complex analysis 3

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Complex Analysis

Math 534 Autumn 2009

Donald E. Marshall

If you are not in my class, you are still welcome to view these notes. My only re-

quirement is that you send me any typos you observe or suggestions for improvement you

might have. The homeworks are given at the end of these Notes. Additional information

is available from our course web page:

http://www.math.washington.edu/∼marshall/math534-09.html

For example, there is a link to a matlab program for viewing analytic functions. There

are also links to color pictures (from an older version of the matlab program) along with

some homework questions related to the pictures.

These notes are subject to change during the quarter.

i

ii

PREFACE

This course is a three quarter graduate level introduction to complex analysis. There

are four points of view for this subject due primarily to Cauchy, Weierstrass, Riemann

and Runge. Cauchy thought of analytic functions in terms of a complex derivative and

through his famous integral formula. Weierstrass instead stressed the importance of power

series expansions. Riemann thought of analytic functions as locally rigid mappings from

one domain to another, a more geometric point of view. Runge showed that analytic

functions are nothing more than limits of rational functions. The seminal modern text

in this area was written by Ahlfors, Complex Analysis, which stresses Cauchy’s point of

view. One aspect of the first year course in complex analysis is that the material has

been around so long that some very slick and elegant proofs have been discovered. The

subject is quite beautiful as a result, but some theorems then may seem mysterious. I’ve

decided instead to start with Weierstrass’s point of view for local behavior. Power series

are elementary and give you many non-trivial functions immediately. In many cases it is a

lot easier to see why certain theorems are true from this point of view. For example, it is

remarkable that a function which has a complex derivative actually has derivatives of all

orders. On the other hand, the derivative of a power series is just another power series and

hence has derivatives of all orders. Cauchy’s theorem is a more global result concerned

with integrals of analytic functions. Why integrals of the form∫

1z−adz are important in

Cauchy’s theorem is very easy to understand using partial fractions for rational functions.

So we will use Runge’s point of view for more global results: analytic functions are simply

limits of rational functions.

As a dydactic device we will use the term “analytic” for local power series expansion

and “holomorphic” for possessing a continuous complex derivative. We will of course prove

that these concepts (and several others) are equivalent eventually, but in the early chapters

the reader should be alert to the different definitions. The connection between analytic

and harmonic functions is deferred until much later in the course. The emphasis in the

beginning is to view analytic functions as behaving like polynomials.

iii

iv

Prerequisites

You should be on friendly terms with the following concepts. If you have only seen the

corresponding proofs for real numbers and real-valued functions, check to see if the same

proofs also work when “real” is replaced by “complex”, after reading the first two sections

of Chapter I. If many of the concepts below are new to you, then I would recommend that

you first take a senior level analysis class.

Let an∞n=0, bn∞n=0 be sequences of real numbers and let fn be a sequence of

real-valued functions defined on some interval I ⊂ R.

1. an converges to a (Notation: an → a) “ε− δ ” version.

2. Cauchy sequence

3.∑

an converges, converges absolutely (Notation:∑ |an| <∞)

4.∑

an converges implies an → 0, but not conversely.

5. lim supn→∞ an, lim infn→∞ an

6. ratio test, comparison test, Weierstrass M-test, root test

7. Rearranging absolutely convergent series gives the same sum, but a similar statement

does not hold for for conditionally convergent series.

8. If∑∞n=0 an = A and

∑∞n=0 bn = B then

A+B =

∞∑

n=0

(an + bn)

cA =∞∑

n=0

can.

If∑

an converges absolutely and cn =∑nk=0 akbn−k then

AB =∞∑

n=0

cn.

9.∞∑

n=0

∞∑

k=0

an,k =∞∑

k=0

∞∑

n=0

an,k

provided at least one sum converges absolutely.

Prerequisites v

10. Continuous function, uniformly continuous function

11. fn(x) → f(x) pointwise, fn(x) → f(x) uniformly.

12. Uniform limit of a sequence of continuous functions is continuous.

13. If fn → f uniformly on a bounded interval I then

lim

∫

I

fn(x)dx =

∫

I

lim fn(x)dx =

∫

I

f(x)dx

14. Corollary:∞∑

n=0

∫

I

fn(x)dx =

∫

I

∞∑

n=0

fn(x)dx,

if the partial sums of∑

fn converge uniformly on the bounded interval I.

15. open set, closed set, connected set, compact set, metric space.

16. f continuous on a compact set X implies f is uniformly continuous on X .

17. X ⊂ Rn is compact if and only if it is closed and bounded.

18. A metric space X is compact if and only if every infinite sequence in X has a limit

point in X . (This can fail if X not a metric space)

19. If f is continuous on a connected set U , then f(U) is connected. If f is continuous on

a compact set K then f(K) is compact.

20. A continuous real-valued function on a compact set has a maximum and a minimum.

21. Green’s theorem. You should look at the proof you learned (if in fact you saw a proof)

and figure out exactly what the hypotheses are for that version. Many undergrad

books prove a special case, then wave their hands.

All of the above can be found in the undergraduate text Rudin, Principles of Math-

ematical Analysis as well as many other sources. Items #15 -#20 are also in Ahlfors,

Complex Analysis, Chapter 3 section 1 (pages 50-61).

vi

I

Preliminaries

§1. Complex Numbers.

The “complex numbers” C consist of pairs of real numbers: (x, y) : x, y ∈ R.The complex number (x, y) can be represented geometrically as point in the plane R2, or

viewed as a vector whose tip has coordinates (x, y) and whose tail has coordinates (0, 0).

The complex number (x, y) can be identified with another pair of real numbers (r, θ), called

the polar coordinate representation. The line from (0, 0) to (x, y) has length r and forms

an angle with the positive x-axis. The angle is measured by using the distance along the

corresponding arc of the circle of radius 1 (centered at (0, 0)). By similarity, the length of

the subtended arc on the circle of radius r is rθ.

θ

rθr

1x

y(x, y)

Figure I.1 Cartesian and polar representation of complex numbers.

Conversion between these two representations is given by

x = r cos θ, y = r sin θ

and

r =√

x2 + y2, tan θ =y

x.

Care must be taken to find θ from the last equality since many angles can have the

same tangent. However, consideration of the quadrant containing (x, y) will give a unique

θ ∈ [0, 2π), provided r > 0 (we do not define θ when r = 0).

1

2 I. Preliminaries

Addition of complex numbers is defined coordinatewise:

(a, b) + (c, d) = (a+ c, b+ d),

and can be visualized by vector addition.

(a, b)

(c, d)

(a+ c, b+ d)

Figure I.2 Addition.

Multiplication is given by:

(a, b) · (c, d) = (ac− bd, bc+ ad)

and can be visualized as follows: The points (0, 0), (1, 0), (a, b) form a triangle. Construct

a similar triangle with corresponding points (0, 0), (c, d), (x, y). Then it is an exercise in

high school geometry to show that (x, y) = (a, b) · (c, d). By similarity, the length of the

product is the product of the lengths and angles are added.

(1, 0)(0, 0)

(c, d)(a, b)

(a, b) · (c, d)

Figure I.3 Multiplication.

The real number t is identified with the complex number (t, 0). With this identifica-

tion, complex addition and multiplication is an extension of the usual addition and multipli-

cation of real numbers. For conciseness when t is real, t(x, y) means (t, 0) · (x, y) = (tx, ty).

The additive identity is 0 = (0, 0) and −(x, y) = (−x,−y). The multiplicative identity

is 1 = (1, 0) and the multiplicative inverse of (x, y) is (x/(x2 + y2),−y/(x2 + y2)). It

§1: Complex Numbers 3

is a tedious exercise to check that the commutative and associative laws of addition and

multiplication hold, as does the distributive law.

The notation for complex numbers becomes much easier if we use a single letter instead

of a pair. It is traditional, at least among mathematicians, to use the letter i to denote

the complex number (0, 1). If z is the complex number given by (x, y), then because

(x, y) = x(1, 0) + y(0, 1), we can write z = x + yi. If z = x + iy then the “real” part of

z is Rez = x and the “imaginary” part is Imz = y. Note that i · i = −1. We can now

just use the usual algebraic rules for manipulating complex numbers together with the

simplification i2 = −1. For example, z/w means multiplication of z by the multiplicative

inverse of w. To find the real and imaginary parts of the quotient, we use the analog of

“rationalizing the denominator”:

x+ iy

a+ ib=

(x+ iy)(a− ib)

(a+ ib)(a− ib)=xa− i2yb+ iya− ixb

a2 + b2=(xa+ yb

a2 + b2)

+(ya− xb

a2 + b2)

i.

Here is some additional notation: if z = x + iy is given in polar coordinates by the

pair (r, θ) then

|z| = r =√

x2 + y2

called the modulus or absolute value of z. Note that |z| is the distance from the complex

number z to the origin 0. The angle θ is called the “argument” of z and written

θ = arg z.

The most common convention is that −π < arg z ≤ π, where positive angles are measured

counter-clockwise and negative angles are measured clockwise. The complex conjugate of

z is given by

z = x− iy.

The complex conjugate is the reflection of z about the real line R.

4 I. Preliminaries

It is an easy exercise to show

|zw| = |z||w|

|cz| = c|z| if c > 0,

z/|z| has absolute value 1,

zz = |z|2,

Rez = (z + z)/2,

Imz = (z − z)/(2i),

z + w = z + w,

zw = z · w,

z = z,

|z| = |z|,

arg zw = arg z + argw modulo 2π,

arg z = − arg z = 2π − arg z modulo 2π.

The statement “modulo 2π” means that the difference between the left and right hand

sides of the equality is an integer multiple of 2π.

The identity a+ (z − a) = z expressed in vector form shows that z − a is (a translate

of) the vector from a to z. Thus |z − a| is the length of the complex number z − a but it

is also equal to the distance from a to z. The circle centered at a with radius r is given by

z : |z − a| = r and the disk centered at a of radius r is given by z : |z − a| < r. The

open disks are the basic open sets generating the standard topology on C. We will use T

to denote the unit circle,

T = z : |z| = 1,

and use D to denote the unit disk,

D = z : |z| < 1.

Complex numbers were around for at least 250 years before good applications were

found. Cardano discussed them in his book Ars Magna (1545). Beginning in the 1800’s,

§2: Estimates 5

and continuing today, there has been an explosive growth in their usage. Now complex

numbers are very important in the application of mathematics to engineering and physics.

It is a historical fiction that solutions to quadratic equations forced us to take complex

numbers seriously. How to solve x2 = mx + c has been known for 2000 years and can be

visualized as the points of intersection of the standard parabola y = x2 and the line

y = mx + c. As the line is shifted up or down by changing c, it is easy to see there are

two, or one, or no (real) solutions. The solution to the cubic equation is where complex

numbers really became important. A cubic equation can be put in the standard form

x3 = 3px+ 2q,

by scaling and translating. The solutions can be visualized as the intersection of the

standard cubic y = x3 and the line y = 3px+ 2q. Every line meets the cubic, so there will

always be a solution. By formal manipulations, Cardano showed that a solution is given

by:

x = (q +√

q2 − p3)13 + (q −

√

q2 − p3)13 .

Bombelli pointed out 30 years later that if p = 5 and q = 2 then x = 4 is a solution,

but q2 − p3 < 0 so the above solution doesn’t make sense. His “wild thought” was to use

complex numbers to understand the solution:

x = (2 + 11i)13 + (2 − 11i)

13 .

He found that (2 ± i)3 = 2 ± 11i and so the above solution actually equals 4. In other

words, complex numbers were used to find a real solution.

§2. Estimates.

Here are some elementary estimates which the reader should check:

−|z| ≤ Rez ≤ |z|

−|z| ≤ Imz ≤ |z|

and

|z| ≤ |Rez| + |Imz|.

Perhaps the most useful inequality in analysis is the

6 I. Preliminaries

Triangle inequality.

|z + w| ≤ |z| + |w|,

and

|z + w| ≥∣

∣|z| − |w|∣

∣.

The associated picture perhaps makes this result clear:

z

wz + w

Figure I.4 Triangle inequality

Analysis is used to give a more rigorous proof of the triangle inequality (and it is good

practice with the notation we’ve introduced):

Proof.|z + w|2 = (z + w)(z + w)

= zz + wz + zw + ww

= |z|2 + 2Re(wz) + |w|2

≤ |z|2 + 2|w||z| + |w|2

= (|z| + |w|)2.

To obtain the second part of the triangle inequality:

|z| = |z + w + (−w)| ≤ |z + w| + | − w| = |z + w| + |w|

and by subtracting |w|,|z| − |w| ≤ |z + w|,

and switching z and w:

|w| − |z| ≤ |z + w|,

§2: Estimates 7

so that∣

∣|z| − |w|∣

∣≤ |z + w|.

These estimates can be used to prove that zn converges if and only if both Reznand Imzn converge. The series

∑

an is said to converge if the sequence of partial sums

Sm =

m∑

n=1

an

converges, and the series converges absolutely if∑ |an| converges. A series is said to

diverge if it does not converge. Absolute convergence implies convergence because Cauchy

sequences converge. We sometimes write∑ |an| < ∞ to denote absolute convergence

because the partial sums are increasing. It also follows that∑

an is absolutely convergent

if and only if both∑

Rean and∑

Iman are absolutely convergent. By comparing the nth

partial sum and the (n − 1)st partial sum, if∑

an converges then an → 0. As with real

series, the converse statement is false.

Another useful estimate is the

Cauchy-Schwarz inequality.

∣

∣

∣

∣

n∑

j=1

ajbj

∣

∣

∣

∣

≤( n∑

j=1

|aj|2)

12( n∑

j=1

|bj|2)

12

.

If v and w are vectors in Cn, the Cauchy-Schwarz inequality says that |〈v, w〉| ≤||v||||w||, where the left side is the absolute value of the inner product and the right side

is the product of the lengths of the vectors.

Proof.

0 ≤ 1

2

n∑

j=1

n∑

i=1

|aibj − ajbi|2

=1

2

n∑

j=1

n∑

i=1

|ai|2|bj |2 + |aj|2|bi|2 − aibjajbi − aibjajbi

=

n∑

i=1

|ai|2n∑

j=1

|bj|2 −( n∑

i=1

aibi

)( n∑

j=1

ajbj

)

8 I. Preliminaries

Thus∣

∣

∣

∣

n∑

j=1

ajbj

∣

∣

∣

∣

2

≤n∑

j=1

|aj |2n∑

j=1

|bj|2.

The proof above also gives the error

1

2

n∑

j=1

n∑

i=1

|aibj − ajbi|2,

and so equality occurs if and only if ai = cbi for all i and some (complex) constant c, or

bi = 0 for all i.

The reader can use Riemann integration to deduce the following, which is also called

the Cauchy-Schwarz inequality.

Corollary. If f and g are continuous complex-valued functions defined on [a, b] ⊂ R then

∣

∣

∣

∣

∣

∫ b

a

f(t)g(t)dt

∣

∣

∣

∣

∣

≤(

∫ b

a

|f(t)|2dt)

12(

∫ b

a

|g(t)|2dt)

12

.

This corollary can also be proved directly by expanding

∫ b

a

∫ b

a

|f(x)g(y)− f(y)g(x)|2dxdy (2.1)

in a similar way, giving a proof for f, g ∈ L2. Moreover, the error term is half of the integral

(2.1) and equality occurs if and only if f = cg, for some constant c, or g is identically zero.

§3: Stereographic Projection 9

§3. Stereographic Projection.

A component of Riemann’s point of view of functions as mappings is that ∞ is like any

other complex number. But we cannot extend the definition of complex numbers to include

∞ and still have the usual laws of arithmetic hold. However, there is another “picture” of

complex numbers that can help us visualize this idea. The picture is called “stereographic

projection”. We identify the complex numbers with the plane (x, y, 0) : x, y ∈ R in R3.

If z = x + iy, let z∗ be the unique point on the unit sphere in R3 which also lies on the

line from the “North Pole” (0, 0, 1) through (x, y, 0). Thus

z∗ = (x1, x2, x3) = (0, 0, 1) + t[(x, y, 0)− (0, 0, 1)].

(x, y, 0)

(x1, x2, x3)

N = (0, 0, 1)

Figure I.5 Stereographic projection.

Then

|z∗| =√

(tx)2 + (ty)2 + (1 − t)2 = 1,

which gives

t =2

x2 + y2 + 1,

where 0 < t ≤ 2, and

z∗ =

(

2x

x2 + y2 + 1,

2y

x2 + y2 + 1,x2 + y2 − 1

x2 + y2 + 1

)

.

The reader is invited to find z = x + iy from z∗ = (x1, x2, x3). The sphere is some-

times called the “Riemann sphere” and is denoted S2 or C∗. It is an explicit one-point

compactification of the complex plane. The North Pole corresponds to “∞”.

10 I. Preliminaries

Theorem 3.1. Under stereographic projection, circles and straight lines in C correspond

precisely to circles on S2 = C∗.

Proof. A circle on S2 is the intersection of a plane with the sphere. Indeed a circle

determines a plane. Conversely the sphere is invariant under rotation about the normal

direction to a plane P , so that the intersection of the sphere and the plane must be a circle.

If the plane is given by

Ax1 +Bx2 + Cx3 = D

and if (x1, x2, x3) corresponds to (x, y, 0) under stereographic projection then

A

(

2x

x2 + y2 + 1

)

+B

(

2y

x2 + y2 + 1

)

+C

(

x2 + y2 − 1

x2 + y2 + 1

)

= D.

Thus

(C −D)(x2 + y2) + 2Ax+ 2By = C +D.

If C = D, then this is the equation of a line, and all lines can be written this way. If

C 6= D, then by completing the square, we get the equation of a circle, and all circles can

be put in this form.

So we will consider a line in C as just a special kind of “circle”.

The sphere S2 inherits a topology from the usual topology on R3 generated by the

balls in R3.

Corollary 3.2. The topology on S2 induces the standard topology on C via stereographic

projection, and moreover a basic neighborhood of ∞ is of the form z : |z| > r.

For later use, we note that the chordal distance between two points on the sphere

induces a metric on C which is given by

χ(z, w) = |z∗ − w∗| =2|z − w|

√

1 + |z|2√

1 + |w|2.

This metric is bounded (by 2).

II

Analytic Functions

§0. Polynomials.

This course is about complex-valued functions of a complex variable. We could think

of such functions in terms of real variables as maps from R2 into R2 given by

f(x, y) = (u(x, y), v(x, y)),

and think of the graph of f as a subset of R4. But the subject becomes more tractable if

we use a single letter z to denote in the independent variable and write f(z) for the value

at z, where z = x+ iy and f(z) = u(z) + iv(z). For example

f(z) = zn

is much simpler to write (and understand) than its real equivalent. Here zn means the

product of n copies of z.

The simplest functions are the polynomials in z:

p(z) = a0 + a1z + a2z2 + . . .+ anz

n, (0.1)

where a0, . . . , an are complex numbers. If an 6= 0, then we say that n is the degree of p.

Note that z is not a (complex) polynomial, and neither is Rez or Imz.

Let’s take a closer look at linear or degree 1 polynomials. For example if b is a (fixed)

complex number, then

g(z) = z + b

translates, or shifts, the plane. If a is a (fixed) complex number then

h(z) = az

11

12 II. Analytic Functions

can be viewed as a rotation and dilation. To see this, by Chapter I and Homework problem

I.3, write z = reiθ where r = |z| > 0 and θ = arg z ∈ (−π, π]. Similarly, write a = Aeiα,

where a > 0. Then

h(z) = Arei(θ+α)

so that h rotates the point z by the angle α and dilates, or scales, by a factor of A. A

linear function

f(z) = az + b

can then be viewed as a dilation and rotation followed by a translation. Equivalently,

writing f(z) = a(z + b/a) we can view f as a translation followed by a rotation and

dilation.

Another instructive example is the function

p(z) = zn = rneinθ.

Each pie slice

Sk = z :

∣

∣

∣

∣

arg z − 2πk

n

∣

∣

∣

∣

<πk

n ∩ z : |z| < r,

k = 0, . . . , n− 1 is mapped to a slit disk

z : |z| < rn \ (−rn, 0).

Angles between straight line segments issuing from the origin are multiplied by n and for

small r, the size of the image disk is much smaller than the “radius” of the pie slice. See

Figure II.1

00

zn

r −rn

Figure II.1 The power map.

The function k(z) = b(z − z0)n can be viewed as a translation by −z0, followed by

the power function, and then a rotation and dilation. To put it another way, k translates

§1: Power Series 13

a neighborhood of z0 to the origin, then acts like the power function zn, followed by a

dilation and rotation by b.

To understand the local behavior of a polynomial (0.1) near a point z0, rewrite it in

the form

p(z) = p(z0) + b1(z − z0) + b2(z − z0)2 + . . .+ bn(z − z0)

n. (0.2)

If b1 6= 0 then p(z) behaves like the linear function p(z0) + b1(z − z0) for z near z0. If

b1 = 0 then near z0, p(z) is closely approximated by p(z0) + bk(z − z0)k, where bk is the

first non-zero coefficient in the expansion (0.2). Indeed, for small ε,

|p(z0 + εeiθ) − [p(z0) + bkεkeikθ]| ≤ Cεk+1,

for some constant C. See Figure II.2.

s

p(z0)

bkεkeikθ

Figure II.1 p(z0 + εeiθ) lies in a small disk of radius s = Cεk+1 < |bk|εk.

