Course Meeting Times Lectures: 2 sessions / week, 1.5 hours / session

Course Objective The aim of the course is to teach the principal techniques and methods of analytic function theory. This is quite different from real analysis and has much more geometric emphasis. It also has significant applications to other fields like analytic number theory.

Prerequisite Analysis I (18.100B) or the equivalent.

Text Ahlfors, Lars V. Complex Analysis: An Introduction to the Theory of Analytic Functions of One Complex Variable. 3rd ed. New York, NY: McGraw-Hill, 1979. ISBN: 9780070006577. This is a text with an attractive geometric flavor. I plan to cover most of the material up to p. 232, but will add two lectures devoted to a complete proof of the Prime Number Theorem, which fits very naturally in with the text material on the Riemann zeta function.

In most of the lectures I will add some material not in the text, partly giving alternative proofs, and partly working through solutions of some of the more interesting problems. In Lec #2, 13, 14, 16, 19, 21, and 22 the treatment is really quite different from the corresponding material in the text. Lec #21 and 22 actually contain a complete proof of the Prime Number Theorem.

Since the lectures deviate so much from the text, regular attendance is strongly recommended.

Homework Assignments Given out every other Thursday, due the next Thursday. This will be selected from problems in the text.

Exams Two in-class tests following Lec #4 and 16. Also a final examination.

Grading The final grade is based on a cumulative point total.

ACTIVITIES PERCENTAGES

Homework 10%

Two in-class tests (20% each) 40%

Final exam 50%

General Remarks Working on the homework problems is very important for your understanding and command of the subject. Be sure you allow plenty of time for this. It is ok for you to discuss the problems with another student but you must give your own exposition of the solution. In the process you will often be lead to a better solution. It is very important to do this exposition in a neat fashion. This is the right place for you to develop a clear mathematical writing style. (Tex is of course most welcome). Sloppy homework might not be accepted.

There are dozens of books available on the topic of complex variables. I urge you strongly to browse in some of these in the library. Caratheodory's books Theory of Functions Vol. I and II are closely related to our text. Also, I would particularly recommend Pólya and Szegö: Problems and Theorems in Analysis. Several chapters there deal with the subject of complex variables. Rudin's book, Real and Complex Analysis is also a valuable reference. Caratheódory, Constantin. Theory of Functions of a Complex Variable. Rhode Island: AMS Chelsea Pub, 2001. ISBN: 9780821828311. Caratheódory, Constantin, and F. Steinhardt. Theory of Functions of a Complex Variable. Vol. 2. New York, NY: Chelsea, 1960. Pólya, George, and Gábor Szego. Problems and Theorems in Analysis. Berlin: Springer, 1978. ISBN: 9780387056722. Rudin, Walter. Real and Complex Analysis. New York, NY: McGraw-Hill, 1986. ISBN: 9780070542341.

Lecture 1: The algebra of Complex numbers

(Text 1-11 & 19-20)

Remarks on Lecture 1

◮ On p.19-20, it is stated that each circle in C

(x − a)2 + (y − b)2 = r 2 (1)

has the form

(α0 − α3)(x 2 + y 2) − 2α1x − 2α2y + α0 + α3 = 0,

so the mapping z 7→ Z maps circles in the plane to circles on S. Solving the equations

a = α1

, b = α2

, r 2 − a 2 − b2 = − α0 + α3

α0 − α3 α0 − α3 α0 − α3

for α0, α1, α2, α3 is disagreeable so we instead determine the image of the curve (1) under the map z 7→ Z. Using the formulas (24)-(26) and

2 1 − x3 = ,

1 + |z|2

formula (1) becomes

1 + r2 − a2 − b2 a2 + b2 − r2 + 1 ax1 + bx2 + x3 = .

2 2

This is a plane which must intersect the sphere so has distance < 1 from 0.

1

= � .

= � �

◮ The formula (28) can be proved geometrically as follows (Exercise 4):

N

This proves (28).

Z

z 0

1

1

Fig. 1-1

Let Z ∈ S lie on the plane

x2 = 0.

The angles at Z are right angles, so by similar triangles:

d(N, Z) 1 =

2 d(N, z) 1

1 + |z|2

Thus N

Z Z'

z'

Fig. 1-2

d(N, Z) 2 d(N, z′) 1 + |z|2 1 + |z ′|2

and by symmetry this is

z d(N, Z ′) .

d(N, z)

Thus the triangles △NZZ ′ and △Nzz ′ are similar, so the above ratio is

d(Z, Z ′) .

|z − z ′|

2

◮ Finally we show that the spherical representation z 7→ Z is conformal. This means that if l and m are two lines in the plane intersecting in z at an angle α, then the corresponding circles C and D through N and Z intersect Z at the same angle α. Consider the tangent plane π to S at the point N . the plane through Z and l intersects π in a line l ′ . Similarly the plane through Z and m intersect π in m ′ . Clearly l ′ and m ′ intersect at N at the same angle α. Since they are tangents to C and D at N , C and D must intersect at the angle α both at N and at Z.

