, and the resulting chain alwayscontains the same number of touching circles.

13. (i) Let P be a sphere resting on the flat surface Q of a table. LetS1, S2, . . . be a string of spheres touching one another successivelyand all the same size as P. If each S-sphere touches both P andQ, show that S6 touches S1, so that we have a closed ’necklace’of six spheres around P .

The right hand side figure in the solution of the previous ex-ercise shows the ’necklace’.

(ii) Let A, B, C be three spheres (not necessarily of equal size) alltouching one another. As in the previous part, let Sl, S2, . . . be astring of spheres (now of unequal size) touching one another suc-cessively, and all touching A, B, C. Astonishingly (cf. previousexercise), S6 will always touch S1, forming a closed ’necklace’ ofsix spheres interlocked with A, B, C. Prove this by first applyingan inversion centred at the point of contact of A and B, thenappealing to part (i).

.???The chain of six spheres in part (ii) is called Soddy’s Hexlet, after theamateur mathematician Frederick Soddy who discovered it (without in-version!). For further information on Soddy’s Hexlet, see Ogilvy [1969].Soddy’s full time job was chemistry-in 1921 he won the Nobel Prize forhis discovery of isotopes!

14. The figure below shows four collinear points a, b, c, d, together withthe (necessarily coplanar) light rays from those points to an observer.Imagine that the collinear points lie in the complex plane, and that theobserver is above the plane looking down. Show that the cross-ratio[a, b, c, d] can be expressed purely in terms of the directions of theselight rays; more precisely, show that [a, b, c, d] = − sinα sin γ

sinβ sin δ .

view point

a b c d

α βγ

δ = α+ β + γ

Suppose the observer now does a perspective drawing on a glass ’canvasplane’ C (arbitrarily positioned between himself and C). That is, for


each point p in C he draws a point p′ where the light ray from p tohis eye hits C. Use the above result to show that although angles anddistances are both distorted in his drawing, cross-ratios of collinearpoints are preserved: [a′, b′, c′, d′] = [a, b, c, d].

[a, b, c, d] = (a−b)(c−d)(a−d)(c−b) . Let the viewpoint be O = (0, h). Then the

sine theorem gives h|a− b| = |O− a| · |O− b| sinα, h|c− d| = |O− c| ·|O−d| sin γ, h|a−d| = |O−a| · |O−d| sin γ, and h|c−b| = |O−c| · |O−b| sin β so that [a, b, c, d] = − (b−a)(d−a)

(d−a)(c−b) = − |O−a|·|O−b| sinα|O−c|·|O−d| sin γ|O−a|·|O−d| sin γ|O−c|·|O−b| sinβ =− sinα sin γ

sinβ sin δ . Further, [a′, b′, c′, d′] = − sinα sin γsinβ sin δ = [a, b, c, d].

15. Show that in both of the figures below, Arg[z, q, r, s] = θ + φ.q

r

s

z

φ

θq

r

s

z φθ

Hence deduce A point p lies on the circle C through q, r, sif and only if =[p, q, r, s] = 0. etc (31)

on p.155.In the left figure Arg[z, q, r, s] = Arg (z−q)(r−s)

(z−s)(r−q) = (Arg(z − q) −Arg(z − s))+ (Arg(r − s)−Arg(r − q)) = θ + φ holds whereas in theright figure actually the same holds when considering the orientation.z on C ⇐⇒ the points z, q, r, s are a cyclic quadrilateral ⇐⇒θ + φ = π (cp. https://en.wikipedia.org/wiki/Cyclic_quadrilateral)⇐⇒ M(p) = [p, q, r, s] ∈ R ⇐⇒ =[p, q, r, s] = 0.

16. As in figure [28] on p.156, think of the cross-ratio [z, q, r, s] as a Möbiustransformation.

(i) Explain geometrically why permuting q, r, s, in [z, q, r, s] yieldssix different Möbius transformations.

Fixing z in the first position there are 3! = 6 permutations ofq, r, s each of which defines another Möbius transform.

(ii) If I(z) is the Möbius transformation that leaves 1 fixed and thatswaps 0 with∞, explain geometrically why I[z, q, r, s] = [z, s, r, q].

I[z, q, r, s] is again a Möbius transform which equals [z, s, r, q]

as

qrs

[z,q,r,s]7−→

01∞

I7−→

∞10

is the same as

srq

[z,s,r,q]7−→

01∞

.

https://en.wikipedia.org/wiki/Cyclic_quadrilateral


(iii) If J(z) is the Möbius transformation that sends 0, 1,∞ to 1,∞, 0,respectively, explain geometrically why J [z, q, r, s] = [z, s, q, r].

J[z, q, r, s] is again a Möbius transform which equals [z, s, q, r]

as

qrs

[z,q,r,s]7−→

01∞

J7−→

1∞0

is the same as

sqr

[z,s,q,r]7−→

01∞

.(iv) Employing the abbreviation χ = [z, q, r, s], explain why the six

Möbius transformations in part (i) can be expressed as χ, I χ,J χ, I J χ, J I χ, I J I χ.

χ = [., q, r, s], I χ = [., s, r, q], J χ = [., s, q, r], I J χ =[., r, q, s], J I χ = [., q, s, r], I J I χ = [., r, s, q], a total ofsix different Möbius transforms.

(v) Show that I(z) = 1/z and J(z) = 1/(1− z).For I(z) = 1/z we have I(0) = ∞, I(1) = 1 and I(∞) = 0.

For J(z) = 1/(1− z) we have J(0) = 1, J(1) =∞ and J(∞) = 0.(vi) Deduce that the six possible values of the cross-ratio are χ, 1

χ,

11−χ , 1− χ, χ

χ−1 ,χ−1χ

.In the order of part (iv) we have χ, I χ = 1

χ, J χ = 1

1−χ ,I J χ = 1− χ, J I χ = J( 1

χ) = 1

1−1/χ = χχ−1 , I J I χ =

I J( 1χ) = I( 1

1−1/χ) = I( χχ−1) = χ−1

χ.

17. Show geometrically that if a and c lie on a circle K, and b and d aresymmetric with respect to K, then the point [a, b, c, d] lies on the unitcircle. [Hints: Draw the two circles through a, b, d and through b, c, d.Now think of [z, b, c, d] as a Möbius transformation.]

With a ∈ K, c ∈ K and d = b, the two pairs of triangles ∆(aqb) ≈∆(dqa) and ∆(bqc) ≈ ∆(cqd) are similar, hence in the first pair |b−q||d−q| =|a−b||a−d| and

|b−q||d−q| = |c−b|

|c−d| in the second pair. Together we have |a−b||a−d| = |c−b||c−d|

implying |[a, b, c, d]| = |a−b|·|c−d||a−d|·|c−b| = 1.

18. The curvature κ of a circle is defined to be the reciprocal of its radius.Let M(z) = az+b

cz+d be normalized, i.e. ad− bc = 1. Use (i) to (iv) of (3)on p.124 to show geometrically that M maps the real line to a circle ofcurvature κ =

∣∣∣2c2=dc

∣∣∣.M(z) is decomposed into four mappings

(i) a translation z → z + dcwhich maps the real line R to R + d

c,

(ii) the inversion z → 1zwhich maps R + d

cto Γ = 1

R+ dc

= F (t) :t ∈ R where F (t) = c

tc+d . Now F (∞) = 0 and F ′(t) = −c(tc+d)2

vanishes in ∞ so that F (t) touches the abszissa in 0. F (−1) =cd−c , F (0) = c

d, F (1) = c

d+c , and for to = −<(cd)cc

we have F (to) =


cd−<(cd)/c = cc

cd−<(cd) so that <(F (to)) = 0. Γ is the circle aroundm = 1

2F (to) ∈ iR with radius r = |m| because for F ′(to) =− c2

(toc+d)2 = − c2(cc)2

(ccd−<(cd))2 = − |c|4|ccd−<(cd)|4 c

2(|c|2d−<(cd))2 =??? ???

(iii) an expansion and rotation z → − 1c2z ???

(iv) and another translation z → z + ac

???

19. Let M(z) = az+bcz+d be normalized, i.e. ad− bc = 1.

(i) Using the decomposition of Möbius transforms (3) on p.124, drawdiagrams to illustrate the successive effects of these transforma-tions on a family of concentric circles. Note that the image circlesare generally not concentric.

-6 -4 -2 0 2 4

x

-3

-2

-1

0

1

2

3

4

5

6

y

three circles and their Möbius transform

K1: unit circleK2: 2*unitcircleK3: 1+unit circlet1(K1)invT1(K1)sRinvT1(K1)M(K1)t1(K2)invT1(K2)sRinvT2(K1)M(K2)t1(K3)invT1(K3)sRinvT1(K3)M(K3)

(ii) Deduce that the image circles are concentric if and only if theoriginal family of circles are centred at q = −d/c. Write downthe centre of the image circles in this case. [Note that this is notthe image of the centre of the original circles: M(q) is the pointat infinity!]

The scalar multiple of a complex circle C(t) = reit by a com-plex scalar c = ρeiτ is cC(t) = cr eit = ρrei(t+τ), again a complexcircle.Let Kr = q + reit be the circle around q = −d

cwith radius r

and any normalized Möbius transformation M(z) = az+bcz+d be rep-

resented by M = T2 SR Inv T1 with T1(z)z + dc, Inv(z) = 1

z,

SR(z) = −ad−bcc2

z = − 1c2z and T2(z) = z + a

c, then M(Kr) = T2

SR Inv(reit) = T2 SR(1re−it) = T2(− 1

c21re−it) = a

c− 1

c21re−it =

ac− 1

ρ2re−i(t+2τ), a circle around a

c.

If on the other hand the image circles are concentric then ??? . ???(iii) Hence show geometrically that the circle IM with equation |cz +

d| = 1 is mapped by M to a circle of equal size. Furthermore,show that each arc of IM is mapped to an image arc of equal size.


For this reason, IM is called the isometric circle of M .Bow, |cz + d| = 1 ⇐⇒ |z − q| = 1

|c| and M(K1/|c|) =ac− |c||c|2re

−i(t+2τ) which is a circle around acwith the same radius

1|c| . In the exponent of the parametrization of M(K1/|c|) thereis just an affine transformation of t so that – the radius beingidentical – each arc of IM is mapped to an image arc of equal size.

For applications of the isometric circle, see Ford [1929] and Katok[1992].

20. (i) Show that every Möbius transformation of the form M(z) = pz+qqz+p

where |p| > |q| can be rewritten in the formM(z) = eiθ z−aaz−1 where

|a| < 1. [Notice that the converse is also true. In other words,the two sets of functions are the same.]

Let a = − qpwith |a| < 1 and for p = r eiϕ we have −p

p=

−ei2ϕ = ei(2ϕ+π) =: eiθ, then M(z) = pz+qqz+p = − p

pz− q

p

− qpz−1 = −p

p

z+ qppp

az−1 =eiθ z−a

az−1 .(ii) Use the matrix representation of the first equation to show that

this set of Möbius transformations forms a group under the oper-ation of composition.

ForMk(z) = pkz+qkqkz+pk

we have [Mk] =(pk qkqk pk

)so that [M1][M2]

=(p1 q1q1 p1

)(p2 q2q2 p2

)=(p1p2 + q1q2 p1q2 + q1p2q1p2 + p1q2 q1q2 + p1p2

)=:

(p3 q3q3 p3

)for p3 = p1p2 + q1q2 and q3 = p1q2 + q1p2 with |p3| > |q3| because

|p3|2 − |q3|2 = (p1p2 + q1q2)(p1p2 + q1q2)− (p1q2 + q1p2)(p1q2 + q1p2)= p1p1p2p2 + p1q1p2q2 + q1p1p2q2 + q1q1q2q2

−p1p1q2q2 − p1q1p2q2 − q1p1p2q2 − q1q1p2p2

= |p1|2(|p2|2 − |q2|2) + |q1|2(|q2|2 − |p2|2)= (|p1|2 − |q1|2)(|p2|2 − |q2|2) > 0

So the setM = [M ] : M(z) = pz+qqz+p , p, q ∈ C, |p| > |q| is closed

under multiplication.

Further |[M ]| =∣∣∣∣∣p qq p

∣∣∣∣∣ = pp − qq = |p|2 − |q|2 > 0 and thus

[M ]−1 =(p qq p

)−1

= 1|[M ]|

(p −q−q p

)=:

(r ss r

)with |r| > |s| so

thatM is closed under inversion also.(iii) Use the disc-automorphism interpretation of these transforma-

tions to give a geometric explanation of the fact that they form a


group.. ???

21. (i) Use the matrix representation to show algebraically that the setof Möbius transformations R(z) = az+b

−bz+a with |a|2 + |b|2 = 1 formsa group under the operation of composition.

For the composition R(z) := (R2 R1)(z) = R2(R1(z)) of twoMöbius transformation of this type R1 and R2 it holds R(z) =a2

a1z+b1−b1z+a1

+b2

−b2a1z+b1−b1z+a1

+a2= a2(a1z+b1)+b2(−b1z+a1)

−b2(a1z+b1)+a2(−b1z+a1) = (a2a1−b2b1)z+(a2b1+b2a1)−(b2a1+a2b1)z+(a2a1−b2b1) .

Also[a2 b2−b2 a2

] [a1 b1−b1 a1

]=[a2a1 − b2b1 a2b1 + b2a1−b2a1 − a2b1 −b2b1 + a2a1

]= [R],

hence [R] = [R2][R1]. As in addition |a2a1 − b2b1|2 + |a2b1 +b2a1|2 = (a2a1 − b2b1)(a2a1 − b2b1) + (a2b1 + b2a1)(a2b1 + b2a1) =a2a1a2a1 − a2a1b2b1 − b2b1a2a1 + b2b1b2b1 + a2b1a2b1 + a2b1b2a1 +b2a1a2b1 + b2a1b2a1 = a2a1a2a1 + a2b1a2b1 + b2a1b2a1 + b2b1b2b1 =(a2a2 + b2b2)(a1a1 + b1b1) = 1 the set R is closed under cmpo-sition. As well the set [R] is closed under multiplication. Wedo not want to iterate pp.150, but [R−1] = [R]−1 so that Rand [R] are closed under inversion. In all, R and [R] aregroups.

(ii) Using the interpretation of these functions The most general ro-tation of the Riemann sphere can be expressed as a Möbius trans-formation of the form R(z) = az+b

−bz+a . (38)on p.162 explain part (i) geometrically.

. ???

22. Let H be the rectangular hyperbola with Cartesian equation x2− y2 =1. Show that z 7→ w = z2 maps H to the line <(w) = 1. What isthe image of this line under complex inversion, w 7→ 1

w? Referring

back to figure [9] on p.62, deduce that complex inversion maps H to alemniscate! [Hint: Think of complex inversion as z 7→

√1/z2.]

First, z2 : z = x + iy ∈ H = x2 − y2 + 2ixy : x2 − y2 = 1 =w ∈ C : <w = 1 = 1 + iR.Second, 1

w: <w = 1 = 1

1+iy : y ∈ R = 1−iy1+y2 : y ∈ R.

<12

=


Obviously, the image is the circle C centered at (12 , 0) with Radius 1

2 ,verified by 1

1+t2 (1,−t) ∈ C ⇐⇒ ( 11+t2 −

12)

2 + ( t1+t2 )

2 = 14 ⇐⇒

(1− 12 −

12t

2)2 + t2 = 1

4(1 + t2)2 ⇐⇒ (1− t2)2 + 4t2 = (1 + t2)2 ⇐⇒1 + 2t2 + t4 = 1 + 2t2 + t4.Third, H = x+ iy : x2 − y2 = 1 z 7→w=z2

−→ 1 + iRw 7→u= 1

w−→ C = x+ iy :(x− 1

2)2 + y2 = 14 = 1

2(1 + eit) : t ∈ R u7→v=√u−→???

23. From the simple fact that z 7→ (1/z) is involutory, deduce that it iselliptic, with multiplier −1.

First, M(z) = 1zhas the two fixpoints ξ± = ±1, then F (z) = z−1

z+1and F inv(z) = − z+1

z−1 so that M(z) = F M F inv(z) = F M(− z+1z−1) =

F(− z−1z+1) = − z−1

z+1−1− z−1z+1 +1 = −(z−1)−(z+1)

−(z−1)+(z+1) = −z. So M is a pure rotation andhence M is elliptic with multiplier −1.Second, .???

24. (i) Use the Symmetry Principle on p.148 to show that the most gen-eral Möbius transformation of the upper half-plane to the unitdisc has the form M(z) = eiθ z−a

z−a where =a > 0..???

(ii) The most general Möbius transformation back from the unit discto the upper half-plane will therefore be the inverse ofM(z). Let’scall this inverse N(z). Use the matrix form of M to show thatN(z) = M−1(z) = az−aeiθ

z−eiθ .

For M(z) = eiθ z−az−a we have [M ] =

(eiθ −aeiθ1 −a

)and thus

[M−1] = [M ]−1 = 1−aeiθ+aeiθ

(−a aeiθ

−1 eiθ

)= 1

a−a

(−ae−iθ a−e−iθ 1

)cor-

responding to M−1(z) = −ae−iθz+a−e−iθz+1 = az−aeiθ

z−eiθ .

(iii) Explain why the Symmetry Principle implies thatN(1/z) = N(z)..???

(iv) Show by direct calculation that the formula for N in part (ii) doesindeed satisfy the equation in part (iii).

N(1/z) = a/z−aeiθ1/z−eiθ = a−aeiθ z

1−eiθ z = ae−iθ−aze−iθ−z = az−ae−iθ

z−e−iθ = N(z).

25. LetM(z) be the general Möbius automorphism of the upper half-plane.

(i) Observing that M maps the real axis into itself, use(w−q)(r−s)(w−s)(r−q) = [w, q, r, s] = [z, q, r, s] = (z−q)(r−s)

(z−s)(r−q) (29)on p.154 to show that the coefficients of M are real.

.???


(ii) By considering =[M(i)], deduce that the only restriction on thesereal coefficients is that they have positive determinant ad−bc > 0.

If M(z) = az+bcz+d for a, b, c, d ∈ R then =M(z) = 1

|cz+d|2=((az +b)(cz + d)) = 1

|cz+d|2=(ac|z|2 + adz + bcz + bd) = 1|cz+d|2=(ad(x+

iy) + bc(x− iy)) > 0 ⇐⇒ ad− bc > 0.(iii) Explain (both algebraically and geometrically) why these Möbius

transformations form a group under composition.If M,N ∈ A are Möbius automorphisms of the upper half-

plane then [M N ] = [M ][N ] and |[M N ]| = |[M ][N ]| =|[M ]|·|[N ]| where the determinant of the product of matrices withpositive determinant is again positive so that alsoMN ∈ A. Theidentity I(z) = 1·z−0

0·z+1 is in A and with M ∈ A also M−1 ∈ A with

[M−1] = 1ad−bc

(d −b−c a

)with |[M−1]| = ad−(−b)(−c)

ad−bc = 1 > 0.

(iv) How many degrees of freedom doesM have? Why does this makesense?

3? ???

26. Reconsider Mα(z) = z−ααz−1 (52) on p.178. normalized? ???

(i) Use M(z) = az+bcz+d is elliptic iff a+ d is real and |a+ d| < 2 (44)

on p.170 to show that Ma is elliptic.Mα(z) = z−α

αz−1 hence a+d = 1−1 = 0 ∈ R and |a+d| = 0 < 2.(ii) Use

√m+ 1/

√m = a+ d (43)

on p.170 to show that both multipliers are given by m = −1.Here,

√m+ 1/

√m = a+ d = 0 implying m = −1.

(iii) Calculate the matrix product [Ma] [Ma], and thereby verify thatMa is involutory.

[Ma][Ma] =(

1 −aa −1

)2

=(

1− aa −a+ aa− a −aa+ 1

)= (1− aa)I. ???

(iv) Use ξ± = (a−d)±√

(a+d)2−42c if is M is normalized (27)

on p.152 to calculate the fixed points of Ma.. ???

(v) Show that the result of the previous part is in accord with figure[38] om p.178.

. ???

27. Use Normalized M(z) = az+bcz+d is (i) elliptic iff a+d is real and |a+d| <

2, (ii) parabolic iff a + d = ±2, (iii) hyperbolic iff a + d is real and|a+ d| > 2 (iv) loxodromic iff a+ d is complex (44)


on p.170 to verify If we define Φ = 2 cos−1 |a|, then Mφa is (i) elliptic

if |φ| < Φ, (ii) parabolic if |φ| = Φ, (iii) hyperbolic if |φ| > Φ. (53)on p.180.

Mφa (z) = eiφ z−a

az−1 = eiφz−aeiφaz−1 is normalized??? −eiφ + aaeiφ =

(aa− 1)eiφ = 1 ??? .???

Chapter 4

Differentiation: The AmplitwistConcept


1. Use the Cauchy-Riemann equations to verify that z 7→ z is not analytic.With f(z) = f(x + iy) = x − iy = u + iv where u(x, y) = x and

v(x, y) = −iy the Cauchy-Riemann equations ∂xu = ∂yv and ∂yu =−∂xv are not fulfilled because ∂xu = 1 6= −i = ∂yv.

2. The mapping z 7→ z3 acts on an infinitesimal shape and the imageis examined. It is found that the shape has been rotated by π, andits linear dimensions expanded by 12. Where was the shape originallylocated? [There are two possibilities.]

Because of f(z) = z3 and f ′(z) = 3z2 we have to identify zo with|3z2

o | = 12 or |zo|2 = 4 and arg(3z2o) = arg(z2

o) = π, hence zo = ±2i.

3. Consider z 7→ Ω(z) = z2/z. By writing z in polar form, find out thegeometric effect of Ω. Using two colours, draw two very small arrowsof equal length emanating from a typical point z: one parallel to z;the other perpendicular to z. Draw their images emanating from Ω(z).Deduce that Ω fails to produce an amplitwist. [Your picture shouldshow this in two ways.]

Ω(z) = Ω(reiφ) = z2/z = r e−i2φ

eiφ= re−i3φ, i.e. different arguments

are rotated by different angles.

83

84 CHAPTER 4. DIFFERENTIATION: THE AMPLITWIST CONCEPT

<

=

<

=

Obviously, neither the lengths are amplified by the same factor nor arethe angles preserved.

4. The picture (on p.211) shows the shaded interior of a curve beingmapped by an analytic function to the exterior of the image curve.If z travels round the curve counterclockwise, then which way does itsimage w travel round the image curve? [Hint: Draw some infinitesimalarrows emanating from z, including one in the direction of motion.]

The function f is analytic in some connected region R with bound-ary curve C = ∂R. It maps R to f(R) = R with C = ∂R = f(C).Now, f maps an infinitesimal vector dv at z ∈ C along C to the in-finitesimal vector dv at z = f(z) ∈ C along C (both in blue) and someinfinitesimal vector dv⊥ at z ∈ C perpendicular to dv and pointing intoR to the infinitesimal vector ˜dv⊥ = f(dv⊥) at z along C (both in red).

R

z

R

z

˜dv⊥ points into the exterior of C and the orientation of ∠(dv, dv⊥) ispreserved. Hence z travels along C clockwise.