For z near z0 then, p(z) behaves like a translation, followed by a power function, a

rotation and dilation, and finally a translation by p(z0).

More complicated functions are found by taking limits of polynomials.

§1. Power Series.

Here is the primary example:∞∑

n=0

zn.

This series is important to understand because its behavior is typical of all power series

(defined shortly) and because it is one of the few series we can actually add up explicitly.

The partial sums

Sm =

m∑

n=0

zn = 1 + z + z2 + . . .+ zm


satisfy

(1 − z)Sm = 1 − zm+1,

as can be seen by multiplying out the left side and cancelling. If z 6= 1 then

Sm =1 − zm+1

1 − z.

Notice that if |z| < 1, then |zm| = |z|m → 0 as m → ∞ and so Sm(z) → 1/(1 − z). If

|z| > 1, then |zm| = |z|m → ∞ and so the sum diverges for these z. If |z| = 1 but z 6= 1

then zn does not tend to 0, so the series diverges. Finally if z = 1 then the partial sums

satisfy Sm = m→ ∞, so we conclude that if |z| < 1 then

∞∑

n=0

zn =1

1 − z, (1.1)

and if |z| ≥ 1, then the series diverges. It is important to note that the left and right sides

of (1.1) are different objects. They agree in |z| < 1, the right side is defined for all z 6= 1,

but the left side is defined only for |z| < 1.

The formal power series

f(z) =∞∑

n=0

an(z − z0)n = a0 + a1(z − z0) + a2(z − z0)

2 + . . .

is called a convergent power series centered (or based) at z0 if there is an r > 0 so

that the series converges for all z such that |z − z0| < r. Note: If we plug z = z0 into the

formal power series, then we always get a0 = f(z0) (more formally, the definition of the

summation notation includes the convention that the n = 0 term equals a0, so that we are

not raising 0 to the power 0.) The requirement for a power series to converge is stronger

than convergence at just the one point z0.

A variant of the primary example is:

1

z − a=

1

z − z0 − (a− z0)=

1

−(a− z0)(1 − ( z−z0a−z0

)).

Substituting

w =z − z0a− z0

§1: Power Series 15

into (1.1) we obtain, when |w| = |(z − z0)/(a− z0)| < 1,

1

z − a=

∞∑

n=0

−1

(a− z0)n+1(z − z0)

n. (1.2)

In other words by (1.1), this series converges if |z− z0| < |a− z0| and diverges if |z− z0| ≥|a− z0|. The region of convergence is an open disk and it is the largest disk centered at z0

and contained in the domain of definition of 1/(z − a). In particular, this function has a

power series expansion based at every z0 6= a, but different series for different base points

z0.

Theorem 1.1 (Weierstrass M-Test). If |an(z − z0)n| ≤ Mn for |z − z0| ≤ r and if

∑

Mn <∞ then∑∞n=0 an(z−z0)n converges uniformly and absolutely in z : |z−z0| ≤ r.

Proof. If M > N then the partial sums Sn(z) satisfy

|SM (z) − SN (z)| =∣

∣

M∑

n=N+1

an(z − z0)n∣

∣≤M∑

n=N+1

Mn.

Since∑

Mn < ∞, we deduce∑Mn=N+1 Mn → 0 as N,M → ∞, and so Sn is a Cauchy

sequence converging uniformly. The same proof also shows absolute convergence.

Note that the convergence depends only on the “tail” of the series so that we need

only satisfy the hypotheses in the Weierstrass M-test for n ≥ n0 to obtain the conclusion.

The primary example (1.1) converges on a disk and diverges outside the disk. The

next result says that disks are the only kind of region in which a power series can converge.

Theorem 1.2 (Root Test). Suppose∑

an(z − z0)n is a formal power series. Let

R = lim infn→∞

|an|−1n =

1

lim supn→∞

|an|1n

∈ [0,+∞].

Then∑∞n=0 an(z − z0)

n

(a) converges absolutely in z : |z − z0| < R,


(b) converges uniformly in z : |z − z0| ≤ r for all r < R, and

(c) diverges in z : |z − z0| > R.

converge

diverge

z0R

Figure II.1 Convergence of a power series.

Proof. The idea is to compare the given series with the example (1.1),∑

zn. If |z− z0| ≤r < R, then choose r1 with r < r1 < R. Thus r1 < lim inf |an|−

1n , and there is an n0 <∞

so that r1 < |an|−1n for all n ≥ n0. This implies that |an(z − z0)

n| ≤ ( rr1 )n. But by (1.1),

∞∑

n=0

(

r

r1

)n

=1

1 − r/r1<∞

since r/r1 < 1. Applying Weierstrass’s M-test to the tail of the series (n ≥ n0) proves

(b). This same proof also shows absolute convergence (a) for each z with |z − z0| < R. If

|z − z0| > R, fix z and choose r so that R < r < |z − z0|. Then |an|−1n < r for infinitely

many n and hence

|an(z − z0)n| >

( |z − z0|r

)n

for infinitely many n. Since (|z − z0|/r)n → ∞ as n→ ∞, (c) holds.

The proof of the Root Test shows that if the terms an(z − z0)n of the formal power

series are bounded when z = z1 then the series converges on z : |z − z0| < |z1 − z0|.The Root Test does not give any information about convergence on the circle of radius

R. The series can converge at none, some, or all points of z : |z−z0| = R, as the following

examples illustrate.

Examples.

(i)

∞∑

n=1

zn

n(ii)

∞∑

n=1

zn

n2(iii)

∞∑

n=1

nzn (iv)

∞∑

n=1

2n2

zn (v)

∞∑

n=1

2−n2

zn

§2: Analytic Functions 17

The reader should verify the following facts about these examples. The radius of

convergence of each of the first three series is R = 1. When z = 1, the first series is the

harmonic series which diverges, and when z = −1 the first series is an alternating series

whose terms decrease in absolute value and hence converges. The second series converges

uniformly and absolutely on |z| = 1. The third series diverges at all points of |z| = 1.The fourth series has radius of convergence R = 0 and hence is not a convergent power

series. The fifth example has radius of convergence R = ∞ and hence converges for all

z ∈ C.

What is the radius of convergence of the series∑

anzn where

an =

3−n if n is even

4n if n is odd?

This is an example where ratios of successive terms in the series does not provide sufficient

information to determine convergence.

§2. Analytic Functions

Definition. A function f is analytic at z0 if

f(z) =∞∑

n=0

an(z − z0)n

where the series converges on z : |z − z0| < r for some r > 0. A function f is analytic

on set Ω if f is analytic at each z0 ∈ Ω.

Note that we do not require one series for f to converge in all of Ω. The example

(1.2), (z − a)−1, is analytic on C \ a and is not given by one series. Note that if f is

analytic on Ω then f is continuous in Ω. Indeed, continuity is a local property. To check

continuity near z0, use the series based at z0. Since the partial sums are continuous and

converge uniformly on a closed disk centered at z0, the limit function f is continuous on

that disk.

A natural question at this point is: where is a power series analytic?


Theorem 2.1. If f(z) =∑

an(z − z0)n converges on z : |z − z0| < r then f is analytic

on z : |z − z0| < r.

Proof. Fix z1 with |z1 − z0| < r. We need to prove that f has a power series expansion

based at z1. By the binomial theorem

(z − z0)n = (z − z1 + z1 − z0)

n =n∑

k=0

(

n

k

)

(z1 − z0)n−k(z − z1)

k.

Hence

f(z) =∞∑

n=0

[ n∑

k=0

an

(

n

k

)

(z1 − z0)n−k(z − z1)

k

]

. (1.3)

Suppose for the moment, that we can interchange the order of summation, then

∞∑

k=0

[ ∞∑

n=k

an

(

n

k

)

(z1 − z0)n−k

]

(z − z1)k

should be the power series expansion for f based at z1. To justify this interchange of

summation, it suffices to prove absolute convergence of (1.3). By the root test

∞∑

n=0

|an||w − z0|n

converges if |w − z0| < r. Set

w = |z − z1| + |z1 − z0| + z0.

Then |w − z0| = |z − z1| + |z1 − z0| < r provided |z − z1| < r − |z1 − z0|.

z0

z1

w

rr − |z1 − z0|z

Figure II.2 Proof of Theorem 2.1.

§2: Analytic Functions 19

Thus if |z − z1| < r − |z1 − z0|,

∞ >

∞∑

n=0

|an||w − z0|n

=

∞∑

n=0

|an|(

|z − z1| + |z1 − z0|)n

=∞∑

n=0

[ n∑

k=0

|an|(

n

k

)

|z1 − z0|n−k|z − z1|k]

as desired.

Another natural question is: Can an analytic function have more than one power

series expansion based at z0?

Theorem 2.2 (Uniqueness). Suppose

∞∑

n=0

an(z − z0)n =

∞∑

n=0

bn(z − z0)n,

for all z such that |z − z0| < r where r > 0. Then an = bn for all n.

Proof. Set cn = an − bn. The hypothesis implies that∑∞n=0 cn(z − z0)

n = 0 and we need

to show that cn = 0 for all n. Suppose cm is the first non-zero coefficient. Set

F (z) ≡∞∑

n=0

cn+m(z − z0)n = (z − z0)

−m∞∑

n=m

cn(z − z0)n.

The series for F converges in 0 < |z − z0| < r because we can multiply the terms of the

series on the right side by the non-zero number (z− z0)−m and not affect convergence. By

the root test, the series for F converges in a disk and hence in |z − z0| < r. Since F is

continuous and cm 6= 0, there is a δ > 0 so that if |z − z0| < δ, then

|F (z) − F (z0)| = |F (z) − cm| < |cm|/2.

If F (z) = 0, then we obtain the contradiction | − cm| < |cm|/2. Thus F (z) 6= 0 when

|z − z0| < δ. But (z − z0)m = 0 only when z = z0, and thus

∞∑

n=0

cn(z − z0)n = (z − z0)

mF (z) 6= 0


when 0 < |z − z0| < δ, contradicting our assumption on∑

cn(z − z0)n.

Notice that the proof of Theorem 2.2 shows that if f is analytic at z0 then for some

δ > 0, either f(z) 6= 0 when 0 < |z − z0| < δ or f(z) = 0 for all z such that |z− z0| < δ. If

f(a) = 0, then a is called a zero of f . Recall that a region is a connected open set.

Corollary 2.3. If f is analytic on a region Ω then either f ≡ 0 or the zeros of f are

isolated.

Proof. Let E denote the set of non-isolated zeros of f . In the proof of the Uniqueness

theorem, we showed that if z0 is a non-isolated zero of f then f is identically zero in a

neighborhood of z0. Thus the set of non-isolated zeros of f is open. Since f is continuous,

the set of zeros and hence the set of non-isolated zeros is closed, because we can find a

union of open disks containing only isolated zeros. By connectedness, either E = Ω or

E = ∅.

There are plenty of continuous functions for which the corollary is false, for example

xsin(1/x). The corollary is true because near z0, f(z) behaves like the first non-zero term

in its power series expansion about z0.

§3. Elementary Operations with Analytic Functions

If f and g are analytic at z0 then so are

f + g, f − g, cf (where c is a constant), and fg.

The first three follow from the fact that the partial sums are absolutely convergent near

z0, together with the associative, commutative and distributive laws applied to the par-

tial sums. Here we have used the fact that absolutely convergent complex series can be

rearranged, which follows from the same statement for real series by considering real and

§3: Elementary Operations 21

imaginary parts. To prove that the product of two analytic functions is analytic, multiply

f(z) =∑

an(z − z0)n and g =

∑

bn(z − z0)n as if they were polynomials to obtain:

∞∑

n=0

an(z − z0)n

∞∑

k=0

bk(z − z0)k =

∞∑

n=0

(

n∑

k=0

akbn−k

)

(z − z0)k, (1.4)

which is called the Cauchy product of the two series. Why is this formal computation

valid? If the series for f and the series for g converge absolutely then because we can

rearrange non-negative convergent series

∞ >

∞∑

n=0

|an||z − z0|n∞∑

k=0

|bk||z − z0|k =

∞∑

n=0

(

n∑

k=0

|ak||bn−k|)

|z − z0|n.

This says that the series on the right-hand side of (1.4) is absolutely convergent and

therefore can be arranged to give the left-hand side of (1.4). To put it another way, the

doubly indexed sequence anbk(z − z0)n+k can be added up two ways: If we add along

diagonals: n + k = m, for m = 0, 1, 2, . . ., we obtain the partial sums of the right-hand

side of (1.4). If we add along partial rows and columns n = m, k = 0, . . . , m, and k = m,

n = 0, . . . , m− 1, for m = 1, 2, . . ., we obtain the product of the partial sums for the series

on the left-hand side of (1.4). Since the series is absolutely convergent (as can be seen by

using the latter method of summing the doubly indexed sequence of absolute values), the

limits are the same.

We can also compose analytic functions where they make sense. Suppose f(z) =∑

an(z−z0)n is analytic at z0 and suppose g(z) =∑

bn(z−a0)n is analytic at a0 = f(z0),

then the composition g f is analytic at z0. To see this, note that

∞∑

m=1

|am||z − z0|m−1 (1.5)

converges in z : 0 < |z − z0| < r for some r > 0 since the series for f is absolutely

convergent, and |z − z0| is non-zero. By the root test, this implies that the series (1.5)

converges uniformly in |z− z0| ≤ r1, for r1 < r, and hence is bounded in |z− z0| ≤ r1.Thus there is a constant M <∞ so that

∞∑

m=1

|am||z − z0|m ≤M |z − z0|,


if |z − z0| < r1, and so

∞∑

m=0

|bm|( ∞∑

n=1

|an||z − z0|n)m

≤∞∑

m=0

|bm|(M |z − z0|)m <∞,

for |z − z0| sufficiently small, by the absolute convergence of the series for g. This proves

absolute convergence for the composed series, and thus we can rearrange the doubly-

indexed series for the composition so that it is a (convergent) power series.

As a consequence, if f is analytic at z0 and f(z0) 6= 0 then composing with the function

1/z which is analytic on C \ 0, we conclude 1/f is analytic at z0. A rational function

r is the ratio

r(z) =p(z)

q(z)

where p and q are polynomials. The rational function r is then analytic on z : q(z) 6= 0.Rational functions and their limits are really what this whole course is about.

Definition. If f is defined in a neighborhood of z then

f ′(z) = limw→z

f(w) − f(z)

w − z

is called the (complex) derivative of f , provided the limit exists.

The next Theorem says that you can differentiate power series term-by-term.

Theorem 3.1. If f(z) =∑

an(z − z0)n converges in B = z : |z − z0| < r then f ′(z)

exists for all z ∈ B and

f ′(z) =∞∑

n=1

nan(z − z0)n−1 =

∞∑

n=0

(n+ 1)an+1(z − z0)n,

for z ∈ B. Moreover the series for f ′ based at z0 has the same radius of convergence as

the series for f .

Proof. If 0 < |h| < r then

f(z0 + h) − f(z0)

h− a1 =

∑∞n=0 anh

n − a0

h− a1 =

∞∑

n=2

anhn−1 =

∞∑

n=1

an+1hn.

§3: Elementary Operations 23

By the root test, the series∑

an+1hn converges in a disk and hence converges uniformly

in h : |h| ≤ r1, if r1 < r. In particular,∑

an+1hn is continuous at 0 and hence

limh→0

∞∑

n=1

an+1hn = 0.

This proves that f ′(z0) exists and equals a1.

By Theorem 2.1, f has a power series expansion about each z1 with |z1−z0| < r given

by∞∑

k=0

[ ∞∑

n=k

an

(

n

k

)

(z1 − z0)n−k

]

(z − z1)k

Therefore f ′(z1) exists and equals the coefficient of z − z1

f ′(z1) =

∞∑

n=1

an

(

n

1

)

(z1 − z0)n−1 =

∞∑

n=1

ann(z1 − z0)n−1.

By the root test and the fact that n1n → 1, the series for f ′ has exactly the same radius of

convergence as the series for f .

Since the series for f ′ has the same radius of convergence as the series for f , we obtain

the following corollary.

Corollary 3.2. An analytic funtion f has derivatives of all orders and

f(z) =∞∑

n=0

f (n)(z0)

n!(z − z0)

n,

if f is analytic at z0.

By definition of the symbols, the n = 0 term in the series is f(z0).

Proof. If f(z) =∑∞n=0 an(z − z0)

n, then we proved in Theorem 3.1 that a1 = f ′(z0) and

f ′(z) =

∞∑

n=1

nan(z − z0)n−1.

Applying Theorem 3.1 to f ′(z), we obtain 2a2 = (f ′)′(z0) ≡ f ′′(z0) and by induction

n!an = f (n)(z0).


If f is analytic in a region Ω with f ′(z) = 0 for all z ∈ Ω then by Corollary 3.2, f

is constant. In fact, by Corollary 2.3, if f is non-constant then the zeros of f ′ must be

isolated.

Corollary 3.3. If f(z) =∑

an(z − z0)n converges in B = z : |z − z0| < r then the

power series

F (z) =∞∑

n=0

ann+ 1

(z − z0)n+1

converges in B and satisfies

F ′(z) = f(z),

for z ∈ B.

The series for F has the same radius of convergence as the series for f , by Corollary

3.1 or by direct calculation.

III

The Maximum Principle.

§1. The Maximum Principle.

The next result is perhaps the most important elementary result in complex analysis.

A region is a connected open set in the plane.

Theorem 1.1 (Maximum Modulus Principle). Suppose f is analytic in a region Ω.

If there exists a z0 ∈ Ω such that

|f(z0)| = supz∈Ω

|f(z)|

then f is constant in Ω.

In particular a non-constant analytic function has no local maximum absolute value.

Proof. If f is analytic at z0, and if ak is the first nonzero coefficient in the series expansion

for f − f(z0) based at z0, then

f(z) − f(z0) = ak(z − z0)k

∞∑

n=k

anak

(z − z0)n−k = ak(z − z0)

kg(z), (1.1)

where g is analytic at z0 and g(z0) = 1. Write z = z0 + εeit. Then eikt traces the unit

circle k times as t increases from 0 to 2π. Thus the image f(z0 + εeit), for ε sufficiently

small, traces a curve that winds k times around f(z0), and which approximately lies on a

circle of radius |ak|εk centered at f(z0), since g is continuous and g(z0) = 1. In particular,

the image of the circle of radius ε contains points with larger absolute value than |f(z0)|.Thus if |f | is maximal at z0, then there is no first non-zero coefficient in the series

expansion about z0 and so f(z) ≡ f(z0) in a neighborhood of z0. If

E = z ∈ Ω : f(z) = f(z0)

25

26 III. The Maximum Principle.

then we have proved E is open. But E is also closed in Ω since f is continuous. Because

E is connected, E = Ω and f is constant.

If f(z0) 6= 0 then the same argument shows that |f(z0)| is not a local minima if f is

non-constant. This fact can also be derived from the statement of the maximum principle

by considering the function 1/f which is analytic off the zeros of f .

Another form of the Maximum Principle is:

Corollary 1.2. If f is a non-constant analytic function in a bounded region Ω and if f is

continuous on Ω then

maxz∈Ω

f(z)

occurs on ∂Ω but not in Ω.

The reader should verify the alternate form: if f is analytic on Ω then

lim supz→∂Ω

f(z) = supΩf(z).

If Ω is unbounded, then ∞ must also be viewed as a point in ∂Ω. The function f(z) = e−iz

is analytic in the upper half plane H = z : Imz > 0, continuous on z : Imz ≥ 0 and

has absolute value 1 on the real line R, but is not bounded by 1 in H.

§2. Consequences of the Maximum Principle.

The maximum principle gives an easy proof of an important result you’ve seen in some

form or another since high school.

Theorem 1.3 (Fundamental Theorem of Algebra). Every non-constant polynomial

has a zero.

This remarkable result says that by extending the real numbers to the complex num-

bers via the solution to the equation z2 + 1 = 0 then every polynomial equation has a

solution.

Proof. Suppose p(z) = anzn + an−1z

n−1 + . . . + a1z + a0 is a polynomial which has no

zeros and for which an 6= 0. Then

1

p(z)=

1

(anzn)(1 +an−1

an

1z + . . . a0

an

1zn )

.

§2: Consequences 27

Since 1/zk → 0 as |z| → ∞, we conclude that given ε > 0, then for |z| = R with R

sufficiently large we have∣

∣

∣

∣

1

p(z)

∣

∣

∣

∣

≤ ε.

By the Maximum Principle

sup|z|≤R

∣

∣

∣

∣

1

p(z)

∣

∣

∣

∣

≤ ε.

Letting ε→ 0 we conclude 1p(z) = 0 for all z, which is a contradiction.

Corollary 1.4. If p is a polynomial of degree n, then there are complex numbers z1, . . . , zn

and a complex constant c so that

p(z) = c

n∏

k=1

(z − zk).

Corollary 1.4 does not tell us how to find the zeros, but it does say that there are

exactly n zeros.

Proof. The proof is by induction. First note that

zn − bn = (z − b)(bn−1 + zbn−2 + . . . zn−2b+ zn−1).