3

Lecture 2: Exponential function & Logarithm for a complex argument

(Replacing Text p.10 - 20)

For b > 1, x ∈ R, we defined in 18.100B,

bx = sup bt

t∈Q, t≤x

(where bt was easy to define for t ∈ Q). Then the formula

bx+y bxby=

was hard to prove directly. We shall obtain another expression for bx making proof easy.

Let � x dt

L(x) = , x > 0. t1

Then L(xy) = L(x) + L(y)

and 1

L ′ (x) = > 0. x

So L(x) has an inverse E(x) satisfying

E(L(x)) = x.

By 18.100B, E ′ (L(x))L ′ (x) = 1,

so E ′ (L(x)) = x.

If y = L(x), so x = E(y), we thus have

E ′ (y) = E(y),

1

It is easy to see E(0) = 1, so by uniqueness,

2 nx xE(x) = 1 + x + + · · · + + · · · and E(1) = e.

2 n!

Theorem 1 bx = E(xL(b)), ∀x ∈ R.

Proof: Let u = L(x), v = L(y), then

E(u + v) = E(L(x) + L(y)) = E(L(xy)) = xy = E(u)E(v),

E(n) = E(1)n = e n ,

and if t = n ,m

E(t)m = E(mt) = E(n) = e n .

so E(t) = e t , t ∈ Q, t > 0.

Since E(t)E(−t) = 1,

So E(t) = e t , t ∈ Q.

Now bn = E(nL(b))

and � �

1 1 mb = E L(b)

m

since both have same mth power.

� �n � �n

� �

1 n 1 n m mb = b = E L(b) = E L(b) ,

m m

so bt = E(tL(b)), t ∈ Q.

Now for x ∈ R,

bx = sup (bt) = sup E(tL(b)) = E(xL(b)) t≤x, t∈Q t≤x, t∈Q

since E(x) is continuous. Q.E.D.

Corollary 1 For any b > 0, x, y ∈ R, we have bx+y = bxby.

2

In particular ex = E(x), so we have the amazing formula

� �x 21 1 x xn 1 + 1 + + · · · + + · · · = 1 + x + + · · · + + · · · .

2! n! 2! n!

The formula for ex suggests defining ez for z ∈ C by

2 nz ze z = 1 + z + + · · · + + · · · .

2! n!

the convergence being obvious.

Proposition 1 ez+w = ezew for all z, w ∈ C.

Proof: Look at the functions

f(t) = e tz+w , g(t) = e tz e w

for t ∈ R. Differentiating the series for etz+w and etz with respect to t, term-by-term, we see that

df dg = zf(t), = zg(t)

dt dt and

f(0) = e w , g(0) = e w .

By the uniqueness for these equations, we deduce f ≡ g. Thus f(1) = g(1). Q.E.D.

Note that if t ∈ R,

e it e −it = 1, and(e it)−1 = e −it .

Thus |e it| = 1.

So eit lies on the unit circle.

Put eit + e−it t2

cos t = = 1− + · · · ,

2 2

eit − e−it t3

sin t = = t − + · · · .

2 3!

Thus we verify the old geometric meaning eit = cos t+i sin t. Note that the eit(t ∈ R) fill up the unit circle. In fact by the intermediate value theorem, {cos t | t ∈ R} fills up the interval [−1, 1], so e it = cos t + i sin t is for a suitable t an arbitrary point on the circle.

3

a

� �

Note that z 7→ ez takes all values w ∈ C except 0. For this note

e z = e x · eiy , z = x + iy.

Choose x with e x = |w|

and then y so that iy w e = ,

|w|

then ez = w.

If z iϕ iψ z = |z|e , w = |w|e ,

w then a zw = |z||w|e i(ϕ+ψ)

Fig.2-1 = |z||w|(cos (ϕ + ψ) + i sin (ϕ + ψ)),

which gives a geometric interpretation

2 /n

of the multiplication.

From this we also have the following very useful formula

(cosϕ+ i sinϕ)n = e inϕ = cosnϕ+ i sinnϕ.

Thus Fig. 2-2

nTheorem 2 The roots of z = 1 are 1, ω, ω2 , · · · , ωn−1 , where

2π 2π ω = cos + i sin .

n n

Geometric meanings for some useful complex number sets:

|z − a| = r ←→ circle

|z − a| + |z − b| = r, (|a − b| < r) ←→ ellipse

|z − a| = |z − b| ←→ perpendicular bisector

{z | z = a + tb, t ∈ R} ←→ line

{z | Imz < 0} ←→ lower half plane

z − a {z | Im < 0} ←→ general half plane

b

4

For x real, x 7→ ex has an inverse. This is NOT the case for z 7→ ez, because

z+2πi z e = e ,

thus ez does not have an inverse. Moreover, for w 6 0, =

z e = w

has infinitely many solutions:

e x = |w|, eiy = w

=⇒ x = log |w|, y = arg(w). |w|

So logw = log |w| + iarg(w)

takes infinitely many values, thus not a function.