5. Consider f(x + iy) = (x2 + y2) + i(y/x). Find and sketch the curvesthat are mapped by f into (a) horizontal lines, and (b) vertical lines.Notice from your answers that f appears to be conformal. Show thatit is not in two ways:

(i) by explicitly finding some curves whose angle of intersection isn’tpreserved;

(a) =f = c in f(x + iy) = x2 + y2 + i yximplies y = cx, i.e. f

maps straight lines y = cx to horizontal lines y = c. (b) <f = c2

implies x2 + y2 = c2, i.e. f maps circle x2 + y2 = c2 around theorigin to vertical lines x = c. Because circles around the originand lines through the origin intersect at right angle as do theirimages under f , f appears to be conformal so far.


Now, f maps the line y = mx into the horizontal line y = m andf maps the vertical line x = xo into the parabola f(xo + iy) =x2o + y2 + i y

xo. The two lines intersect in (xo,mxo) and the two

image curves intersect in (d2(1 +m2),m).

<

=

(x2o(1 +m2),m)

<

=

(xo,mxo)

The angle between the two lines is fixed for any xo whereas theangle between the two image curves varies between π

2 and 0 forxo →∞.

(ii) by using the Cauchy-Riemann equations.With f(z) = f(x + iy) = u(x, y) + iv(x, y) where u(x, y) =

x2 + y2 and v(x, y) = y/x the Cauchy-Riemann equations ∂xu =2x 6= 1

x= ∂yv and ∂yu = 2y 6= y

x2 = −∂xv are violated.

6. Continuing from the previous exercise, show that no choice of v canmake f(x+ iy) = (x2 + y2) + iv analytic.

The Cauchy-Riemann equations with ∂yv = 2x imply v(x, y) =2xy + c(x). With ∂xv = −2y they imply v(x, y) = −2xy + d(y) whichis a contradiction.

7. (i) If g(z) ≡ 3 + 2i then explain geometrically why g′(z) ≡ 0.Because every infinitesimal vector ε emanating from some z is

mapped to the zero vector, the amplification factor necessarily is0 and thus g′(z) = 0 anywhere.

(ii) Show that if the amplification of an analytic function is identicallyzero (i.e., f ′(z) ≡ 0) on some connected region, then the functionis constant there.

. ???(iii) Give a simple counterexample to show that this conclusion does

not follow if the region is instead made up of disconnected com-ponents.

. ???

8. Use pictures to explain why if f(z) is analytic on some connected region,each of the following conditions forces it to reduce to a constant.

(i) <f(z) = 0All images lie on the imaginary axis – also the image of two


infinitesimal vectors inclosing/confining a nonzero angle. The im-ages can only preserve the angle if all infinitesimal vectors aremapped to zero. Hence with part (ii) of the previous exercise fmust be constant.

(ii) |f(z)| = const|f(z)|2 = u2 + v2 = const.???

(iii) Not only is f(z) analytic, but f(z) is too.If f locally preserves angles while f must rotate an arrow at

angle φ by −2φ to obtain its image arrow at angle −φ, cp p.200.Hence f cannot be analytic unless f and f are constant.

9. Use the Cauchy-Riemann equations to give rigorous computationalproofs of the results of the previous two exercises.

(i) <f(z) = 0From the Cauchy-Riemann equations we get ∂xu = 0 = ∂yv

and ∂yu = 0 = −∂xv so that ∂xv = 0 = ∂yv and hence v(x, y) = c.(ii) |f(z)| = const

|f(z)|2 = u2 + v2 = const implies u∂xu + v∂xv = 0 and

u∂yu + v∂yv = 0. On one hand we get uv∂xu+ v2∂xv= 0u2∂yu+ uv∂xu= 0, by

(u2 + v2)∂xv = 0 implying −∂yu = ∂xv = 0, on the other hand

analogously u2∂xu+ uv∂xv= 0−uv∂xv + v2∂yu= 0, by (u2 + v2)∂yv = 0 implying

∂xu = ∂yv = 0. Because all partial derivatives vanish f must beconstant.

(iii) Not only is f(z) analytic, but f(z) is too.If f = u + iv then f = u − iv. f being analytical gives

∂xu = ∂yv and ∂yu = −∂xv whereas f being analytical gives∂xu = −∂yv and ∂yu = ∂xv. Again, all partial derivatives vanish,so that f must be constant.

10. Instead of writing a mapping in terms of its real and imaginary parts(i.e. f = u+ iv), it is sometimes more convenient to write it in terms oflength and angle: f(z) = ReiΨ where R and Ψ are functions of z. Showthat the equations that characterize an analytic f are now ∂xR = R∂yΨand ∂yR = −R∂xΨ.

First, ∂xR = ∂√x2+y2

∂x= x√

x2+y2and R∂yΨ =

√x2 + y2 ∂ arctan(y/x

∂y=

√x2 + y2 1/x

(y/x)2+1 =√x2 + y2 x

x2+y2 .

Second, ∂yR = ∂√x2+y2

∂y= y√

x2+y2and −R∂xΨ = −

√x2 + y2 ∂ arctan(y/x)

∂x

= −√x2 + y2 −y/x2

(y/x)2+1 =√x2 + y2 y

x2+y2 = y√x2+y2

.


11. Let’s agree to say that f = u+ iv satisfies the Cauchy-Riemann equa-tions" if u and v do. Show that if f(z) and g(z) both satisfy theCauchy-Riemann equations, then their sum and their product do also.

We have ∂x<f = ∂y=f and ∂y<f = −∂x=f as well as ∂x<g = ∂y=gand ∂y<g = −∂x=g, hence ∂x<(f+g) = ∂x<f+∂x<g = ∂y=f+∂y=g =∂y=(f + g) and ∂y<(f + g) = −∂x=(f + g) analogously.∂x<(fg) = ∂x(<f<g−=f=g) = <g∂x<f+<f∂x<g−=g∂x=f−=f∂x=gand ∂y=(fg) = ∂y(<f=g + =f<g) = =g∂y<f + <f∂y=g + <g∂y=f +=f∂y<g = −=g∂x=f + <f∂x<g + <g∂x<f − =f∂x=g and ∂y<(fg) =−∂x=(fg) analogously.

12. For nonzero z, let f(z) = <z=zz

= f(x+ iy) = xyx+iy .

(i) Show that f(z) approaches 0 as z approaches any point on thereal or imaginary axis, including the origin.

Let z = x + iy → xo 6= 0, then especially y → 0, hencelimz→xo f(z) = limz→xo

xyx+iy = limz→xo

x2yx2+y2 − i limz→xo

xy2

x2+y2 =0− 0i. The case z = x+ iy → iyo is handled analogously.

(ii) Having established that f = 0 on both axes, deduce that theCauchy-Riemann equations are satisfied at the origin.

∂xu|(0,0) = ∂xx2yx2+y2

∣∣∣(0,0)

= ddx

0∣∣∣x=0

= 0 and similarly ∂yv|(0,0) =

∂yxy2

x2+y2

∣∣∣(0,0)

= ddy

0∣∣∣y=0

= 0.

∂yu|(0,0) = ∂yx2yx2+y2

∣∣∣(0,0)

= ddy

0∣∣∣y=0

= 0 and similarly − ∂xv|(0,0) =

− ∂x xy2

x2+y2

∣∣∣(0,0)

= ddx

0∣∣∣x=0

= 0.

(iii) Despite this, show that f is not even differentiable at 0, let aloneanalytic there! To do so, find the image of an infinitesimal arrowemanating from 0 and pointing in the direction eiφ. Deduce thatwhile f does have a twist at 0, it fails to have an amplificationthere.

Let z = reiφ be some infinitesimal vector, then f(z) = r2 cosφ sinφreiφ

= 12 sin(2φ)re−iφ. There cannot be any amplification because it

would depend on φ by the factor sin(2φ).

13. Verify that z 7→ ez satisfies the Cauchy-Riemann equations, and find(ez)′.

For f(x+ iy) = ex(cos y+ i sin y) = u+ iv we have u = ex cos y andv = ex sin y, hence ∂xu = ex cos y = ∂yv as well as ∂yu = −ex sin y =−∂xv. Further, (ez)′ = ∂xu + i∂xv = ex cos y + iex sin y = ex(cos y +i sin y) = ez.

14. By sketching the image of an infinitesimal rectangle under an analyticmapping, deduce that the local magnification factor for area is the


square of the amplification. Re-derive this fact by looking at the de-terminant of the Jacobian matrix.

Some analytical function f maps infinitesimal vectors dv and dv⊥at z to infinitesimal vectors dv = f ′(z)dv and ˜dv⊥ = f ′(z)dv⊥ atz = f(z). The area of the rectangle spanned by dv and dv⊥ is |dv×dv⊥|.The area of the rectangle spanned by dv and ˜dv⊥ is |dv × ˜dv⊥| =|f ′(z)|2 |dv × dv⊥|, i.e. the local magnification factor for area is thesquare of the amplification.

The determinant of the Jacobian J =(∂xu ∂yu∂xv ∂yv

)=(∂xu −∂xv∂xv ∂xu

)is

|J | = (∂xu)2 + (∂xv)2 = |∂xu+ i∂xv|2 = |f ′(z)|2.

15. Let us define S to be the square region given by a − b ≤ <z ≤ a + band −b ≤ =z ≤ b.

(i) Sketch a typical S for which b < a. Now sketch its image S underthe mapping z 7→ ez.

<

=

1S<

=

1S

(ii) Deduce the area of S from your sketch, and write down the ratioΛ = area of S

area of S .The area of the square is 4b2 whereas the area of the annulus

sector is 2b2ππ(e

2(a+b)−e2(a−b)) = b(e2(a+b)−e2(a−b)) = 2be2a sinh b,hence Λ = b(e2(a+b)−e2(a−b))

4b2 = e2a sinh b2b .

(iii) Using the results of the previous two exercises, what limit shouldΛ approach as b shrinks to nothing?

Because of limx→0sinhxx

= limx→0coshx

1 = 1 we have limb→0 Λ =12e

2a.

16. Consider the complex inversion mapping I(z) = 1/z. Since I is confor-mal, its local effect must be an amplitwist. By considering the imageof an arc of an origin-centred circle, deduce that |(1/z)′| = 1/|z|2.

Let z = reit and z + dz = rei(t+dt), then dz = z(ei dt − 1). Thearguments z and z + dz, respectively are mapped to I(z) = w = 1

re−it

and I(z + dz) = w + dw = 1z+dz = 1

re−i(t+dt), respectively, then dw =

w(e−i dt − 1), hence |f ′(z)| =∣∣∣dwdz

∣∣∣ = 1|z|2

∣∣∣ e−i dt−1ei dt−1

∣∣∣ ≈ 1|z|2

∣∣∣−i dti dt

∣∣∣ = 1|z|2 .

17. Consider the complex inversion mapping I(z) = 1/z.


(i) If z = x+ iy and I = u+ iv, express u and v in terms of x and y.From I(x + iy) = 1

x+iy = x−iyx2+y2 we get u(x, y) = x

x2+y2 andv(x, y) = −y

x2+y2 .

(ii) Show that the Cauchy-Riemann equations are satisfied every-where except the origin, so that I is analytic except at this point.

Firstly, ∂xu = (x2+y2)−2x2

(x2+y2)2 = −x2+y2

(x2+y2)2 and ∂yv = −(x2+y2)+2y2

(x2+y2)2 =−x2+y2

(x2+y2)2 . Secondly ∂yu = −2xy(x2+y2)2 and −∂xv = −2xy

(x2+y2)2 .

(iii) Find the Jacobian matrix, and by expressing it in terms of polarcoordinates, find the local geometric effect of I.

For z = x+ iy = reiφ:(ux uyvx vy

)= 1

(x2+y2)2

(−x2 + y2 −2xy

2xy −x2 + y2

)=

r2

r4

(− cos2 φ+ sin2 φ −2 cosφ sinφ

2 cosφ sinφ − cos2 φ+ sin2 φ

)= 1

r2

(− cos(2φ) − sin(2φ)sin(2φ) − cos(2φ)

)=

1r2

(−1 00 −1

)(cos(−2φ) − sin(−2φ)sin(−2φ) cos(−2φ)

). Locally I is an amplification

by 1r2 and a twist by π − 2φ.

(iv) Use f ′ = ∂xu+ i∂xv = ∂xf (7)and f ′ = ∂yv − i∂yu = −i∂yf (8)on p.210 to show that the amplitwist is −1/z2, just as in ordinarycalculus, and in accord with the previous exercise. Use this toconfirm the result of part (iii).

f ′(z) = −x2+y2

(x2+y2)2 + i 2xy(x2+y2)2 and on the other hand − 1

z2 =− 1x2+i2xy−y2 = − x2−y2−i2xy

(x2−y2)2+(2xy)2 = −x2+y2+i2xy(x2+y2)2 = −x2+y2

(x2+y2)2 +i 2xy(x2+y2)2 .

18. Recall Ex. 19, p.186, where you showed that a general Mobius trans-formation M(z) = az+b

cz+d maps concentric circles to concentric circles ifand only if the original family (call it F) is centred at q = −d

c. Let

ρ = |z − q| be the distance from q to z, so that the members of F areρ = const.

(i) By considering orthogonal connecting vectors from one memberof F to an infinitesimally larger member of F , deduce that theamplification of M is constant on each circle of F . Deduce that|M ′| must be a function of ρ alone.

With ∆ = ad − bc we have M ′(z) = a(cz+d)−c(az+b)(cz+d)2 = ∆

(cz+d)2

so that |M ′(q+ ρeit| =∣∣∣ ∆(−d+cρeit+d)2

∣∣∣ = |∆||c|2ρ2 which is constant for

all t and is a function of ρ alone.(ii) By considering the image of an infinitesimal shape that starts far

from q and then travels to a point very close to q, deduce that atsome point in the journey the image and preimage are congruent.

. ???


(iii) Combine the above results to deduce that there is a special mem-ber IM of F such that infinitesimal shapes on IM are mapped tocongruent image shapes on the image circle M(IM). Recall thatIM is called the isometric circle of M .

.???(iv) Use the previous part to explain why M(IM) has the same radius

as IM ..???

(v) Explain why IM−1 = M(IM)..???

(vi) Suppose that M is normalized. Using the idea in Ex. 16, showthat the amplification of M is M ′(z) = 1

c2ρ2 ..???

19. Consider the mapping I(z) = z4, illustrated below. On the left is aparticle p travelling upwards along a segment of the line x = 1, whileon the right is the image path traced by I(p).

u

v

A

B

p

x

y

1

−1

1

p b1

b2

(i) Copy this diagram, and by considering the length and angle of pas it continues its upward journey, sketch the continuation of theimage path.

With u(x, y) = x4 − 6x2y2 + y4 and v(x, y) = 4x3y − 4xy3

we have limy→±∞ u(1, y) = +∞ and with d u(1,y)dy

= 4y(y2 − 3)then u(1, y) has the three critical points (−

√3,−8), (0, 1) and

(√

3,−8), corresponding to the three points −8 − i8√

3, 1 + 0iand −8 + i8

√3 on the curve z(t) = (1 + it)4.

Further limy→±∞ v(1, y) = limy→±∞ 4y(1 − y2) = ∓∞ and withd v(1,y)dy

= 4(1−3y2) then v(1, y) has the two critical points −√

3/3and√

3/3. Then B = (1± i)4 = −4 + 0i, C = (1 + 0i)4 = 1 + 0i,D1,2 = (1 ∓

√3/3)4 = −8

9(1 ± i√

3) and E1,2 = (1 ± i√

3)4 =−8± i8

√3.


u−1−2−3 1

v

A1

A2

BC

D1

D2

E1

E2

(ii) Show that A = i sec4 π8 which in turn is A = i 1

cos4 π8

= i 4(1+cos π4 )2 =

i 4(1+√

2/2)2 = i 83+2√

2 = i8(3− 2√

2) because cos2 θ2 = 1

2(1 + cos θ).I(z) = I(x + iy) = u + iv with u = x4 − 6x2y2 + y4 and

v = 4x3y − 4xy3. u(1, y∗) = 0 ⇐⇒ y∗1,2,3,4 = ±√

3± 2√

2 giv-ing exactly the four intersections ±i4

√3± 2

√2(1 − 3 ∓ 2

√2) =

∓i8√

3± 2√

2|1±√

2| of z(t) = (1 + it)4 with the ordinate, sym-metric to the origin. The intersection on the positive ordinateclosest to the origin is i8

√3− 2

√2(√

2 − 1) = i8(3 − 2√

2) ≈1.372583i.

(iii) Find and mark on your picture the two positions (call them b1and b2) of p that map to the self-intersection point B of the imagepath.

For z = 1± i we have z4 = (1± i)4 = 1± 4i− 6∓ 4i+ 1 = −4.(iv) Assuming the result I ′(z) = 4z3, find the twist at b1 and also at

b2.I ′(bj) = 4(1± i)3 = 4(1± 3i∓ 3∓ i) = 8(−1± i).

(v) Using the previous part, show that (as indicated at B) the imagepath cuts itself at right angles.

Obviously, −1+ i and −1− i are perpendicular as the two realvectors (−1, 1) and (−1,−1) with (−1, 1)(−1,−1)> = 0 show.

20. The figure on p.215 is a copy of [9], p.62.

(i) Show geometrically that if z moves distance ds along the lemnis-cate, increasing θ by dθ, then w = z2 moves a distance 4dθ alongthe circle.

. ???


(ii) Using the fact that (z2)′ = 2z, deduce geometrically that ds =2dθ/r.

.???(iii) Using the fact that r2 = 2 cos(2θ), show by calculation that r dr =

2√

1− r4/4 dθ.r2 = 2 cos(2θ) implies d r2

dθ= 2r dr

dθ= −4 sin(2θ), hence r dr =

−2 sin(2θ) dθ. Now, with sin(2θ) =√

1− cos2(2θ) =√

1− r4/4we have r dr = −2

√1− r4/4 dθ Vorzeichen????

(iv) Let s represent the length of the segment of the lemniscate con-necting the origin to the point z. Deduce from the previous twoparts that s =

∫ ro

dr√1−r4/4

, hence the name, lemniscatic integral.

s =∫ zo ds =

∫ θo

2rdθ =

∫ ro

dr√1−r4/4

.???

21. (i) By extending the argument given in the text, show that in three-dimensional space the effect of a linear transformation is to stretchspace in three mutually perpendicular directions (generally bythree different factors), then to rotate it.

.???(ii) Deduce that a mapping of three-dimensional space to itself is

locally a three-dimensional amplitwist if and only if it maps in-finitesimal spheres to infinitesimal spheres.

.???(iii) Deduce that inversion in a sphere preserves the magnitude of the

angle contained by two intersecting curves in space..???

(iv) Deduce that stereographic projection is conformal..???

Remark: In stark contrast to the bountiful conformal mappings of theplane, Liouville and Maxwell independently discovered that the onlyangle-preserving transformation of space is an inversion, or perhapsthe composition of several inversions.

Chapter 5

Further Geometry ofDifferentiation

5.6 Visual Differentiation of the Power Func-tion

On p.230 let w = f(z) = 3√z so that f has three branches f(r eiθ) =

3√rei(θ+2πZ)/3 which differ by the factor ei2π/3.

<1

w = 3√reiθ/3

θ/3

φ

w = 3√rei(θ/3+2π/3) φ

w = 3√rei(θ/3+4π/3)

φ

<

z = reiθ

θ

12

φ

f−→

5.9 An Application of Higher Derivatives:Curvature*

5.9.3 Complex Curvature

On p.241 to show E = ξ ×K ???

93

94 CHAPTER 5. FURTHER GEOMETRY OF DIFFERENTIATION

5.10 Celestial Mechanics*

5.10.1 Central Force Fields

On p.241 let the particle p in a central force field at o have initial positionpo ∈ R3 and initial velocity vo ∈ R3. Then the orbit of p lies in the planethrough o and po which contains vo.

On p.241 the angular momentum h = m~r × ~r of particle p with mass mand areal speed A = d ~A

dt= ~r×~v

2 in a central force field is h = m~r× ~r = 2mA.

5.10.2 Two Kinds of Elliptical Orbit

On p.242 let ` be the length of the string and let ϕ be the displacementangle resulting in the horizontal displacement r = ` sinϕ. Then the forcemg = −mg eπ/2 is split in the part −mg cosϕ eπ/2−ϕ along the string (takenby the string) and the part mg sinϕ e−ϕ tangent to the circular trajectory

(accelerating p). Check −eπ/2 =(

01

)= − cosϕ

(sinϕcosϕ

)+ sinϕ

(cosϕ− sinϕ

)=

− cosϕ(

cos(π/2− ϕ)sin(π/2− ϕ)

)+sinϕ

(cos(−ϕ)sin(−ϕ)

)= − cosϕ eπ/2−ϕ+sinϕ e−ϕ. Now,

for |ϕ| 1 we have mg sinϕ e−ϕ ≈ mg`rex, i.e. a (central) force proportional

to the displacement r.

On p.243 the potential energy is the work needed to pull the particle awayfrom the origin out to a distance of r, i.e. Epot =

∫ r0 F (ρ) dρ =

∫ r0 ρ dρ = 1

2r2.

5.10.5 An Explanation

On p.246 put f(z) = z2 into κ = 1|f ′(p)|(=[

f ′′(p)ξf ′(p) ] + κ) (23)

on p.234 to get κ = 12|p|

(=[ 2ξ

2p ] + κ)

= 12(=(pξ)r3 + κ

r ) = 12(

cos γr2 + κ

r ) because=(pξ) = =(ip(−iξ)) = <(pn) = 〈p, n〉 = |p| |n| cos∠(p, n) = r cos γ wheren = −iξ is one of the two unit normal vector at the curve in p.

5.11. ANALYTIC CONTINUATION* 95

5.11 Analytic Continuation*

5.11.5 Analytic Continuation via Reflections

On p.253 let f be the rotation of φ, i.e. f(z) = z eiφ, then f ∗(z) = f(z) =z eiφ = z e−iφ is the rotation of −φ.On p.254 Show: If f maps a line-segment L (not necessarily real) to anotherline segment L, then we can analytically continue from one side of L to theother by using the fact that points symmetric in L map to points symmetricin L. ???Let L = p+ tc : t ∈ R and L = p+ tc : t ∈ R with unit vectors c and c.Then h(z) := (z − p)c maps L to R because p+ tc 7→ tc 7→ tcc = t as well ash(z) := (z − p)¯c maps L to R because p + tc 7→ tc 7→ tc¯c = t. Define g byf = hinv g h, i.e. g = h f hinv with hinv(w) = p + w

c= p + wc. Then

obviously g maps R to L to L to R. With f , h and h also g is analytical.Now, g by g∗(z) = g(z) has an analytic continuation across the real axis.As h and h map the half spaces specified by L and L respectively to theupper or the lower half plane, their analytical continuations h∗ and h∗ doso correspondingly. Hence f ∗(z) = f(z) = hinv g h(z) = hinv g(h(z)) =hinv(g∗ h∗(z)) = hinv

∗ g∗ h∗(z) is the analytical continuation across L.On p.254 if f maps part of the unit circ1e to part of the real axis andis only known inside the circle then f †(z) := f(1/z) defines the analyticalcontinuation to the exterior of the unit circle coinciding on part of it.On p.256 let a ellipse E be given by x2

a2 + y2

b2= 1. With x = z+z

2 andy = z−z

2i we have z2+2zz+z2

a2 − z2−2zz+z2

b2= 4 ⇐⇒ b2(z2 + 2zz + z2) −

a2(z2 − 2zz + z2) = 4a2b2 ⇐⇒ (b2 − a2)(z2 + z2) + (a2 + b2)2zz = 4a2b2

⇐⇒ z2 + a2+b2b2−a2 2zz + z2 = 4a2b2

b2−a2 . Solving for z gives z = −a2+b2b2−a2 z ±

1b2−a2

√(a2 + b2)2z2 − (b2 − a2)2z2 + 4a2b2(b2 − a2) = −a2+b2

b2−a2 z± 2abb2−a2

√z2 + b2 − a2.