So if p(b) = 0, then

q(z) ≡ p(z)

z − b=p(z) − p(b)

z − b=

n∑

k=1

ak(

k−1∑

j=0

bk−1−jzj). (1.2)

The coefficient of zn−1 in (1.2) is an so q is a polynomial of degree n− 1. Repeating this

argument n times proves the Corollary.

Corollary 1.5. If p(z) =∑nk=0 akz

k is a polynomial with real coefficients ak then p

can be factored into a product of linear and quadratic factors:

p(z) = C

p∏

k=1

(z − xk)

q∏

j=1

(z2 + bjz + cj),


where C, xk, bj, cj ∈ R and the quadratic factors z2 + bjz + cj are non-zero on R.

Proof. Since the coefficients of p are real, if p(a) = 0 then p(a) = p(a) = 0. The quadratic

factors come from grouping the non-real zeros in pairs: (z−a)(z−a) = z2 +(Rea)z+ |a|2.

For example the polynomial zn− 1 has n zeros. If zn = 1 then 1 = |zn| = |z|n so that

|z| = 1. Write z = cos t + i sin t = eit, by a homework problem. Then zn = eint = 1 so

that nt = 2πk for some integer k, and thus t = 2πk/n. The n distinct zeros of zn are then

ei2πk/n, k = 0, 1, . . . , n− 1 which are equally spaced around the unit circle.

Corollary 1.6. A non-constant analytic function defined on a region is an open map.

In other words, if f is analytic and non-constant on a region Ω and if U ⊂ Ω is open

then f(U) is an open set.

Proof. Suppose f has a power series expansion which converges on z : |z − z0| < R.Pick r < R and set

δ = inf|z−z0|=r

|f(z) − f(z0)|.

Since the zeros of f − f(z0) are isolated, we may suppose that δ > 0 by decreasing r if

necessary. If |w − f(z0)| < δ/2 and if f(z) 6= w for all z such that |z − z0| ≤ r, then

1/(f − w) is analytic in |z − z0| ≤ r and∣

∣

∣

∣

1

f(z) − w

∣

∣

∣

∣

≤ 1

|f(z) − f(z0)| − |w − f(z0)|<

1

δ − δ/2=

2

δ

on |z − z0| = r. By the maximum principle the inequality persists in |z − z0| < r. But

evaluating this expression at z0 we obtain the contradiction 2/δ < 2/δ.

Corollary 1.7 (Liouville’s Theorem). If f is analytic in C and bounded, then f is

constant.

Proof. Suppose |f | ≤M <∞. Set g(z) = (f(z)−f(0))/z. Then g is analytic and |g| → 0

as |z| → ∞. By the maximum principle g ≡ 0 and hence f ≡ f(0).


Corollary 1.8 (Schwarz’s Lemma). Suppose f is analytic in D = z : |z| < 1 and

suppose |f(z)| ≤ 1 and f(0) = 0. Then

|f(z)| ≤ |z|, (1.3)

for all z ∈ D and

|f ′(0)| ≤ 1. (1.4)

Moreover, if equality holds in (1.3) or (1.4) then f(z) = cz where c is a constant with

|c| = 1.

Proof. By the first Homework problem, the function g given by

g(z) =

f(z)

zif z ∈ D \ 0

f ′(0) if z = 0

is analytic in D and

sup|z|=r

|g(z)| ≤ 1

r.

Fix z0 ∈ D, then for r > |z0|, the maximum principle implies |g(z0)| ≤ 1r, so that, letting

r → 1, we obtain (1.3) and (1.4). If equality holds in (1.3) or (1.4) then g(z) has a

maximum at z0 and hence is constant.

Corollary 1.9 (Invariant form of Schwarz’s Lemma). Suppose f is analytic in D =

z : |z| < 1 and suppose |f(z)| ≤ 1. If a, b ∈ D then

∣

∣

∣

∣

f(b) − f(a)

1 − f(a)f(b)

∣

∣

∣

∣

≤∣

∣

∣

∣

b− a

1 − ab

∣

∣

∣

∣

, (1.5)

and|f ′(z)|

1 − |f(z)|2 ≤ 1

1 − |z|2 . (1.6)

Proof. If |c| < 1, the function

Tc(z) =z − c

1 − cz


is analytic except at z = 1/c. Moreover for z = eit ∈ ∂D,

|Tc(eit)| =

∣

∣

∣

∣

eit − c

1 − ceit

∣

∣

∣

∣

=|eit − c||e−it − c| = 1.

By the maximum principle (or direct computation), |Tc| ≤ 1 on D. Setting c = f(a), the

composition Tc f is analytic on D and bounded by 1. Furthermore

Tc f(z)

Ta(z)=

(

f(z) − f(a)

1 − f(a)f(z)

)(

1 − az

z − a

)

is analytic on D and

lim sup|z|→1

∣

∣

∣

∣

Tc f(z)

Ta(z)

∣

∣

∣

∣

= lim sup|z|→1

|Tc f(z)| ≤ 1,

by the maximum principle. This proves (1.5). Inequality (1.6) follows by dividing both

sides of (1.5) by b− a and letting b→ a.

Corollary 1.10. If f is analytic on D, |f | ≤ 1, and f(zj) = 0, for j = 1, . . . , n then

f(z) =n∏

j=1

(

z − zj1 − zjz

)

g(z),

where g is analytic in D and |g(z)| ≤ 1 on D.

Proof. If f(a) = 0, then by the proof of Corollary 1.9, g(z) = f(z)/Ta(z) is analytic in

the disk and bounded by 1. Repeating this argument n times proves the Corollary.

By the Uniqueness theorem, if f is analytic on D, then the zeros of f do not cluster

in D, unless f is identically zero. If we have more restrictions on f , such as boundedness,

then the zeros cannot approach the unit circle too slowly by the next Corollary.

Corollary 1.11. If f is non-constant, bounded and analytic in D and if zj are the zeros

of f then∑

j

(1 − |zj |) <∞.


Proof. We may suppose |f | ≤ 1, by dividing f by a constant if necessary. If f(0) 6= 0,

then using the notation of the proof of Corollary 1.10,

|f(0)| = (

n∏

j=1

|zj |)|g(0)| ≤n∏

j=1

|zj |,

so that by taking logarithms (base e),

log1

|f(0)| ≥n∑

j=1

log1

|zj |≥

n∑

j=1

(1 − |zj |).

If f(0) = 0, then write f(z) = zkh(z) where h(0) 6= 0. Applying the preceeding argument

to h, we obtainn∑

j=1

(1 − |zj |) ≤ log1

|h(0)| + k.

The Corollary follows by letting n→ ∞.

Much of what we’ll do in this class takes place on D or on C, which look like rather

special domains, but we know a power series converges on a disk, and by scaling the domain,

we can assume it is D or C. Also in the third quarter, we’ll prove the uniformization theorem

which says in some sense the only analytic functions we need to understand can be defined

on D or C.

As an aside, there is one application of the Fundamental Theorem of Algebra I’d like

to cover because you may use it when you teach differential equations. The setting is

solving differential equations using the Laplace Transform. The Laplace Transform of a

function f defined on [0,∞) is given by

L(f)(z) =

∫ ∞

0

f(t)e−ztdt.

An integration by parts shows that

L(f ′) = zL(f) − f(0).

The logarithm was developed because it converted a hard problem, multiplying two 5 digit

numbers, to a simpler problem of addition:

log(ab) = log(a) + log(b).


To multiply a and b, first take their logs, then add the logs, then use the inverse of

the logarithm to find ab. The Laplace Transform acts similarly, converting a differential

equation to an algebraic equation. We solve the algebraic equation then need to convert

back to the original context by taking the inverse of the Laplace transform. For example,

to solve

y′′ − 3y′ + 2y = f(t), with y(0) = 1 and y′(0) = 2

apply the Laplace transform to obtain

z2L(y) − z − 2 − 3zL(y) + 3 + 2L(y) = L(f),

where we have used L(y′′) = zL(y′) − y′(0), and L(y′) = zL(y) − y(0). Thus

L(y) =z − 1 + L(f)

z2 − 3z + 2.

For simple electrical circuits for example, L(f) is a rational function and hence so is

L(y). To compute the inverse Laplace transform of a rational function, it is best to simplify

the rational function using partial fractions. By the Fundamental Theorem of Algebra, a

rational function can be written in the form

r(z) =p(z)

∏Nj=1(z − zj)nj

,

where the zj are distinct.

Corollary 1.12 (Partial Fraction Expansion). If p is a polynomial then there is a

polynomial q and constants ck,j so that

p(z)∏Nj=1(z − zj)nj

= q(z) +N∑

j=1

nj∑

k=1

ck,j(z − zj)k

. (1.7)

Proof. There are two initial cases to consider: If p is a polynomial then

p(z)

z − a=p(z) − p(a)

z − a+

p(a)

z − a. (1.8)


and as in (1.2), (p(z) − p(a))/(z − a) is a polynomial. Secondly, we can write

1

(z − a)(z − b)=

A

z − a+

B

z − b, (1.9)

for some constants A and B. To see that this is true, multiply both sides by (z− a)(z− b)

to obtain a linear equation:

1 = A(z − b) +B(z − a),

when z 6= a or b. Since a linear polynomial has only one zero, the coefficient A + B of z

must be equal to 0. This implies 1 = −Ab−Ba, and solving we obtain A = 1/(a− b) and

B = 1/(b − a). These steps are reversible, so that (1.9) is correct. The full theorem now

follows by induction. Suppose the Corollary is true if the degree of the denominator is at

most d. If we have an equation of the form (1.7) then we can divide the equation by z−a.The right side consists of lower degree terms to which the induction hypothesis applies,

with one exception: if the denominator of the left side of (1.7) is (z − b)d. If b = a there

is nothing to prove because dividing each term by (z − a) puts it in the correct form. If

b 6= a, then we could have applied the inductive assumption to the decomposition of

p(z)

(z − b)d−1(z − a)

and then divided the result by z − b.

Now that we know the form of the solution, we can try to determine the coefficients.

The brute force way is to multiply both sides of (1.7) by the denominator of (1.7) and solve

a (big) system of equations. This usually leads to errors. There is a faster way, called the

cover-up method. Here is an illustration. Suppose we’d like to decompose

z2

(z − a)(z − b)(z − c),

where a, b, and c are distinct. Then by the Corollary

z2

(z − a)(z − b)(z − c)=

A

z − a+

B

z − b+

C

z − c. (1.9)


If we multiply the left side by z − a then let z → a, the left side tends to

a2

(a− b)(a− c).

If we multiply each term on the right side by z − a we just get A for the first term. The

other two terms will have then a factor of z − a which tend to 0 as z → a. Thus

a2

(a− b)(a− c)= A.

The same process works for determining B and C. This is called the “cover-up method”

because we can write down the coefficients immediately with little writing in the following

way: Write out (1.9) but simply leave blanks in place of the coefficients A, B and C.

Multiplying the left side by z − a, is accomplished by putting your hand over the factor

z − a on the left. Now the limit as z → a in the remaining portion of the left side can

be immediately written down as a (possibly complex) fraction, and it must be the same

as the coefficient of 1/(z − a) on the right and hence can be written in the appropriate

blank spot above z − a. The same works for the remaining coefficients. If you try doing

the same problem by solving three equations and three unknowns, you’ll immediately see

the benefit. The same process works for an arbitrary number of distinct factors. Notice

that we did not say to multiply by z−a then set z = a, for then we would have multiplied

by 0, but rather we let z → a.

What happens if the roots are not distinct? Again we’ll illustrate with an example:

(z − d)

(z − a)(z − b)2=

A

z − a+

B

z − b+

C

(z − b)2.

We can determine the coefficient A using exactly the same cover-up method. It doesn’t

take long to discover that we can find C by similarly multiplying both sides by (z − b)2

and letting z → b. But what about B? Here’s how to find B using the cover-up method.

It is to simply apply the idea of the proof of the partial fraction expansion: First solve

z − d

(z − a)(z − b)=

D

z − a+

E

z − b

using the cover-up method. Now divide both sides by z − b. We obtain

(z − d)

(z − a)(z − b)2=

D

(z − a)(z − b)+

E

(z − b)2.


The first term on the right can be further decomposed by the cover-up method, and the

last is already in the desired form. This can be repeated to cover all cases.

The cover-up method isn’t always the fastest method, but if you’ve ever done any

computer programming, you might see that it wouldn’t be hard to implement the method.

Solving many equations in many unknowns is in general a delicate process, not only difficult

by hand but requiring more sophisticated numerical methods than elimination to avoid loss

of information. In the case where you know all coefficients except one, for example, it might

be better to simply evaluate both sides of the expression at one value of z and solve for

the one unknown.

An objection I’ve seen to the cover-up method is that partial fractions that occur in

Engineering problems generally all have real-coefficients. It is better many times to give

the partial fraction decomposition without complex numbers in the expression. In other

words, allow irreducible quadratics and their powers in the partial fraction expansion.

You can of course factor the denominator using complex roots, find the partial fraction

expansion, then combine the terms that correspond to complex conjugate pairs of zeros of

the denominator. A more straight-forward method is to just apply the idea of the cover-up

method directly. Again we’ll use an example to illustrate the idea. It is not hard to prove

that if a, b, c, d and e are real, then

z2 + dz + e

(z − a)((z − b)2 + c2)=

A

z − a+B(z − b) +D

(z − b)2 + c2, (1.10)

where A, B, and D are real. Notice that we’ve written the numerator of the last term as

B(z− b) +D. We could have used just Bz+D, but there are two reasons for the form we

chose: first is that if we want to take the inverse Laplace Transform of the right side using

familiar functions, we would need to change to Bz+D to B(z−b)+E anyways. Secondly,

the form B(z − b) + D is actually easier to find using the cover-up method. Indeed, we

find the coefficient A by the usual cover-up method. Then to find B and D, we multiply

both sides of (1.10) by (z − b)2 + c2 and let it tend to 0. Thus z → b ± ic. Cover up the

quadratic factor in the denominator of the left side of (1.10) and let z → b + ic. On the

right side, when we multiply by the quadratic factor, the first term will tend to 0, and the

denominator of the second term will be cancelled and B(z − b) +D will tend to Bic+D.


Thus the real part of the result on the left equals D and the imaginary part equals Bc

and we can then write down immediately the coefficients B and D. Try this process for

two irreducible quadratic factors in the denominator, and you’ll see how much faster and

accurate it is than solving many equations with many unknowns.

The one case we haven’t treated is when the degree of the numerator is larger than

the degree of the denominator. In this case, we can use long division to reduce the degree.

How would you use the cover-up method to find the coefficients of the polynomial q on the

right side of (1.7)?

The technique above for polynomials with real coefficients can be taught to your

differential equation students once you remind them how to multiply and divide complex

numbers.

§3. Boundary Behavior.

We conclude this chapter with some examples and a theorem about boundary behavior

of analytic functions on the unit disk.

The first example is

I(z) = ez+1z−1 .

Since I is the composition of an analytic function on C \ 1 and the exponential function,

which is analytic in C, I is analytic on C \ 1. By a homework problem, |ez| = eRez so

that by a computation

|I(z)| = e|z|2−1

|z−1|2 .

Thus |I(z)| ≤ 1 on D \ 1. On the unit circle I(eit) = e−i cot(t/2), for 0 < t < 2π. In

particular if ζ ∈ ∂D \ 1, then

limz→ζ

I(z)

exists and has absolute value 1. However, for 0 < r < 1, I(r) = er−1r+1 → 0 as r → 1. On

the unit circle I(eit) is spinning rapidly as t → 0. Hence I(z) does not have a limit as

z → 1. The proper way to take limits in the disk is through cones. For ζ ∈ ∂D and α > 1,

define

Γα(ζ) = z ∈ D : |z − ζ| ≤ α(1 − |z|).

§3: Boundary Behavior 37

to be the Stolz angle or cone at ζ.

0

zΓα(ζ)

2 sec−1 α

1 − |z|

|z − ζ|

ζ

Figure III.1 Stolz angle Γα(ζ).

The precise shape of Γα is not important except that it is symmetric about the line

segment [0, ζ] and forms an angle less than π at ζ. For z ∈ Γα(1) we have

|I(z)| = e−(1−|z|)

|z−1|(1+|z|)|z−1| ≤ e−

1α|z−1| → 0

as z ∈ Γα(1) → 1. Thus I(z) → 0 in every cone, and ∪αΓα(1) = D, but there is still no

limit as z → 1.

Another example is the function given by

L(z) =

∞∑

n=0

z2n

.

This series converges uniformly and absolutely in D. It is sometimes called a lacunary

series since the spacing between nonzero coefficients increases as n increases. The function

L does not extend continuously to any point of ∂D. For example if ζ2k

= 1 then

L(rζ) =

k−1∑

n=0

(rζ)2n

+

∞∑

n=k

r2n

.

The second sum is clearly positive and increasing to ∞ as r → 1. Since the 2k roots of 1

are evenly spaced around ∂D, if eit ∈ ∂D, then we can find a ζ as close to eit as we like

with ζ2k

= 1. Thus in any neighborhood of eit, f is unbounded.


The next Theorem gives a connection between Fourier series and analytic functions

in D.

Theorem 1.13 (Abel’s Limit Theorem). If ζ = eit ∈ ∂D and if∑∞n=0 ane

int converges,

then f(z) =∑∞

n=0 anzn converges in D and if Γ = Γα(ζ) is any Stolz angle at ζ then

limz∈Γ→ζ

f(z) =∞∑

n=0

aneint.

The convergence in Theorem 1.13 is called nontangential convergence. Abel’s

Theorem says that you get what you would expect for a limit, but only nontangentially.

Note that the limit does not depend on which Stolz angle you choose.

Proof. Replacing f(z) by f(ζz), we may suppose ζ = 1 and by subtracting a constant

from a0 we may suppose∑∞n=0 an = 0. By the root test, the series for f converges in D.

Suppose z ∈ Γα(1). Then

f(z)

1 − z=

∞∑

n=0

zn∞∑

k=0

akzk =

∞∑

n=0

( n∑

k=0

ak

)

zn.

Set sk =∑nk=0 ak. Then

|f(z)| ≤ |1 − z|∣

∣

∣

∣

N−1∑

n=0

snzn

∣

∣

∣

∣

+|1 − z|∣

∣

∣

∣

∞∑

n=N

snzn

∣

∣

∣

∣

.

Given ε > 0, there exists N <∞ so that |sn| < ε for n ≥ N . Thus

|f(z)| ≤ |1 − z|N−1∑

n=0

|sn| + |1 − z|ε∞∑

n=N

|z|n = |1 − z|N−1∑

n=0

|sn| +|1 − z||z|Nε

1 − |z| .

We can choose δ > 0 so that if z ∈ Γα(1) and |z − 1| < δ,

|f(z)| ≤ ε+ αε.

Since ε > 0 was arbitrary,

limz∈Γα→1

|f(z)| = 0.

§3: Boundary Behavior 39

A Fourier Series is a function of the form

F (t) =

∞∑

n=−∞

aneint.

If F (t) converges then

f(z) =

∞∑

n=0

anzn and g(z) =

∞∑

n=0

a−nzn

converge and are analytic on D. By Abel’s Theorem, f(reit) + g(reit) converges to F (t)

at each t where the series for F converges. Thus the harmonic function f + g “extends” F

to D, and the infinitely differentiable functions defined on [0, 2π] by

fr(t) = f(reit) and gr(t) = g(reit)

satisfy

fr + gr → F

as r → 1, at each t where the series for F converges.

Fourier Series arose from attempting to solve certain differential equations. Each

square integrable function on [0, 2π] has a Fourier Series, and it was a famous problem for

many years to prove that the Fourier series converges almost everywhere on [0, 2π]. The

proof that was eventually found remains perhaps one of the hardest proofs in analysis, but

it used this connection between Fourier Series and analytic functions on the disk, and the

relation between convergence on the circle and nontangential convergence.

40

IV

Integration Along Curves

§1. Definitions.

Definition. A curve is a continuous mapping of an interval I ⊂ R into C.

Different curves can have the same image. For example if γ(t) : [0, 1] → C then

γ(t2) : [0, 1] → C and both curves have the same image. Many times we will write

γ

Figure IV.1 A curve γ.

and call the curve γ. But we really mean a particular choice (though unstated) of the

parameterization. The arrows show how γ(t) traces the image as t ∈ I increases.

Definition. A curve γ is called an arc if it is two-to-two.

“Two-to-two” means two points go to two points (if a 6= b then f(a) 6= f(b)). A less

enlightened, but unfortunately widespread terminology is “one-to-one” which really should

be the definition of a function.

Definition. A curve γ : [a, b] → C is called closed if γ(a) = γ(b).

Definition. A closed curve γ : [a, b] → C is called simple if γ restricted to [a, b) is

two-to-two (one-to-one).

A simple closed curve γ : [0, 2π] → C can also be viewed as a two-to-two mapping of

the unit circle given by ψ(eit) = γ(t).

41

42 IV. Integration Along Curves

Definition. A curve γ = x+ iy is called piecewise continuously differentiable if

γ′(t) = x′(t) + iy′(t)

exists except for finitely many t and x′ and y′ have one-sided limits (from both sides) at

the exceptional points.