Then SE(z) = 1a2−b2 ((a

2 + b2)z − 2ab√z2 + b2 − a2)


1. Show that if f = u + iv is analytic then (∇u) · (∇v) = 0, where ∇ isthe ’gradient operator’ of vector calculus. Explain this geometrically.

(∇u) · (∇v) =(∂xu∂yu

)·(∂xv∂yv

)= ∂xu∂xv + ∂yu∂yv = −∂xu∂yu +

∂yu∂xu = 0 due to the Cauchy-Riemann equations.Geometrie?(∇u) · (∇v) = 0 ⇐⇒ ∇u ⊥ ∇v ???


2. Show that the real and imaginary parts of an analytic function are har-monic, i.e., they both automatically satisfy Laplace’s equation ∆Φ = 0,where ∆ (which is often instead written ∇2) is defined by ∆ = ∂2

x + ∂2y

and is called the Laplacian. [In Chapter 12 we will see that this equa-tion represents a crucial link between analytic functions and physics.]

Let f = u + iv, then ∆u = ∂2xu + ∂2

yu = ∂x∂yv − ∂y∂xv = 0 and∆v = ∂2

xv + ∂2yv = −∂x∂yu+ ∂y∂xu = 0 .

3. Use the previous exercise (not calculation) to show that each of thefollowing functions is ’harmonic’.

(i) f(x, y) = ex cos y∂2xf = ∂2

xf + ∂2yf = ∂xe

x cos y− ∂yex sin y = ex cos y− ex cos y = 0.

(ii) g(x, y) = ex2−y2 cos(2xy)

∂2xg = ∂2

xg + ∂2yg

= ∂xex2−y2(2x cos(2xy)−2y sin(2xy))−∂yex

2−y2(2y cos(2xy)+2x sin(2xy))= ex

2−y2 (4x2 cos(2xy)− 4xy sin(2xy) + 2 cos(2xy)− 4xy sin(2xy)− 4y2 cos(2xy))−

ex2−y2 (−4y2 cos(2xy)− 4xy sin(2xy) + 2 cos(2xy)− 4xy sin(2xy) + 4x2 cos(2xy)

)= 0.

(iii) h(x, y) = ln |f(x+ iy)|, where f(z) is analytic.∂2xh = (∂2

x + ∂2y) ln

√u2 + v2 = ∂x

u∂xu+v∂xvu2+v2 + ∂y

u∂yu+v∂yvu2+v2

= (∂xu∂xu+u∂2xu+∂xv∂xv+v∂2

xv)(u2+v2)−(u∂xu+v∂xv)2(u∂xu+v∂xv)(u2+v2)2

+ (∂yu∂yu+u∂2yu+∂yv∂yv+v∂2

yv)(u2+v2)−(u∂yu+v∂yv)2(u∂yu+v∂yv)(u2+v2)2

= ((∂xu)2+(∂xv)2+(∂yu)2+(∂yv)2)(u2+v2)−2(u∂xu+v∂xv)2−2(u∂yu+v∂yv)2

(u2+v2)2

= −(u∂xu)2+(u∂xv)2−(u∂yu)2+(u∂yv)2−4uv∂xu∂xv(u2+v2)2

+ (v∂xu)2−(v∂xv)2+(v∂yu)2−(v∂yv)2−4uv∂yu∂yv(u2+v2)2 = 0 due to the Cauchy-

Riemann equations.

4. What is the most general function u = ax2 + bxy + cy2 that is thereal part of an analytic function? Construct this analytic function, andexpress it in terms of z.

First, ∂xu = ∂x(ax2 + bxy + cy2) = 2ax + by = ∂yv implies v =g(x)+2axy+ 1

2by2 for some function g = g(x). Second, ∂yu = ∂y(ax2 +

bxy + cy2) = bx + 2cy = −∂xv implies v = −12bx

2 − 2cxy + h(y) forsome function h = h(y). Comparison of coefficients gives a = −c andthus u = −cx2 + bxy + cy2 and v = −1

2bx2 − 2cxy + 1

2by2.

f(z) = f(x + iy) = −(c + i12b)z

2 = −(c + i12b)(x

2 − y2 + i2xy) =−cx2 + cy2 − i2cxy − i1

2bx2 + i1

2by2 + bxy = u+ iv.

5. Which of the following functions are analytic?

(i) f(x+iy) = e−y(cosx+i sin x) with u = e−y cosx and v = e−y sin x∂xu = −e−y sin x and ∂yv = −e−y sin x as well as ∂yu =


−e−y cosx and ∂xv = e−y cosx so that both Cauchy-Riemannequations are fulfilled.

(ii) f(z) = f(x + iy) = cosx − i sin y = z with u = cosx and v =− sin y

∂xu = − sin x and ∂yv = − cos y so that the first Cauchy-Riemann equation is violated and f is not analytic.

(iii) f(z) = f(reiθ) = r3 + i3θ.The Cauchy-Riemann equation in polar-cartesian form ∂θf =

ir∂rf here is violated: ∂θf = 3i 6= ir3r2 = ir∂rf as long as r 6= 1.

(iv) f(z) = f(reiθ) = [rer cos θ]ei(θ+r cos θ).∂θf = ( − r sin θ + i(1 − r sin θ))f and ∂rf = (1

r+ cos θ +

i cos θ))f so that ir∂rf = (i + ir cos θ − r cos θ))f . Together∂θf 6= ir∂rf implies that f is not analytic.

6. Solve the Polar CR equations given that ∂θv = 0. Express your answerin terms of a familiar function, and interpret everything you have donegeometrically.

∂θv = 0 implies v = V (r) and due to the Cauchy-Riemann equation∂ru = 0 implies u = U(θ). But ∂θu = dU(θ)

dθ= −r d V (r)

dr= −r∂rv as on

p.221 implies dU(θ)dθ

= A = −r d V (r)dr

. Integration gives U(θ) = Aθ + Cθwith constant Cθ ∈ R and V (r) = −A ln r + Cr with constant Cr ∈ R,hence f(reiθ) = Aθ+Cθ−i(A ln r−Cr) = −i(A ln r−Cr+i(Aθ+Cθ)) =−iA(ln r + iθ) +B = −iA log(z) +B for constants A ∈ R and B ∈ C.∂θv = 0, i.e. f maps circles to horizontal lines, and ∂ru = 0, i.e. f mapscircles to vertical lines. Hence as on p.220/221 we conclude f must bethe complex logarithm scaled by real A and rotated by −π

2 save forsome additive complex constant.

7. Use the Cartesian CR equations to show that the only analytic map-ping that sends parallel lines to parallel lines is the linear mapping.[Hint: Begin with the case of horizontal lines being mapped to hori-zontal lines. How does this translate into ’Equationspeak’? Now solveCR.]

If horizontal lines are mapped to horizontal lines then f(x+ iyo) =u(x, yo) + iv(x, yo) with v(x, yo) = yo so that v = v(y), hence ∂yu =−∂xv = 0 so that u = u(x). Now, from d u(x)

dx= ∂xu = a = ∂yv = d v(y)

dy

for a real constant A we get u(x) = ax + Cu and v(y) = ay + Cvwith real constants Cu = Cu(yo) and Cv = Cv(yo). Reinserting weget ∂yu = C ′u(y) = −∂xv = 0 so that Cu(y) is constant, and likewise∂xu = a = ∂yv = a + C ′v(y) so that Cv(y) is constant. Generalizationby conformality? ???


8. Calculate, then draw on a picture, a possible location for log(1 + i).Draw a small shape at 1 + i. Use the amplitwist of log(z) to draw itsimage. Verify this using your computer.

log(1 + i) = log |1 + i|+ i arg(1 + i) = log√

2 + iπ4 .

u

v

1

ilog(1 + i)

x

y

1

i 1 + i

At 1+ i we have log(1+ i+dz) ≈ log(1+ i)+ log′(1+ i) dz =√

2+ iπ4 +1−i

2 dz, e.g. for dz = a cos t + ib sin t then log(1 + i + dz) ≈√

2 + iπ4 +12(a cos t+ b sin t) + i1

2(b sin t−a cos t) ,√

2 + iπ4 + 12

(1 1−1 1

)(a cos tb sin t

)where multiplication by 1−i

2 amounts to a twist by −π4 and to an am-

plification by√

22 . The resulting image curves are represented dashed.

9. Derive the quotient rule in an analogous way to the product rule (seepage 226). [Hint: Multiply top and bottom of A/B by b− ξg′.]

Let a = f(z), b = g(z), A = a + ξf ′ and B = b + ξg′. Then in afirst order approximation A

B= a+ξf ′

b+ξg′ = (a+ξf ′)(b−ξg′)b2−(ξg′)2 ≈ (a+ξf ′)(b−ξg′)

b2−(ξg′)2 ≈ab+ξf ′b−aξg′

b2= a

b+ ξ f

′b−ag′b2

so that (fg )′(z) = f ′g−fg′

g2 (z).

10. Consider the polynomial P (z) = (z − a1)(z − a2) . . . (z − an).

(i) Show that the critical points of P (z) are the solutions of 1z−a1

+1

z−a2+ . . .+ 1

z−an = 0. (no critical point is a zero of P .)P ′(z∗) = ∑n

k=1∏n`=1,` 6=k(z∗−a`) = P (z∗)∑n

k=11

z∗−ak= 0 ⇐⇒∑n

k=11

z∗−ak= 0.

(ii) Let K be a circle with centre p. By considering the conjugate ofthe equation in (i), deduce that p is a critical point if and only ifit is the centre of mass of the inverted points IK(aj).

Let K be a circle around p with radius R, then IK(z) = R2

z−p .Hence p is a critical point ⇐⇒ 1

n

∑nk=1

1p−ak

= 0 by part (i) ⇐⇒1n

∑nk=1 IK(ak) = 1

n

∑nk=1 (

R2

ak−p+ p) = 0 + p.

(iii) Show that the equation in (i) is equivalent to ∑nk=1

z−ak|z−ak|2

= 0and by interpreting the LHS as a (positively) weighted sum of


the vectors from z to the roots of P (z), deduce Lucas’ Theo-rem: The critical points of a polynomial in C must all lie withinthe convex hull of its zeros. This is a complex generalization ofRolle’s Theorem in ordinary calculus. [Hint: Use the fact thatThe centroid Z must lie in the interior of the convex hull H. (32)on p.104 is still valid even if the masses are not equal.]

First,∑nk=1

1z−ak

= 0 ⇐⇒ ∑nk=1

1z−ak

= 0 = ∑nk=1

z−ak|z−ak|2

. Sec-ond, any critical point p thus fulfills p∑n

k=11

|p−ak|2= ∑n

k=1p

|p−ak|2=∑n

k=1ak

|p−ak|2or equivalently p = ∑n

k=1wkak with weights wk =1∑n

`=1|p−ak|2

|p−a`|2

where all wk ≥ 0 and ∑nk=1wk = ∑n

k=11∑n

`=1|p−ak|2

|p−a`|2

= 1∑n

`=11

|p−a`|2

∑nk=1

1|p−ak|2

= 1. Hence p is a convex sum of ak andas such lies in the convex hull of the zeroes of P .

11. Use (ez)′ = ez to show that the derivatives of all the trig functions aregiven by the familiar rules of real analysis.

cos′ z = (12(eiz+e−iz))′ = 1

2(ieiz−ie−iz) = 1

2i(−eiz+e−iz) = − sin z

and sin′ z = ( 12i(e

iz − e−iz))′ = 12i(ie

iz + ie−iz) = 12(e

iz + e−iz) = cos z.

12. Provided it is properly interpreted, show that (zµ)′ = µzµ−1 is still trueeven if µ is complex.

(zµ)′ = (eµ ln z)′ = eµ ln zµ1

z= µ zµ 1

z= µ zµ−1.

13. (i) If a is an arbitrary constant, show that the series f(z) = 1 + az+a(a−1)

2! z2 + a(a−1)(a−2)3! z3 + . . . converges inside the unit circle.

The ratio test on p.74 gives the radius R of convergence as R =limn→∞ |cn/cn+1| = limn→∞

∣∣∣a(a−1)·...·(a−n+1) (n+1)!n! a(a−1)·...·(a−n)

∣∣∣ = limn→∞

∣∣∣n+1a−n

∣∣∣= limn→∞

|1+1/n||a/n−1| = 1.

(ii) Show that (1+z)f ′(z) = a f(z). (We define(ak

)= a(a−1)...(a−k+1)

k! .)(1 + z)f ′ = (1 + z)(

∑∞k=0

(ak

)zk)′ = (1 + z)∑∞k=1

(ak

)kzk−1 =∑∞

k=1a(a−1)...(a−k+1)

(k−1)! zk−1 +∑∞k=1

a(a−1)...(a−k+1)(k−1)! zk =∑∞

k=0a(a−1)...(a−k)

k! zk +∑∞k=1

a(a−1)...(a−k+1)(k−1)! zk =

a+∑∞k=1

a(a−1)...(a−k+1)(k−1)!

a−k+kk

zk = a+ a∑∞k=1

(ak

)zk = af .

(iii) Deduce that [(1 + z)−af ]′ = 0.[(1 + z)−af ]′ = −a(1 + z)−a−1f + (1 + z)−a−1(1 + z)f ′ =

(1 + z)−a−1(−a+ a)f = 0.(iv) Conclude that f(z) = (1 + z)a.

Because of part (iii) the function h(z) = (1 + z)−af(z) mustbe constant, all the more we have h(z) = h(0) = 1 so that f(z) =(1 + z)a.


14. As we pointed out in Chapter 3, stereo graphic projection has a verypractical use in drawing a conformal map of the world. Once we havethis map we can go on to generate further conformal maps, simply byapplying different analytic functions to it. One particularly useful onewas discovered (using other means) by Gerhard Mercator in 1569. Wecan describe it (though he could not have) as the result of applyinglog z to the stereo graphic map.

(i) Look up both a stereographic map and a Mercator map in anatlas, and make sure you can relate the changes in shape you seeto your understanding of the complex logarithm.

.???(ii) Imagine plotting a straight-line course on a Mercator map and

then actually travelling it on the high seas. Show that as you sail,the reading of your compass never changes.

.???

15. (i) By noting that the unit tangent (in the counterclockwise direc-tion) to an origin-centred circle can be written as ξ = iz/|z|, showthat the formula κ = 1

|f ′(p)|(=[f ′′(p)ξ/f ′(p)] + κ) (23)on p.234 for the curvature of the image of such a circle can bewritten as κ = 1

|zf ′|(1 + <[zf ′′/f ′]).Obviously |ξ| = 1 and ξ ⊥ z as <(ξz) = <(i|z|) = 0. Now,

κ = 1|f ′|(=[f ′′i z|z|

1f ′

] + κ) = 1|zf ′|(=[f ′′iz 1

f ′] + |z|κ) = 1

|zf ′|(<[ zf ′′f ′

] +|z|κ) = 1

|zf ′|(1 + <[ zf ′′f ′

]) (were evaluation in z is assumed) as|z|κ = |z| 1

|z| = 1.Because f is analytic, it is conformal and thus it is sufficient tocheck on the curvature of exactly one curve, namely the circlecentered at the origin through z.

(ii) What should this formula yield if f(z) = log z? Check that itdoes.

As f ′(z = 1zand f ′′(z) = − 1

z2 we have κ = 1|zf ′|(1 + <[z f ′′

f ′]) =

1+<[z−1z2

11/z ] = 1+<(−1) = 0 as to be expected as the logarithm

maps circles centered at the origin to vertical lines.(iii) What should this formula yield if f(z) = zm? Check that it does.

What is the significance of the negative value of κ when m is neg-ative? [Hint: Which way does the velocity complex number ofthe image rotate as z travels counterclockwise round the originalcircle?]

As f ′(z) = mzm−1 and f ′′(z) = m(m − 1)zm−2 we haveκ = 1

|zf ′|(1 + <[zf ′′/f ′]) = 1|mzm|(1 + <[zm(m−1)zm−2

mzm−1 ]) = 1+m−1|mzm| =


sgn(m)|zm| . For positive m the result is to be expected because f maps

circles centered at the origin with radius r and κ = 1rto such cir-

cles with radius rm and κ = 1rm

= κm.For negative m the orientation is inverted resulting in a neg-ative curvature, see also e.g. https://en.wikipedia.org/wiki/Curvature.

16. As illustrated below, a region is called convex if all of it is visible froman arbitrary vantage point inside. Let an analytic mapping f act onan origin-centred circle C to produce a simple image curve f(C), theinterior of which is convex.

(i) From the formula of Ex.15, deduce that if f maps the interior ofC to the interior of f(C), then the following inequality holds atall points z of C: <[zf ′′/f ′] ≥ −1

Assume <[zf ′′/f ′] < −1. With Ex.15 (i) then κ < 0 indicatingthat f maps the interior of C to the exterior of f(C).

(ii) What is the analogous inequality when f maps the interior of Cto the exterior of f(C). [Hint: Ex.4 on p.211 of chapter 4.]

Ex.4 on p.211 of chapter 4 states: If a point z travels along Ccounterclockwise then f(z) travels along f(C) clockwise. Henceκ < 0 implying <[zf ′′/f ′] < −1.

17. Let S be a directed line-segment through a point p = x+ iy.

(i) Let f(z) = ez. Without calculation, decide which direction of Syields an image f(S) having vanishing curvature at f(p).

As in [17] on p.235 κ = 0 when S is horizontal. If S = t+ iy :t ∈ R then the image f(S) = et eiy : t ∈ R is a straight linewhose curvature vanishes.

(ii) The complex curvature K = if ′′

f ′|f ′| must therefore point in one ofthe two orthogonal directions. Which? By considering the imageof S when it points in this direction, deduce the value of |K|, andthereby conclude that K(p) = ie−x.

Geometrisch? ???(iii) Use K = if ′′

f ′|f ′| (28) on p.238 to verify this formula.K = i

|ez | = iex

= ie−x.

(iv) Repeat as much as possible of the above analysis in the casesf(z) = log z and g(z) = zm, where m is a positive integer.[In neither of these cases will you be able to see the exact valueof |K(p)|.]

As f ′(z) = 1zand f ′′(z) = − 1

z2 we have K = −i/z2

|1/z|/z = −i|z|z

=

https://en.wikipedia.org/wiki/Curvature

https://en.wikipedia.org/wiki/Curvature


−i z|z| . ???As g′(z) = mzm−1 and g′′(z) = m(m − 1)zm−2 we have K =im(m−1)zm−2

mzm−1|mzm−1| = i(m−1)zm|z|m−1 = m−1

miz

|z|m+1 = m−1m

i|z|m

z|z| .???

(v) According to the geometric reasoning in Ex.18 on p.213, the am-plification of a Möbius transformation M(z) = az+b

cz+d is constanton each circle centred at −d

c. Thus the complex curvature of M

should be tangent to these concentric circles. Verify this by cal-culating K.

M ′(z) = ad−bc(cz+d)2 and M ′′(z) = −2c(ad−bc)

(cz+d)3 so that K = if ′′

f ′|f ′| =i2c(bc−ad)(cz+d)3

ad−bc(cz+d)2

| ad−bc(cz+d)2

|= −i2c|cz+d|2

(cz+d)|ad−bc| = −i2c(cz+d)|ad−bc| . Now, for z = −d

c+ reiφ

on such a circle we have cz + d = creiφ and that K = −i2ccreiφ|ad−bc| =

2|c|2|ad−bc|e

i(φ+π/2+π) is orthogonal to z.(vi) Use a computer to verify figure [21] on p.240 for all four mappings

above..???

18. Let two curves C1 and C2 emerge from a point p in the same direction.Two examples are illustrated below. Although in both cases the angleat p is zero, there is a great temptation to say that the curves on theright meet at a smaller ’angle’ than those on the left.

C1

C2

C3

C4

Any putative definition of such an ’angle’ Θ should (presumably) beconformally invariant: if the curves are mapped to C1 and C2 by amapping f that preserves ordinary angles (i.e., an analytic mapping),then the new ’angle’ Θ should equal the old ’angle’ Θ.

(i) Newton [1670] attempted to define such a Θ as the difference ofthe curvatures of C1 and C2 at p: Θ = κ1 − κ2.Use κ = K · ξ+ κ

|f ′| (30) on p.239 to show that this definitionis not quite conformally invariant: Θ = Θ/|f ′(p)|.

Θ = κ1 − κ2 = (K · ξ + κ1|f ′|)− (K · ξ + κ2

|f ′|) = κ1|f ′| −

κ2|f ′| = Θ

|f ′| .(ii) Consider an infinitesimal disc D (radius ε) centred at p. Let c1

and c2 be the centres of curvature of C1 and C2, and let D be


the difference between the angular sizes of D as seen from c1 andc2. Show that D = εΘ. If a conformal mapping f is applied toC1, C2, and D, deduce that D = D. [Of course this is not whatwe were after: D is (a) infinitesimal, and (b) not defined by thecurves alone. The discovery of a true conformal invariant had toawait Kasner [1912]. See Ex.10 on p.571.]

. ???

19. In more advanced work on Mobius transformations (e.g., Nehari [1952]and Beardon [1984]), an important role is played by the so-called Schwar-zian derivative f(z), z := (f

′′

f ′ )′− 1

2(f ′′

f ′ )2 of an analytic function f(z)

with respect to z. (Another notation is (Sf)(z) = f(z), z.)

(i) Show that the Schwarzian derivative may also be written asf(z), z = f ′′′

f ′− 3

2(f ′′

f ′ )2.

(f′′

f ′ )′ − 1

2(f ′′

f ′ )2 = f ′′′ f ′−(f ′′)2

(f ′)2 − 12(

f ′′

f ′ )2 = f ′′′

f ′− 3

2(f ′′

f ′ )2.

(ii) Show that az + b, z = 0 = (1/z), z.First, for f(z) = az + b we have f ′(z) = a, f ′′(z) = 0 = f ′′′(z)

and thus az+ b, z = 0a− 3

2(0a)

2 = 0 and second, for g(z) = 1zwe

have g′(z) = − 1z2 , g′′(z) = 2

z3 , g′′′(z) = − 6z4 and thus g(z), z =

g′′′

g′− 3

2(g′′

g′ )2 = −6z2

−z4 − 32(

2z2

−z3 )2 = 6

z2 − 32

4z2 = 0.

(iii) Let f and g be analytic functions, and write w = f(z). Show thatthe Schwarzian derivative of the composite function g[f(z)] =g[w] is given by the following "chain rule":g(w), z = [f ′(z)]2g(w), w+ f(z), z.

With (g f)′ = (g′ f)f ′ and (g f)′′ = (g′′ f)(f ′)2 + (g′ f)f ′′ we have (gf)′′

(gf)′ = (g′′f)(f ′)2

(g′f)f ′ + (g′f)f ′′(g′f)f ′ = (g′′f)f ′

(g′f) + f ′′

f ′. Then

g(w), z = ( (gf)′′(gf)′ )

′− 12(

(gf)′′(gf)′ )

2 = ( (g′′f)f ′(g′f) )

′+(f′′

f ′ )′− 1

2((g′′f)f ′

(g′f) )2−

(g′′f)f ′(g′f)

f ′′

f ′ −12(

f ′′

f ′ )2 = (g

′′

g′ f)′f ′ + (g

′′

g′ f)f′′ − 1

2(g′′

g′ f)2(f ′)2 −

(g′′

g′ f)f′′ + f(z), z = (g

′′

g′ f)′f ′ − 1

2(g′′

g′ f)2(f ′)2 + f(z), z =

(f ′)2((g′′

g′ )′ − 1

2(g′′

g′ )2) f + f(z), z = [f ′(z)]2g(w), w+ f(z), z.

(iv) Use the previous two parts to show that all Möbius transforma-tions have vanishing Schwarzian derivative. [Hint: Recall thatthe mappings in part (ii) generate (via composition) the set of allMöbius transformations.] Remark: Ex.19 on p.424 shows that theconverse is also true: If f(z), z = 0 then f is a Möbius trans-formation. Thus Möbius transformations are completely charac-terized by their vanishing Schwarzian derivative.

As a Möbius transformation is a composition of affine trans-formations and inversion. With part (iii) the Schwarzian of any


Möbius transformation vanishes.(v) Use the previous two parts to show that the Schwarzian deriva-

tive is ’invariant under Möbius transformations’, in the followingsense: if M is a Möbius transformation, and f is analytic, thenM [f(z)], z = f(z), z.

With parts (iii) we have M [f(z)], z = [f ′(z)]2M(w), w +f(z), z where the first summand vanishes due to part (iv) sothat M [f(z)], z = f(z), z follows.

20. Think of the real axis as representing time t, and let a moving particlew = f(t) (where f(z) is analytic) trace an orbit curve C. The velocityis then v = w = f ′(t).

(i) Use κ = 1|f ′(p)|(=

f ′′(p)ξf ′(p) + κ) (23)

on p.234 to show that the curvature of C is κ = 1|v|=(v/v).

As κ = 0 for the real line and with f ′′ = v and ξ = v|v| = v

|f ′(t)|

we have κ = 1|v|=

vv/|v|v

= 1|v|=

v|v| .???

(ii) Argue that this result does not in fact depend on C being pro-duced by an analytic mapping, but is instead true of any motionfor which the velocity v and acceleration v are well-defined.

.???(iii) Show that the formula may be rewritten as κ = =(vv)

|v|3 .κ = 1

|v|=(v/v) = 1|v|3=(|v2|v/v) = 1

|v|3=(vvv/v) = 1|v|3=(vv).

(iv) Deduce that it may also be written vectorially as κ = |v×v||v|3 .

Let v = vx + ivy and a = v = ax + iay, then on one hand=(av) = =[(ax + iay)(vx − ivy)] = ayvx − axvy and on the other

hand |v×a| = |

∣∣∣∣∣∣∣ex ey ezvx vy 0ax ay 0

∣∣∣∣∣∣∣ | = |(0, 0, vxay−axvy)| = |vxay−axvy|so the two expressions for the curvature are identical except forthe orientation.

By considering C to be the "osculating plane" (see Hilbert [1932]) ofa curve in 3-dimensional space, we see that this formula holds in thatcase also.

21. In 3-dimensional space, let (X, Y, Z) be the coordinates of a movingparticle. If X = a cos(ωt), Y = a sin(ωt), Z = bt, then the path tracedby the particle is a helix.

(i) Give interpretations for the numbers a, ω, and b.a is the radius of the cylinder on which the particle moves,


ω with |v| = |(−aω sin(ωt), aω cos(ωt), b)| =√a2ω2 + b2 is the

angular frequency of the rotation of the projection into the realplane. Let s = ωt. Then Z = bt = b

ωs so that b

ωis the slope on

the cylinder.(ii) If a and ω remain fixed, what does the helix look like in the two

limiting cases of b becoming very small or very large? What if aand b remain fixed while ω becomes very small or very large?

If a and ω are fixed, then the helix for very small b approachesa circle and for very large b approaches a line.If a and b are fixed and ω becomes very small or very large thenthe slope on the cylinder becomes infinity (the helix approaches avertical line) or zero (the helix approaches a circle), respectively.

(iii) What limiting values would you anticipate for the curvature ofthe helix for each of the limiting cases considered in (ii)?

If a and ω are fixed and b becoming very small or very largethen κ approaches 1

aor 0 respectively.

If a and b are fixed and ω becomes very small or very large thenκ approaches 0 or some constant, respectively.

(iv) Use Ex.20 (iv) to show that the curvature of the helix is κ =aω2

b2+a2ω2 and use this to confirm your hunches in (iii).With v = (−aω sin(ωt), aω cos(ωt), b) and |v| =

√a2ω2 + b2,

v× v =

∣∣∣∣∣∣∣ex ey ez

−aω sin(ωt) aω cos(ωt) b−aω2 cos(ωt) −aω2 sin(ωt) 0

∣∣∣∣∣∣∣ =

abω2 sin(ωt)

abω2 cos(ωt)a2ω3

and

|v× v| =√a2b2ω4 + a2ω6 = aω2√b2 + a2ω2 we have κ = |v×v|

|v|3 =aω2√b2+a2ω2√a2ω2+b23 = aω2

b2+a2ω2 .If a and ω are fixed then limb→0 κ = limb→0

aω2

b2+a2ω2 = 1aand

limb→∞aω2

b2+a2ω2 = 0. If a and b are fixed then limω→0aω2

b2+a2ω2 = 0and limω→∞

aω2

b2+a2ω2 = 1.

22. Continuing from Ex.20, take f(z) to be a general Möbius transfor-mation: w = f(t) = at+b

ct+d where ∆ = (ad − bc) 6= 0. Show that thecurvature of this path is κ =

∣∣∣2c2∆ =dc

∣∣∣ in agreement with Ex.18 on p.186.The fact that this is constant provides a new proof that the image is acircle, for only circles have constant curvature.

As f ′(t) = a(ct+d)−(at+b)c(ct+d)2 = ∆

(ct+d)2 and f ′′(t) = −2 c∆(ct+d)3 we have κ =

1|v|

∣∣∣= vv

∣∣∣ = |ct+d|2|∆|

∣∣∣=( −2c∆(ct+d)3

(ct+d)2

∆ )∣∣∣ = 2|ct+d|2

|∆|

∣∣∣= cct+d

∣∣∣ = 2|∆|

∣∣∣=(c(ct+ d))∣∣∣

= 2|∆|

∣∣∣=(cd)∣∣∣ = 2

|∆|

∣∣∣=(cd)∣∣∣ = 2|c|2

|∆|

∣∣∣= dc

∣∣∣ = 2|c|2|∆|

∣∣∣=(dc)∣∣∣ =∣∣∣2c2∆ =

dc

∣∣∣.23. As another continuation of Ex.20, let us see how the Schwarzian deriva-


tive f(z), z of Ex.19 arises rather naturally in the context of curva-ture.

(i) Show that d κdt

= 1|f ′|=f(z), z. [This formula was discovered by

G. Pick. For elegant applications, see Beardon [1987]. For an-other connection between curvature and Schwarzian derivatives,see Ex.28 (iii).]

With Ex.20 (i) κ = 1|v|=

vvso that d κ

dt= d

dt(1|f ′|=

f ′′

f ′ ) =???1|f ′|=f(z), z = 1

|f ′|=(f ′′′

f ′− 3

2(f ′′

f ′ )2) =.

(ii) Use Ex.19, part (iv) to deduce that if f(z) is a Möbius transfor-mation then d κ

dt= 0. Why is this result geometrically obvious?

Ex.19, part (iv) gives f(z), z = 0 implying d κdt

= 0.f maps the real line to a circle. Its curvature is constant (cp. theprevious exercise) so that the derivative of its curvature vanishes.

24. Let the position at time t of a moving particle in C be z(t) = r(t)eiθ(t).

(i) Show that the acceleration of the particle is z = [r − rθ2]eiθ +[2rθ + rθ]ieiθ

From z = [r+irθ]eiθ we get z = [r+irθ+irθ+(r+irθ)iθ]eiθ =[r − rθ2]eiθ + [2rθ + rθ]ieiθ.

(ii) What are the radial and transverse components of the accelera-tion?

Obviously, [r − rθ2]eiθ is the radial and [2rθ + rθ]ieiθ thetransversal component of the acceleration.

(iii) If the particle is moving in a central force field, with the centreof force at the origin, deduce that the areal speed A = 1

2r2θ is

constant. For a beautiful geometric proof of this fact, see Newton[1687, p.40].

.???

25. Sometimes the circle of convergence of a power series is so denselypacked with singularities that it becomes a genuine barrier for thegeometric mapping, beyond which it cannot be continued. This iscalled a natural boundary. An example of this is furnished by f(z) =z + z2 + z4 + z8 + z

6 + . . . which converges inside the unit circle. Showthat every point of |z| = 1 is either a singularity itself, or else has sin-gularities arbitrarily near to it. [Hint: What is f(1)? Now note thatf(z) = z+ f(z2), and deduce that f is singular when z2 = 1. Continu-ing in this manner, show that the 2n-th roots of unity are all singular.]

f(z) = ∑∞k=0 z

2k diverges for z = 1 so 1 is a singularity of f .As z + f(z2) = z + ∑∞

k=0 (z2)

2k = z + ∑∞k=0 z

2k+1 = f(z) all second


roots of unity are singularities. The same idea gives z + z2 + f(z4) =z + z2 + z + ∑∞

k=0 z2k+2 = f(z) so that all fourth roots of unity are

singularities. With the 2n roots of unity for all n ∈ N the singularitiesare dense on the unit circle.

26. Unlike inversion in a circle, show that Schwarzian reflection in an ellipseE (see figure [31]) does not interchange the interior and the exterior.Indeed, how does RE(z) behave for large values of |z|?

. ???

27. (i) If L is a line passing through the real point X, and making anangle σ with the horizontal, then show that its Schwarz functionis SL(z) = ze−i2σ +X(1− e−i2σ).

The line L is given by z(t) = X + eiσt. Then SL(z(t)) =(X + eiσt)e−i2σ + X(1 − e−i2σ) = X + e−iσt = z(t). As affinetransformation SL(z) is also analytic and hence the only functionwith SL(z(t)) = z(t). Uniqueness?

(ii) If C is a circle with centre p and radius r, show that its Schwarzfunction is SC(z) = p+ r2

z−p .Again SC(z) is analytic and for z(t) = p + reit we have

SC(z(t)) = p+ r2

reit= p+ re−it = z(t). Uniqueness?

(iii) Verify the claim that in both these cases z → R(z) is the ordinaryreflection, even if z is far from the curve.

. ???

28. Let a be a point on a (directed) curve K having Schwarz function S(z).

(i) Show that the curvature of K at a is κ ≡ φ = i2S′′(a)

[S′(a)]3/2 where φis the angle in [30] on p.255, and the dot denotes differentiationwith respect to distance ` along K (in the given sense). Deducethat |κ| = |S ′′/2|. [Hints: Since S is analytic, so is S ′. Thusto calculate S ′′ = dS′

dzwe need only find the change dS ′ in S ′

produced by an infinitesimal movement dz of z, taken in anyonedirection of our choosing. At a let us choose dz along K, so thatdz = eiφd`. The corresponding change in S ′ is then determinedsolely by the shape of K, for the values of S ′ on K are given byS ′ = e−2iφ.]

. ???(ii) Deduce that the centre of curvature of K at a is a+ 2 S′(a)

S′′(a) .. ???

(iii) Show that the rate of change of the curvature of K is given bythe "Schwarzian derivative" [Ex.19] of the Schwarz function: κ =


i2S′S(z), z.

.???

29. Check the result of Ex.28 (i) by applying it to the results of Ex.27.The line L in Ex.27 (i) has curvature 0 and its Schwarz function

is SL(z) = ze−i2σ + X(1 − e−i2σ), then S ′′ ≡ 0 is consistent with κ =i2S′′(a)

[S′(a)]3/2 of Ex.28 (i).

30. Let a be a point on a curveK having Schwarz function S(z). By the stillunproven result on the infinite differentiability of analytic functions,S(z) may be expanded into a Taylor series in the vicinity of a: S(z) =S(a) + S ′(a)(z − a) + 1

2!S′′(a)(z − a)2 + 1

3!S′′′(a)(z − a)3 + . . ..

(i) Show that the Schwarz function of the tangent line to K at a isgiven by the first two terms of the series above. This reconfirmssomething we saw in [30] on p.255: very close to a, reflection inthe tangent is a good approximation to Schwarzian reflection.

.???(ii) It is natural to suspect that a better approximation to RK(z)

would be inversion in the circle of curvature (call it C) of K ata. Let’s verify this. Use Ex.28 (ii) and Ex.27 (ii) to find SC , andshow that it may be written as where it is understood that thederivatives are all evaluated at a. Show that the first three termsin the binomial expansion of SC agree with those of S, but thatthey generally differ thereafter. [Hint: You will need the fact that(S ′/S ′′) = −(S ′)2/S ′′ on K. Prove this.]

.???(iii) If the curvature κ of K were constant then K would be identical

to its circle of curvature. The fact that S and SC disagree beyondthe third term thus reflects the fact that κ does change. Oneis thus led to guess that the faster κ changes, the greater thediscrepancy between RK and inversion in C. Continuing fromthe last part, use Ex.28 (iii) to verify this hunch in the followingprecise form: SC(z)− S(z) ≈ i

3 [S ′]2κ(z − a)3..???

31. Let C and D be intersecting circles. Let us say that "D is symmetricin C" if reflection (inversion) in C maps D into itself. We know thisoccurs if and only if D is orthogonal to C, so "D is symmetric in C⇐⇒ C is symmetric in D". Briefly, we may simply say that "C and Dare symmetric". Let’s see what happens if we generalize C and D tointersecting arcs possessing Schwarz functions, and generalize inversionto Schwarzian reflection.


(i) Explain why the statement "D is symmetric in C" is the same as"if the point d lies on D then RD[RC(d)] = RC(d)". Must thearcs be orthogonal?

. ???(ii) If D is symmetric in C, deduce that the mappings RD RC andRC RD are equal at points of D.

. ???(iii) Using the fact that these two mappings are analytic (why?), de-

duce that C must also be symmetric in D. Thus, as with circles,we may simply say that C and D are symmetric.

. ???


Chapter 6

Non-Euclidean Geometry*


1. Draw a geodesic triangle ∆ on the surface of a suitable fruit or veg-etable. Now draw a geodesic segment from one of the vertices to anarbitrary point of the opposite side. This divides ∆ into two geodesictriangles, say ∆1 and ∆2. Show that the angular excess function E isadditive, i.e. E(∆) = E(∆1) + E(∆2). By continuing this process ofsubdivision, deduce that (5) on p.275 implies (6) on p.275 wherefor an infinitesimal triangle ∆ of area dA E(∆) = k(p) dA (5)for any triangle T E(T ) =

∫∫T k(p) dA (6)

Let ∆ = ∆(ACB) so that ∆1 = ∆(ACD) and ∆2 = ∆(BDC), thenE(∆) = ∠(BAC) +∠(CBA) +∠(ACB)− π = ∠(DAC) +∠(CBD) +(∠(ACD)+∠(DCB))−π+(∠(CDA)+∠(BDC))−π = (∠(DAC)+∠(ACD) +∠(CDA)− π)+ (∠(CBD) +∠(BDC) +∠(DCB)− π) =E(∆1) + E(∆2).

111

112 CHAPTER 6. NON-EUCLIDEAN GEOMETRY*

xy

z

A

B

C D

Continuing in that fashion any triangle T can be thought of as (infinite)sum of infinitesimal triangles, hence (5) implies (6) on p.275.

2. Explain Reflection of Σ in a line induces reflection (inversion)of C in the stereographic image of that line. (18)

on p.287 by generalizing the argument that was used to obtain the spe-cial case Inversion of C in the unit circle induces a reflection

of the Riemann sphere in its equatorial plane, C. (17)on p.143. That is, think of reflection of the sphere in terms reflectionof space in a plane Π, as in [8] on p.280. Also, think of stereographicprojection as the restriction to the sphere of the three-dimensional in-version IK , where K is the sphere of radius

√2 centred at the north

pole of Σ (see [13b] on p.285). Now let a be a point on Σ, and considerthe effect of IK on a, Π, and RΠ(a).

For N = (0, 1) and a = (|a|, 0) and a = (xo, yo) we have a =( 2|a|+1/|a| , 1 −

xo|a|), then |Na|2|Na|2 = (x2

o + (yo − 1)2)(|a|2 + 1) =x2o(1 + 1

|a|2 )(|a|2 + 1) = 4

(|a|+1/|a|)2(|a|2+1)2

|a|2 = 4 = (√

22)

2 so that stereo-graphic projection actually is identical to IK .

O

N

C

Σ

K

a

a

√2 1


Further, IK conformally transforms circles to circles (cp p.142), espe-cially the great circle CΠ in which the plane Π intersects the sphere Σand the great circle C⊥Π through a and RΠ(a) in Π⊥ orthogonal Π sothat CΠ and C⊥Π are orthogonal also. Hence also IK(CΠ) and IK(C⊥Π )are orthogonal and as such inverse to each other1. It remains to showthat IK(a) and IK(RΠ(a)) are inverse w.r.t. IK(CΠ). ???

3. Let C be a circle in C, and let C be its stereographic image on Σ. If Cis a great circle, then Reflection of Σ in a line induces reflection

(inversion) of C in the stereographic image of that line. (18)on p.287 says that IC stereographically induces reflection of Σ in C, butwhat transformation is induced if C is an arbitrary circle? Generalizethe argument of figure [14] on p.287 to show that IC becomes projectionfrom the vertex v of the cone that touches Σ along C. That is, ifw = IC(z) then w is the second intersection point of Σ with the line inspace that passes through the vertex v and the point z. Explain how(18) may be viewed as a limiting case of this more general result.

. ???

4. Use F M = M F for F (z) = z−ξ+z−ξ− or M(z)−ξ+

M(z)−ξ− = m z−ξ+z−ξ− (41)

on p.166 to show that if the Mobius transformation M(z) has fixedpoints ξ±, and the multiplier associated with ξ+ is m, then [M ] =[1 −ξ+1 −ξ−

]−1 [√m 00 1/

√m

] [1 −ξ+1 −ξ−

]. By putting ξ+ = a, m = e−iψ,

and ξ− = − 1a, deduce [Rψ

a ] =[eiψ/2|a|2+e−iψ/2 2ia sin(ψ/2)

2ia sin(ψ/2) e−iψ/2|a|2+eiψ/2](19)

on p.288. [Hint: Remember that you are free to multiply a Mobiusmatrix by a constant.]

[M ] =[1 −a1 1/a

]−1 [e−iψ/2 0

0 eiψ/2

] [1 −a1 1/a

]=[1/a a−1 1

] [e−iψ/2 −e−iψ/2aeiψ/2 eiψ/2/a

]=[e−iψ/2/a+ eiψ/2a (−e−iψ/2 + eiψ/2)a/a−e−iψ/2 + eiψ/2 e−iψ/2a+ eiψ/2/a

]=[eiψ/2|a|2 + e−iψ/2a a(eiψ/2 − e−iψ/2)a(eiψ/2 − e−iψ/2) e−iψ/2|a|2 + eiψ/2

]=[eiψ/2|a|2 + e−iψ/2a 2ia sin(ψ/2)

2ia sin(ψ/2) e−iψ/2|a|2 + eiψ/2

]with multiplication by a

1/a+a .

5. Show that the Mobius transformations Rψa (z) = Az+B

−Bz+A (20)on p.288 satisfy the differential equation f ′(z) = 1+|f(z)|2

1+|z|2 (17)on p.286.

f ′(z)(1 + |z|2) = A(−Bz+A)+B(Az+B)(−Bz+A)2 (1 + |z|2) = 1 + Az+B

−Bz+AAz+B−Bz+A =

1 + |f(z)|2 ⇐⇒ AA+BB−Bz+A (1 + |z|2) = (−Bz+A)(−Bz+A)+(Az+B)(Az+B)

−Bz+A =(AA+BB)(1+zz)

−Bz+A ⇐⇒ −Bz + A = −Bz + A ???1cp. T. Lewis, Univ. of Alberta www.math.ualberta.ca/~tlewis/343_10/03sec.pdf

www.math.ualberta.ca/~tlewis/343_10/03sec.pdf


6. (i) The conjugate V of a quaternion V = v + V is defined to bethe conjugate transpose V∗ of the corresponding matrix. Showthat V := V∗ = v −V, and deduce that V is a pure quaternion(analogous to a purely imaginary complex number) if and only ifV = −V.

If V = v + V = v + (v1, v2, v3) = v1 + v1I + v2J + v3K =[v + iv3 iv1 − v2iv1 + v2 v − iv3

]then V =

[v − iv3 −iv1 + v2−iv1 − v2 v + iv3

]= v − V.

Now, V by definition is a pure quaternion ⇐⇒ v = 0 ⇐⇒v = −v ⇐⇒ V = V = −(−V) = −V.

(ii) The length |V| of V is defined (by analogy with complex numbers)by |V|2 = VV. Show that |V|2 = v2 + |V|2 = |V|2.|V|2 = VV = (v + V)(v −V) = (v1 + v1I + v2J + v3K)(v1−

v1I− v2J− v3K) = (v21− vv1I− vv2J− vv3K) + (vv1I + v211−

v1v2K + v1v3J) + (vv2J + v1v2K + v221− v2v3I) + (vv3K− v1v3J +

v2v3I+v231) = (v2+v2

1 +v22 +v2

3)1 = v2+|V|2 = v2+|−V|2 = |V|2.(iii) If |V| = 1, then V is called a unit quaternion. Verify that Rψ

v [seeRψ

v = cos(ψ/2) + v sin(ψ/2) with |v| = 1 (28)on p.292] is a unit quaternion, and that Rψ

v = [Rψv ]∗ = R−ψv .

|Rψv |2 = cos2(ψ/2)+|v|2 sin2(ψ/2) = cos2(ψ/2)+sin2(ψ/2) = 1

since |V| = 1, and Rψv = cos(ψ/2) + v sin(ψ/2) = cos(ψ/2) −

v sin(ψ/2) = cos(−ψ/2) + v sin(−ψ/2) = R−ψv .(iv) Show that VW = WV and deduce that |VW| = |V| |W|. Thus,

for example, the product of two unit quaternions is another unitquaternion.

VW = (v1+v1I+v2J+v3K)(w1+w1I+w2J+w3K) = (vw1+vw1I+vw2J+vw3K)+(v1wI−v1w11+v1w2K−v1w3J)+(v2wJ−v2w1K − v2w21 + v2w3I) + (v3wK + v3w1J − v3w2I − v3w31) =(vw−v1w1−v2w2−v3w3)1+(vw1 +v1w+v2w3−v3w2)I+(vw2−v1w3 + v2w + v3w1)J + (vw3 + v1w2 − v2w1 + v3w)K and henceVW = (vw−v1w1−v2w2−v3w3)1− (vw1 +v1w+v2w3−v3w2)I−(vw2−v1w3+v2w+v3w1)J−(vw3+v1w2−v2w1+v3w)K andWV =(w1−w1I−w2J−w3K)(v1− v1I− v2J− v3K) = (vw1− v1wI−v2wJ− v3wK) + (−vw1I− v1w11 + v2w1K− v3w1J) + (−vw2J−v1w2K− v2w21 + v3w2I) + (−vw3K + v1w3J− v2w3I− v3w31) =(vw − v1w1 − v2w2 − v3w3)1 + (−v1w − vw1 + v3w2 − v2w3)I +(−v2w − v3w1 − vw2 + v1w3)J + (−v3w + v2w1 − v1w2 − vw3)K.Now, |VW|2 = VWVW = VWWV = V|W|2V = |V|2|W|2.