From now on, all curves will be assumed to be piecewise continuously differentiable

unless stated otherwise. In particular, by the Fundamental Theorem of Calculus if γ

is piecewise continuously differentiable then

γ(t2) − γ(t1) = (x(t2) − x(t1)) + i(y(t2) − y(t1)) =

∫ t2

t1

x′(t)dt+ i

∫ t2

t1

y′(t)dt

Definition. A curve ψ : [c, d] → C is called a reparameterization of a curve γ : [a, b] →C if there exists an increasing function α : [a, b] → [c, d] such that ψ(α(t)) = γ(t).

For example ψ(t) = t2 + it4, for 0 ≤ t ≤ 1, is a reparameterization of γ(t) = t + it2,

for 0 ≤ t ≤ 1, with α(t) = t12 . If σ : [0, 1] → C is a curve then the curve β, defined

by β(t) = σ(1 − t), is not a reparameterization of σ because 1 − t is decreasing. If

ψ is a reparameterization of γ with ψ(α(t)) = γ(t), where α is piecewise continuously

differentiable then ψ′(α(t))α′(t) = γ′(t), by the chain rule applied to the real and imaginary

parts, or by taking limits of difference quotients.

Definition. If γ : [a, b] → C is a curve and if f is a continuous complex-valued function

defined on (the image of) γ then

∫

γ

f(z)dz ≡∫ b

a

f(γ(t))γ′(t)dt.

The reader can check that a reparameterization of γ will not change the integral by

the chain rule, and so the integral really depends on the image of γ, not the choice of

parameterization and for that reason we use the notation∫

γfdz. Note, however, that the

direction of the image curve is important. If ψ(−t) = γ(t), then∫

ψ

f(z)dz = −∫

γ

f(z)dz.

§1: Definitions. 43

Another reason for the notation∫

γfdz is the following.

Suppose a = t0 < t1 < t2 < . . . < tn = b and set γ(tj) = zj . Then

n−1∑

j=0

f(zj)(zj+1 − zj) =

n−1∑

j=0

f(γ(tj))[γ(tj+1) − γ(tj)]

≈n−1∑

j=0

f(γ(tj))γ′(tj)[tj+1 − tj ],

and the latter is a Riemann sum using the independent variable z for

∫ b

a

f(γ(t))γ′(t)dt =

∫

γ

f(z)dz.

Integration is linear∫

γ(f(z) + g(z))dz =

∫

γf(z)dz +

∫

γg(z)dz, and if C is constant

then∫

γCf(z)dz = C

∫

γf(z)dz.

Definition. If γ : [a, b] → C is a curve and if f is a continuous complex-valued function

defined on (the image of) γ then

∫

γ

f(z)|dz| =

∫ b

a

f(γ(t))|γ′(t)|dt.

Thus∣

∣

∣

∣

∫

γ

f(z)dz

∣

∣

∣

∣

=

∣

∣

∣

∣

∫ b

a

f(γ(t))γ′(t)dt

∣

∣

∣

∣

≤∫ b

a

|f(γ(t))||γ′(t)|dt =

∫

γ

|f(z)||dz|.

Note that if zj = γ(tj) then∑

f(zj)|zj+1 − zj |

is a Riemann sum for∫

γf(z)|dz|.

Definition. If γ : [a, b] → C is a curve then the length of γ is defined to be

`(γ) = |γ| =

∫

γ

|dz|.


Note that we are assuming all curves are piecewise C1 on a closed interval in R so that all

curves have finite length.

The following important estimate follows immediately from the definitions:

∣

∣

∣

∣

∫

γ

f(z)dz

∣

∣

∣

∣

≤(

supγ

|f(z)|)

`(γ). (1.1)

Consequently, if fn converges uniformly to f on γ and if `(γ) <∞ then

limn

∫

γ

fn(z)dz =

∫

γ

f(z)dz.

To prove this just use the estimate:

|∫

γ

(fn − f)dz| ≤(

supγ

|fn − f |)

`(γ).

Note that all curves are assumed to be piecewise continuously differentiable, and hence

have finite length.

§2. Equivalence of Analytic and Holomorphic.

Definition. A complex valued function f is said to be holomorphic on a set S if it is

defined in an open set U ⊃ S and

f ′(z) = limw→z

f(w) − f(z)

w − z

exists and is continuous on U .

An analytic function is holomorphic by Theorem II.3.1.

The Chain Rule for complex differentiation says that if γ is a piecewise continuously

differentiable curve and if f is holomorphic on a neighborhood of γ then f γ is a piecewise

continuously differentiable curve and

d

dtf(γ(t)) = f ′(γ(t))γ′(t).

§2: Equivalence of Analytic and Holomorphic. 45

The chain rule follows from the real-variables version or can be proved directly using

difference quotients. By the Fundamental Theorem of Calculus and the chain rule, if f is

holomorphic in a neighborhood of γ : [a, b] → C then∫

γ

f ′(z)dz =

∫ b

a

f ′(γ(t))γ′(t)dt =

∫ b

a

d

dtf(γ(t))dt = f(γ(b))− f(γ(a)).

For example the line segment from z to ζ can be parameterized by γ(t) = z+ t(ζ−z),for t ∈ [0, 1], so that if f is holomorphic in a neighborhood of γ, then

f(ζ) − f(z) =

∫ 1

0

f ′(z + t(ζ − z))(ζ − z)dt. (2.1)

Corollary 2.1. If γ : [a, b] → C is a closed curve and if f is holomorphic in a neighborhood

of γ then∫

γ

f ′(z)dz = f(γ(b))− f(γ(a)) = 0.

The next example will be key to understanding integrals of analytic functions, as we

shall see later. If r > 0 set

Cr = z0 + reit : 0 ≤ t ≤ 2π,

then we have

Proposition 2.2.

1

2πi

∫

Cr

1

z − adz =

1 if |a− z0| < r

0 if |a− z0| > r.

Proof. Suppose |a− z0| < r. Then C′r(t) = ireit and

1

2πi

∫

Cr

1

z − adz =

1

2πi

∫ 2π

0

1

reit − (a− z0)ireitdt

=1

2π

∫ 2π

0

1

1 − (a−z0reit )

dt

=1

2π

∫ 2π

0

∞∑

n=0

(

a− z0reit

)n

dt

=

∞∑

n=0

(a− z0)n

rn1

2π

∫ 2π

0

e−intdt = 1.


Interchanging the order of summation and integration is justified since |(a−z0)/(reit)| < 1

implies uniform and absolute convergence of the series.

If |a− z0| > r, then write

reit

reit − (a− z0)=

(

reit

z0 − a

)

1

1 − reit

a−z0

= − reit

a− z0

∞∑

n=0

rneint

(a− z0)n= −

∞∑

n=1

rneint

(a− z0)n,

so that

1

2πi

∫

Cr

1

z − adz =

1

2π

∫ 2π

0

reit

reit − (a− z0)dt = −

∞∑

n=1

rn

(a− z0)n1

2π

∫ 2π

0

eintdt = 0.

An immediate consequence of Corollary 2.1 and Proposition 2.2 is that there is no

function f defined in a neighborhood of ∂D satisfying f ′(z) = 1/z.

Theorem 2.3. If f is holomorphic on z : |z − z0| ≤ r then for |z − z0| < r,

f(z) =1

2πi

∫

Cr

f(ζ)

ζ − zdζ.

where Cr is the circle of radius r centered at z0, parameterized in the counter-clockwise

direction.

Theorem 2.3 shows that it is possible to find the values of an analytic function inside

a disk from the values on the bounding circle.

Proof. By (2.1) and Corollary 2.1

∫

Cr

f(ζ) − f(z)

ζ − zdζ = lim

ε→0

∫

Cr

∫ 1

ε

f ′(z + t(ζ − z))dtdζ

= limε→0

∫ 1

ε

∫

Cr

f ′(z + t(ζ − z))dζdt

= limε→0

∫ 1

ε

∫

Cr

d

dζf(z + t(ζ − z))dζ

dt

t= 0.

Thus1

2πi

∫

Cr

f(ζ)

ζ − zdζ = f(z) · 1

2πi

∫

Cr

dζ

ζ − z= f(z),

§2: Equivalence of Analytic and Holomorphic. 47

by Proposition 2.2.

A consequence of Theorem 2.3 is the converse to Theorem II.3.1.

Corollary 2.4. A complex-valued function f is holomorphic on a region Ω if and only if

f is analytic on Ω. Moreover, the series expansion for f based at z0 ∈ Ω converges on the

largest open disk centered at z0 and contained in Ω.

Proof. If f is analytic in Ω then f is holomorphic in Ω, by Theorem II.3.1. To prove the

converse, suppose f is holomorphic on z : |z − z0| ≤ r. If |z − z0| < r, then by II.1.2

f(z) =1

2πi

∫

Cr

f(ζ)

ζ − zdζ =

1

2πi

∫

Cr

( ∞∑

n=0

1

(ζ − z0)n+1(z − z0)

n

)

f(ζ)dζ

=

∞∑

n=0

(

1

2πi

∫

Cr

f(ζ)

(ζ − z0)n+1dζ

)

(z − z0)n.

Interchanging the order of the summation and integral is justified by the uniform and ab-

solute convergence of the series for z fixed. Thus f has a power series expansion convergent

in z : |z − z0| ≤ r, provided this closed disk is contained in Ω.

In particular, if f is analytic in C then f has a power series expansion which converges in

all of C. Such functions are called entire. From now on, we will use the words analytic

and holomorphic interchangably.

The proof of Corollary 2.4 yields a bit more information. Not only can we find the

values of an analytic function inside a disk from it values on the boundary, but we also

have a formula for each of its derivatives in the disk.

Corollary 2.5 (Cauchy’s Estimate). If f is analytic in z : |z− z0| ≤ r and Cr(z0) =

z0 + reit : 0 ≤ t ≤ 2π, then

f (n)(z0)

n!=

1

2πi

∫

Cr(z0)

f(ζ)

(ζ − z0)n+1dζ, (2.2)


and∣

∣

∣

∣

f (n)(z0)

n!

∣

∣

∣

∣

≤supCr(z0)

|f |rn

. (2.3)

Proof. Equation (2.2) follows from Corollary II.3.2, the proof of Corollary 2.4, and The-

orem II.2.2. Inequality (2.3) follows from (2.2) by using inequality (1.1).

§3. Approximation by Rational Functions.

In this section we will show that Theorem 2.3 also holds if the circle Cr is replaced by

the boundary of a square, and then use it to prove Runge’s theorem that analytic functions

can be uniformly approximated by rational functions.

Corollary 3.1. If f is analytic in an open disk B and if γ ⊂ B is a closed curve, then

∫

γ

f(z)dz = 0.

Proof. By Corollary 2.4, f has a power series expansion which converges on B. By

Corollary II.3.3, there is an analytic function F so that F ′ = f on B. Now apply Corollary

2.1 to F ′.

Definition. If γ1 : [a, b] → C and γ2 : [c, d] → C with γ1(b) = γ2(c), then γ1 + γ2 is the

curve defined on the interval [a, b+ d− c] by

γ1 + γ2(t) =

γ1(t) if t ∈ [a, b]

γ2(t+ c− b) if t ∈ [b, b+ d− c].

γ1

γ2

γ1(b) = γ2(c)

Figure IV.2 The curve γ1 + γ2.

§3: Approximation by Rational Functions. 49

Definition. If γ : [a, b] → C, then −γ : [−b,−a] → C is the curve defined by

−γ(t) = γ(−t).

The curve −γ has the same geometric image as γ, but it is traced in the opposite

direction. If F is continuous on γj , j = 1, 2, then

∫

γ1+γ2

F (z)dz =

∫

γ1

F (z)dz +

∫

γ2

F (z)dz,

and∫

−γ1

F (z)dz = −∫

γ

F (z)dz.

For example, the integral around two adjacent squares, each in the counter-clockwise

direction is equal to the integral around the boundary of the union of the squares.

R

S1 S2

∂(S1 ∪ S2)

Figure IV.3 Integrals around squares.

In the left side of Figure IV.3, the boundary of the squares, ∂S1 and ∂S2, are param-

eterized in the counter-clockwise direction and

∫

∂(S1∪S2)

F (z)dz =

∫

∂S1

F (z)dz +

∫

∂S2

F (z)dz,

for every continuous function F defined on ∂S1 ∪ ∂S2. This can be seen by writing the

integrals around each square as the sum of integrals on the bounding line segments. The

common boundary edge is traced in opposite directions, so the corresponding integrals will


cancel. A similar argument applies to a finite union of squares, so that after cancellation,

the sum of the integrals around the boundaries of all the squares is equal to the integral

around the boundary of the union of the squares.

Proposition 3.2. If S is an open square with boundary ∂S parameterized in the counter-

clockwise direction then

1

2πi

∫

∂S

1

z − adz =

1 if a ∈ S

0 if a ∈ C \ S.

Proof. If a ∈ C \ S, then we can find a disk B which contains S and does not contain a.

aB S

Figure IV.4 If a is outside S.

By II.1.2 and Corollary 3.1,

∫

∂S

1

z − adz = 0.

If a ∈ S, then let C be the circumscribed circle to ∂S parameterized in the clockwise

direction.


s1

s2

s3

s4

c1

c2

c3

c4

B1a

Figure IV.5 The square S and its circumscribed circle C.

Then we can write

∂S = s1 + s2 + s3 + s4,

where sj , j = 1, . . . , 4, are the sides of ∂S and

C = c1 + c2 + c3 + c4,

where cj , j = 1, . . . , 4 are the arcs of C subtended by the corresponding sides of ∂S. Then

sj + cj is a closed curve contained in a disk Bj with a /∈ Bj , for j = 1, . . . , 4. By Corollary

3.1,∫

sj+cj

1

z − adz = 0, (3.1)

for j = 1, . . . , 4. Adding the four integrals (3.1) we obtain

∫

∂S

1

z − adz +

∫

C

1

z − adz =

∫

∂S+C

1

z − adz = 0.

But we also have∫

C

1

z − adz = −2πi,

because −C is the circle parameterized in the counter-clockwise direction as in Proposition

2.2. This proves Proposition 3.2.

Proposition 3.2 can also be proved by explicit computation, but we chose this proof

because the idea will be used later to compute the integral of 1/(z − a) for other curves.


Theorem 3.3. If f is analytic in a neighborhood of the closure S of an open square S,

then for z ∈ S,

f(z) =1

2πi

∫

∂S

f(ζ)

ζ − zdζ,

where ∂S is parameterized in the counter-clockwise direction.

Proof. The proof of Theorem 3.3 is exactly like the proof of Theorem 2.3 except that

Proposition 3.2 is used instead of Proposition 2.2.

Theorem 3.4 (Runge). If f is analytic on a compact set K and if ε > 0 then there is a

rational function r so that

supz∈K

|f(z) − r(z)| < ε.

Proof. Suppose f is analytic on U open, with U ⊃ K. Let

d = dist(∂U,K) = inf|z − w| : z ∈ ∂U, w ∈ K.

Construct a grid of squares with side length d/2.

Figure IV.6 A union of closed squares covering K and contained in U .


Shade each square in the grid whose closure intersects K. Note that none of the

shaded squares intersect ∂U because they have diameter d/√

2. Let Sk be the collection

of shaded squares and let Γ denote the boundary of the union of the closed shaded squares

Γ = ∂

(

∪jSj)

,

formed from ∪∂Sj , each parameterized in the counter-clockwise direction, by cancelling

edges which are traced in opposite directions. If z is in the interior of one of the squares,

Sj0 , then f(ζ)/(ζ − z) is analytic as a function of ζ on Sj , for j 6= j0. Apply Theorem 3.3

to Sj0 and Corollary 3.1 to all of the other squares, then cancel edges traced in opposite

directions to obtain

f(z) =1

2πi

∫

Γ

f(ζ)

ζ − zdζ. (3.2)

Both sides of (3.2) are continuous functions of z, for z /∈ Γ, so that equality holds for all z

in the region bounded by Γ.

Fix z0 ∈ K. Choose a Riemann sum for the integral on the right side of (3.2) to obtain

∣

∣

∣

∣

f(z0) −m∑

j=1

f(ζj)(ζj+1 − ζj)

ζj − z0

∣

∣

∣

∣

< ε.

We can in fact choose the partition so that the inequality remains true for all refinements

of the partition. By continuity, this inequality remains true for all z in a disk containing

z0. Cover K by finitely many such disks, and take a common refinement.

Definition. If r is a rational function, by the Fundamental Theorem of Algebra we write

r(z) = p(z)/q(z) where p and q are polynomials with no common zeros. The zeros of q are

called the poles of the rational function r.

If b is a pole of r(z) then |r(z)| → ∞ as z → b. A rational function is analytic

everywhere in the plane, except at its poles.

It is possible to improve the statement of Runge’s theorem by restricting where the

poles of the rational function need to be placed, using the next Lemma.


Lemma 3.5. Suppose U is open and connected, and suppose b ∈ U . Then a rational

function with poles only in U can be uniformly approximated on C \ U by a rational

function with poles only at b.

Proof. Suppose a ∈ U and suppose |a−b| < 13dist(a, ∂U). Note that C\U ⊂ z : |z−b| >

2|a− b| ≡ V . Then for z ∈ V

1

z − a=

1

z − b− (a− b)=

1

(z − b)(1 − (a−bz−b

))=

∞∑

n=0

(a− b)n

(z − b)n+1.

The sequence of partial sums approximate 1/(z − a) uniformly on V and hence on C \ U .

By taking products we can also approximate (z − a)−n for n ≥ 1 on C \ U , and by taking

finite linear combinations, we can uniformly approximate on C \ U any rational function

with poles only at a by rational functions with poles only at b.

Write a ∈ Rb if every rational function with poles only at a can be uniformly approx-

imated on C \ U by rational functions with poles only at b. This relation is transitive: if

a ∈ Rb and b ∈ Rd then a ∈ Rd. Set E = b ∈ U : a ∈ Rb. By transitivity and the

argument above, if b ∈ E then E contains a disk centered at b with radius dist(b, ∂U)/3.

Thus E is open. Moreover if bn ∈ E converges to b0 ∈ U , then b0 lies in a disk of radius

less than dist(bn, ∂U)/3 centered at bn, for n large and hence b0 ∈ E. This proves E is

closed in U and by connectedness E = U .

Now suppose that r is rational with poles only in U and fix b ∈ U . Each term 1/(z−c)k

in the partial fraction expansion of r can be approximated by a rational function with poles

only at b. Adding the approximations, gives an approximation of r by a rational function

with poles only at b.

Corollary 3.6. Suppose U is connected and open and suppose z : |z| > R ⊂ U for some

R <∞. Then a rational function with poles only in U can be uniformly approximated on

C \ U by a polynomial.

Proof. By Lemma 3.5 we need only prove that if |b| > R, then a rational function with


poles at b can be uniformly approximated by a polynomial on C \ U . But

1

z − b=

1

−b(1 − zb )

= −1

b

∞∑

n=0

(

z

b

)n

,

where the sum converges uniformly on |z| ≤ R. As in the proof of Lemma 3.5, we

can approximate (z − b)−n for n ≥ 1 and by taking finite linear combinations, we can

approximate any rational function with poles only at b by a polynomial, uniformly on

z : |z| ≤ R ⊃ C \ U .

Theorem 3.4, Lemma 3.5 and Corollary 3.6 combine to give the following improvement

of Runge’s Theorem.

Theorem 3.7 (Runge). Suppose K is a compact set. Choose one point an in each

bounded component Un of C \ K. If f is analytic on K and ε > 0, then we can find a

rational function r with poles only in the set an such that

supz∈K

|f(z) − r(z)| < ε.

If C \K has no bounded components, then we may take r to be a polynomial.

For example ifK1 andK2 are disjoint compact sets such that C\(K1∪K2) is connected

and ε > 0 then we can find a polynomial p so that |p| < ε on K1 and |p − 1| < ε on K2

because the function which is equal to 0 on K1 and equal to 1 on K2 is analytic on K1∪K2.

Corollary 3.8. If f is analytic on an open set Ω then there is a sequence of rational

functions rn with poles in C \ Ω so that rn converges to f uniformly on compact subsets

of Ω.

Proof. Set

Kn = z ∈ Ω : dist(z, ∂Ω) ≥ 1

nand |z| ≤ n.

Then Kn is compact, ∪Kn = Ω and each component of C \Kn contains a point of C \ Ω.

By Theorem 3.7, we can choose the rational functions approximating f to have poles only

in C \ Ω.


We complete this chapter by showing that a uniform limit of rational functions is

always analytic.

Theorem 3.9 (Weierstrass). Suppose fn is a collection of analytic functions on a

region Ω such that fn → f uniformly on compact subsets of Ω. Then f is analytic on Ω.

Moreover f ′n → f ′ uniformly on compact subsets of Ω.

Lemma 3.10. If G is continuous on γ and then

g(z) ≡∫

γ

G(ζ)

ζ − zdζ

is analytic in C \ γ and

g′(z) =

∫

γ

G(ζ)

(ζ − z)2dζ.