(v) Show that V is a pure, unit quaternion if and only if V2 = −1.V2 = (v + v1I + v2J + v3K)2 = (v − v2

1 − v22 − v2

3)1 + 2v(v1I +v2J + v3K). Now, if V is pure and of unit length then v = 0 and


v21 + v2

2 + v23 = 1, hence V2 = −1. If on the other hand V2 = −1

then the assumption v 6= 0 implies v1 = v2 = v3 = 0 and thus thecontradiction V2 = v2 > 0 > −1. Hence v = 0 and v2

1 +v22 +v2

3 = 1so that V is pure and of unit length.

(vi) Show that any quaternion V can be expressed as |V|Rψu for some

unit vector u and some ψ.Let V = v+ V = |V|( v√

v2+v21+v2

2+v23

+ V√v2+v2

1+v22+v2

3) with V =

(v1, v2, v3) so that V = |V|( v√v2+v2

1+v22+v2

3+ V|V|

|V|√v2+v2

1+v22+v2

3) =

|V|( cosφ+u sinφ) = |V|Rψu with cosφ := v√

v2+v21+v2

2+v23implying

sin2 φ = 1−cos2 φ = v2+v21+v2

2+v23−v

2√v2+v2

1+v22+v2

32 = |V|2√

v2+v21+v2

2+v23

2 , and u := V|V|

and φ = ψ2 .

(vii) Suppose we generalize the transformation P = RψvPR−ψv (29)

on p.292 where Rψv rotates P = (x, y, z) into P = (x, y, z) and P

and P is represented by P = xI + yJ + zK and P = xI + yJ + zKrespectively to V 7→ V = WVW, where W is an arbitrary quater-nion. When interpreted in this way, deduce that W representsa dilative rotation of space, and the product of two quaternionsrepresents the composition of the corresponding dilative rotations.[This confirms the claim at the end of Chapter 1.]

Since by part (vi) any quaternion W can be represented asW = |W|Rψ

u and W = |W|R−ψu the mapping V 7→ V = WVW =|W|Rψ

uV|W|R−ψu = |W|2 RψuVR−ψu is a dilatation and a rotation. ???

7. The following proof of P = RψvPR−ψv (29)

on p.292 is based on a paper of H.S.M. Coxeter [1946].

(i) Use VW = −V ·W + V×W for pure V and W (27)on p.291 to show that the pure quaternions V and W are orthog-onal if and only if VW + WV = 0.

Two quaternions V and W are orthogonal iff their scalar prod-uct in R4 vanishes. https://www.researchgate.net/post/How_do_I_calculate_the_smallest_angle_between_two_quaternionsTwo pure quaternions V = V and W = W are orthogonal ifftheir scalar product V ·W in R3 vanishes iff VW = −WV iffVW + WV = 0.

(ii) If V has unit length, so that V2 = −1 deduce that the previousequation may be expressed as W = VWV.

VW + WV = 0 ⇐⇒ −VW = WV ⇐⇒ W = VWV.(iii) Now keep the pure, unit quaternion V fixed, but let W represent

an arbitrary pure quaternion. Let ΠV denote the plane with nor-

https://www.researchgate.net/post/How_do_I_calculate_the_smallest_angle_between_two_quaternions

https://www.researchgate.net/post/How_do_I_calculate_the_smallest_angle_between_two_quaternions


mal vector V that passes through the origin, so that its equationis W ·V = 0. Consider the transformation W 7→W′=VWV (49)Show that (a) W′ is automatically pure, and |W′| = |W|, so that(49) represents a motion of space; (b) every point on ΠV remainsfixed; (c) every vector orthogonal to ΠV is reversed. Deduce that(49) represents reflection RΠV of space in the plane ΠV.

(a) W′ is by exercise 6 (i) pure iff W′ = −W′ where W′ =VWV = VWV = (−V)(−W)(−V) = −W′ by exercise 6 (i) sinceV and W are pure quaternions. By part (iv) of exercise 6 wehave |W′| = |W| since V has unit length. So the transformationW 7→W′ preserves lengths and thus is a motion.(b) Let W be in ΠV, then W ·V = 0. Now, identify W with purequaternion W, then W′ = (VW)V = ( −V ·W + V ×W)V =(0+V×W)V = −(V×W)·V+(V×W)×V = − det(V,W,V)+(V ·V)W− (W ·V)V = −0 + 1W− 0V = W.(c) Every vector W orthogonal to ΠV is a multiple of V, sayW = λV, so that W′ = V(λV)V = (λV)V2 = −λV = −W.

(iv) Deduce that if the angle from ΠV to a second plane ΠW is ψ/2,and the unit vector along the intersection of the planes is U,then the rotation Rψ

U = RΠW RΠV is given by P 7→ P =(WV)P(VW) = (−WV)P(−WV).

By part (iii) RΠV maps P to VPV which in turn is mappedby RΠW to WVPVW. Now, V and W are pure so that V = −Vand V = −V which implies VW = VW = WV. Mapping P toWVPVW = (−WV)P(−WV) by Ex.7 (iv) represents a dilativerotation. The two reflections???

(v) Use VW = −V ·W + V×W for pure V and W (27)on p.291 to show that −WV = cos(ψ/2) + V sin(ψ/2), therebysimultaneously proving (29) and (28).

.???

8. Here is another proof of P = RψvPR−ψv (29)

on p.292. As in the text, we shall assume that P is a unit vector with itstip at the point p on the unit sphere. If we represent the stereographicimages p and p of p and ˆp by their homogeneous coordinate vectors ~pand ~p in C2, then we know that the rotation is represented as ~p 7→ ~p =Rψ

v~p where Rψv is being thought of as a 2× 2 matrix.

(i) Show that in homogeneous coordinates,X + iY = 2z

1+|z|2 = 2(x+iy)1+x2+y2 and Z = |z|2−1

|z|2+1 (20)on p.146, becomes X + iY = 2p1p2

|p1|2+|p2|2 and Z = |p1|2−|p2|2|p1|2+|p2|2 .

In homogeneous coordinates ~p = [p1, p2]> we have z = p1p2,


cp pp157, hence X + iY = 2z1+|z|2 = 2p1/p2

1+|p1|2/|p2|2 = 2p1p2|p1|2+|p2|2 and

Z = |z|2−1|z|2+1 = |p1|2−|p2|2

|p1|2+|p2|2 .(ii) To simplify this, recall that all multiples of ~p describe the same

point p in C. We can therefore choose the ¨length¨ of ~p to be√

2:〈~p,~p〉 = |p1|2 + |p2|2 = 2.With this choice, show that the above equations can be written

as[

1 + Z X + iYX − iY 1− Z

]=[p1p1 p1p2p2p1 p2p2

]=[p1p2

][p1, p2] = ~p~p∗.

Part (i) implies X + iY = p1p2 and X − iY = X + iY =p1p2 = p2p1 and 1 + Z = |p1|2+|p2|2+|p1|2−|p2|2

|p1|2+|p2|2 = p1p1 and 1 − Z =|p1|2+|p2|2−|p1|2+|p2|2

|p1|2+|p2|2 = p2p2.

(iii) Verify that[

1 + Z X + iYX − iY 1− Z

]= 1− iP.

1 − iP = 1 − i(XI + Y J + ZK) =[1 00 1

]− iX

[0 ii 0

]−

iY

[0 −11 0

]− iZ

[i 00 −i

]=[

1 + Z X + iYX − iY 1− Z

].

(iv) Deduce that 1− iP = Rψv(1− iP)[Rψ

v ]∗ = 1− iRψvPR−ψv from which

(29) follows immediately.. ???

9. (i) Figure [40a] on p.324 gave a two-step process for carrying a pointz in the Poincaré disc to the corresponding point z′ in the Kleinmodel, pp.301. Explain why the net mapping z 7→ z′ of the disc toitself is the one shown in figure [a] below, where C is an arbitrarycircle passing through z and orthogonal to the unit circle U .

Σ

Cz

z′

|z′|

a

b

[b]

C

U

z′

z

[a]

. ???(ii) Figure [b] is a vertical cross section of [40a] on p.324 through z

and z′. Deduce that |z′|a

= 1band a

|z| = 2b. By multiplying these


two equations, deduce that z′ = 2z1+|z|2 . [Thus we have a geometric

explanation of the result X + iY = 2z1+|z|2 and Z = |z|2−1

|z|2+1 (20)on p.146.]

The upper dashed triangle at the north pole and the lowerdashed triangle at the south pole are similar, so |z′|

a= long cathetus

hypotenuse

= 1b. Further we have a

2 = short cathetushypotenuse = |z|

bimplying a

|z| = 2b.

Together we have |z′|a

a|z| = 1

b2band hence |z′| = 2|z|

b2= 2|z|

12+|z|2 . zand z′ pointing into the same direction implies z′ = 2z

12+|z|2 .Zu zeigen Z = |z|2−1

|z|2+1???

(iii) This formula can be derived directly from figure [a], without theassistance of the hemisphere. Redraw the figure with C chosenorthogonal to 0z. Explain geometrically why the centre of C maybe viewed as either IU(z′), or as the midpoint of z and IU(z).Conclude that 1

z′= IU(z′) = 1

2 [z + IU(z)] = 12 [z + 1

z], from which

the result follows immediately..???

10. Think of the sphere as the surface of revolution generated by a semi-circle. Construct a conformal map of the sphere by strict analogy withthe construction of the map of the pseudosphere in [20] on p.297. Showthat this is the Mercator map that you obtained in Ex.14 on p.259.

.???

11. (i) In the hyperbolic plane, show that the h-circumference of an h-circle of h-radius ρ is 2π sinh ρ. [Hint: Represent the h-circle asan origin-centred Euclidean circle in the Poincaré disc.]

.???(ii) Let the inhabitants of the sphere of radius R draw a circle of

(intrinsic) radius ρ. Use elementary geometry to show that thecircle’s circumference is 2πR sin ρ

R. Show that if we take the radius

of the sphere to be R = i, then this becomes the formula in part(i)! [Compare this with Ex.14.]

On the sphere the radius is a geodesic, i.e. a section of a greatcircle.

Rr ρ

α


With ρ = αR and hence α = ρR

and sinα = rR

we have for thereal circumference 2πr = 2π R sinα = 2π sin ρ

R. And for R = i we

have 2πR sin ρR

= 2πi sin(−iρ) = 2π(−i sin(iρ)) = 2π sinh ρ dueto sinh z = −i sin(iz).


Chapter 7

Winding Numbers andTopology


1. A "simple" loop can get very complicated (see diagram). However, if weimagine creating this complicated loop by gradually deforming a circle,it is clear that it will wind round its interior points precisely once. LetN(p) be the number of intersection points of the simple loop with a rayemanating from p (cf. [4] on p.341). What distinguishes the possiblevalues of N(interior point) from those of N(exterior point)? In placeof the crossing rule (1) on p.340, you now possess (for simple loops) amuch more rapid method of determining whether a point is inside oroutside.

Exercises 369

IX Exercises

1 A "simple" loop can get very complicated (see diagram). However, if we imagine creating this complicated loop by gradually deforming a circle, it is clear that it will wind round its interior points precisely once. Let N (p) be the number of intersection points of the simple loop with a ray emanating from p (cf. [4 D. What distinguishes the possible values of N (interior point) from those of N (exterior point)? In place of the crossing rule 0), you now possess (for simple loops) a much more rapid method of determining whether a point is inside or outside.

You can use this result to playa trick on a friend F: (1) So that foul play cannot be suspected, get F to draw a very convoluted simple loop for himself; (2) choose a random point in the thick of things and ask F if it's inside or not, i.e., starting at this point, can one escape through the maze to the outside?; (3) after F has been forced to recognize the time and effort required to answer the question, get him to choose a point for you; (4) choosing a ray in your mind's eye, scan along it and count the intersection points. Amaze F with your virtually instantaneous answer!

2 Reconsider the mapping lin (4) of the unit circle to itself, and the associated graph of <I> (()) in [7]. If <1>' (a) > 0 then the graph is rising above the point () = a, and small movement of z will produce a small movement of the w having the same sense. We say that <I> is orientation-preserving at a and that the topological multiplicity v (a) of z = e ia as a preimage of w = ei<l>(a) is +1. Similarly, if <1>' (a) < 0 then the mapping is orientation-reversing and v (a) = -1. In other words,

v (a) = the sign of <I>'(a).

Compare this with the 2-dimensional formula (9).

(i) In [7], explain how the complete set of preimages of w = eiA can be found by drawing the family of horizontal lines <I> = A, A ± 21f, A ± 41f, etc.

(ii) If the set of preimages is typical in the sense that <1>' f=. 0 at any of them, what do we obtain if we sum their topological multiplicities? Thus to say that the degree of l (the winding number of L) is v is essentially to say that lis v-to-one. [Hint: In [4], consider a ray as describing the location of w.]

You can use this result to play a trick on a friend F : (1) So that foulplay cannot be suspected, get F to draw a very convoluted simple loopfor himself; (2) choose a random point in the thick of things and ask F ifit’s inside or not, i.e., starting at this point, can one escape through themaze to the outside?; (3) after F has been forced to recognize the timeand effort required to answer the question, get him to choose a pointfor you; (4) choosing a ray in your mind’s eye, scan along it and countthe intersection points. Amaze F with your virtually instantaneous

121

122 CHAPTER 7. WINDING NUMBERS AND TOPOLOGY

answer!If N(p) is odd then p is inside else outside.

2. Reconsider the mapping w = L(eiθ) = eiΦ(θ) (4)on p.343 of the unit circle to itself, and the associated graph of Φ(θ)in [7] on p.344. If Φ′(a) > 0 then the graph is rising above the pointθ = a, and a small movement of z will produce a small movement of thew having the same sense. We say that Φ is orientation-preserving at aand that the topological multiplicity ν(a) of z = eia as a preimage ofw = eiΦ(a) is +1. Similarly, if Φ′(a) < 0 then the mapping is orientation-reversing and ν(a) = −1. In other words, ν(a) is the sign of Φ′(a).Compare this with the 2-dimensional formula ν(a) = sgn(|J(a)|) (9)on p.349.

(i) In [7] on p.344, explain how the complete set of preimages ofw = eiA can be found by drawing the family of horizontal linesΦ = A,A± 2π,A± 4π, etc.

Any intersection of Φ(θ) with any line in A+2πZ correspondsto some Φ(θ) = A+ 2πk for some k ∈ Z so that .???

(ii) If the set of preimages is typical in the sense that Φ′ 6= 0 at any ofthem, what do we obtain if we sum their topological multiplicities?Thus to say that the degree of L (the winding number of L) isν is essentially to say that L is ν-to-one. [Hint: In [4] on p.341,consider a ray as describing the location of w.]

.???

3. For each of the following functions f(z), find all the p-points lyinginside the specified disc, determine their multiplicities, and by using acomputer to draw the image of the boundary circle, verify the ArgumentPrinciple.

(i) f(z) = e3πz and p = i, for the disc |z| ≤ 43 .

f(z) = e3πz = e3πxe3πi y = i ⇐⇒ e3πx = 1 and e3πi y = i ⇐⇒x = 0 and y ∈ 1

6 + 23Z. For any k ∈ Z we have z = ei(1+4k)/6 with

|z| = 1 that all obviously lie in the disc |z| ≤ 43 . Actually there

are the three p-points z0 = 16i, z1,2 = ±5

6i. Because f′(zj) 6= 0 for

j = 0, 1, 2 each p-point is simple.


-2 0 2

105

-2

-1

0

1

2

105

f(z) = e3 z

for |z| = 4/3for t [0 6.2832]

0 2 4 6

104

-1

-0.5

0

0.5

1105

f(z) = e3 z

for |z| = 4/3for t [0.3927 5.8905]

0 10000 20000

-2

-1

0

1

2

104

f(z) = e3 z

for |z| = 4/3for t [0.58905 5.6941]

-6000 -4000 -2000 0

-2000

0

2000

f(z) = e3 z

for |z| = 4/3for t [0.7854 5.4978]

-400 -200 0

-500

0

500

f(z) = e3 z

for |z| = 4/3for t [0.98175 5.3014]

0 5 10

-2

-1

0

1

2

f(z) = e3 z

for |z| = 4/3for t [1.3744 4.9087]

So the Argument Principle 3 = ν(f(L), i) with L = 43eit : t =

0..2π is verified.(ii) f(z) = cos z and p = 1, for the disc |z| ≤ 5.

f(z) = cos z = 12(e

iz + e−iz) = 1 ⇐⇒ eiz + e−iz = 2 ⇐⇒e−y(cosx+i sin x)+ey(cosx−i sin x) = 2 ⇐⇒ (e−y+ey) cosx = 2and (e−y−ey) sin x = 0 ⇐⇒ cosh y cosx = 1 and sinh y sin x = 0⇐⇒ x ∈ 2πZ and y = 0 so that only zo = 0 we have f(zo) = 1and |zo| ≤ 5. Because of cos z − cos 0 = − 1

2!z2 + 1

4!z4 ∓ . . . zo = 0

is a double p-point.

-20 0 20 40 60-60

-40

-20

0

20

40

60cos z for z=5*eit, p = 1

-25 -20 -15 -10 -5 0

-6

-4

-2

0

2

4

6

cos z for z = 5 eit, p = 1for |t| 0.9

So the Argument Principle 2 = ν(f(L), 1) with L = 5eit : t =0..2π is verified. ???

(iii) f(z) = sin z4 and p = 0, for the disc |z| ≤ 2.f(z) = 1

2i(eiz4− e−iz4) = 0 ⇐⇒ eir

4e4iθ = e−ir4e4iθ for z = reiθ

with 0 ≤ r ≤ 2 ⇐⇒ eir4(cos(4θ)+i sin(4θ)) = eir

4(− cos(4θ)−i sin(4θ))

⇐⇒ e−r4 sin(4θ) = er

4 sin(4θ) and eir4 cos(4θ) = e−ir4 cos(4θ) ⇐⇒ (r =

0 or θ ∈ π4Z) and r

4 cos(4θ) ∈ πZ ⇐⇒ r = 0 or r = 4√π ≈ 1.3313,


4√

2π ≈ 1.5832, 4√

3π ≈ 1.7521, 4√

4π ≈ 1.8828, 4√

5π ≈ 1.9908 andθ ∈ 0,±1

4π,±12π,±

34π,−π = k π4 : k = 0, 1, . . . , 7.

Because of sin z4 = ∑∞k=0

(−1)k(2k+1)!(z

4)2k+1 = z4 − 13!z

12 ± . . . =z4(1 − 1

3!z8 ± . . . ) the p-point z = 0 has multiplicity 4. The

Taylor expansion of f around e.g. zo = 4√πeiπ/4 is f(z) = sin(z4

o)+4z3o cos z4

o

1! (z−zo)+ . . . = sin(−π)+4 4√π

3ei3π/4 cos(−π)(z−zo)+ . . ..

The first derivative is non-vanishing, so that zo is a simple p-point.The same holds for the other 39 p-points. So in total there are 44p-points in the disk |z| ≤ 2.

-4 -2 0 2 4

106

-4

-3

-2

-1

0

1

2

3

4

106

f(z) = sin(z4)

for z = 2 eit, t [0,2 ]

-2 -1 0 1 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1|f(t ei /4)| with t [-2,2]

The left figure is deceiving because e.g. sin ((2ei0)4) = sin 16 ≈−0.2879 and sin ((2eiπ/4)4) = sin(−16) ≈ 0.2879.So the Argument Principle ? = ν(f(L), 0) with L = 2eit : t =???0..2π is verified.

4. Reconsider [8] on p.345.

(i) Use a computer to draw the image under a cubic mapping f(z) =(z−a)(z−b)(z−c) of an expanding circle Γ, and observe the man-ner in which the winding number increases as Γ passes throughthe roots a, b, and c. In particular, observe that the shape thatmarks the birth of a new loop is this: ≺.

Dashed curves are the preimages, solid curves are the images.


-1.5 -1 -0.5 0 0.5 1 1.5 2

x

-1.5

-1

-0.5

0

0.5

1

y

f(z) = (z-a)(z-b)(z-c) for |z| = 0.2136 0.57786 1.0565a, b, c = 0.4+0.15i 0.2-0.5i -0.75+0.6i

a

b

c

Each time the family of circles Γ pass through a root, the numberof self intersections of f(Γ) is incremented. cp Ch_7Ex4_animated.m ???

(ii) If f ′(p) 6= 0 then a little piece of Γ passing through p is merelyamplitwisted to another almost straight piece of curve throughf(p). Deduce that ≺ shapes can only occur when Γ hits a criticalpoint. Explain why the particular shape ≺ is consistent with acritical point of order 1.

For f(z) = z2


-0.1 -0.05 0 0.05 0.1-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

preimages a*ei a-a*ei t for parameter a

0 5 10 15 20

10-3

-0.025

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

images (a*ei a-a*ei t)2 for parameter a

???.

(iii) Observe that there are only two points in the evolution of Γ atwhich a ≺ shape is produced. Explain this algebraically in termsof the degree of f ′.

.???

(iv) Let T be the triangle with vertices a, b, and c. There are manyellipses which can be inscribed in T so as to touch all three sides,but show that there is only one (call it E) that touches T at themidpoints of the sides.

A general ellipsis E is given by the quadratic form Ax2 +2Bxy+Cy2+2Dx+2Ey+F = 0 or with homogeneous coordinates

and matrix M =

A B DB C ED E F

by

xy1

>

M

xy1

= 0. The tangent

in point po =

xoyo1

to E fulfills p>o M

xy1

= 0. So we have three

equations for the three midpoints to be points of E and threeequations for these points to be contact points – each up to acommon scalar multiple – to determine the six parameters A, B,C, D, E and F up to a common scalar multiple, c.p. e.g. SAGE:


(v) [Hard] Show that the two critical points of f are the foci of E !The critical points are the two solutions of (z−b)(z−c)+(z−

a)(z−c)+(z−a)(z−b) = 3z2−(a+b+c)z+ab+ac+bc = 0, namelyz1,2 = a+b+c

3 ± 13

√a2 + b2 + c2 − ab− ac− bc for a, b, c ∈ C. . ???

5. As in the text, let ξa, ηa, φa denote the two perpendicular expansionfactors and the rotation angle used to describe the local linear trans-formation at a produced by a mapping. By considering the case of arotation by π

4 , for which J is constant, show that ξ and η are generallynot the eigenvalues λ1 and λ2 of the Jacobian J . However, confirm forthis example that det(J) = λ1λ2 = ξη.

A rotation at a by π4 is defined by z 7→ eiπ/4(z−a)+a = 1+i√

2 (z−a)+a =

√2

2 (1+i)z+√

22 (√

2−1−i)a =√

22 ((x−y+(

√2−1)a)+i(x+y−a)) =

u(x, y) + i v(x, y) where z = x+ iy and a = b+ ic. Then the Jacobian

J =(∂xu ∂yu∂xv ∂yv

)=√

22

(1 −11 1

)is independent of a. Eigenvalues λ1,2

of J are the zeroes of p(λ) = |J − λI| = ( 1√2 − λ)2 + 1

2 = λ2−√

2λ+ 1,namely λ1,2 =

√2

2 (1± i). Summarizing we get |J | = 2 (√

2)2

4 = 1 = λ1λ2as well as . cp. p.349 ???

6. Even in three or more dimensions the local linear transformation in-duced by a mapping f at a point a can still be represented by theJacobian matrix J(a), and if a is not a critical point then its topologi-cal multiplicity ν(a) as a preimage of f(a) is still given by

ν(a) = sgn(|J(a)|) (9)on p.349. If n is the number of real negative eigenvalues of J(a),counted with their algebraic multiplicities, show that ν(a) = (−1)n.[Hint: Since the characteristic equation det[J(a) − λI] = 0 has realcoefficients, any complex eigenvalues must occur in conjugate pairs.]

. ???

7. Consider the nonanalytic mapping h(z) = |z|2 − iz.


(i) Find the roots of h.h(z) = 0 ⇐⇒ zz = |z|2 = iz ⇐⇒ z = 0 or z = i.

(ii) Calculate the Jacobian J , and hence find det(J).h(z) = x2+y2−i(x−iy) = x2+y2−y−ix so that u(x, y) = x2+

y2−y and v(x, y) = −x, hence J =(∂xu ∂yu∂xv ∂yv

)=(

2x 2y − 1−1 0

),

hence det[J ] = 0 + 2y − 1.(iii) Use ν(a) = sgn(|J(a)|) (9)

on p.349 to calculate the multiplicities of the roots in (i).ν(0) = sgn(2 · 0− 1) = −1 and ν(i) = sgn(2 · 1− 1) = 1.

(iv) Find the image curve traced by h(z) as z = 2eiθ traverses thecircle |z| = 2, and confirm the prediction of the Topological Ar-gument PrincipleThe total number of p-points inside a simple loop Γ (counted withtheir topological multiplicities) is equal to the winding number ofh(Γ) round p. (10)on p.350.

Let Γ be the circle |z| = 2, then h(2eiθ) = 4 − i2e−iθ =4 − i2(cos θ − i sin θ) = 4 − 2 sin θ − i2 cos θ is obviously anotherparametric representation of the circle around 4 ∈ C with ra-dius 2 as a simple curve, so ν(h(Γ), p) = 1 for p??? TopologicalArgument Principle states that ??????

(v) Gain a better understanding of the above facts by observing thath(z) = z(z − i), and then mimicking the analysis of [8].

h(z) = |z|2 − iz = zz − iz = z(z − i)..???(vi) Use the insight of the previous part to find ν(1

2i), which cannotbe done with (9).

.???

8. Let Q(t) be a real function of time t, subject to the (linear homoge-neous) differential equation ∑n

k=0 ckdk Qd tk

= 0. Recall that one solvesthis equation by taking a linear superposition of special solutions ofthe form Qj(t) = esjt. Substitution into the previous equation showsthat the sj’s are the roots of the (characteristic) polynomial F (s) =cns

n + cn−1sn−1 + ... + c1s + co. Note that Qj(t) will decay with time

if sj has a negative real part. The issue of whether or not the generalsolution of the differential equation decays away with time therefore re-duces to the problem of determining whether or not all n roots of F (s)lie in the half-plane <(s) < O. Let R be the net rotation of F (s) as straverses the imaginary axis from bottom to top. Explain the followingresult:


The general solution of the differential equation will die away if andonly if R = nπ.This is called the Nyquist Stability Criterion, and an F that satisfiesthis condition is called a Hurwitz polynomial. [Hints: Apply the Argu-ment Principle to the loop consisting of the segment of the imaginaryaxis from −iR to +iR, followed by one of the two semicircles havingthis segment as diameter. Now let R tend to infinity.]

. ???

9. Referring to the previous exercise, consider the differential equationd3 Qd t3−Q = 0

(i) Find R for this equation. Does it satisfy the Nyquist StabilityCriterion?

F (s) = s3− s is the characteristic polynomial of d3Qd t3−Q = 0.

Consider F (it) = −it3 − it = i(t− t3) for t = −∞..∞.

t

1iF (it) = t− t3

1√3

29

√3

F (it) for t = −∞..∞ moves from +i∞ on the imaginary axisdown to −i2

9

√3, then back to i2

9

√3 and then down to −i∞. ???

(ii) Confirm your conclusion by explicitly solving the differential equa-tion.

The characteristic polynomial F (s) = s3 − s = s(s2 − 1) =(s + 1)s(s− 1) of d3Q

d t3−Q = 0 has the three zeroes −1, 0, 1. So

the solution is spanned by e−t, e0 and et, in which the two basicsolutions e0 and et are not decaying.

10. If a is real and greater than 1, use Rouché’s Theorem on p.354 to showthat the equation znea = ez has n solutions inside the unit circle Γ.

With znea = ez ⇐⇒ zn−ez−a = 0 let f(z) := zn and g(z) := ez−a,then |g(z)| = ecos t−a ≤ e1−a < 1 = |zn| = |f(z)| for |z| = 1, hence byRouché’s Theorem f + g has as many zeroes inside Γ as f , namely n.


11. (i) Applying Rouche’s Theorem to f(z) = 2z5 and g(z) = 8z − 1,show that all five solutions of the equation 2z5 + 8z− 1 = 0 lie inthe disc |z| < 2.

Let f(z) = 2z5 − 1 and g(z) = 8z, then |g(z)| = |8z| ≤ 16 on|z| = 2. By Rouche’s Theorem (f + g)(z) = 2z5 + 8z − 1 has asmany zeroes within |z| = 2 as f has. Thus all five zeroes of f + glie in |z| = 2 because f(z) = 2z5 − 1 = 0 ⇐⇒ z5 = 1

2 has fivesolutions within |z| = 2.

(ii) By reversing the roles of f and g, show that there is only oneroot in the unit disc. Deduce that there are four roots in the ring1 < |z| < 2.

For f(z) = 8z − 1 and g(z) = 2z5 we have |g(z)| = |2z5| =2 < 7 ≤ |8z − 1| = |f(z)| on |z| = 1. By Rouche’s Theorem(f + g)(z) = 2z5 + 8z − 1 has as many zeroes within |z| = 1as f has, namely one. So there must be four roots in the ring1 < |z| < 2.

12. We can formalize, and slightly generalize, our explanation of Rouché’sTheorem as follows:

(i) If p(z) and q(z) are nonzero on a simple curve Γ, and Γ is theimage curve under z 7→ p(z)q(z), show that ν[Γ, 0] = ν[p(Γ), 0] +ν[q(Γ), 0].

.???(ii) Write f(z) +g(z) = f(z)(1 + g(z)

f(z)) =: f(z)H(z). If |g(z)| < |f(z)|on Γ, sketch a typical H(Γ). Deduce that ν[H(Γ), 0] = 0. Usingthe previous part, obtain Rouché’s Theorem.

Let g(z) = zn, f(z) = z− 3 and Γ be the unit circle. On Γ wehave |g(eit)| = |eint| = 1 < 2 ≤ |eit − 3| = |f(eit)| for t ∈ R. Now,for n = 2 and H(z) := 1 + g(z)

f(z) the image curve H(Γ) looks like


which makes obvious that 0 lies outside H(Γ) because <H(z) >0. In general, H(z) = 1 + r eiθ with 0 ≤ r = r(z) < 1 andθ = θ(z) = arg g(z) − arg f(z), hence <H(z) = 1 + r cos θ > 0implying ν(H(Γ), 0) = 0.With part (i) we have ν((f + g)(Γ), 0) = ν((f · H)(Γ), 0) =ν(f(Γ), 0) + 0 which together with the Argument Principle (6)on p.345 gives Rouché’s Theorem.

(iii) If we only stipulate that |g(z)| ≤ |f(z)| on Γ, then parts of H(Γ)could actually coincide with the circle |z − 1| = 1, rather thanlying strictly inside it, and ν[H(Γ), 0] might not be well-defined.However, show that if we further stipulate f + g 6= 0 on Γ, thenν[H(Γ), 0] = 0 as before. Deduce that ν[(f+g)(Γ), 0] = ν[f(Γ), 0].

. ???

13. Let w = f(z) be analytic inside and on a simple loop Γ, and supposethat f(Γ) is an origin-centred circle.

(i) If ∆ is an infinitesimal movement of z along Γ and φ is the cor-respondingly infinitesimal rotation of w, show geometrically thatf ′∆f

= iφ.. ???

(ii) As z traverses Γ, explain why ν[∆, 0] = 1 and ν[iφ, 0] = 0.. ???

(iii) Referring to part (i) of the previous exercise, show that ν[f(Γ), 0] =ν[f ′(Γ), 0] + 1.

. ???(iv) Deduce from the Argument Principle that f has one more root

inside Γ than f ′ has. This is sometimes called Macdonald’s The-orem, though I believe its essence goes back as far as Riemann.

. ???(v) From this we deduce, in particular, that f has at least one root

inside Γ. Derive this fact directly by considering the portion ofthe modular surface lying above Γ and its interior.

. ???


Chapter 8

Complex Integration: Cauchy’sTheorem


1. Thinking of x as representing time, z(x) = x + if(x) is a parametricdescription of the ordinary graph, y = f(x),

(i) Show that the complex velocity is v = 1 + i tan θ, where θ is theangle between the horizontal and the tangent to the graph. Also,show that complex acceleration is a = if ′′.

v ,

(<z=z

)=(

1f ′

), 1+i tan θ and a ,

(<z=z

)=(

0f ′′

), if ′′.

(ii) Recall from Ex.20 on page 262 that the curvature of the orbit isκ = 1

|v|3=(av). Deduce from (i) that κ = sec3 θf ′′(x).κ = 1

|v|3=(av) = 1√1+tan2 θ

3=(if ′′(1− i tan θ)) = cos3 θf ′′.

(iii) From (ii), deduce that the error equationarea(ABCD) = 1

8κ∆3 sec3 θ (3)on p.382 can be written as area(ABCD) = 1

8f′′(x)∆3.

area(ABCD) = 18(κ sec3 θ)∆3 = 1

8f′′(x)∆3.

2. In [6] on p.383, show that lim∆→0area between the chordAB and the curve

area between the tangentCD and the curve = 2.In other words, the midpoint Riemann sum RM is twice as accurate asthe Trapezoidal formula.

. ???

3. In the integration of an ordinary real function f(x), let L denote thelength of the integration range, and let M denote the maximum sizeof f ′′(x) in this range. From the previous two exercises, deduce the

133

134 CHAPTER 8. COMPLEX INTEGRATION: CAUCHY’S THEOREM

standard result, total Trapezoidal error < 112LM∆2. Likewise,

deduce the somewhat less familiar result, totalRM error < 124LM∆2.

Let L = n∆. Because the error of the trapezoidal rule is to thatone of the midpoint rule as 2 to 1, the total trapezoidal error is becauseEx.1 less than n2

3 ·18f′′∆3 ≤ 1

12LM ∆2. The total error of RM is halfof it: i.e. total RM error is less than 1

24LM ∆2.

4. Write down the values of∮C

1zdz for each of the following choices of C,

then confirm the answers the hard way, using parametric evaluation.

(i) |z| = 1∮C

1zdz = limt→2π Logeit−Log1 = 2πi and with z(t) = eit and

z = iz we have∮C

1zdz =

∫ 2π0

1eitieit dt = 2πi.

(ii) |z − 2| = 1∮C

1zdz = limt→2π Log(2 + eit)−Log3 = 0. With z(t) = 2 + eit

and z = ieit we have∮C

1zdz =

∫ 2π0

ieit dt2+eit = i

∫ 2π0

eit(2+e−it) dt(2+eit)(2+e−it) =

i∫ 2π

02 cos t+1+i2 sin t

5+4 cos t dt = 0 as ( t2−arctan (13 tan t

2))′ = 1

2−13 (1+tan2 t

2 ) 12

1+ 19 tan2 t

2

= 12−

3(1+tan2 t2 )

2(9+tan2 t2 ) = 1

2−3(cos2 t

2 +sin2 t2 )

2(9 cos2 t2 +sin2 t

2 ) = 12−

32

12(1+8 cos2 t

2= 1+8 cos2 t

2−32(1+8 cos2 t

2 )

= 4 cos2 t2−1

1+8 cos2 t2

= 4(1+cos t)/2−11+8(1+cos t)/2 = 2+2 cos t−1

1+4+4 cos t = 1+2 cos t5+4 cos t . Now we have∫ 2π/3

02 cos t+15+4 cos t dt = −

∫ π2π/3

2 cos t+15+4 cos t dt as

(t2 − arctan (1

3 tan t2))∣∣∣2π/3

0=

π3 − arctan (1

3 tan π3) = π

3 − arctan√

33 = π

3 −π6 = π

6 as well as(t2 − arctan (1

3 tan t2))∣∣∣π

2π/3= π

3−π2−arctan (1

3 tan π3)+arctan(∞)

= −π6 −arctan

√3

3 + π2 = −π

6 −π6 + π

2 = π6 and because on the other

hand sin t5+4 cos t is an odd 2π-periodic function so that

∫ π−π

sin t dt5+4 cos t

vanishes.

(iii) |z − 1| = 2For = |z − 1| = 2 we have

∮Cdzz

= 2πi ν(C, 0) = 2πi. Withz(t) = 1 + 2eit and z(t) = 2ieit we have

∮Cdzz

=∫ 2π

02ieit dt1+2eit =∫ 2π

02i dte−it+2 =

∫ 2π0

2i dt2+cos t−i sin t = 2i

∫ 2π0

(2+cos t+i sin t) dt(2+cos t)2+sin2 t

= <I + i=Iwhere the real part <I =

∫ π−π−2 sin t dt5+4 cos t vanishes because the in-

tegrand −2 sin t5+4 cos t is odd and 2π-periodic. In the imaginary part

=I =∫ π−π

2(2+cos t) dt5+4 cos t the integrand has anti-derivative F (t) = t

2 +arctan (1

3 tan t2) as F ′(t) = 1

2 +13 (1+tan2 t

2 ) 12

19 tan2 t

2 +1 = tan2 t2 +9+3(1+tan2 t

2 )2(tan2 t

2 +9) =6+2 tan2 t

2tan2 t

2 +9 = 6+2 1−cos t1+cos t

1−cos t1+cos t+9 = 6+6 cos t+2−2 cos t

1−cos t+9+9 cos t = 8+4 cos t10+8 cos t = 2(2+cos t)

5+4 cos t sothat =I = F (t)|π−π = π

2 − −π2 + arctan(+∞) − arctan(−∞) =

π + π2 −−

π2 = 2π.


5. Evaluate parametrically the integral of 1/z round the square with ver-tices ±1± i, and confirm that the answer is indeed 2πi .

Let C be the square with vertices ±1± i.∮Cdzz

= I1 + I2 + I3 + I4

In I1 the real integrand is odd, hence I1 =∫ 1+i

1−ii dt

1+it =∫ 1−1

i(1−it) dt1+t2 =∫ 1

−1i dt

1+t2 = i arctan(t)|1−1 = π2 i, I2 =

∫−1+i1+i

dzz

=∫−1

1dtt+i =

∫−11

(t−i)dtt2+1 =

−i arctan(t)|−11 = π

2 i, I3 =∫−1−i−1+i

dzz

=∫−1

1i dt−1+it =

∫−11−i(1+it)dt

1+t2 =∫−11−i dt1+t2 = −i arctan(t)|−1

1 = π2 i, I4 =

∫ 1−i−1−i

dzz

=∫ 1−1

dtt−i =

∫ 1−1

(t+i) dtt2+1 =∫ 1

−1i dt

1+t2 = i arctan(t)|1−1 = π2 i implying

∮Cdzz

= I1 + I2 + I3 + I4 = 2πi.

6. Confirm by parametric evaluation that the integral of zm round anorigin-centred circle vanishes, except when m = −1.

Let C be parameterized by z(t) = reit with z(t) = ireit, then∮C z

m dz =∫ 2π

0 rmeimt ireit dt = irm+1 ∫ 2π0 ei(m+1)t dt= rm+1

m+1 ei(m+1)t

∣∣∣2π0

= 0.

7. Hold a coin (of radius A) down on a flat surface and roll another one(of radius B) round it. The path traced by a point on the rim of therolling coin is called an epicycloid, and it is a closed curve if A = nB,where n is an integer.

(i) With the centre of the fixed coin at the origin, show that theepicycloid can be represented parametrically as z(t) = B[(n +1)eit − ei(n+1)t].

When the outer circle makes a full revolution the point ofcontact covers 2πB on the inner circle with radius A = nB.

Hence the centre of the outer circle covers 2πBA+BA

= 2π n+1nB

on a circle with radius A + B = (n + 1)B. Movement on theepicycloid consists of revolution around a rotating center: whenz(t) = (A+B)eit +Beit = (n+ 1)Beit ???


(ii) By evaluating the integral in∮C z dz = 2i(area enclosed) (8)

on p.394 parametrically, show that the area enclosed by the epicy-cloid is πB2(n+ 1)(n+ 2).∮

C z dz =∫ 2π

0 B((n+1)e−it−e−i(n+1)t)iB(n+1)(eit−ei(n+1)t) dt=∫ 2π

0 iB2(n + 1)2 dt −∫ 2π

0 iB2(n + 1)((n + 1)eint − e−int) dt +∫ 2π0 iB2(n + 1) dt = 2iπB2(n + 1)2 − 0 − 0 + 2iπB2(n + 1) =

2iπB2(n+ 1)(n+ 2).

8. The figure below shows four simple loops, and in each case we haveindicated how much shaded area is enclosed. Use parametric evaluationto verify equation

∮C z dz = 2i(area enclosed) (8)

on p.394 for each of the four loops.

(i) C = Reit : 0 ≤ t ≤ 2π is a circle with enclosed area πR2∮C z dz =

∫ 2π0 Re−itiReit dt =

∫ 2π0 iR2 dt = 2iπR2.

(ii) C = a cos t + ib sin t : 0 ≤ t ≤ 2π is an ellipsis with enclosedarea πab∮

C z dz =∫ 2π

0 (a cos t− ib sin t)(−a sin t+ ib cos t) dt=∫ 2π0 (−a2 cos t sin t+ iab cos2 t+ iab sin2 t+ b2 cos t sin t) dt

= (b2 − a2)12∫ 2π

0 sin(2t) dt+ 2iπab = 0 + 2iπab.(iii) C = Reit : 0 ≤ t ≤ π

2 ∪ it : 1 ≥ t ≥ 0 ∪ t : 0 ≤ t ≤ 1 is aquarter circle in the first quadrant with enclosed area 1

4πR2∮

C z dz =∫ π/2

0 Re−itiReit dt +∫ 0R(−it)i dt +

∫ R0 t dt = i1

2πR2 +

t2

2

∣∣∣0R

+ t2

2

∣∣∣R0

= 2i14πR

2.

(iv) C = R+ it : 0 ≤ t ≤ R ∪ t+ iR : R ≥ t ≥ 0 ∪ t+ i(R− t) :0 ≤ t ≤ R is a triangle with enclosed area 1

2R2∮

C z dz =∫ R

0 (R−it)i dt+∫ 0R(t−iR) dt+

∫ R0 (t−iR+it)(1−i) dt =

(iR2 + 12R

2) + (−12R

2 + iR2) +∫ R

0 (t − it − iR − R + it + t) dt =2iR2 +

∫ R0 (2t−R− iR) dt = 2iR2 + (R2 −R2 − iR2) = iR2.

9. What is the generalization of∮L z dz = 2i(area enclosed) (8)

on p.394 to the case where the contour is not closed?To shed light on

∮L z dz for some curve C we use [15] on p.393.

Conjugation 393

In the analytic case, provided that the special point z = 0 was not enclosed, the loop integral vanished. Even when the integral of (liz) did not vanish, its possible values were still neatly quantized in units of 2rr i; one unit for each time the special point z = 0 was enclosed by the loop. As we will see later, this behaviour is typical, although a more general mapping may well possess several special points (at which analyticity breaks down) dotted about in the plane. Once again, the integral is not sensitive to the precise shape of the loop. Provided that none of the special points are enclosed by the loop, then the integral vanishes. However, if some of the points are enclosed, then each one makes its own distinctive contribution (generally not 2rri) to the integral, one unit for each time it's encircled. The value ofthe integral is just the sum of these discrete contributions.

Contrast all this with our non-analytic example. The area of the loop (and hence the integral of z) will almost never vanish. Furthermore, instead of being determined by stable topological properties, the value of the integral is sensitive to the detailed geometry of the loop. Finally, the value is not neatly quantized, but instead varies continuously as the loop changes shape.

2 Area Interpretation Let us now verify the area interpretation of the integral of Z. Recall from Chapter 1 that 1m ( a b) is just twice the area of the triangle spanned by a and b. As z traces the loop Lin [15a], think of the area it sweeps out as being decomposed into triangular elements, as illustrated. Thus

o

Figure [IS]

2 (element of area) = Im[(z + ~)z] = Im[Z~].

Adding these elements together, we obtain the imaginary part of the Riemann sum corresponding to the integral of Z. Thus we conclude that

1m t z dz = 2 (area enclosed).

This result can be further simplified by noticing that z and (lIz) both point in

O

A

B

O

A

B

For C from A to B where the origin is always on the left when travelingfrom A to B the shaded area is just 1

2∮C z dz. In the second situation


the origin is on the left and on the right when traveling from A to B.Here the area on the right is to be taken negative.

10. Use K = ∑j νjCj (23) on p.416 (where νj is the winding number

of K about Dj and Cj = ∂Dj) to verify∮L z dz = 2i∑inside νjAj (9)

on p.395 (where Aj denotes the area of Dj).With L = ∑

j νjCj and Cj = ∂Dj we have∮L z dz = ∑

j νj∮Cjz dz =

2i∑inside νjAj.

11. The perfect symmetry of figure [18] on p.397 results from integrationround the unit circle. Roughly how would this figure look if we insteadused a somewhat larger circle?

. ???

12. Let K be the contour in [21] on p.400.

400 Complex Integration: Cauchy's Theorem

fez) z+i z-i

Applying (7) therefore yields

i fez) dz = 2n[v(C, i) - v(C, -i)].

Assuming (as in [20a]) that L encloses both singularities, use this formula to verify (12) for this particular function.

5 Residues Since we now possess a fairly complete understanding of the loop integrals of power functions, it is relatively easy to integrate simple rational functions: we need only find the decomposition into so-called partial fractions, and then integrate term by term. Indeed, this is precisely what we did in the example of the last paragraph.

Figure [21]

Here is a slightly more complicated example: the integral of f (z) = z5 / (z + 1)2 taken round the contour Kin [21]. By writing the numerator as [(z + 1) - 1]5, we quickly find that

fez) =

But we know that the loop integral of powers other than -1 is zero, and so only the complex inversion term [in square brackets] can contribute. In detail,

Ii fez) dz = 5· 2ni v(K, -1) = -20ni.