Proof. There are at least two ways to prove this Lemma. One way is to write out a

power series expansion for 1/(ζ − z) based at z0, where z0 /∈ γ, then interchange the order

of summation and integration to obtain a power series expansion for g based at z0. The

derivative of g can be found by differentiating the series term by term. The second proof

is to write

g(z + h) − g(z)

h−∫

γ

G(ζ)

(ζ − z)2dζ =

∫

γ

G(ζ)h

(ζ − z)2(ζ − (z + h))dζ,

which → 0 as h→ 0. The proof shows that

g′(z) =

∫

γ

G(ζ)

(ζ − z)2dζ.

Proof of Theorem 3.9. Analyticity is a local property, so to prove the first statement

we may suppose D is a disk with D ⊂ Ω. Then by Theorem 2.3, if z ∈ D,

fn(z) =1

2πi

∫

∂D

fn(ζ)

ζ − zdζ.

Set

F (z) =1

2πi

∫

∂D

f(ζ)

ζ − zdζ.


Then

|fn(z) − F (z)| → 0

for each z ∈ D, because fn → f uniformly on ∂D. Thus F = f on D and by Lemma 3.10,

F is analytic on D. By Theorem 2.3 and Lemma 3.10

f ′n(z) =

1

2πi

∫

∂D

fn(ζ)

(ζ − z)2dζ.

and

f ′(z) = F ′(z) =1

2πi

∫

∂D

f(ζ)

(ζ − z)2dζ.

Again, since fn → f uniformly on ∂D, we have that f ′n converges uniformly to f ′ on

compact subsets of D. Thus f ′n converges uniformly to f ′ on closed disks contained in Ω.

Given a compact subset K of Ω, we can cover K by finitely many closed disks and hence

f ′n converges uniformly on K to f ′.

58

V

Cauchy’s Theorem and Applications

§1. Cauchy’s Theorem.

The reader can verify by induction that

d

dzzk = kzk−1,

and by the Chain Ruled

dz

1

(z − a)k=

−k(z − a)k+1

,

when k ≥ 1.

(COMMENT: I should have pointed this out earlier!)

By Corollary IV.2.1, if γ is a closed curve then

∫

γ

q(z)dz = 0

for all polynomials q and∫

γ

1

(z − a)kdz = 0,

when k ≥ 2 and a /∈ γ, because the integrand in each case is the derivative of an analytic

function. Suppose

r(z) =

N∑

k=1

nk∑

j=1

ck,j(z − pk)j

+ q(z)

is a rational function and q is a polynomial. If γ is a closed curve which does not intersect

any of the poles of r, then

∫

γ

r(ζ)dζ =N∑

k=1

ck,1

∫

γ

1

(ζ − pk)dζ. (1.1)

Definition. A cycle is a finite union of disjoint closed curves.

59

60 V. Cauchy’s Theorem

Theorem 1.1 (Cauchy). Suppose γ is a cycle contained in a region Ω and suppose∫

γ

dζ

ζ − a= 0 (1.2)

for all a /∈ Ω. If f is analytic on Ω then∫

γ

f(ζ)dζ = 0.

Proof. By Runge’s Theorem, we can find a sequence of rational function rn with poles in

C \ Ω so that rn converges to f uniformly on the compact set γ ⊂ Ω. By (1.1) and (1.2),∫

γrn(z)dz = 0. But then

|∫

γ

f(z)dz| = |∫

γ

(f(z) − rn(z))dz| ≤ supγ

|f − rn|`(γ) → 0.

We can use Cauchy’s Theorem to extend Theorems 2.3 and 3.3.

Theorem 1.2 (Cauchy’s Integral Formula). Suppose γ is a cycle contained in a region

Ω and suppose∫

γ

dζ

ζ − a= 0

for all a /∈ Ω. If f is analytic on Ω then

1

2πi

∫

γ

f(ζ)

ζ − zdζ = f(z) · 1

2πi

∫

γ

1

ζ − zdζ.

Proof. For each z ∈ Ω, the function g(ζ) = (f(ζ) − f(z))/(ζ − z) extends to be analytic

on Ω and by Cauchy’s Theorem has integral over γ equal to 0. Theorem 1.2 follows by

splitting the integral of g along γ into two pieces.

γ

z

Figure V.1 Integration on a cycle.

§2: Winding Number. 61

The cycle γ in Figure V.1 consists of two circles, parameterized in opposite directions

as indicated. If f is analytic on the closed region bounded by the two circles, then by

Proposition 2.2, Cauchy’s Integral Formula applies, and

f(z) =1

2πi

∫

γ

f(ζ)

ζ − zdζ,

when z is between the two circles, again using Proposition 2.2. When z is outside the

larger circle or inside the inner circle, the integral is equal to 0 by Cauchy’s Theorem,

because f(ζ)/(ζ − z) is an analytic function of ζ on the region between the two circles, if

z is not in this region.

§2. Winding Number.

The important integrals (1.2) have a geometric interpretation which we will next

explore.

Lemma 2.1. If γ is a cycle and a /∈ γ, then

1

2πi

∫

γ

1

ζ − adζ

is an integer.

Proof. Without loss of generality we may suppose γ is a closed curve parameterized by

γ(t) : [0, 1] → C. Define

h(x) =

∫ x

0

γ′(t)

γ(t) − adt.

Then h′(x) exists and equals γ′(x)/(γ(x)− a), except at finitely many points x. Then

d

dxe−h(x)(γ(x) − a) = −h′(x)e−h(x)(γ(x)− a) + e−h(x)γ′(x)

= −γ′(x)e−h(x) + γ′(x)e−h(x) = 0,

except at finitely many points. Since e−h(x)(γ(x) − a) is continuous, it must be constant.

Thus

e−h(1)(γ(1)− a) = e−h(0)(γ(0) − a) = 1 · (γ(1) − a).


Since γ(1) − a 6= 0, e−h(1) = 1 and h(1) = 2πki, where k is an integer. Thus

1

2πi

∫

γ

dζ

ζ − adζ =

h(1)

2πi= k,

an integer.

Definition. The index or winding number of γ with respect to a (or about a) is

n(γ, a) =1

2πi

∫

γ

dζ

ζ − a,

for a /∈ γ.

Note the following properties of n(γ, a):

(a) n(γ, a) is an analytic function of a, for a /∈ γ, by Lemma 3.10. In particular it

is continuous and integer-valued by Lemma 2.1. Thus n(γ, a) is constant in each

component of C \ γ.

(b) n(γ, a) → 0 as a→ ∞. Thus n(γ, a) = 0 in the unbounded component of C \ γ.

(c) n(−γ, a) = −n(γ, a).

(d) n(γ1 + γ2, a) = n(γ1, a) + n(γ2, a).

(e) If γ(t) = eikt, for 0 ≤ t ≤ 2π, then

n(γ, 0) =1

2πi

∫

γ

dz

z=

1

2πi

∫ 2π

0

ikeikt

eiktdt = k.

The curve γ in (e) “winds” k times around 0.

If a ∈ C, let

La = z : Imz = Ima

be the horizontal line through a. Suppose that γ1 is an arc in the half-plane z : Imz >

Ima with endpoints b and c on La such that Reb < Rea < Rec. See Figure V.2a.

ab c

γ1

γ2

γ3 γ4La

Figure V.2a The Winding of an arc in a half-plane.


For δ > 0 small, let γ2 be the semi-circular arc in the half-plane centered at a and

radius δ. Let γ3 be the segment on La between b and γ2 and let γ4 be the segment on

La between γ2 and c. Then we can choose directions for γ2, γ3, and γ4 so that γ =

γ1 + γ2 + γ3 + γ4 is a closed curve. Note that a is in the unbounded component of C \ γ(indeed, we can draw a vertical line from a to ∞ which does not meet γ) and hence

∫

γ

dζ

ζ − a= 0 (2.1)

by (b).

a b c

γ1

La

Figure V.2b The Winding of an arc in a half-plane.

If γ1 is an arc in the half-plane z : Imz > Ima with endpoints b and c on La such that

Rea < Reb ≤ Rec, then we can add the line segment between b and c to form a closed curve

γ. See Figure V.2b. Then a is in the unbounded component of C \ γ so that (2.1) holds.

A similar construction can be made if the endpoints b and c of γ satisfy Reb ≤ Rec < Rea.

Similar constructions can be performed starting with an arc γ1 contained in the half-plane

z : Imz < Ima.Now suppose that γ is a closed curve and a /∈ γ. Then γ \ La is a finite union of arcs

Aj . Each Aj is contained in the upper half-plane or the lower half-plane. Choose δ > 0

so small that the disk of radius δ centered at a does not intersect γ. For each arc Aj form

the corresponding curve Bj contained in the lower or upper half-plane as described above.

Then∑

Bj = γ + σ + `,

where σ is a closed curve contained in Ca = z : |z − a| = δ, consisting of semi-circular

arcs and ` is the sum of line segments. The integral of 1/(ζ − a) over the line segments is

zero. One way to see this is to keep track of the segments added each time the curve meets


the half-line z : Imz = Ima,Rez > Rea. The added segment ends at the beginning of

the previous segment, and the integral over these segments is zero because the integration

over all these line segments ends where it begins. The same thing happens to the segments

on z : Imz = Ima,Rez < Rea If α is a semicircular arc contained in Ca then∫

α

dζ

ζ − adζ = ±1

2,

where the sign is + if α is traced in the counter-clockwise direction and otherwise the sign

is −. The winding number of σ about a can then be found by counting +12

or −12

for each

semi-circular arc depending on the direction it is parameterized and by (c) and (d)

n(γ, a) = −n(σ, a).

More informally, if we trace γ starting at a point b ∈ La with Reb > Rea, then we can

compute n(γ, a) by counting the “winding” of γ about a adding ±12

if the endpoints are

on opposite sides of a in La, depending on the direction of travel, and no change to the

count if the endpoints are on the same side of a on La.

i

−121

Figure V.3 Calculating the winding number.

In Figure V.3 n(γ,−1) = 0, n(γ, 1) = 2, n(γ, 2) = 0 and n(γ, i) = 1.

σ

0 1

Figure V.4 n(γ, 0) = n(γ, 1) = 0.


Figure V.4 shows a cycle γ such that n(γ, 0) = 0 and a closed curve σ contained in

Ω0 = C \ 0, 1 with the property that n(σ, a) = 0 for all a /∈ Ω0. If you are familiar with

homotopy (we will treat this subject later in the course), this curve shows that homotopy

and homology are different since this curve cannot be shrunk to a point while remaining

in Ω0. By Cauchy’s theorem if f is analytic on Ω0, then∫

σf(z)dz = 0.

If Ω is a region in C bounded by piecewise differentiable curves, then we can pa-

rameterize ∂Ω so that as you trace each boundary component, the region Ω lies on the

left. In otherwords iγ′(t) is the “inner normal”, rotated counter-clockwise by π/2 from the

tangential direction γ′(t). It is an exercise to show that n(∂Ω, a) = 1 for all a ∈ Ω and

n(∂Ω, a) = 0 for all a /∈ Ω. We call this the positive orientation of ∂Ω.

Definition. A region Ω ⊂ C is called simply connected if S2 \ π(Ω) is connected in S2,

where π : C → S2 is the stereographic projection.

Simply connected essentially means “no holes”. For example, the unit disk D is simply

connected. The vertical strip z : 0 < Rez < 1 is simply connected. The punctured plane

C \ 0 is not simply connected. More generally, a connected open subset of S2 is called

simply connected, if its complement in S2 is connected. Thus the set that corresponds to

C \ D together with ∞ (the “North Pole” on the sphere) is simply connected, but C \ D

is not simply connected. The open set inside a “figure 8” curve is not simply connected

because it is not connected.

Theorem 2.2. A region Ω ⊂ C is simply connected if and only if n(γ, a) = 0 for all cycles

γ ⊂ Ω and all a /∈ Ω.

The point of Theorem 2.2 is that simply connected is a geometric condition which is

sufficient for Cauchy’s Theorem to apply.

Proof. Suppose Ω is simply connected and suppose γ is a cycle contained in Ω and suppose

a /∈ Ω. Then a is in the unbounded component of C \ γ, and so n(γ, a) = 0 by (b).

Conversely, suppose that S2 \π(Ω) = A∪B where A and B are non-empty closed sets

in S2 with A∩B = ∅. Without loss of generality ∞ ∈ B. Since A is closed, a neighborhood


of ∞ does not intersect A and hence π−1(A) is bounded. Pick a0 ∈ A. We’ll construct a

curve γ0 ⊂ Ω such that n(γ0, a0) 6= 0, proving Theorem 2.2.

The construction is the same construction used to prove Runge’s theorem. Let

d = dist(π−1(A), π−1(B)) = inf|a− b| : a ∈ π−1(A), b ∈ π−1(B) > 0.

Pave the plane with squares of side d/2 such that a0 is the center of one of the squares.

Orient the boundary of each square in the positive, or counter-clockwise direction (like all

storms in the Northern hemisphere). Shade each square Sj with Sj ∩A 6= ∅. Let γ0 denote

the curve obtained from ∪∂Sj after performing all possible cancellations. Then γ0 ⊂ Ω

since γ0 does not intersect either A or B, and n(γ0, a0) = 1.

Note that in the proof of Theorem 2.2, the curve γ0 satisfies n(γ0, a) = 0 or 1 for each

a /∈ γ, and equals 1 for all a ∈ A. This is not hard to see for a in the interior of one of

the squares Sj , and as in the proof of Runge’s Theorem, it holds at all other points of

∪∂Sj \ γ0 by continuity. Since none of the Sj intersect B, n(γ0, b) = 0 for all b ∈ B.

Definition. Closed curves γ1 and γ2 are homologous in a region Ω if n(γ1 − γ2, a) = 0

for all a /∈ Ω and we write

γ1 ∼ γ2.

Homology is an equivalence relation on the curves in Ω. A closed curve γ is homologous

to 0 in Ω if n(γ, a) = 0 for all a /∈ Ω. In this case we write γ ∼ 0.

Cauchy’s Theorem says that if γ ∼ 0 in Ω and if f is analytic in Ω then∫

γ

f(z)dz = 0,

and if γ1 is homologous to γ2 in Ω, then∫

γ1

f(z)dz =

∫

γ2

f(z)dz.

The most common application of Cauchy’s Integral Formula is when γ ⊂ Ω with γ ∼ 0

and n(γ, z) = 1. Then for f analytic on Ω,

f(z) =1

2πi

∫

γ

f(ζ)

ζ − zdζ.

§3: Riemann’s Theorem and Laurent Series. 67

§3. Riemann’s Theorem and Laurent Series.

The first result in this section is a generalization of a problem from the first homework.

Corollary 3.1 (Riemann’s Theorem on Removable Singularities). Suppose f is

analytic in Ω = z : 0 < |z − a| < δ and suppose

limz→a

(z − a)f(z) = 0.

Then f extends to be analytic in z : |z − a| < δ.

In other words, we can “remove” the “singularity” of f at a.

Proof. Fix z ∈ Ω and choose ε and r so that 0 < ε < |z − a| < r < δ. Let Cε and

Cr denote the circles of radius ε and r centered at a, oriented in the counter-clockwise

direction as in Figure V.1. The cycle Cr−Cε is homologous to 0 in Ω so that by Cauchy’s

Integral Formula

f(z) =1

2πi

∫

Cr

f(ζ)

ζ − zdζ − 1

2πi

∫

Cε

f(ζ)

ζ − zdζ.

But∣

∣

∣

∣

∫

Cε

f(ζ)

ζ − zdζ

∣

∣

∣

∣

≤ maxζ∈Cε

|f(ζ)| 1

|z − a| − ε2πε.

But if ζ ∈ Cε then |f(ζ)|ε = |f(ζ)||ζ − a| → 0 as ε→ 0 and hence

f(z) =1

2πi

∫

Cr

f(ζ)

ζ − zdζ. (3.1)

By Lemma IV.3.10, the right side of (3.1) is analytic in z : |z−a| < r. Thus if we define

f(a) as the value of the right side of (3.1) when z = a, then this extension is analytic at a

and hence we have extended f to be analytic in z : |z − a| < δ.

The most important special case of Riemann’s Removable Singularity Theorem is:

if f is bounded and analytic in neighborhood of a then f extends to be analytic in a

neighborhood of a.


We say that a compact set E has one-dimensional Hausdorff measure 0 if for

every ε > 0 there are finitely many disks Dj with radius rj so that

E ⊂ ∪jDj

and∑

j

rj < ε.

Corollary 3.2 (Painleve). Suppose E is a compact set with one-dimensional Hausdorff

measure 0. If f is bounded and analytic on U \ E, where U is open and E ⊂ U , then f

extends to be analytic on U .

Proof. As in the proof of Runge’s Theorem and Theorem 2.2, we can find a curve γ ⊂ U\Eso that n(γ, b) = 1 for all b ∈ E, and n(γ, b) = 0 for b ∈ C \ U . Cover E by finitely many

disks Dj of radius rj so that∑

rj < ε. For small ε, the disks Dj will not intersect γ. Let

V = z : n(γ, z) = 1 and let σ = ∂(

∪Dj)

, which we orient in the positive sense with

respect to V \ ∪Dj . Then γ + σ ∼ 0 in U \ ∪Dj , so that by Cauchy’s Theorem

f(z) =1

2πi

∫

γ

f(ζ)

ζ − zdζ +

1

2πi

∫

σ

f(ζ)

ζ − zdζ,

for z ∈ V \ ∪Dj . Since `(σ) ≤ `(∪Dj) < 2πε and since f is bounded, the second integral

tends to 0 as ε→ 0, exactly as in the proof of Riemann’s Theorem. Thus

1

2πi

∫

γ

f(ζ)

ζ − zdζ

provides an analytic extension of f to E, by Lemma IV.3.10.

Painleve asked for geometric conditions on a compact set E to be removable for

bounded analytic functions in 1888. Removable means that the second sentence of Corol-

lary 3.2 holds. A major accomplishment in complex analysis within the last ten years was

to solve this problem, though there is some debate about whether the solution is really

“geometric”.


An annulus is the region between two concentric circles. If f is analytic on the

annulus A = z : r < |z − a| < R then by Runge’s Theorem, we can approximate f by

a rational function with poles only at a. The Laurent Series is another version of this

result, similar to a power series expansion.

Theorem 3.3 (Laurent Series). Suppose f is analytic on A = z : r < |z − a| < R.Then

f(z) =

∞∑

n=−∞

an(z − a)n,

where the series converges uniformly and absolutely on compact subsets of A. Moreover

an =1

2πi

∫

Cs

f(ζ)

(ζ − a)n+1dζ, (3.2)

where Cs is the circle centered at a with radius s, r < s < R, oriented counter-clockwise.

Proof. Without loss of generality, a = 0. By shrinking A slightly we may suppose f is

analytic on A. By the Cauchy Integral Formula,

f(z) =1

2πi

∫

∂A

f(ζ)

ζ − zdζ (3.3)

for z ∈ A, where ∂A has positive orientation with respect to A. Set

fs(z) =1

2πi

∫

Cs

f(ζ)

ζ − zdζ

where Cs = seit : 0 ≤ t ≤ 2π. Then fs is analytic off Cs and by Cauchy’s Integral

Formula fs(z) does not depend on s so long as |z| < s, since Cs1 ∼ Cs2 with respect to A if

r < s1 < s2 < R. Expanding 1ζ−z in a power series expansion about 0, and interchanging

the order of summation and integration, as we have done before, we conclude that fR has

a power series expansion

fR(z) =∞∑

n=0

anzn,

where an satisfies (3.2). Likewise fs does not depend on s so long as |z| > s. Expanding

1ζ−z in a power series expansion about ∞, i.e. in powers of 1/z, and interchanging the

order of summation and integration we get that fr has a power series expansion

fr(z) = −∞∑

n=1

a−nz−n,


where a−n satisfies (3.2). By (3.3) f = fR − fr.

For example the function

r(z) =z

z2 + 4+

2

(z − 3)2− 1

z − 4

is analytic in C except at ±2i, 3, 4 and hence is analytic on four regions centered at 0:

A1 = z : |z| < 2, A2 = z : 2 < |z| < 3, A3 = z : 3 < |z| < 4, and A4 = z : 4 < |z|.Of course A1 is a disk, so that r has a power series expansion in A1. We can use

1

z − a=

1

−a(1 − za )

=

∞∑

n=0

−1

an+1zn,

provided |z| < |a|, and

1

z − a=

1

z(1 − az )

=∞∑

n=1

an−1

zn,

provided |z| > |a|. These formulae give the Laurent expansion of 1/(z − 4) in each region.

Differentiating a series of this form with a = 3 will give the series for 1/(z − 3)2. Setting

a = −4 and replacing z with z2 will give the series for 1/(z2 +4), and multiplying by z will

give the series for z/(z2 + 4). If a rational function is given as a ratio of two polynomials,

p/q, then use the techniques of Section III.2 to find its partial fraction expansion, and then

use the series expansion for 1/(z − a) as above. Sometimes it might be easier to find the

partial fraction expansion for 1/q, then multiply the resulting series by the polynomial p

and collect together similar powers of z.

Laurent Series are useful for analyzing the behavior of an analytic function near an

isolated singularity. We say that f has an isolated singularity at b if f is analytic in

0 < |z − b| < ε for some ε > 0. Write

f(z) =

∞∑

n=−∞

an(z − b)n.

Note the following:

(i) If an = 0 for n < 0 then f is analytic at b. In this case we say that f has a removable

singularity at b.