Thus the value of the integral has been determined by just two factors: the winding number of the loop, and the amount (i.e. coefficient) of complex inversion contained in the decomposition of the mapping. Because this latter number is the only part of the function that remains after we integrate, it is called the residue of the function

(i) Evaluate∮K

z dzz2−iz−1−i by factoring the denominator and putting

the integrand into partial fractions.∮K

z dzz2−iz−1−i =

∮K

z dz[z−1−i][z+1] = 1−i

2−i∮K

dzz−(1+i) + 1

2−i∮K

dzz−(−1) =

2πi ν(K, 1 + i)g(1 + i) + 2πi ν(K,−1)h(−1) = 0− 4πi2−i = 4πi

i−2 withν(K, 1 + i) = 0, g(z) = 1−i

2−i and ν(K,−1) = −2, h(z) = 12−i .

(ii) Write down the Laurent series (centred at the origin) for cos zz11 .

Hence find∮K

cos zz11 dz.

f(z) = cos zz11 = 1

z11∑∞k=0

(−1)k(2k)! z

2k = 1z11− 1

2!1z9 + 1

4!1z7− 1

6!1z5 + 1

8!1z3−

110!

1z± . . . implies Res(f, 0) = −1

10! and∮K

cos zz11 dz = 2πi ν(K, 0) ·

Res(f, 0) = 2πi10! .

13. This exercise illustrates how one type of difficult real integral may beevaluated easily using a complex integral.Let L be the straight contour along the real axis from −R to R, andlet J be the semi-circular contour (in the upper half plane) back fromR to −R. The complete contour L+ J is thus a closed loop.


(i) Using the partial fraction idea of the previous exercise, show thatthe integral

∮L+J

dzz4+1 vanishes if R < 1, and find its value if

R > 1.z4 + 1 = (z− a1)(z− a2)(z− a3)(z− a4) with ak = ei(2k−1)π/4,

i.e. a = a1 = eiπ/4, a2 = ei3π/4 = ia, a3 = ei5π/4 = −a anda4 = ei7π/4 = −ia so that 1

z4+1 = −a1/4z−a1

+ −a2/4z−a2

+ −a3/4z−a3

+ −a4/4z−a4

=−a/4z−a + −ia/4

z−ia + a/4z+a + ia/4

z+ia by SAGE.If R < 1 then L+J lies within the unit circle so that all fourth unitroots are outside, hence by Cauchy’s Theorem

∮L+J

dzz4+1 vanishes.

If R > 1 then C := L+ J encloses the two fourth unit roots a =eiπ/4 and ia = ei3π/4 and excludes the other two, hence

∮C

dzz4+1 =

−a4∮C

dzz−a + −ia

4∮C

dzz−ia + a

4∮C

dzz+a + ia

4∮C

dzz+ia = −a

4 2πiν(C, a) +−ia

4 2πiν(C, ia) + 0 + 0 = (−a2 + −ia2 )πi = π

2 (1− i)a.(ii) Using the fact that z4 + 1 is the complex number from −1 to z4,

write down the minimum value of |z4 + 1| as z travels round J .Now think of R as large, and use inequality

|∫K f(z) dz| ≤ |K|maxz∈K |f(z)| (5)

on p.387 to show that the integral round J dies away to zero asR grows to infinity.

Because of∣∣∣∫J dz

1+z4

∣∣∣ =∣∣∣∫ π0 iReit dt

1+R4ei4t

∣∣∣ ≤ πRmax0≤t≤πR

|1+R4ei4t| =πR2

R4−1 we have limR→∞∫J

dz1+z4 = 0.

(iii) From the previous parts, deduce the value of∫∞−∞

dxx4+1 .∫∞

−∞dxx4+1 = limR→∞

∮L

dzz4+1 = limR→∞

∮L+J

dzz4+1 = π

2 (1− i)a =π2

√2

2 ((1 + i)− (i− 1)) = π√

22 = π√

2 .

14. (i) The integral∫∞−∞

dxx2+1 is easily found by ordinary means, but eval-

uate it instead by the method of the previous exercise.Ordinarily,

∫∞−∞

dxx2+1 = arctan x|∞−∞ = π

2 − −π2 = π, Now,

write f(z) = 1z2+1 = 1

(z+i)(z−i) = i2

1z+i −

i2

1z−i and let L be the

straight contour along the real axis from −R to R, and let J bethe semi-circular contour in the upper back from R to −R. Thecomplete contour L+J is thus a closed loop. On one hand we have∮L+J f(z) dz =

∫L f(z) dz +

∫J f(z) dz =

∫ R−R

dtt2+1 +

∫ π0

iReit dtR2ei2t+1 →∫∞

−∞dtt2+1 for R → ∞ as

∣∣∣∫ π0 iReit dtR2ei2t+1

∣∣∣ ≤ πM → 0 for R → ∞as M = max0≤t≤π

R|R2ei2t+1| ≤

RR2−1 → 0 for R → ∞. On the

other hand due to Cauchy’s theorem we have∮L+J f(z) dz =

i2∫L+J

dzz+i −

i2∫L+J

dzz−i = 0− i

22πiν(L+ J, i) = π.

(ii) Likewise, evaluate∫∞−∞

dx(x2+1)2 by ordinary means and then by con-

tour integration.[Hint: The quickest way to find the partial fraction decomposition


for this function is to square the decomposition of 1/(z2 + 1).]Ordinarily,

∫∞−∞

dx(x2+1)2 = 1

2 ( arctan(x) + xx2+1)

∣∣∣∞−∞

= π2 . Now,

on one hand∮L+J

dz(z2+1)2 = (

∫L +

∫J )

dz(z2+1)2 =

∫ R−R

dt(t2+1)2 because

for J = Reit : t = 0...π we know∫J

dz(z2+1)2 =

∫ π0

iReit dt(R2ei2t+1)2

to vanish for R → ∞ since∣∣∣∫ π0 iReit dt

(R2ei2t+1)2

∣∣∣ ≤ πRM with M =max0≤t≤π

R|R2ei2t+1|2 ≤

R(R2−1)2 so that πRM → 0 for R→∞.

With 1(z2+1)2 = ( i/2z+i −

i/2z−i)

2 = 1/2(z+i)(z−i) −

1/4(z+i)2 − 1/4

(z−i)2 and1/2

(z+i)(z−i) = i/4z+i −

i/4z−i on the other hand Cauchy’s theorem gives∮

L+Jdz

(z2+1)2 = − i4∮L+J

dzz−i −

14∮L+J

dz(z−i)2 = − i

42πi ν(L + J, i) −14−1z−i

∣∣∣zozo

= π2 + 0 (cp. p.406) for any zo ∈ L+ J .

15. (i) Use the Fundamental Theorem to evaluate∫ a+ib0 ez dz.

Because (ez)′ = ez we have∫ a+ib0 ez dz = ea+ib − 1.

(ii) Equate the answer with the one obtained by parametric evaluationalong the straight contour from 0 to (a + ib), and deduce thatC :=

∫ 10 e

ax cos(bx) dx = a(ea cos b−1)+bea sin ba2+b2 and

S :=∫ 1

0 eax sin(bx) dx = b(1−ea cos b)+aea sin b

a2+b2 .With z(t) = t(a+ ib) and z(t) = a+ ib we verify

∫ a+ib0 ez dz =∫ 1

0 et(a+ib)(a+ ib) dt = (a+ ib) 1

a+ibet(a+ib)

∣∣∣10

= ea+ib − 1.C = <

∫ 10 e

t(a+ib) dt = <( 1a+ib

∫ 10 e

t(a+ib)(a + ib) dt) = <( ea+ib−1a+ib ) =

<((ea cos b−1+iea sin b)(a−ib))a2+b2 = aea cos b−a+bea sin b

a2+b2 .S = =

∫ 10 e

t(a+ib) dt = =( 1a+ib

∫ 10 e

t(a+ib)(a + ib) dt) = =( ea+ib−1a+ib ) =

=((ea cos b−1+iea sin b)(a−ib))a2+b2 = b−bea cos b+aea sin b

a2+b2 .(iii) Prove the results in (ii) by ordinary methods.

Both for C as for S we do product integration twice.C =

∫ 10 e

ax cos(bx) dx = 1aeax cos(bx)

∣∣∣10− b

a

∫ 10 e

ax sin(bx) dx =1a(e

a cos b−1)+ ba(

1aeax sin(bx)

∣∣∣10− ba

∫ 10 e

ax cos(bx) dx) = ea cos b−1a

+bea sin ba2 − b2

a2

∫ 10 e

ax cos(bx) dx ⇐⇒ a2+b2a2 C = ea cos b−1

a+ bea sin b

a2 ⇐⇒C = aea cos b−a+bea sin b

a2+b2 .S =

∫ 10 e

ax sin(bx) dx = eax

asin(bx)

∣∣∣10− ba

∫ 10 e

ax cos(bx) dx = ea sin ba−

ba(

1aeax cos(bx)

∣∣∣10+ b

a

∫ 10 e

ax sin(bx) dx) = ea sin ba− b

a2 (ea cos b−1)−b2

a2

∫ 10 e

ax sin(bx) dx ⇐⇒ a2+b2a2 S = ea sin b

a− b2

a2 ea cos b + b

a2 ⇐⇒S = aea sin b+b−bea cos b

a2+b2 .

16. (i) Show that when integrating a product of analytic functions, wemay use the ordinary method of "integration by parts".


The Fundamental theorem (15) on p.404 together with theproduct rule of Ex.9 in chapter 5 give

∫ ba f′(z)g(z) dz = f(z)g(z)|ba

−∫ ba f(z)g′(z) dz for a, b ∈ C.

(ii) Let L be a contour from the real number −θ to +θ. Show that∫L ze

iz dz = 2i(sin θ− θ cos θ), and verify this by taking L to be aline-segment and integrating parametrically.∫

L zeiz dz = z 1

ieiz∣∣∣θ+i0−θ+i0

− 1i

∫L e

iz dz = −i(θeiθ + θe−iθ) −1ieiz∣∣∣θ+i0−θ+i0

= −2iθ cos θ + i(eiθ − e−iθ) = 2i(sin θ − θ cos θ).

17. Let f(z) = 1z(z + 1

z)n where n is a positive integer.

(i) Use the Binomial Theorem to find the residue of f at the originwhen n is even and when n is odd.

f(z) = 1z

∑nk=0

(nk

)zkz−(n−k) = ∑n

k=0

(nk

)z2k−n−1 implies

Res(f, 0) =(nn/2

)if n is even and Res(f, 0) = 0 if n is odd.

(ii) If n is odd, what is the value of the integral of f round any loop?If n is odd then

∮L f(z) dz = 2πi ν(L, 0)Res(f, 0) = 0 for any

loop L.

(iii) If n = 2m is even and C is a simple loop winding once round theorigin, deduce from part (i) that

∮C f(z) dz = 2πi (2m)!

(m!)2 .∮C f(z) dz = 2πi ν(C, 0)Res(f, 0) = 2πi

(nn/2

)= 2πi

(2mm

)=

2πi (2m)!(m!)2 .

(iv) By taking C to be the unit circle, deduce the following result dueto Wallis:

∫ 2π0 cos2m θ dθ = (2m)!

22m−1(m!)2π.From (iii) 2πi (2m)!

(m!)2 =∮C f(z) dz =

∫ 2π0

1eit (e

it + e−it)2mieit dt =

i∫ 2π

0 (2 cos t)2m dt we get π (2m)!(m!)2 = 22m−1 ∫ 2π

0 cos2m θ dθ.

(v) Similarly, by considering functions of the form zkf(z) where k isan integer, evaluate

∫ 2π0 cosn θ cos(kθ) dθ and

∫ 2π0 cosn θ sin(kθ) dθ.∮

C zkf(z) dz =

∮C z

k−1(z + 1/z)n dz =∫ 2π

0 ei(k−1)tieit(eit +e−it)n dt = i

∫ 2π0 eikt2n cosn t dt = i2n

∫ 2π0 cos(kt) cosn t dt+2n

∫ 2π0 sin(kt) cosn t dt

so that∫ 2π0 cos(kt) cosn t dt =??? and 2n

∫ 2π0 sin(kt) cosn t dt =??????

18. Let E be the elliptical orbit z(t) = a cos t + ib sin t, where a and b arepositive and t varies from 0 to 2π. By considering the integral of 1

z

round E, show that∫ 2π

0dt

a2 cos2 t+b2 sin2 t= 2π

ab.

2πi =∮Edzz =

∫ 2π0−a sin t+ib cos ta cos t+ib sin t dt =

∫ 2π0

(−a sin t+ib cos t)(a cos t−ib sin t)a2 cos2 t+b2 sin2 t

dt =∫ 2π0

(b2−a2) cos t sin ta2 cos2 t+b2 sin2 t

dt+∫ 2π

0iab dt

a2 cos2 t+b2 sin2 thence

∫ 2π0

dta2 cos2 t+b2 sin2 t

= 2πab .


19. Let us verify the claim of Ex.19 on p.262, that if a function has vanish-ing Schwarzian derivative, then it must be a Möbius transformation.Following Beardon [1984, p.77], suppose that f(z), z = 0, and defineF ≡ (f ′′/f ′).

(i) Show that 1F (z) −

1F (w) = −1

2(z − w).From f(z), z = (f

′′

f ′ ) −12(

f ′′

f ′ )2 = F ′ − 1

2F2 = 0 we get the

separable differential equation F ′ = 12F

2 with solution F (z) = 2a−z

for some constant a. Then we have 1F (z) −

1F (w) = a−z

2 −a−w

2 =−1

2(z − w).(ii) Deduce that d

dzlog f ′(z) = − 2

z−a , for some constant a.ddz

log f ′(z) = f ′′(z)f ′(z) = F (z) = − 2

z−a .(iii) Perform two further integrations to conclude that f(z) is a Möbius

transformation.Let g(z) = log f ′(z). From g′(z) = − 2

z−α we conclude g(z) =log f ′(z) = −2 log(z − α) + β = log 1

(z−α)2 + β implying f ′(z) =γ

(z−α)2 so that f(z) = −γz−α + δ = −γ+δ(z−α)

z−α = az+bcz+d for a, b, c, d ∈ C.

20. In [27] on p.411 consider the white fragments of squares sandwichedbetween K and C. (The side length of each square is ε.)Cauchy's Theorem 411

Figure [27]

C. However, as we let E shrink, the shaded region fills the interior of C ever more completely, and K follows C ever more precisely. Thus, in order to see whether or not the integral of a mapping f along C vanishes, it is sufficient to instead investigate the behaviour of the integral of f along K, as E shrinks to zero. [This is justified in greater detail in Ex. 20.]

Next we seek to relate this integral along K to the behaviour of the mapping inside the shaded region that it bounds. Consider the sum of all the integrals of f taken counterclockwise round each of the infinitesimal shaded squares. This counterclockwise sense of integration is illustrated in [27] for two adjacent squares. When we add the integrals from these two squares, their common edge is traversed twice, once in each direction, and hence the integrals along it cancel. But this is true of every edge that lies in the shaded region, so that when we sum the integrals for all the shaded squares, the only edges that do not self-destruct in this manner are those that make up K:

1:Kf (Z)dZ = L 1: f(z)dz. h shaded squares fo

(18)

The investigation of the integral of f along C has thus been reduced to the study of the local effect of f on infinitesimal squares in the interior region.

It should be stressed that the discussion thus far is equally applicable to nonanalytic and analytic mappings. For example, with f(z) = Z, (18) simply says [see (8)] that the area inside K is the sum of the areas of the shaded squares. In order to understand Cauchy's Theorem, we must specialize to the case (illustrated in [27]) where the local effect of f is an amplitwist throughout the interior of C. First, though, let us try to guess how the magnitude of a typical integral in the above summation will depend on E (as the squares shrink) for a general mapping.

Experience with real integration, as well as the inequality (5), might lead one to guess that the integral round an infinitesimal square would die away at the same rate as its perimeter, that is, as E. This is false. The fact that the square is a

(i) Show that the sum of the integrals round these fragments equalsthe difference between the integrals round C and K.

As the integral around K equals the sum of the integralsaround each shaded square because those parts of the integralscovering adjacent sides are cancelling (as explained on p.411), thesum of the integrals round these fragments equals the differencebetween the integrals round C and K.

(ii) As ε shrinks, what is the approximate size of each term in theabove series?∮

C f(z) dz ≈∮K f(z) dz = ∑

shaded∮f(z) dz where

∮f(z) dz

≈ ε(p+ iq) = O(ε2). ???


(iii) Roughly how many terms are there in the series?Using a bounding box or refering to (fixed area inside C, di-

vided by the area of each square) on p.412 the number of terms isO( 1

ε2).

(iv) From the previous parts, what do you conclude about the dif-ference between the integrals round C and K, as ε shrinks tonothing?

limε→0∮K f(z) dz =

∮C f(z) dz.

21. Let K be the straight contour from a− (ε/2) to a+ (ε/2), where ε is ashort complex number in an arbitrary direction.

(i) Use the Fundamental Theorem to integrate z2 along K, and thenwrite down the value obtained by using a single term in RM . Showthat the error induced by RM is 1

12ε3.∫

K z2 dz = 1

3z3∣∣∣a+ε/2

a−ε/2= 1

3(a+ ε2)3 − 1

3(a− ε2)3 = 1

3(ε3

8 + 3a ε24 +3a2 ε

2 +a3)− 13(−

ε3

8 +3a ε24 −3a2 ε2 +a3) = 2

3(ε3

8 +3a2 ε2) = ε3

12 +a2ε

so that∫K z

2 dz −RM = ε3

12 + a2ε− a2ε = ε3

12 .(ii) As in part (i), find both the exact value and the RM value for

the integral of ez along K. By expanding eε/2 as a power series,deduce that the error in this case is roughly 1

24eaε3.∫

K ez dz = ez|a+ε/2

a−ε/2 = ea+ε/2 − ea−ε/2 = ea(eε/2 − e−ε/2) andwith RM = eaε we have

∫K e

z dz − RM = ea(1 + ε2 + ε2

2·4 + ε3

6·8 +. . .− (1− ε

2 + ε2

2·4 −ε3

6·8 ± . . .)− ε) ≈ ea(ε+ ε3

3·8 − ε) = ε3

24ea.

(iii) Repeat the error analysis of the previous parts for the non-analyticfunction z2. [You will need to use parametric evaluation to findthe exact value of the integral.]∫

K z2 dz =

∫ 1/2−1/2(a−tε)2ε dt = ε

−3ε(a−tε)3∣∣∣1/2−1/2

= 13((a+ε/2)3−

(a − ε/2)3) = 13(a

3 + 3a2 ε2 + 3a ε24 + ε3

8 − (a3 − 3a2 ε2 + 3a ε24 −

ε3

8 )) = 23(3a

2 ε2 + ε3

8 ) = a2ε + ε3

12 and with RM = a2ε we have∫K z

2 dz −RM = a2ε+ ε3

12 − a2ε = (a2 − a2)ε+ ε3

12 .???

22. Let K be the short contour of the previous exercise. Suppose that f(z)possesses a Taylor series centred at a that converges at points of K:f(a + h) = f(a) + f ′(a)h + f ′′(a)

2! h2 + f ′′′(a)3! h3 + . . .. [The existence of

such a series for any analytic function is derived in the next chapter.]

(i) By integrating this series along K, show that the difference be-tween the exact integral and the RM value is roughly 1

24f′′(a)ε3.

Verify that the results of the first two parts of the previous exer-cise are in accord with this finding.


We switch summation and integration in order to get∫K f(z) dz

=∫K (f(a) + f ′(a)

1 (z−a) + f ′′(a)2 (z−a)2 + f ′′′(a)

6 (z−a)3 + ...) dz

= (f(a)z + f ′(a)2 (z−a)2 + f ′′(a)

6 (z−a)3 + f ′′′(a)24 (z−a)4 + ...)

∣∣∣a+ε/2

a−ε/2.

Here, all summands with even powers of z − a vanish so that∫K f(z) dz = 2f(a) ε2 + 2f ′′(a)

6ε3

8 + . . . ≈ f(a)ε + f ′′(a)24 ε3. With

RM = f(a)ε we have∫K f(z) dz−RM ≈ f(a)ε+ f ′′(a)

24 ε3− f(a)ε =f ′′(a)

24 ε3.(ii) Use the series to show that the complex number from the image of

the midpoint of K to the midpoint of the images of the ends of Kis roughly 1

4f′′(a)ε2. As ε shrinks, are these two types of midpoint

distinguishable under the magnifying lens that produces figure[28] on p.412?

Image of the midpoint ofK is f(a), the images of the endpointsof K are f(a±) where a± = a± ε

2 , respectively. Thenf(a−)+f(a+)

2 −f(a) = 1

2(f(a)−f ′(a) ε2 + f ′′(a)2!

ε2

4 −f ′′′(a)

3!ε3

8 +− . . .+f(a)+f ′(a) ε2 +f ′′(a)

2!ε2

4 + f ′′′(a)3!

ε3

8 + . . . )− f(a) ≈ f(a) + f ′′(a)2!

ε2

4 − f(a) = f ′′(a)2!

ε2

4 .Lenses? ???

(iii) From the Fundamental Theorem, deduce that the existence ofsuch a series implies the vanishing of the integral of f round loopswithin the disc of convergence.

Interchanging summation and integration gives∮L f(z) dz =∮

L

∑∞k=0

f (k)(a)k! (z − a)k dz = ∑∞

k=0f (k)(a)k!

∮L(z − a)k dz = 0.

23. Let f(z) be analytic throughout a region which contains a triangle withvertices a, b, c, and hence with edges A = c− b, B = a− c, C = b− a.Given a pair of point p and q, let us define wpq as a kind of average off(z) along the line-segment pq: wpq = 1

q−p∫ qp f(z) dz. Show that this

complex average mapping sends the sides of the triangle ∆(abc) to thevertices wab, wbc, wca of a similar triangle! We merely rediscovered thisresult, which is apparently due to Echols [1923].[Hint: Show that Awbc +Bwca + Cwab = 0, and use A+B + C = 0.]∮

∆(abc) f(z) dz = 0 = c−bc−b

∫ cb f(z) dz+ a−c

a−c∫ ac f(z) dz+ b−a

b−a∫ ba f(z) dz =

Awbc+Bwca+Cwab implying wbc(b−c) = wca(a−b+b−c)−wab(a−b)⇐⇒ wbc(b− c)−wca(b− c) = wca(a− b)−wab(a− b) ⇐⇒ wbc−wca

a−b =wca−wabb−c .

24. Let K be a closed contour, and let ν be its winding number aboutthe point a. Show that

∮K

ez dzz−a = 2πi νea. [Hint: Write ez as eaez−a,

and expand ez−a as a power series.] This is a special case of Cauchy’sIntegral Formula (explained in the next chapter), which states that iff is analytic inside K, then

∮K

f(z)z−a dz = 2πi νf(a).


∮K

ez dzz−a = ea

∮K

ez−a dzz−a = ea

∮K (

1z−a + 1 + 1

2!(z − a) + 13!(z − a)2 +

. . . ) dz = ea(∮K

dzz−a+

∮K dz+ 1

2!∮K(z−a) dz+ 1

3!∮K(z−a)2 dz+. . . ) dz =

ea∮K

dzz−a = 2πi νea.

25. Consider the image of the disc |z| ≤ R under the mapping z 7→ kzm. Asthe radius sweeps round the disc once, its image sweeps m times roundthe image disc of radius |k|Rm. Thus we may sensibly define the area ofthe image to be mπ(|k|Rm)2. With this understanding, show that if amapping has a convergent power series f(z) = a+ bz+ cz2 + dz3 + . . .,then the area of the image is just the sum of the areas of the im-ages under each of the separate terms of the series: area of image= π(|b|2R2 + 2|c|2R4 + 3|d|2R6 + . . .). This is Bieberbach’s Area The-orem. Hint: Recall that the local area expansion factor is |f ′|2, so theimage area is

∫∫|z|≤R |f ′|2 dxdy =

∫ R0 (

∫ 2π0 f ′(reiθ)f ′(reiθ) dθ)r dr.

If area(kzm) = mπ(|k|Rm)2 = mπ|k|2R2m then area(f(z)) = area(a)+area(bz1) + area(cz2) + area(dz3) + . . . = 0 + 1π|b|2R2 + 2π|c|2R4 +3π|d|2R6 + . . ..???

26. (i) Show that if f is an analytic function without singularities or p-points on a loop L, then ν[f(L), p] = 1

2πi∮L

f ′(z)f(z)−p dz.

When substituting w = f(z) − p with dw = f ′(z) dz we get1

2πi∮L

f ′(z)f(z)−p dz = 1

2πi∮f(L)−p

dww

= ν(f(L)− p, 0) = ν(f(L), p).

(ii) Now let f(z) = (z−a1)A1 (z−a2)A2 ...(z−an)An(z−b1)B1 (z−b2)B2 ...(z−bm)Bm and by considering (log f)′,

find f ′

f.

log f(z) = ∑nk=1Ak log(z − ak) −

∑mk=1Bk log(z − bk) so that

f ′(z)f(z) = (log f)′(z) = ∑n

k=1Akz−ak−∑m

k=1Bkz−bk

.

(iii) In part (i) put p = 0 and take L to be a simple loop containingthe roots a1 to ar and containing the poles b1 to bs . Therebyobtain a calculation proof of the Generalized Argument Principlein the case of rational functions: ν[f(L), 0] = ∑r

j=1 Aj−∑sj=1Bj =

(number of interior roots)−(number of interior poles).Part (i) gives ν(f(L), 0) = 1

2πi∮Lf ′(z)f(z) dz = 1

2πi∑rj=1

∮LAj dz

z−aj −1

2πi∑sj=1

∮LBj dz

z−bj = ∑rj=1Aj−

∑sj=1Bj = (number of interior roots)

− (number of interior poles).

Chapter 9

Cauchy’s Formula and ItsApplications


1. If C is the unit circle, show that∫ 2π

0dt

1+a2−2a cos t =∮C

i dz(z−a)(az−1) . Use

Cauchy’s Formula to deduce that if 0 < a < 1, then∫ 2πo

dt1+a2−2a cos t =

2π1−a2 .

Let z = z(t) = eit, then we have∮C

i dz(z−a)(az−1) =

∫ 2π0

−eit dt(eit−a)(aeit−1) =∫ 2π

0−dt

aeit−1−a2+ae−it =∫ 2π0

−dt1+a2+2a cos t . Now, if 0 < a < 1 then 1

a> 1 and

f(z) = iaz−1 is analytic in and on C so that Cauchy’s Formula gives∮ f(z) dz

z−a = 2πif(a) = −2πa2−1 = 2π

1−a2 .

2. Let f(z) be analytic on and inside a circle K defined by |z − a| = ρ,and let M be the maximum of |f(z)| on K.

(i) Use f (n)(a) = n!2πi∮L

f(z) dz(z−a)n+1 (6)

on p.432 to show that |f (n)(a)| ≤ n!Mρn

.If z on K then |z − a| = ρ and |K| = 2πρ and

∣∣∣f (n)(a)∣∣∣ =

n!2π

∣∣∣∮K f(z) dz(z−a)n+1

∣∣∣ ≤ n!2π

Mρn+1 |K| = n!M

ρn.

(ii) Suppose that |f(z)| ≤M for all z ∈ C, whereM is some constant. ???By putting n = 1 in the above inequality, rederive Liouville’sTheorem [p.360].

Let f be bounded by M and analytic on C, then the modulusof each coefficient in the Taylor series f(z) = ∑∞

n=0f (n)(a)n! (z − a)n

of f is bounded by Mρn

and hence has to vanish for ρ→∞. So fmust be constant.

145

146 CHAPTER 9. CAUCHY’S FORMULA AND ITS APPLICATIONS

(iii) Suppose |f(z)| ≤ M |z|no for all z ∈ C, where no is some positive???integer. Show that f (no+1)(z) ≡ 0, and deduce that f(z) must bea polynomial whose degree does not exceed no.

Let z(t) = a+ρeit, as in part (ii) then |f (n)(a)| ≤ n!2π

Mρno

ρn+1 2πρ =n!Mρn−no

which for ρ→∞ tends to 0 for any n ≥ no + 1.

3. (i) Show that if C is any simple loop round the origin, then(nr

)=

12πi∮C

(1+z)n dzzr+1 .

For f(z) = (1 + z)n we have f (r)(z) = n(n − 1) · ... · (n −r + 1)(1 + z)n−r and 1

2πi∮C

(1+z)n dzzr+1 = 1

2πi∮C

f(z) dz(z−0)r+1 = f (r)(0)

r! =n(n−1)·...·(n−r+1)

r! = n!r! (n−r)! =

(nr

).

(ii) By taking C to be the unit circle, deduce that(

2nn

)≤ 4n.(

2nn

)=∣∣∣ 12πi∮C

(1+z)2n dzzn+1

∣∣∣ = 12π

∣∣∣∮ 2π0

(1+eit)2ni dz(eit)n

∣∣∣ ≤ 12π22n(2π) =

4n.For other interesting applications of complex analysis to problems in-volving binomial coefficients, see Bak: and Newman [1982, Chap. 11].

4. The Legendre polynomials are defined by Pn(z) = 12nn!

dn

dzn[(z2 − 1)n].

These polynomials are important in many physical problems, includingthe quantum mechanical description of the hydrogen atom.

(i) Calculate P1(z) and P2(z), and explain why Pn(z) has degree n.P1(z) = 1

2ddz

[z2 − 1] = 122z = z and P2(z) = 1

8d2

dz2 [(z2 − 1)2] =14ddz

[2z(z2 − 1)] = 12(z2 − 1 + 2z2) = 1

2(3z2 − 1).

(ii) Use (6) on p.432 to show that Pn(z) = 12πi∮K

(Z2−1)n dZ2n(Z−z)n+1 where K

is any simple loop round z.Using (6) Pn(z) = 1

2nn!dn

dzn[(z2 − 1)n] = 1

2nn!n!

2πi∮K

(Z2−1)n dZ(Z−z)n+1 =

12πi∮K

(Z2−1)n dZ2n(Z−z)n+1 .

(iii) By takingK to be a circle of radius√|z2 − 1| centred at z, deduce

that Pn(z) = 1π

∫ π0 (z +

√z2 − 1 cos θ)n dθ.

For Z = z +√|z2 − 1|eit we have Pn(z) = 1

2πi∮K

(Z2−1)n dZ2n(Z−z)n+1 =

12π∫ 2π

0((z+√|z2−1|eit)2−1)n dt

2n(√|z2−1|eit)n

= 12π∫ 2π

0 (z2+2z√|z2−1|eit+|z2−1|ei2t−12√|z2−1|eit

)ndt =

12π∫ 2π

0 (z+ z2−12√|z2−1|

e−it +√|z2−1|

2 eit)ndt which under the assump-

tion z2−1√|z2−1|

=√|z2 − 1| equals = 1

2π∫ 2π

0 (z +√|z2 − 1| cos t)n dt.???

???(iv) Check that this last formula yields the same P1(z) and P2(z) as

you obtained in part (i).


P1(z) = 1π

∫ π0 (z +

√z2 − 1 cos θ) dθ = z +

√z2−1π

∫ π0 cos θ dθ =

z +√z2−1π

sin θ∣∣∣π0

= z and P2(z) = 1π

∫ π0 (z +

√z2 − 1 cos θ)2 dθ =

1π

∫ π0 (z

2+2z√z2−1 cos θ+(z2−1)1+cos(2θ)

2 ) dθ = z2+ 2√z2−1π

sin θ∣∣∣π0

+ z2−12π π + z2−1

2πsin(2θ)

2

∣∣∣π0

= z2 + 12(z2 − 1) = 1

2(3z2 − 1).

5. If C denotes the unit circle, show∫ 2π

0sin2 θ dθ5−4 cos θ = − i

4∮C

(z2−1)2 dzz2(z−2)(2z−1) = π

4 .Let z = eit, then − i

4∮C

(z2−1)2 dzz2(z−2)(2z−1) = − i

4∫ 2π

0(ei2t−1)2ieit dt

(eit)2(eit−1)(2eit−1) =14∫ 2π

0(eit−e−it)2 dt

(eit−1)(2−e−it) = 14∫ 2π

0(2i sin t)2 dt

−5+2(eit+e−it) =∫ 2π

0sin2 t dt5−4 cos t . Now, let f(z) =

(z2−1)2

z2(z−2)(2z−1) with a pole in 0 of second order and a pole in 12 of first order.

Then, on one hand Res(f, 0) = 1(2−1)!

ddz

[z2f(z)]∣∣∣z=0

= ddz

(z2−1)2

(z−2)(2z−1)

∣∣∣z=0

=4(z2−1)z(z−2)(2z−1)−(z2−1)2[2z−1+2(z−2)]

(z−2)2(2z−1)2

∣∣∣z=0

= 54 and on the other hand

Res(f, 12) = (z − 1

2)f(z)∣∣∣z=1/2

= 12

(z2−1)2

z2(z−2)

∣∣∣z=1/2

= 12

(−3/4)2

(1/4)(−3/2) = −34 ,

hence − i4∮C

(z2−1)2 dzz2(z−2)(2z−1) = − i

42πi(54 −

34) = π

212 = π

4 .

6. Let f(z) be an analytic function with no poles on the real axis, andsuch that |f(z)| < M

|z|2 for sufficiently large |z|. By integrating f(z)eizalong the contour (L+ J) shown in [4a], deduce that∫∞−∞ f(x) cosx dx+ i

∫∞−∞ f(x) sin x dx = 2πi∑Imz>0 Res(f(z)eiz).

[Hint: First show that if y > 0, then |eiz| < 1.]Let L = z(t) = Rt : t ∈ [−1, 1] and J = z(t) = Reit : t ∈ [0, π],

and for y > 0 we have |ei(x+iy)| = |eix| e−y < 1, hence |∫J f(z)eiz dz| ≤

(πR) maxz∈J |f(z)eiz| ≤ πMR

maxz∈J |eiz| ≤ πMR

which for R → ∞tends to 0. Thus by (9), the Cauchy Integral Formula on p.435 wehave

∫∞−∞ f(x) cosx dx+ i

∫∞−∞ f(x) sin x dx = limR→∞

∮L+J f(z)eiz dz =

2πi∑=z>0 Res(f(z)eiz).

7. Use the result of the previous exercise to do the following problems, inwhich we assume that a > 0.

(i) Show that∫∞−∞

cosx dxx2+a2 = π

ae−a.

The integrand is an odd function so that∫∞−∞

sinx dxx2+a2 = 0.

For f(x) = 1x2+a2 we have

∫∞−∞ f(x) cosx dx =

∫∞−∞ f(x) cosx dx+

i∫∞−∞ f(x) sin x dx = 2πiRes(f(z)eiz, ia) = 2πiRes( eiz

(z+ia)(z−ia) , ia)where ia is the only simple pole in the upper half plane andRes(f(z)eiz, ia) = (z − ia)f(z)|z=ia = eiz

z+ia

∣∣∣z=ia

= e−a 12ia so that∫∞

−∞ f(x) cosx dx = 2πie−a 12ia = π

ae−a follows.

(ii) Evaluate∫∞−∞

x sinx dx(x2+a2)2 .

x cosx(x2+a2)2 is an odd integrand so that

∫∞−∞

x sinx dx(x2+a2)2 vanishes, hence

i∫∞−∞

x sinx dx(x2+a2)2 =

∫∞−∞

x cosx dx(x2+a2)2 + i

∫∞−∞

x sinx dx(x2+a2)2 = 2πiRes(f(z)eiz, ia)


with f(z) = z(z2+a2)2 = z

(z−ia)2(z+ia)2 with exactly one pole ia inthe upper half plane. ia is a double pole. Now, Res(f(z)eiz, ia) =ddz

(z − ia)2f(z)eiz∣∣∣z=ia

= ddz

zeiz

(z+ia)2

∣∣∣ia

= (1+iz)eiz(z+ia)2−zeiz2(z+ia)(z+ia)4

∣∣∣ia

= eiz (1+iz)(z+ia)−2z(z+ia)3

∣∣∣ia

= eiz −z+ia+iz2−az(z+ia)3

∣∣∣ia

= e−a−2ia2

−8ia3 = 14ae−a so

that∫∞−∞

x sinx dx(x2+a2)2 = 2π 1

4ae−a = π

2ae−a.

8. Let Fn(z) = 11+zn , where n = 2m is even.

(i) Use Res(P (z)Q(z) , a) = P (a)

Q′(a) (11)on p.435 to show that if p is a pole of Fn then Res[Fn, p] = − p

n.

Fn(z) = P (z)Qn(z) with P (z) = 1 and Qn(z)1 + zn. If 1 + pn = 0

or pn = −1 we have Res(P (z)Q(z) , p) = 1

npn−1 = pnpn

= − pn.

(ii) With the help of part (i), show that the sum of the residues of Fnin the upper half-plane is a geometric series with sum 1

i n sin(π/n) .zn = −1 ⇐⇒ zk = ei(2k+1)π/n. zk for k = 0, 1, . . . , n − 1 are

simple poles with z0, z1, . . . , zm−1 lying in the upper half-plane.Now, 1 + zn = (z − zk)

∑n−1`=0 z

`kz

n−1−` implies Res(f(z), zk) =z−zk1+zn

∣∣∣z=zk

= 1∑n−1`=0 z

`kzn−1−`

∣∣∣∣z=zk

= zk∑n−1`=0 z

nk

= − 1nzk. Thus we

have ∑m−1k=0 Res(f(z), zk) = − 1

n

∑m−1k=0 zk = − 1

nzo∑m−1k=0 (z2

o)k =− zo

n1−z2m

o

1−z2o

= zon

2z2o−1 = 1

in2i

zo−1/zo = 1in

1sin(π/n) .

(iii) By applying the Residue Theorem to the contour (L+ J) shownin [4a] on p.437, deduce that

∫∞0

dx1+xn = π

n sin(π/n) . (15)The integrand is even so that 2

∫∞o

dx1+xn =

∫∞−∞

dx1+xn . Now,

on one hand∣∣∣∫J dz

1+zn

∣∣∣ =∣∣∣∫ π0 iReit dt

1+Rneint

∣∣∣ ≤ πmax0≤t≤πR

|1+Rneint| =π RRn−1 so that limR→∞

∫J

dz1+zn = 0 and with part (ii)

∫∞o

dx1+xn =

12 limR→∞

∫L+J

dz1+zn = 1

22πi 1i n sin(π/n) = π

n sin(π/n) .

(iv) Although the above derivation breaks down when n is odd [why?],use a computer to verify that (15) is nevertheless still true.

If n is odd then the pole at −1 lies on L so the Residue The-orem is not applicable.SAGE can solve the integral for even n only: f = 1/(1+x^(2*m));then integral(f,x,-oo,oo); returns pi/(m*sin(1/2*pi/m)) elsejust integrate(1/(x^(2*m+1)+1), x, -Infinity, +Infinity).With assume(n,’integer’); and assumeAlso(n>0); MATLAB’sSMT on int(f,x,-inf,inf) returns limit(x*hypergeom([1,1/n],[1/n+1],-x^n),x==Inf) - limit(x*hypergeom([1,1/n],[1/n+1],-x^n), x==-Inf).


On integrate 1/(1+x^n) dx from -infty to +infty WolframAlpha returns Standard computation time exceeded... ???

9. Continuing from the previous question, consider the wedge-shaped con-tour K with K = K1 +K2 +K3 and K1 = t : t = 0..R, K2 = Reit :t = 0..2π/n], K3 = tei2π/n : t = R..0 and R > 1 so that zo = eiπ/n

is within K.

(i) Use the Residue Theorem to show that if n = 2, 3, 4, . . . and R > 1then

∮K

dz1+zn = −2πi

neiπ/n.∮

Kdz

1+zn = 2πiRes(f(z), zo) with only (simple) pole zo = eiπ/n

and Res(f(z), zo) = z−zo1+zn

∣∣∣z=zo

= 1∑n−1`=0 z

`ozn−1−`

∣∣∣∣z=zo

= 1∑n−1`=0 z

n−1o

=1

nzn−1o

= − 1nzo so that

∮K

dz1+zn = −2πi

nzo = −2πi

neiπ/n.

(ii) Show that limR→∞∮K

dz1+zn = [1− ei2π/n]

∫∞0

dx1+xn .

First, z = z(t) = t on K1 so that∫K1

dz1+zn =

∫ R0

dt1+tn , second

z = z(t) = Reit on K2 so that∣∣∣∫K2

dz1+zn

∣∣∣ =∣∣∣∫ 2π/n

0iReit dt

1+Rneint

∣∣∣ ≤2πn

RRn−1 → 0 for R → ∞ and third z = z(t) = tz2

o on K3

so that∫K3

dz1+zn =

∫ 0R

z2o dt

1+tnz2no

= −z2o

∫ R0

dt1+tn . Together we have

limR→∞∮K

dz1+zn = (1− z2

o)∫∞

0dt

1+tn = [1− ei2π/n]∫∞0

dx1+xn .

(iii) Deduce that∫∞0

dx1+xn = π

n sin(π/n) . (15)of Ex.8 (iii) is indeed valid for odd n as well as even n.

For any n = 2, 3, ... we have∫∞

0dx

1+xn = 11−ei2π/n limR→∞

∮K

dz1+zn

= 1ei2π/n−1

2πineiπ/n = π

n2i

eiπ/n−e−iπ/n = πn sin(π/n) .

10. Use If f(z) analytic and |f(z)| ≤ M|z|2 for sufficiently large |z|

then ∑∞n=−∞ f(n) = −π∑poles off Res[f(z) cot(πz)] (13)

on p.441 to show that ∑∞n=11n4 = 1

90π4.

The Taylor series is cot(πz) ≈ 1πz− 1

3(πz) − 145(πz)3 − . . . (cp.

https://de.wikipedia.org/wiki/Tangens_und_Kotangens) so that withf(z) = 1

z4 we have Res[f(z) cot(πz), 0] = Res[ cot(πz)z4 ] = − 1

45π3. With

(13) we get ∑∞n=11n4 = 1

2∑∞n=−∞,n 6=0 f(n) = −π

2 Res[f(z) cot(πz), 0] =190π

4. We excluded the pole in 0 from the LHS, cp. Ex.11.

11. If f(z) is an analytic function such that |f(z)| < M|z|2 for sufficiently

large |z|, then ∑∞n=−∞(−1)nf(n) = −π∑poles off(z) Res[f(z)cosec(πz)].In this formula, it is understood that if any of the poles of f(z) happento be integers, then these values of n are excluded from the LHS.

cosec(πz) = 1πz

+ 16(πz)+ 7

360(πz)3 + . . . (cp. https://en.wikipedia.org/wiki/Trigonometric_functions). cp. p.??? ???

12. Use the result of the previous question to do the following:

https://de.wikipedia.org/wiki/Tangens_und_Kotangens

https://en.wikipedia.org/wiki/Trigonometric_functions

https://en.wikipedia.org/wiki/Trigonometric_functions


(i) Show that ∑∞n=1(−1)n+1

n2 = 1− 14 + 1

9 −116 ± . . . = π2

12 .Let f(z) = z2. Then f(z)cosec(πz) = 1

πz3 + 16πz

+ . . . so that∞∑n=1

(−1)n+1

n2 = −12

∞∑n=−∞n 6=0

(−1)nf(n) = π2 Res[f(z)cosec(πz), 0] = π2

12 .

(ii) Find the sum of the series ∑∞n=1(−1)n+1

n2+1 = 12 −

15 + 1

10 −117 ± . . ..

Let f(z) = 11+z2 = 1

1−(iz)2 = ∑∞k=0(iz)2k = ∑∞

k=0(−1)kz2k =1−z2+z4∓. . .. Then f(z)cosec(πz) = (1−z2±...)(1/π

z+π

6 z+...) so

that Res[f(z)cosec(πz), 0] = 1π. With Ex.11 we have

∞∑n=1

(−1)n+1

n2+1 =

−12

∞∑n=−∞n 6=0

(−1)nf(n) = π2 Res[f(z)cosec(πz), 0] = 1

2 .

13. (i) Show that ∑∞n=−∞1

z2−n2 = π cot(πz)z

.Fix z ∈ C and let f(w) = 1

z2−w2 . Using (13) on p.441we get ∑∞n=−∞

1z2−n2 = ∑∞

n=−∞ f(n) = −πRes[ cot(πw)z2−w2 , w = z] −

πRes[ cot(πw)z2−w2 , w = −z] = −π cot(πw) (w−z)

(z+w)(z−w)

∣∣∣w=z− π cot(πw) (z+w)

(z+w)(z−w)

∣∣∣w=−z

= π cot(πz)2z − π cot(−πz)

2z = π cot(πz)z

.(ii) Show that part (i) can be rewritten as cot z = 1

z+∑∞

n=12z

z2−n2π2 .Starting from ∑∞

n=−∞1

w2−n2 = π cot(πw)w

and with z = πw weget π2

zcot z = 1

w2 +∑∞n=1

2w2−n2 = π2

z2 +∑∞n=1

2π2

z2−n2π2 ⇐⇒ cot z =1z

+∑∞n=1

2zz2−n2π2 .

(iii) Show that the previous equation can be rewritten as ddz

[ln(sin z/z)]= ∑∞

n=1ddz

ln(z2 − n2π2).On one hand d

dz[ln(sin z/z)] = z

sin zz cos z−1·sin z

z2 = z cos z−1·sin zz sin z =

cot z− 1z, on the other hand ∑∞n=1

ddz

ln(z2−n2π2) = ∑∞n=1

2zz2−n2π2

implying ddz

[ln(sin z/z)] = ∑∞n=1

ddz

ln(z2 − n2π2).(iv) By integrating along any path from 0 to z that avoids integers, and

then exponentiating both sides of the resulting equation, deducethat sin z = z

∏∞n=1 (1− z2

n2π2 ). [Hint: Recall that limz→0sin zz

= 1.]This famous formula is due to Euler, who used it to evaluate∑∞n=1

1n2 .

Integration gives ln sin zz

= ∑∞n=1 ln(z2−n2π2)+C for some con-

stant C ∈ C. Due to limz→0sin zz

= 1 we get 0 = ∑∞n=1 ln(−n2π2)+

C which implies ln sin zz

= ∑∞n=1 ( ln(z2 − n2π2) − ln(−n2π2)) =∑∞

n=1 ln z2−n2π2

−n2π2 = ∑∞n=1 ln (1 − z2

n2π2 ). Exponentiation then givessin zz

= ∏∞n=1 (1− z2

n2π2 ).

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Tristan Needham: Visual Complex Analysis Clarendon · PDF fileTristan Needham: Visual Complex Analysis Clarendon Oxford University Press z.B. umv.science.upjs.sk/hutnik/NeedhamVCA.pdf