(ii) If an = 0 for n < n0 with n0 > 0 and an06= 0, then we can write

f(z) = (z − b)n0

∞∑

n=0

an0+n(z − b)n = an0(z − b)n0 + an0+1(z − b)n0+1 + . . . .

In this case b is called a zero of order n0.

(iii) If an = 0 for n < −n0 with n0 > 0 and a−n06= 0 then we can write

f(z) = (z − b)−n0

∞∑

n=0

a−n0+n(z − b)n =an0

(z − b)n0+

an0+1

(z − b)n0−1+ . . . .

In this case b is called a pole of order n0, and |f(z)| → ∞ as z → b.

In each of the above cases there is a unique integer k so that

limz→b

(z − b)kf(z)

exists and is non-zero, and

(z − b)kf(z)

is analytic and non-zero in a neighborhood of b.

(iv) If an 6= 0 for infinitely many negative n, then b is called an essential singularity.

For example

f(z) = e−1

z2 =

∞∑

n=0

(−1)n

n!z−2n

has an essential singularity at 0.

Definition. A zero or pole is called simple if the order is 1.

Definition. If f is analytic in a region Ω except for isolated poles in Ω then we say

that f is meromorphic in Ω. A meromorphic function in C is sometimes just called

meromorphic.

If f is meromorphic in Ω, then 1/f is meromorphic in Ω, and a zero of order k

becomes a pole of order k, and a pole of order k becomes a zero of order k. If f and g are

meromorphic in Ω then the order of a zero or pole for the meromorphic function f/g at b


can be found by factoring the appropriate power of z − b out of f and g then taking the

difference of these powers.

The next result gives an idea of the behavior near an essential singularity.

Theorem 3.4. If f is analytic in U = z : 0 < |z − b| < δ and if 0 is an essential

singularity for f then f(U) is dense in C.

In other words, every (punctured) neighborhood of an essential singularity has a dense

image.

Proof. If not there exists A ∈ C and ε > 0 so that |f(z) − A| > ε for all z ∈ U . Then

1

f(z) −A

is analytic and bounded by 1/ε on U . By Riemann’s Theorem, 1/(f(z) − A) extends to

be analytic in U ∪ b and so f(z) − A is meromorphic in U ∪ b. Thus the Laurent

expansion for f has at most finitely terms with a negative power of z − b, contradicting

the assumption that b is an essential singularity.

§4. The Argument Principle.

The next result is useful for locating zeros and poles of meromorphic functions.

Theorem 4.1(Argument Principle). Suppose f is meromorphic in a region Ω with

zeros zj and poles pk. Suppose γ is a cycle with γ ∼ 0 in Ω and suppose zj ∩ γ = ∅and pk ∩ γ = ∅. Then

n(f(γ), 0) =1

2πi

∫

γ

f ′(z)

f(z)dz =

∑

j

n(γ, zj) −∑

k

n(γ, pk). (4.1)

In the statement of the Argument Principle, if f has a zero of order k at z, then z occurs

k times in the list zj, and a similar statement holds for the poles. For example, if γ

is a simple closed curve in Ω which is homologous to 0 in Ω, then the number of zeros

“enclosed” by γ minus the number of poles “enclosed” by γ is equal to the winding number

of the image curve f(γ) about zero.

§4: The Argument Principle. 73

Proof. The first equality in (4.1) follows from the change of variables w = f(z). Note that

γ ∼ 0 and γ ⊂ Ω implies that n(γ, a) = 0 if a is sufficiently close to ∂Ω. Thus n(γ, zj) 6= 0

for only finitely many zj . Similarly n(γ, pk) 6= 0 for only finitely many pj because there

are no cluster points of zj or pk in Ω. This implies that the sums in (4.1) are finite.

If b is a zero or pole of f then we can write

f(z) = (z − b)kg(z)

where g is analytic in a neighborhood of b and g(b) 6= 0. Then

f ′(z) = k(z − b)k−1g(z) + (z − b)kg′(z)

and

f ′(z)

f(z)=

k

z − b+g′(z)

g(z).

Since g(b) 6= 0, g′/g is analytic in a neighborhood of b and hence f ′/f − k(z − b)−1 is

analytic near b. Thus

f ′(z)

f(z)−∑ 1

z − zj−∑ 1

z − pk(4.2)

is analytic at all points z where n(γ, z) 6= 0. In the sums, we repeat zj and pk according

to their multiplicity. By Cauchy’s Theorem integrating (4.2) over γ gives (4.1).

At this point in the course, we will examine pictures of some meromorphic functions.

See

http://www.math.washington.edu/∼marshall/math 534/534functions.html

Be sure to read the text at the bottom of each picture. Try to answer the questions.

Better pictures will eventually be available here.

Corollary 4.2 (Rouche). Suppose γ is a closed curve in a region Ω with γ ∼ 0 in Ω and

n(γ, z) = 0 or = 1 for all z ∈ Ω. If f and g are analytic in Ω and satisfy

|f(z) + g(z)| < |f(z)| + |g(z)| (4.2)


for all z ∈ γ, then f and g have the same number of zeros enclosed by γ.

Equation (4.2) says that strict inequality holds in the triangle inequality. The points

“enclosed” by γ are those z ∈ Ω for which n(γ, z) = 1. The number of zeros of f and g

are counted according to their multiplicity.

Proof. The function fg is meromorphic in Ω and satisfies

∣

∣

∣

∣

f

g+ 1

∣

∣

∣

∣

<

∣

∣

∣

∣

f

g

∣

∣

∣

∣

+1. (4.3)

By (4.2), f 6= 0 and g 6= 0 on γ, so that the hypotheses of the Argument Principle are

satisfied.

w

01−1

|w|

γ

|w + 1|

Figure V.5 Proof of Rouche’s Theorem.

The left side of (4.3) is the distance from w = f(z)/g(z) to −1. But |w − (−1)| = |w| + 1

if and only if w ∈ [0,∞). See Figure V.5. Thus the assumption (4.2) implies that fg (γ)

omits the half-line [0,∞) and hence does not wind around 0. By the argument principle,

the number of zeros of fg

equals the number of poles and hence the number of zeros of f

equals the number of zeros of g, counting multiplicity.

Example. f(z) = z9 − 2z6 + z2 − 8z − 2.

How many zeros does f have in |z| < 1? The biggest term is −8z, so comparing f and

−8z:

|f(z) + 8z| = |z9 − 2z6 + z2 − 2| ≤ 1 + 2 + 1 + 2 = 6 < |8z|

§5: Local behavior. 75

on |z| = 1. By Rouche’s Theorem, f and 8z have the same number of zeros in |z| < 1,

namely one. How many zeros does f have in |z| < 2? In this case we compare f with z9:

|f(z) − z9| = | − 2z6 + z2 − 8z − 2| ≤ 128 + 4 + 16 + 2 = 150 < |z9| = 512.

Therefore f and z9 have the same number of zeros in |z| < 2, namely 9. Thus 8 of the

zeros of f lie in 1 < |z| < 2, and the remaining zero lies in |z| < 1.

To find the number of zeros of z4 − 4z + 5 in |z| < 1 we compare with the constant

function 5 on |z| = 1:

|z4 − 4z + 5 − 5| = |z4 − 4z| ≤ 5 ≤ |5| + |z4 − 4z + 5|.

If equality holds in the first inequality, then 5 = |z4 − 4z| = |z3 − 4|. But then z3 = −1,

and so z4 − 4z+ 5 = −z− 4z+ 5 = 5(−z+ 1). Since z 6= 1, |z4 − 4z+ 5| > 0 and we have

|z4 − 4z + 5 − 5| < |5| + |z4 − 4z + 5|.

By Rouche’s Theorem, z4 − 4z + 5 and 5 have the same number of zeros in |z| < 1.

A more elaborate process for locating zeros of polynomials is in the appendix.

§5. Local Behavior of Analytic Functions.

The next Corollary to the Argument Principle will give us a better picture of the local

behavior of an analytic function.

Corollary 4.3. Suppose f is analytic at z0 and suppose f−f(z0) has a zero of order n at

z0. If ε is sufficiently small, then there exists δ > 0 so that f(z)−w has exactly n distinct

roots in z : 0 < |z − z0| < ε, provided 0 < |w − f(z0)| < δ.

In other words, f(Bε(z0)) covers Bδ(w0) \ w0 exactly n times, where Br(ζ) is the

ball centered at ζ with radius r. If f has a pole of order n > 0 at z0, then 1/f has a

zero of order n at z0 and so by Corollary 1.1, f(z) − w has exactly n distinct roots in

z : 0 < |z − z0| < ε, provided |w| is sufficiently large.


Proof. Choose ε > 0 so that f is analytic on z : |z − z0| < ε, and f ′(z) 6= 0 and

f(z) − f(z0) 6= 0 in z : 0 < |z − z0| ≤ ε. If γ is the circle centered at z0 with positive

orientation, then n(f(γ), w) = n(f(γ), f(z0)) if |w − f(z0)| is sufficiently small, since the

winding number is constant in each component of C \ f(γ). By the Argument Principle,

f − w has the same number of zeros as f − f(z0), counting multiplicity. Since f ′ 6= 0 in

0 < |z − z0| < ε, all the roots of f(z) − w have order 1.

An immediate consequence of Corollary 4.3 is that analytic functions are open maps.

Corollary 4.4. If f is analytic at z0 and f ′(z0) 6= 0, then f is one-to-one in a neighborhood

of z0 and has a local inverse g which is analytic at w0. Conversely if f is analytic and

one-to-one in a neighborhood of z0, then f ′(z0) 6= 0.

Proof. If f ′(z0) 6= 0 the by Corollary 4.3, f is a one-to-one map of a sufficiently small

neighborhood of z0 onto a neighborhood of w0. Thus g = f−1 is defined in a neighborhood

of w0 and since analytic maps are open, g is continuous. Moreover, if w is near f(z0) and

|h| sufficiently small, we can find z1 and z2 near z0 so that f(z1) = w and f(z2) = w + h.

Theng(w + h) − g(w)

h=

z2 − z1f(z2) − f(z1)

→ 1

f ′(z1),

as h→ 0. Thus g has a complex derivative at w and it equals 1/f ′(g(w)). This derivative

is continuous, so that g is holomorphic and hence analytic. Conversely if f ′(z0) = 0 then

f − f(z0) has a zero of order at least 2 and by Corollary 4.3, f is not one-to-one near z0.

We can give a formula for the inverse of f . Suppose f is analytic on |z − z0| ≤ ε and

suppose f(z)−w has exactly one zero in |z−z0| < ε and no zeros on Cε = z : |z−z0| = ε.Then

limz→f−1(w)

(z − f−1(w))f ′(z)z

f(z) − w= f−1(w).

By Riemann’s Removable Singularity Theorem

h(z) =f ′(z)z

f(z) − w− f−1(w)

z − f−1(w)


is analytic in z : |z − z0| ≤ ε, and by Cauchy’s Theorem∫

Cε

h(z)dz = 0,

where Cε is given the usual postive orientation. Since n(Cε, w) = 1, we obtain

f−1(w) =1

2πi

∫

Cε

f ′(z)z

f(z) − wdz.

For example f(z) = ez =∑∞n=0

zn

n! satisfies f ′(z) = ez by term-by-term differentiation

and f ′(z) 6= 0. The inverse function is called log z. The inverse is defined locally by

Corollary 4.4, andd

dzlog z = 1/z.

Note that

eRe log z = |elog z| = |z|,

so that Re log z = log |z|. Throughout this course log denotes the natural log, not base 10.

We also deduce thatz

|z| = eiIm log z

so that Im log z = arg z, the direction of z. Since ez+2πi = ez , arg z is defined only modulo

2π. There is no function h analytic in a neighborhood of the unit circle with eh(z) = z,

for if so then h′(z) = 1/z and∫

|z|=1h′(z)dz = 2πi, which contradicts the Fundamental

Theorem of Calculus. Some books call log z a “multiple-valued function” which is a bit of

a contradiction in terms. We will only consider log z on domains where it can be defined

as a function.

We can now give a picture of the local correspondence given by an analytic function.

If f is analytic at z0, we can write

f(z) − f(z0) = an(z − z0)ng(z),

where g is analytic at z0 and g(z0) = 1. Choose a so that an = an. The function log z is

analytic in a neighborhood of 1 so that

g1(z) = e1n

log g(z)


is analytic at z0 and thus

f(z) − f(z0) =

[

a(z − z0)g1(z)

]n

.

By Corollary 4.4, a(z− z0)g1(z) is one-to-one in a neighborhood of z0 so that we have the

following picture.

a(z − z0)g1(z) zn + f(z0)

z0 0 f(z0)

Figure V.6 Local Behavior of an Analytic Function.

In Figure V.6, the case when n = 3 is illustrated. The composed function is equal to

f(z) near z0. Each of the three regions in the left-hand figure is mapped one-to-one onto

the slit disk in the right-hand figure. Asymptotically (as the radius tends to 0) the map

“looks” like z3, translated and dilated. Note that Figure V.6 was contructed right-to-left.

The left side is the preimage of the right side.

If f ′(z0) 6= 0 and z = z0 + reit then

f(z) ≈ f(z0) + reitf ′(z0).

Thus for small r, map f approximately dilates by the factor |f ′(z0)| and rotates by

arg f ′(z0).

αα

σ

γ

f

f(γ)

f(σ)

z0

f(z0)

Figure V.7 Conformality.


To put it another way, if γ and σ are two curves passing through z0 with angle α from

γ to σ, then f(γ) and f(σ) will be curves passing through f(z0) and the angle from f(γ)

to f(σ) will also be equal to α.

Definition. We say that f is conformal at z0, if it preserves angles (including direction)

between curves.

A one-to-one analytic map on a region Ω is conformal at each point of Ω. An analytic

function can be conformal at each point of Ω without being one-to-one, however modern

usage of the phrase “conformal map” usually means a one-to-one and analytic map. The

meaning should be clear from the context.

In Section VI.X.X (various characterizations of analytic functions), we will show that

conformal maps are analytic.

If f is analytic in z : |z| > R, then f(1/z) has a isolated singularity at 0, and we

say that f has an isolated singularity at ∞. We classify this singularity at ∞ as a zero,

pole or essential singularity if f(1/z) has a zero, pole or (respectively) essential singularity

at 0. In terms of the Laurent expansion of f in |z| > R,

f(z) =∞∑

n=−∞

bnzn,

f has an essential singularity at ∞ if bn 6= 0 for infinitely many positive n. The reader can

supply the corresponding statement for zeros and poles and their orders.

80

VI

Behavior in the Large

In this chapter we will study global properties of analytic functions. The emphasis

will be on the behavior of linear fractional transformations and the elementary power,

trigonometric, and exponential functions related to the familiar functions of a real variable.

These functions are all built from linear functions a + bz, ez =∑

zn/n!, and its locally

defined inverse log z using algebraic operations and composition.

§1. Linear Fractional Transformations.

Linear Fractional Transformations (or LFTs) are non-constant rational functions

of the form

T (z) =az + b

cz + d, (1.1)

where a, b, c, d are complex constants. The four basic types of LFTs are:

Translation: T (z) = z + b,

Rotation: T (z) = eiθz,

Dilation: T (z) = az for a > 0, and

Inversion: T (z) =1

z.

The translation above shifts every point by the vector b. The rotation rotates the

plane by an angle θ. The dilation expands (if a > 1) or contracts (if a < 1). The inversion

is best understood by writing z = reit. Then the argument of 1/z is −t and the length of

1/z is the reciprocal of the length of z.

81

82 VI. Behavior in the Large

An LFT can be build out of these examples using composition. If c = 0 in (1.1) then

T =a

dz +

b

d. (1.2)

In this case T is a dilation by∣

∣

∣

ad

∣

∣

∣, a rotation by arg a

d followed by a translation by bd . If

c 6= 0, then we can rewrite (1.1) as

T (z) =bc− ad

c21

(z + dc )

+a

c. (1.3)

In this case, T is a translation by dc , an inversion, a dilation by

∣

∣

∣

bc−adc2

∣

∣, a rotation by

arg bc−adc2

followed by a translation by ac.

Note that by (1.2) and (1.3), T is non-constant if and only if bc− ad 6= 0.

Proposition 1.1. The LFTs form a group under composition.

Proof. The composition of any one of the four basic types of LFTs with another LFT is

an LFT. By (1.1) and (1.2), LFTs are closed under composition. The inverse of (1.1) is

easily found to be

z =dw − b

−cw + a.

If T is given by (1.1) with c 6= 0, then T is meromorphic in C with a simple pole at

−dc . Moreover T extends to be analytic at ∞ with T (∞) = a

c . Thus T extends to be a

one-to-one map of the extended plane, or the Riemann sphere onto itself.

Proposition 1.2. If f is entire and one-to-one, then f(z) = az + b, where a and b are

constant.

Proof. The function f =∑∞

n=0 anzn cannot have an essential singularity at ∞, since if

otherwise, by Theorem 3.4, there is a sequence zn → ∞ with f(zn) ∈ f(D), since f(D) is

open. But then there is a ζn ∈ D with f(zn) = f(ζn), contradicting the assumption that

f is one-to-one. Thus f is a polynomial. By the Fundamental Theorem of Algebra, the

degree of f is 1, since f is one-to-one.

§1: Linear Fractional Transformations. 83

We remark that entire one-to-one functions must map C onto C by Proposition 1.2.

Proposition 1.3. If f is analytic on C \ z0 and one-to-one then f is an LFT.

Proof. . Without loss of generality z0 = 0. By the proof of Proposition 1.2, f cannot

have an essential singularity at ∞. But if f has a pole of order n at ∞ then 1/f has a

zero of order n at ∞ and by Corollary V.4.3 f is n-to-one near ∞. Since f is one-to-one,

we must have n ≤ 1. By the same argument, f has at worst a simple pole at 0. Thus the

Laurent expansion of f is

f(z) =a

z+ b+ cz.

If a 6= 0 and c 6= 0 then w = az + b + cz can be simplified to a quadratic equation in z,

contradicting the assumption that f is one-to-one. Thus a = 0 or c = 0, but not both,

since f is non-constant.

The LFTs are characterized geometrically by the next Proposition. A “circle” or

generalized circle is a circle or a straight line. As we saw in Section I.3, “circles”

correspond precisely to the circles on the Riemann sphere. Lines lift to circles through the

north pole. Similarly a “disk” is a region (in C∗) bounded by a “circle”.

Proposition 1.4. LFTs map “circles” onto “circles” and “disks” onto “disks”.

Proof. We need only check this for the four basic types of LFTs given in Section 1. If

|z− c| = r then |az− ac| = |a|r, and |(z− b)− (c− b)| = r so that rotations, dilations and

translations map circles to circles. The equation of a straight line is given by

Re(c(z − b)) = 0,

since we can translate the line so it passes through 0 then rotate it to correspond to the

imaginary axis. Rotations, dilations and translations map lines to lines exactly as in the

case of circles. To check that inversions preserve “circles”, suppose |z − c| = r and set


w = 1/z. Multiply out | 1w − c|2 = r2. If r2 = |c|2, then Re2cw = 1, the equation of a line.

If r2 6= |c|2, then by completing the square we obtain

∣

∣

∣

∣

w − c

r2 − |c|2∣

∣

∣

∣

2

=

(

r

r2 − c2

)2

,

which is the equation of a circle. The similar reasoning for the image of a line is left to the

reader to verify.

The equation of a disk is found by replacing the equal sign in the equation for a circle

with < or >, so that the proof of the statement for “disks” follows in a similar way.

The most common example is the conformal map from the upper half-plane H onto

the disk D, sometimes called the Cayley Transform:

C(z) =z − i

z + i.

Notice that the distance from x ∈ R to i is equal to the distance to −i, so that C maps

R∪ ∞ onto ∂D. Since C(i) = 0, the image of H must be D by Proposition 1.4. Another

way to see that the image of H is D is to note that if z ∈ H then the distance from z to

i is less than the distance from z to −i, so that |C(z)| < 1 and similarly |C| > 1 on the

lower half-plane. Since C maps C∗ onto iC∗, the image of H must be D.

The Cayley transform can be used, for example, to transform an integral on R to an

integral on ∂D and vice-versa.

Conformality can also be used to determine the image of an LFT. For example the

LFT given by T (z) = (z − 1)/(z + 1) is real-valued on R so it maps the unit circle to a

“circle” which is orthogonal to R and passes through ∞, the image of −1. Thus it maps the

unit circle to the imaginary axis, and since T (0) = −1, it maps D onto the left half-plane

Rez < 0.

The proof of the next Proposition is useful for constructing LFTs.

Proposition 1.5. Given z1, z2, z3 distinct points in C∗, and w1, w2, w3 distinct points in

C∗, there is a unique LFT, T , such that

T (zi) = wi, (1.4)

§2: log f. 85

for i = 1, 2, 3.

In the statement of Proposition 1.4, C∗ is the extended plane, so that ∞ is included

in the possibilities.

Proof. First suppose that w1 = 0, w2 = ∞ and w3 = 1. Set

T (z) =

(

z − z1z − z2

)(

z3 − z2z3 − z1

)

.

Then T (zi) = wi, i = 1, 2, 3. For the general case, choose LFTs R and S so that R(z1) =

S(w1) = 0, R(z2) = S(z2) = ∞ and R(z3) = S(w3) = 1, then

T = S−1 R

satisfies (1.4). If U is an LFT satisfying (1.4) then

S U R−1 =az + b

cz + d

maps 0 to 0 so that b = 0. It maps ∞ to ∞ so that then c = 0, and it maps 1 to 1 so that

a = d, and U(z) = S−1 R(z).

One additional property of LFTs that is sometimes useful for determining their image

is: if T is an LFT and if z1, z2, z3 are three points on the boundary of a “disk” D such

that D lies to the left of ∂D as ∂D is traced from z1 to z2 to z3, then T (D) lies to the left

of T (∂D) as it is traced from T (z1) to T (z2) to T (z3). For example, the unit disk D lies to

the left of the unit circle as it is traced from 1 to i to −1. If T (z) = (z − 1)/(z + 1) then

T (1) = 0, T (i) = i and T (−1) = ∞. Thus the image of ∂D is the imaginary axis, since it

is a circle through 0, i,∞ and the image T (D) must lie to the left of the imaginary axis as

it is traced from 0 to i to ∞. Thus T (D) is the left half-plane z : Rez < 0. The reason

for this is that conformal maps preserve angles (including direction) between curves and

LFTs are conformal everywhere, in particular on the boundary of D.

§2. log f.


We have already discussed the function

ez = exeiy = ex(cos y + i sin y).

It maps the horizontal line y = c onto the ray arg z = c from 0 to ∞, and it maps each

segment of length 2π in the vertical line x = c onto the circle |z| = ec.

We begin with a fundamental consequence of the Argument Principle.

Corollary 2.1. Suppose f is analytic on a simply connected domain Ω and f(z) 6= 0 for

all z ∈ Ω. Then we can define g(z) = log f(z) to be analytic on Ω.

The conclusion of Corollary 2.1 is that there is an analytic function g such that

f(z) = eg(z). (2.1)

Note that g(z) + 2πi is another solution. We are not claiming that log z can be defined

on the range f(Ω) of f on Ω. The function g is locally the composition of a function log z

and f , but there might not be a function h defined on all of f(Ω) such that z = eh(z). For

example, the LFT (z − 1)/(z + 1) maps the disk D onto the left half-plane Rez < 0 and

so the function f(z) = e(z−1)/(z+1) maps the D onto D \ 0 and is non-vanishing. Thus

f satisfies the hypotheses of Corollary 2.1 and indeed g = (z − 1)/(z + 1) works. But we

cannot define log z on f(D) = D \ 0.

Proof. If g satisfies (2.1) then g′ = f ′/f . Fix z0 ∈ Ω and define

g(z) =

∫

γz

f ′(w)

f(w)dw,

where γz is any curve in Ω beginning at z0 and ending at z. If γ1 and γ2 are curves in Ω

from z0 to z, then γ1 − γ2 is a closed curve in Ω which is homologous to 0 by Theorem

V.2.2. Since f 6= 0 on Ω, f ′/f is analytic on Ω and by Cauchy’s Theorem

0 =

∫

γ1−γ2

f ′(w)

f(w)dw =

∫

γ1

f ′(w)

f(w)dw −

∫

γ2

f ′(w)

f(w)dw.

§2: log f. 87

Thus g(z) does not depend on the choice of γz. If z1 ∈ Ω and |h| is small then we may

take γz1+h = γz1 + σh, where σh ⊂ Ω is the straight line segment from z1 to z1 + h. Thus

∣

∣

∣

∣

g(z1 + h) − g(z1)

h− f ′(z1)

f(z1)

∣

∣

∣

∣

=

∣

∣

∣

∣

1

h

∫

σh

(

f ′(w)

f(w)− f ′(z1)

f(z1)

)

dw

∣

∣

∣

∣

≤ supw∈σh

∣

∣

∣

∣

f ′(w)

f(w)− f ′(z1)

f(z1)

∣

∣

∣

∣

`(σh)

|h| < ε

if |h| is small, since f ′/f is continuous. Thus g′ exists and g′ = f ′/f . By Corollary IV.2.4

g is analytic.

To compare f and eg, set

F = fe−g.

Then

F ′ = f ′e−g − fg′e−g = f ′e−g − f ′e−g = 0.

This implies F is a constant, and evaluating at z0, F (z0) = f(z0). Thus

f(z) = f(z0)eg(z) = eg(z)+a,

where a is any complex number with ea = f(z0), proving Corollary 2.1.

Note that if g and h satisfy f = eg = eh on a region Ω and if g(z0) = h(z0) for some

z0 ∈ Ω, then g = h in a neighborhood of z0 because log z has a unique inverse in a (small)

neighborhood of f(z0) with log f(z0) = g(z0). By the Uniqueness Theorem, g = h in Ω.

For example, the function z is non-zero on the simply connected domain C \ (−∞, 0].

Then log z, with log 1 = 0, is the function given by

log z = log |z| + i arg z, (2.2)

where −π < arg z < π. If instead we specified that log 1 = 2πi then (2.2) holds with

π < arg z < 3π. However, if Ωγ = C \ γ where γ is the spiral given in polar coordinates by

r = eθ, −∞ < θ < ∞, then Ωγ is simply connected and Im log z is unbounded on Ωγ . In

this case we can still specify, for example, log(−1) = πi and this uniquely determines the

function log z on Ωγ .


Notice that with the definition (2.2), log(zw) 6= log(z) + log(w) if arg(z) + arg(w) is

not in the interval (−π, π).

The hypothesis that Ω is simply connected is essential in Corollary 2.1. If Ω is not

simply connected, then there exists a point a ∈ C \Ω and a curve γ ⊂ Ω so that n(γ, a) =

12πi

∫

γdzz−a 6= 0. The function f(z) = z−a is non-zero on Ω and if log(z−a) could be defined

as an analytic function, then f ′(z) = 1/(z − a) by the chain rule. By the Fundamental

Theorem of Calculus∫

γf ′(z)dz = 0, contradicting n(γ, a) 6= 0.

§3. zα.

If Ω is a simply connected region not containing 0, and if α ∈ C, we define

zα = eα log z.

where log z can be specified by giving its value at one point z0 ∈ Ω. Then zα is an analytic

function on Ω. For example suppose 0 < α < 1, suppose Ω = C \ (−∞, 0] and define

log 1 = 0. If z = reit, where −π < t < π, then zα = rαeiαt.

2βzα

2αβ00

Figure VI.1 zα.

The image of the sector z : | arg z| < β is the sector z : | arg z| < αβ. The map zα is

conformal in Ω, but it is not conformal at 0. Indeed angles are multiplied by α at 0.

If we define log 1 = 2πi instead of 0, then the image sector is rotated by the angle

2πα. How would you define log z on C \ [0,+∞) and what would be the image of this

region by the map z12 ? Test your understanding by showing that there are exactly two

definitions of z12 on this region. In fact, there are exactly two possible definitions of z

12 on

any simply connected region not containing 0.

§4:1

2(z +

1

z). 89

A slightly more complicated function is

z1+i2

defined on Ω0 = C \ (−∞, 0]. If log(1) = 0, then the function ϕ(z) = log z is analytic on

Ω0 and has image equal to the horizontal strip Ω1 = z : |Imz| < π. The image of Ω1 by

the map (1+i)2 z is the strip

Ω2 = x+ iy : x− π < y < x+ π.

The function ez maps the line y = x onto a spiral S given in polar coordinates by r = eθ.

The image of y = x+ c is the rotation of S by the angle c. Thus ez is analytic and one-to-

one on Ω2 with image Ω3 = C\−S. The composition of these maps is z1+i

2 , which is then

a one-to-one analytic map of Ω0 onto Ω3. Notice that z1+i

2 does not extend continuously to

(−∞, 0]. The function zα has been used to explain one of M.C. Escher’s lithographs. The

image of a small square by the conformal map zα is approximately a small square since the

map is approximately linear wherever it is conformal. Thus a picture in C \ (−∞, 0] can

be transferred to a distorted picture in C \ −S by redrawing the portion in each square

of a fine grid in C \ (−∞, 0] on the corresponding approximate square in the image region.

§4. 1

2

(

z +1

z

)

.

The next function we will consider in this chapter is

w(z) =1

2

(

z +1

z

)

,

which is analytic in C \ 0. By the quadratic formula

z = w ±√

w2 − 1 (3.1)

so that w is two-to-one unless w2 = 1. Since w(z) = w(1/z), the two roots in (3.1) are

reciprocals of each other, one inside D and one outside D or else complex conjugates of

each other on ∂D. If we write z = reit, then

w =1

2

(

z +1

z

)

=1

2

(

r +1

r

)

cos t+i

2

(

r − 1

r

)

sin t.


If we also write w = u+ iv then(

u12(r + 1

r )

)2

+

(

v12(r − 1

r )

)2

= 1,

which is the equation of an ellipse, unless r = 1. Thus for each r 6= 1, the circles of radius

r and 1/r are mapped onto the same ellipse. The circle of radius r = 1 is mapped onto

the interval [−1, 1]. We leave as an exercise for the reader to show that the image of a ray

from the origin to ∞ which is not on a coordinate axis is a branch of a hyperbola which is

perpendicular to each ellipse given above, by conformality.

The function w is a one-to-one analytic map of C\D onto C\ [−1, 1], and a one-to-one

analytic map of D\0 onto C\[−1, 1]. The function w is also analytic on H = z : Imz > 0and one-to-one, since Im1/z < 0 if z ∈ H. The image of H by the map w is the region

Ω2 = C \ (−∞,−1] ∪ [1,∞).The inverse of w is given by (3.1) in each case, but we must make the correct choice

for the square root on the image region. If U is any simply connected domain with ±1 /∈ U

then we can define log(w2 − 1) so as to be analytic in Ω by Corollary 2.1, and thus

w +√

w2 − 1 = w + e12 log(w2−1)

is analytic and one-to-one on U with inverse function 12(z + 1/z).

For example if U = C∗ \ [−1, 1], write

√

w2 − 1 = w

√

1 − 1

w2.

The function 1 − 1/w2 is analytic and non-zero on U . Define ϕ(w) =√

1 − 1/w2 so that

ϕ(∞) = 1. More concretely, the image of U by the map 1−1/w2 is contained in C\(−∞, 0]

so if log ζ = log |ζ| + i arg ζ with −π < arg ζ < π then√

1 − 1/w2 is the composition of

ζ = 1 − 1/w2 and exp( 12 log ζ), and therefore analytic. Then

ψ(w) = w

(

1 +

√

1 − 1

w2

)

is an inverse to (z + 1/z)/2 on U . By the preceeding discussion, there are two inverses to

(z+1/z)/2 on U . Since ψ(w) → ∞ as w → ∞, ψ must be the inverse to (z+1/z)/2 with

range C \ D. The inverse with range D \ 0 is

ψ1(w) = w

(

1 −√

1 − 1

w2

)

§5: Trigonometric Functions. 91

which tends to 0 as w → ∞. The reader is invited to test their understanding by finding

the inverses on C \ (−∞,−1] ∪ [1,+∞).

In the next section we will use the function (z+1/z)/2 and the closely related function

(z − 1/z)/2 which can be understood as a composition

1

2

(

z − 1

z

)

= −i · 1

2

(

(iz) +1

(iz)

)

.

It is the composition of rotation by π/2 followed by 12(z + 1/z) followed by rotation by

−π/2. Thus circles and lines through 0 are mapped to ellipses and orthogonal hyperbolas.

The ellipses have semi-major axis along the imaginary axis. The unit circle is mapped to

the interval [−i, i].

§5. Trigonometric Functions.

We define

cos z =eiz + e−iz

2and sin z =

eiz − e−iz

2i.

Then cos z and sin z are entire functions satisfying

cos z + i sin z = eiz,

(cos z)2 + (sin z)2 = 1,

d

dzcos z = − sin z and

d

dzsin z = cos z.

These functions agree with their usual calculus definitions when z is real. However, we

know by Liouville’s Theorem that they cannot be bounded in C. The function cos z is best

understood by viewing it as the composition of the map 12 (z + 1/z) and the function eiz.

For example, the vertical strip z : |Rez| < π is mapped onto C \ (−∞, 0] by the map

eiz. Vertical lines are mapped to rays from 0 to ∞ and horizontal lines are mapped to

circles. The images of rays and circles by the map 12(z + 1/z) are branches of hyperbolas

and ellipses, as we saw in section 3. Other trigonometric functions are defined using sin

and cos, for example

tan z =sin z

cos z.


Hyperbolic trigonometric functions are also defined using the exponential function:

cosh z =ez + e−z

2and sinh z =

ez − e−z

2.

The inverse trigonometric functions can be found by working backward. For example

to find arccos z, set z = (eiw + e−iw)/2, multiply by eiw and obtain a quadratic equation

in eiw, so that by the quadratic formula

eiw = z ±√

z2 − 1.

Thus

arccos z = w = −i log(z ±√

z2 − 1).

If Ω is a simply connected domain such that ±1 /∈ Ω then f(z) = z±√z2 − 1 is an analytic

function, as seen in Section 4. If z ±√z2 − 1 = 0 then z2 = z2 − 1 which is impossible.

Thus f is a non-vanishing function on Ω and by Corollary 2.1, we can define log f as an

analytic function on Ω. Thus arccos is analytic on any simply connected region which does

not contain ±1. To find the arccos with arccos(0) = π2 , it is best to write it in the form

arccos z = w = −i log(z + i√

1 − z2).

with√

1 = 1 and log(i) = iπ/2.

§6. Constructing Conformal Maps.

In this section we will use the functions we’ve studied in this chapter to construct

conformal maps. In modern usage, the phrase conformal map means a one-to-one ana-

lytic map. The entire function f(z) = ez is conformal everywhere, since its derivative is

everywhere non-zero, but it is not a conformal map on C because it is not one-to-one. It

is a conformal map on z : |Imz| < π, for example. The list below is not exhaustive, but

rather is meant to illustrate the techniques which can be used at this point in the course.

1. As we saw in Homework #3, the conformal maps of D onto D are given by

ϕ(z) = c

(

z − a

1 − az

)

,

§6: Constructing Conformal Maps. 93

where a and c are constants with |a| < 1 and |c| = 1.

2. The conformal maps of the upper half-plane H onto H are given by

ϕ(z) =

az

1 − bz+ c, or

−az

+ b.(6.1)

where a > 0, and b and c are real. If ϕ satisfies (6.1) then ϕ is a linear fractional

transformations which is real on R, so by Proposition 1.4, ϕ maps H onto a “disk” and

maps the extended real line onto itself. LFTs are homeomorphisms of the extended plane

C∗ so ϕ is one-to-one and the image of H must either be H or C\H. f ϕ(z) = az/(1−bz)+c

then ϕ′(0) = a > 0, so there is no rotation at 0 and thus ϕ is a conformal map of H onto

H. If ϕ(z) = −a/z + b, then ϕ is the composition of the map −1/z, which has positive

imaginary part on H, and the linear map az + b which maps H onto H since a > 0 and b

is real. Thus ϕ is a conformal map of H onto H. Conversely, if ϕ is a conformal map of H

onto H, then

C ϕ C−1,

where C(z) = (z− i)/(z+ i) is the Cayley transform, is a conformal map of D onto D. By

Example 1 above and Proposition 1.1, ϕ is a linear fractional transformation

ϕ(z) =az + b

cz + d.

If d = 0, then ϕ(z) = a/c − (b/c)(−1/z). Then a/c = ϕ(∞) ∈ R. Since −1/z maps H

onto H, −b/c > 0 for otherwise ϕ would rotate H near 0 and thus the second case in (6.1)

holds. If d 6= 0, then

ϕ(z) =b

d+

(

ad−bcd2

)

z

1 + cdz

.

Then b/d = ϕ(0) ∈ R, and bc−add2

= ϕ′(0) > 0, and c/d must be real since the pole of ϕ

must lie on the extended real line R∗, proving (6.1).

3. A conformal map of the disk centered at c and radius r onto D is given by

ϕ(z) = (z − c)/r.


4. A conformal map of a sector Ω = z : a < arg z < b onto D can be constructed in

steps.

f(z) = zα = eα log z

where α = π/(b− a) will map Ω onto a sector with opening π, a half-plane. The choice of

log z is already given in the description of Ω. A rotation z → eitz will map the half-plane

onto H, and the Cayley transform (z− i)/(z+ i) will map H onto D. It is usually sufficient

to describe a conformal map as a composition of a sequence of simpler conformal maps.

5. If Ω is the intersection of two disks, then in order to map Ω onto D, find the two points

c, d where the bounding circles meet. The map

z − c

z − d

will map each disk onto a “disk” with 0 and ∞ on its boundary, and hence the image

of Ω is the intersection of two half-planes forming a sector at 0. Now apply Example 4.

Note that the region outside the union of two disks is the intersection of two “disks” in

the extended plane if we add the point at ∞.

0c

d

Figure VI.2

6. The region Ω = H\I, where I is the segment [0, i] on the imaginary axis, can be mapped

onto H by the map√

z2 + 1,

where√−1 = i. Indeed the image of Ω by the map z2 + 1 is the slit plane C \ [0,+∞).

The “branch” of the square root is uniquely determined by the requirement that√−1 = i.

It can be given more explicitly as exp( 12 log z) where 0 < arg z = Im log z < 2π.


000 11 −1

i z2 + 1 √z

Figure VI.3

7. The region Ω = H\A, where A is an arc lying on a circle C which is orthogonal to R at

0 can be mapped to D by first applying a conformal map az/(1− z/b) of H onto H where

b is the other point of intersection of C and R, and a > 0. The image of A must lie on a

“circle” through ∞ and be perpendicular to R since LFTs are conformal. Thus the image

of A is an interval [0, ic] on the imaginary axis. We can choose a > 0 so that c = 1 and

then Example 6 applies.

)

0 b

Figure VI.4

8. To map H onto the region between two branches of a hyperbola, first map H onto a

sector symmetric about the y-axis using eitzα with the proper choice of the rotation angle

t and opening α. Then apply 12(z + 1/z). See Section 4.

0

Heitzα

12(z + 1/z)

Figure VI.5


9. To map C \ D onto the exterior of an ellipse, apply the map z → rz, then 12(z + 1/z).

See Section 4.

12 (rz + 1/(rz))

Figure VI.6

10. To map H onto the region below the parabola y = x2 first apply the map z → −iaz+b

where a > 0 and b > 0. The image of H is the half-plane Rez > b. The image of this

region under the map z2 is the exterior of a parabola. Applying the map z → ciz + id

with c < 0 will result in a map to the region below a parabola. The reader can check that

a proper choice of a, b, c will give the desired map.

Figure VI.7

To map D or H onto any of regions bounded by a conic section other than those given

in 8, 9, and 10 will be covered later in this chapter.

11. To map the strip z : 0 < Rez < 1 onto D, first apply the map eπiz. The image of

Rez = c, 0 < c < 1, is the ray reiπc, r > 0, so the image of the strip is H. Now apply

the Cayley transform.

12. To map the half-strip z : 0 < Imz < π, Rez < 0 onto D, first apply the map ez

which has image D ∩ H. Now apply Example 5, or use the inverse of the map 12 (z + 1/z)

and the Cayley transform.


Figure VI.8

Many other examples can be constructed by using combinations of the above ideas.

The conformal map in each of the examples above is a composition of a sequence of simpler

conformal maps. The inverse map can be found by composing the inverses of the simpler

functions in the reverse order. To find the conformal map of D onto a region Ω, it is usually

easier to discover the map from Ω onto D, then compute its inverse.

A natural question at this point is how unique are these maps? A conformal map of

Ω onto D can be composed with the linear fractional transformations of the form given in

Example 1 and still map D onto D.

Proposition 6.1. If there exists a conformal map of a region Ω onto D, then given any

z0 ∈ Ω, there exists a unique conformal map f of Ω onto D such that

f(z0) = 0 and f ′(z0) > 0.

Proof. If g is a conformal map of Ω onto D, set a = g(z0). Then

h(z) = cz − a

1 − az

maps D onto D if |c| = 1, and f = h g maps Ω onto D with h(z0) = 0. Then

f ′(z0) = cg′(z0)/(1 − |a|2),

so that the proper choice of the argument of c will give h′(z0) > 0. If k maps Ω onto D

with k(z0) = 0 and k′(z0) > 0, then H = k h−1 maps D onto D with H(0) = 0. By

Schwarz’s Lemma, |H(z)| ≤ |z| and |H−1(z)| ≤ |z| so that |H(z)| = |z| and H(z) = cz,

with |c| = 1. Since H ′(0) = k′(0)/h′(0) > 0, we must have c = 1 and h = k.


For example, to find a conformal map ϕ of H onto D such that ϕ(z0) = 0 and ϕ′(z0) >

0, first apply the Cayley transform, then apply a linear fractional transformation of the

form given in Example 1. In this example, it is actually easier to first apply the map

f(z) =z − z0z − z0

which maps H onto D because |f(x)| = dist(x, z0)/dist(x, z0) = 1, when x ∈ R, so that

the image of H is either D or C∗ \ D. But f(z0) = 0, so f maps H onto D. Now find the

correct rotation so that ϕ(z) = eitf(z) has positive derivative at z0.

Another natural question is: what regions can be mapped conformally onto D? The

next Proposition gives a necessary condition. We shall see later in the course that it is

also a sufficient condition.

Proposition 6.2. If ϕ is a conformal map of a region Ω onto D, then Ω must be simply

connected.

Proof. Suppose γ is a closed curve contained in Ω and suppose a ∈ C \ Ω. Let f = ϕ−1,

which is analytic by Corollary 4.4. Then

n(γ, a) =1

2πi

∫

γ

dw

w − a=

1

2πi

∫

ϕ(γ)

f ′(z)

f(z) − adz = 0,

by Cauchy’s Theorem since f ′/(f − a) is analytic in D. By Theorem V.2.2 Ω is simply

connected.

§7. Symmetry and Conformal Maps.

In this section we will explore the connection between symmetry in a simply connected

domain and symmetry of a conformal map from the region to the disk D.

We say that a region Ω is symmetric about R provided z ∈ Ω if and only if z ∈ Ω.

If Ω is symmetric about R and if f is analytic on Ω and real-valued on Ω∩ R then f(z) is

analytic on Ω and equal to f on Ω∩R. Indeed, the power series expansion for f(z) based at

§7: Symmetry and Conformal Maps. 99

b has coefficients which are the complex conjugates of the coefficients for the power series

for f at based at b. So f(z) − f(z) does not have isolated zeros and hence

f(z) = f(z) (7.1)

for all z ∈ Ω. In fact, it suffices to assume that f is real-valued on a non-empty interval

(a, b) ⊂ R ∩ Ω to conclude that f(z) = f(z) for all z ∈ Ω, and hence f is real-valued on

all of R∩Ω. Geometrically, (7.1) says that f maps symmetric points to symmetric points.

For one-to-one functions we have the following related result.

Proposition 7.1. If f is a conformal map of a simply connected symmetric region Ω onto

D such that f(x0) = 0 and f ′(x0) > 0 for some x0 ∈ R ∩ Ω then f(z) = f(z).

Proof. Apply Proposition 6.1 to the functions f and f(z).

Proposition 7.1 says that if the region is symmetric then we can take the mapping

function to be symmetric. We’d like to use this idea to help construct conformal maps.

We first prove a converse to Cauchy’s Theorem.

Theorem 7.2 (Morera). If f is continuous on a region Ω and∫

γf(z)dz = 0 for all closed

polygonal curves γ ⊂ Ω then f is analytic on Ω.

Proof. Fix z0 ∈ Ω. Define

g(z) =

∫

γz

f(ζ)dζ, (7.2)

where γz is a polygonal curve from z0 to z in Ω. The function g(z) does not depend on

the choice of the curve γz for if σz were another such curve, γz − σz is closed and so by

assumption∫

γz

f(ζ)dζ −∫

σz

f(ζ)dζ =

∫

γz−σz

f(ζ)dζ = 0.

By the proof of Corollary 2.1, g is analytic in Ω with g′ = f .


Theorem 7.3 (Schwarz Reflection Principle). Suppose Ω is a region which is sym-

metric about R. Set Ω+ = Ω∩H and Ω− = Ω∩ (C \H). If f is analytic on Ω+, continuous

on Ω+ ∪ (∂Ω+ ∩ R) and real-valued on (∂Ω+ ∩ R) then the function defined by

h(z) =

f(z) for z ∈ Ω \ Ω−

f(z) for z ∈ Ω−

is analytic on Ω.

Proof. We need only prove h is analytic on Ω ∩ R. Analyticity is a local property, so

without loss of generality, Ω is a disk centered on R. Suppose that γ is a closed polygonal

curve in Ω. Then by adding and subtracting intervals on ∂Ω+ ∩ R we can write

∫

γ

h(z)dz =

∫

γ+

h(z)dz +

∫

γ−

h(z)dz,

where γ+ and γ− are closed curves with γ+ ⊂ Ω+ ∪ (∂Ω+ ∩R) and γ− ⊂ Ω− ∪ (∂Ω− ∩R).

By Cauchy’s Theorem and the continuity of h,

∫

γ+

h(z)dz = 0 and

∫

γ−

h(z)dz = 0.

Thus∫

γh(z)dz = 0 and by Morera’s Theorem, h is analytic.

Corollary 7.4. If f satisfies the hypotheses of the Schwarz Reflection Principle and also

is one-to-one in Ω+ with Imf > 0 on Ω+, then h is also one-to-one on Ω.

Proof. By definition h is one-to-one on Ω− with Imh < 0 on Ω−. So if h(z1) = h(z2)

then z1, z2 ∈ R. Since h is open, it maps a small disk centered at zj onto a neighborhood

of h(zj) for j = 1, 2. Thus there are two points ζ1, ζ2 ∈ Ω+ near z1, z2 (respectively) with

h(ζ1) = h(ζ2), contradicting the assumption that h is one-to-one on Ω−.

As an application we outline how to find a conformal map of H to the region Ω =

z = x+ iy : y > x2. First map H onto the half-strip S = z = x+ iy : 0 < x < b, y > 0

§8: The Geodesic Algorithm. 101

so that f1(∞) = ∞ (i.e. lim|z|→∞ |f1(z)| = ∞), where b is chosen as in Example 10 of

Section 6. Then the function f2(z) = −iz2 is one-to-one on S with image equal to half

of the desired region. Then the composed map f = f2 f1 satisfies f(∞) = ∞ and by

translating f if necessary, we can arrange that f(0) = 0. Thus the image of the negative

real axis is the positive imaginary axis. By the Schwarz Reflection Principle (applied to

if), f can be extended to be a conformal map of C \ [0,∞) onto Ω. Then f(z2) is the

desired map. This same idea can be used to construct a conformal map from H to the

region outside one branch of a hyperbola. The map to the inside of an ellipse cannot be

done with the elementary maps we’ve constructed so far.

§8. The Geodesic Algorithm.

How do you find a conformal map of the upper half plane H to a complicated region?

Rather few maps can be given explicitly by hand, so that a computer must be used to

find the map approximately. One reasonable way to describe a region numerically is to

give a large number of points on the boundary. One way to say that a computed map

defined on H is “close” to a map to the region is to require that the boundary of the

image be uniformly close to the polygonal curve through the data points. Indeed, the only

information we may have about the boundary of a region are these data points.

The geodesic algorithm is based on the simple map fa : H \ γ −→ H where γ is an

arc of a circle from 0 to a ∈ H which is orthogonal to R at 0. This map can be realized by

a composition of a linear fractional transformation, the square and the square root map

as illustrated in Figure VI.9. The orthogonal circle also meets R orthogonally at a point

b = |a|2/Rea and is illustrated by a dashed curve in Figure VI.9.


H \ γ H

γa

b

c−c

ic

c20

0

0

0

fa

z

1 − z/b

z2 + c2

√z

Figure VI.9. The basic map fa.

In Figure VI.9, c = |a|2/Ima. Observe that the arc γ is opened to two adjacent intervals

at 0 with a, the tip of γ, mapped to 0. The inverse f−1a can be easily found by composing

the inverses of these elementary maps in the reverse order.

Now suppose that z0, z1, . . . , zn are points in the plane. The basic maps fa can be

used to compute a conformal map of H onto a region Ωc bounded by a Jordan curve which

passes through the data points as illustrated in Figure VI.10.


z0

z1z2z3zn

ζ2

ζ3

ζ4

ϕ1 = i√

(z − z1)/(z − z0)

ϕ2 = fζ2

ϕ3 = fζ3 ϕn = fζn

ϕn+1 = −(

z

1 − z/ζn+1

)2

H

0

0

0

0

0

Ωc

ζn+1

Figure VI.10. The Geodesic Algorithm.

The complement in the extended plane of the line segment from z0 to z1 can be

mapped onto H with the map

ϕ1(z) = i

√

z − z1z − z0

and ϕ1(z1) = 0 and ϕ1(z0) = ∞. Set ζ2 = ϕ1(z2) and ϕ2 = fζ2 . Repeating this process,

define

ζk = ϕk−1 ϕk−2 . . . ϕ1(zk)

and

ϕk = fζk.

for k = 2, . . . , n. Finally, map a half-disc to H by letting

ζn+1 = ϕn . . . ϕ1(z0) ∈ R


be the image of z0 and set

ϕn+1 = ±(

z

1 − z/ζn+1

)2

The + sign is chosen in the definition of ϕn+1 if the data points have negative winding

number (clockwise) around an interior point of ∂Ω, and otherwise the − sign is chosen.

Set

ϕ = ϕn+1 ϕn . . . ϕ2 ϕ1

and

ϕ−1 = ϕ−11 ϕ−1

2 . . . ϕ−1n+1.

Then ϕ−1 is a conformal map of H onto a region Ωc such that zj ∈ ∂Ωc, j = 0, . . . , n.

The portion γj of ∂Ωc between zj and zj+1 is the image of the arc of a circle in the upper

half plane by the analytic map ϕ−11 . . .ϕ−1

j . In more picturesque language, after applying

ϕ1, we grab the ends of the displayed horizontal line segment and pull, splitting apart or

unzipping the curve at 0. The remaining data points move down until they hit 0 and then

each splits into two points, one on each side of 0, moving further apart as we continue to

pull.

Note that ϕ is a conformal map of the complement of Ωc, C∗ \Ωc, onto the lower half

plane, C \ H where C∗ denotes the extended plane. Simply follow the unshaded region in

H in Figure VI.10. Finally, we remark that it is easier to use circular arcs in the right-half

plane instead of in the upper-half plane when coding the algorithm, because most computer

languages adopt the convention −π2 < arg

√z ≤ π

2 .

The geodesic algorithm can be applied to any sequence of data points z0, z1, . . . , zn,

unless the points are out of order in the sense that a data point zj belongs to the (computed)

arc from zk−1 to zk, for some k < j. We will next give a simple condition on the data

points z0, z1, . . . , zn which is sufficient to guarantee that the curve computed by the geodesic

algorithm is close to the polygon with vertices zj.

Definition. A disc-chain D0, D1, . . . , Dn is a sequence of pairwise disjoint open discs

such that ∂Dj is tangent to ∂Dj+1, for j = 0, . . . , n − 1. A closed disc-chain is a

disc-chain such that ∂Dn is tangent to ∂D0.


Any simple closed polygon P , for example, can be covered by a closed disc-chain with

arbitrarily small radii and centers on P . There are several ways to accomplish this, but

one straightforward method is the following: Given ε > 0, find pairwise disjoint discs Bjcentered at each vertex, and of radius less than ε. Then

P \⋃

j

Bj =⋃

Lk

where Lk are pairwise disjoint closed line segments. Cover each Lk with a disc-chain

centered on Lk tangent to the corresponding Bj at the ends, and radius less than half the

distance to any other Li, and less than ε.

Figure VI.11. Disc-chain covering a polygon.

Another method for constructing a disc-chain is to use a Whitney decomposition

of a simply connected domain. Suppose Ω is a simply connected domain contained in

the unit square. The square is subdivided into 4 equal squares. Each of these squares is

subdivided again into 4 equal squares, and the process is repeated. If Q is a square, let

2Q denote the square with the same center, and sides twice as long. In the subdivision

process, if a square Q satisfies 2Q ⊂ Ω, then no further subdivisions are made in Q. Let

Un be the union of all squares Q obtained by this process with side length at least 2−n

for which 2Q ⊂ Ω. If z0 ∈ Ω, let Ωn be the component of the interior of Un containing

z0. Then ∂Ωn is a polygonal Jordan curve. Note that ∂Ωn consists of sides of squares Q

with length 2−n. Thus we can form a disc chain by placing a disc of radius 2−n/2 at each

vertex of ∂Ωn. The points of tangency are the midpoints of each square with edge length

2−n on ∂Ωn.


Yet another method for constructing a disc-chain would be to start with a hexagonal

grid of tangent discs, all of the same size, then select a sequence of these discs which form

a disc-chain. Disc-chains can be used to approximate the boundary of an arbitrary simply

connected domain.

Definition. If f is a conformal map of D onto a simply connected region Ω and if I is a

segment contained in a diameter of D, then f(I) is called a geodesic in Ω.

Using linear fractional transformations of D onto D it is not hard to see that I is a

geodesic if and only if I is an arc on a circle which is orthogonal to ∂D.

Later in the course we will study the hyperbolic geometry in plane domains. The

shortest curve between two points in the hyperbolic metric on a simply connected domain

is a geodesic. We will not use that fact here, but we do need the following Lemma.

Lemma 7.5 (Jørgensen). If ∆ is an open disc contained in a simply connected domain

Ω and if J is a geodesic, then J ∩ ∆ is connected and if non-empty, then J is not tangent

to ∂∆ in Ω.

Proof. If A ∈ ∂∆, let B ∈ ∂∆ such that AB is a diameter of ∆. Then z ∈ ∆ if and only

if

ReB − z

z − A> 0. (8.1)

Suppose ϕ is a conformal map of D onto Ω with ϕ(0) = A and ϕ′(0)i/(B − A) > 0. The

geodesics in Ω through A lie on images of diameters of D by the map ϕ. By replacing ϕ(z)

with ϕ(rz), we may suppose that ϕ is analytic on D and without loss of generality we may

assume that ∆ is still a subset of ϕ(D). Note that

Re

[

(B −A)

ϕ′(0)

(

1

z+ z

)]

= 0 (8.2)

on ∂D ∪ (−1, 1). Then

g(z) =B − ϕ(z)

ϕ(z) −A− (B − A)

ϕ′(0)

(

1

z+ z

)


is analytic on D since the singularity at 0 is removable by Riemann’s Theorem. By (8.1)

and (8.2), Reg ≤ 0 on ∂D since ∂Ω ∩ ∆ = ∅. By the maximum principle Reg ≤ 0 on

(−1, 1). By (8.2) again, ReB−ϕϕ−A

≤ 0 on (−1, 1). By (8.1) the geodesic G = ϕ(

(−1, 1))

is does not enter ∆ and thus is tangent to ∆ at A. The unit disk D is divided into two

components D+ and D− by (−1, 1) so that ϕ(D+) and ϕ(D−) are disjoint components of

Ω \G. Each diameter I of D, except (−1, 1) has one half in D+ and the other half in D−.

Thus a geodesic J which intersects ∆ and passes through A must thereafter remain in a

component of Ω \G which does not contain ∆. Each of these geodesics is not tangent to

G at A by conformality, proving the Lemma.

One interpretation of Jørgensen’s Lemma is that disks are convex in the metric on a

simply connected domain given by the geodesics.

If D0, D1, . . . , Dn is a closed disc-chain, set

zj = ∂Dj ∩ ∂Dj+1,

for j = 0, . . . , n, where Dn+1 ≡ D0.

Theorem 7.6. If D0, D1, . . . , Dn is a closed disc-chain, then the geodesic algorithm ap-

plied to the data z0, z1, . . . , zn produces a conformal map ϕ−1c from the upper half plane

H to a region bounded by a C1 curve γ with

γ ⊂n⋃

0

(Dj ∪ zj).

Proof. An arc of a circle which is orthogonal to R is a hyperbolic geodesic in the upper half

plane H. Let γj denote the portion of the computed boundary, ∂Ωc, between zj and zj+1.

Since hyperbolic geodesics are preserved by conformal maps, γj is a hyperbolic geodesic in

C∗ \ ∪j−1

k=0γk.

For this reason, we call the algorithm the “geodesic” algorithm.


Using the notation of Figure VI.9, each map f−1a is analytic across R \ ±c, where

fa(±c) = 0, and f−1a is approximated by a square root near ±c. If f−1

b is another basic

map, then f−1b is analytic and asymptotic to a multiple of z2 near 0. Thus f−1

b f−1a

preserves angles at ±c. The geodesic γj then is a smooth arc which meets γj−1 at zj with

angle π. Thus the computed boundary ∂Ω is C1. The first arc γ0 is a chord of D0 and

hence not tangent to ∂D0. Since the angle at z1 between γ0 and γ1 is π, γ1 must enter

D1, and so by Jørgensen’s Lemma

γ1 ⊂ D1,

and γ1 is not tangent to ∂D1. By induction

γj ⊂ Dj ,

j = 0, 1, . . . , n.

Figure VI.12. Tenerife.

Figure VI.12 shows the conformal map of a grid on the disc to both the interior and

exterior of the island Tenerife (Canary Islands). The center of the interior is the volcano

Teide.

Appendix

§1. Locating zeros of a polynomial.

Algorithm:

Suppose p(z) = anzn + an−1z

n−1 + . . .+ a0 is a polynomial with an 6= 0.

Set

p∗(z) = znp

(

1

z

)

= a0zn + a1z

n−1 + . . .+ an

and let q(z) = a0p(z) − anp∗(z).

Case (a): If |an| < |a0| set p1 = q.

Case (b): If |an| > |a0|, set p1 = q∗/z = (a0p∗ − anp)/z.

Case (c): If |an| = |a0| and q 6= 0 on |z| = 1, set p1 = q.

Let N(p) denote the number of zeros of p in D = z : |z| < 1.

Then

N(p) =

N(p1) in Cases (a) and (c)

N(p1) + 1 in Case (b)

If Case (a), (b), or (c) applies to p1, repeat the process. Since the degree of p1 is less

than the degree of p, the process eventually stops because the degree is zero or because

the first and last coefficients have the same absolute value and q has a zero on |z| = 1.

Proof. Notice first that |p(eiθ)| = |p∗(eiθ)| for all θ. Secondly note that if ζ is a zero of p

on the circle ∂D, then ζ is also a zero of p∗ and hence a zero of q, counting multiplicity.

This implies q/p and q∗/(zp) are analytic in a neighborhood of the circle, ∂D.

109

110 A. Appendix

Case (a): Suppose |an| < |a0|. Then on |z| = 1

∣

∣

∣

∣

∣

p1

a0p− 1

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

−anp∗a0p

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

ana0

∣

∣

∣

∣

∣

< 1.

By Rouche’s theorem, the number of zeros of p equals the number of zeros of p1 in D.

Case (b): Suppose |an| > |a0|. Then on |z| = 1

∣

∣

∣

∣

∣

zp1

anp+ 1

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

a0p∗

anp

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

a0

an

∣

∣

∣

∣

∣

< 1.

By Rouche’s theorem, the number of zeros of p equals the number of zeros of zp1(z) in D,

which equals 1 +N(p1).

Case (c): Suppose |an| = |a0|. Then on |z| = 1

∣

∣

∣

∣

∣

p1

a0p− 1

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

−anp∗a0p

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

ana0

∣

∣

∣

∣

∣

= 1.

If p1 = q 6= 0 on |z| = 1, then p 6= 0 on |z| = 1 and

∣

∣

∣

∣

∣

p1

a0p− 1

∣

∣

∣

∣

∣

< 1 +

∣

∣

∣

∣

∣

p1

a0p

∣

∣

∣

∣

∣

,

so by Rouche’s Theorem, p and p1 have the same number of zeros in |z| < 1.

Remark: If |an| = |a0| and if q has a zero on |z| = 1, then see Cohn, A. Math. Zeit. vol 14

(1922) 110-148. (see also Marden, Morris, The geometry of the zeros of a polynomial in

a complex variable, AMS (1949) 148ff). Cohn shows that the algorithm can be continued

by modifying the sequence. Another approach when |a0| = |an| is to replace p(z) by p(rz)

for r near 1. If r < 1 then Case (a) applies. The algorithm can be used to find the number

of zeros in any disc by composing the polynomial with a linear map Az +B.

Remark: In case (b), p1 = q∗/z. If p1 vanishes to a higher order at the origin, you can

reduce the degree of p1 by dividing by a higher power of z before applying the algorithm

again (and add to the count of the number of zeros).

§1: Locating zeros. 111

Example:p(z) = 2z4 − 3z3 + z2 + z − 1

p∗(z) = −z4 + z3 + z2 − 3z + 2

p1(z) = 3z3 − 5z2 + 3z − 1

p∗1(z) = −z3 + 3z2 − 5z + 3

p2(z) = 8z2 − 12z + 4

p∗2(z) = 4z2 − 12z + 8

p3(z) = 12z − 12

p∗3(z) = −12z + 12

If p has k zeros in D, then p1 has k− 1 zeros in D, p2 has k− 2 zeros in D amd p3 has

k− 3 zeros in D. At this point the algorithm stops since the first and last coefficients have

equal modulus. But p3 clearly has no zeros in D and hence k− 3 = 0 and p has 3 zeros in

D. Note that zeros of p which are symmetric about the unit circle are also zeros of each

pk. Since p3(1) = 1, then p(1) = 0 and its remaining zeros are in the disk.

As an exercise, write down a polynomial and apply the process.

The following is a problem that has been open since the 1930’s and is of current

interest in ergodic theory. Suppose p is a polynomial with integer coefficients

p(z) = anzn + . . .+ a0

with |an| = |a0| = 1. Let z1, z2, . . . , zk be the roots in |z| < 1. If p has at least one root in

|z| < 1, how big can∏ |zj | be? It is conjectured that

k∏

p=1

|zj | ≤ .8537662...

This value is achieved for the polynomial

p(z) = z10 + z9 − z7 − z6 − z5 − z4 − z3 + z + 1.

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complex analysis 3