Define

Arg(w) � principal argument of w in interval − π < Arg(w) < π

and define the principal value of logarithm to be

Log(w) � log |w| + iArg(w),

which is defined in slit plane (removing the negative real axis).

We still have log z1z2 = log z1 + log z2

in the sense that both sides take the same infinitely many values. We can be more specific:

Theorem 3 In slit plane,

Log(z1z2) = Log(z1) + Log(z2) + n · 2πi, n = 0 or ± 1

and n = 0 if −π < Arg(z1) + Arg(z2) < π.

In particular, n = 0 if z1 > 0.

Proof: In fact, Arg(z1), Arg(z2) and Arg(z1z2) are all in (−π, π), thus

−π − π − π < Arg(z1) + Arg(z2)− Arg(z1z2) < π + π + π,

but Arg(z1) + Arg(z2)− Arg(z1z2) = n · 2πi,

5

thus

|n| ≤ 1.

If |Arg(z1) + Arg(z2)| < π,

since |Arg(z1z2)| < π,

they must agree since difference is a multiple of 2π. Q.E.D.

6

�

Lecture 3: Analytic Functions; Rational Functions

(Text 21-32)

Remarks on Lecture 3

� Formula (14) on p.32 was proved under the assumption that R(≡) = ≡. On the other hand, if R(≡) is finite, then (12) holds with G � 0. Then we use the previous proof on R(�j + 1 � ) and we still get the representation (14).

� For theorem 1 on page 29, we have the following stronger version:

Theorem 1 (Stronger version) The smallest convex set which contains all the zeros of P (z) also contains the zeros of P �(z).

Proof: Let �1, · · · , �n be the zeros of P , so

P (z) = an(z − �1) · · · (z − �n).

Then P �(z) 1 1

= + · · · + . P (z) z − �1 �n

If z0 is a zero of P �(z) and z0 =≥ each �i, then this vanishes for z = z0; conjugating the equation gives

z0 − �1 z0 − �n + · · · + = 0,

|z0 − �1|2 |z0 − �n|2

so z0 = m1�1 + · · · + mn�n,

where n

mi � 0 and mi = 1. i=1

We now only need to prove the following simple result:

1

� �

�

�

�

�

�

Proposition 1 Given a1, · · · , an ≤ C, the set

n n

{ miai | mi � 0, mi = 1} (1) i=1 i=1

is the intersection C of all convex sets containing all ai (which is called the convex hull of a1, · · · , an).

n

Proof: We must show that each point aimi in (1) is contained in each convex set i=1

containing the ai and thus in C. We may assume it has the form

p

x = miai i=1

where mi > 0 for 1 � i � p

and mj = 0 for j > p.

We prove x ≤ C by induction on p. Statement is clear if p = 1. Put

p−1

� = mi i=1

and p−1 � m1

a = ai. i=1

By inductive assumption, a ≤ C. But

p

x = miai = �a + (1 − �)ai i=1

where 0 � � � 1. So x ≤ C as stated. Q.E.D.

2

Solution to 4 on p.33

Suppose R(z) is rational and

|R(z)| = 1

for |z| = 1. Then |R(e i�)| � 1 λ ≤ R.

Let S(z) be the rational functions obtained by conjugating all the coefficients in R(z), then

R(e i�)S(e −i�) = R(e i�)R(ei�) = 1.

So 1

R(z)S( ) = 1 on |z| = 1. z

Clearing denominators we see this relation

1 R(z)S( ) = 1

z

holds for all z ≤ C.

Since a polynomial has only finitely many zeroes, let

�1, · · · , �p

be all the zeroes of R(z) which are not equal to 0 or ≡. Then

1 1 , · · · ,

�1 �p

are the poles of S(z) which are not equal to 0 or ≡. So

1 1 , · · · ,

�̄1 �̄p

are the poles of R(z) which are not equal to 0 or ≡ because of the definition of S. Then

� �

−1 z − �1 z − �p

R(z) · · · 1 − �̄1z 1 − �̄pz

has no poles or zeros except possibly 0 and ≡. Hence

R(z) = Czl z − �1

· · · z − �p

1 − �̄1z 1 − �̄pz

where C is constant with |C| = 1, l is integer.

Conversely, such R has |R(z)| = 1 on |z| = 1.

3

�

� �

Lecture 4: Power Series

(Text 33-42)

Remarks on Lecture 4

Problem 8 on p.41

∞ nWe know 0 w converges only for |w| < 1. Otherwise the terms do not

converge to 0. Now put 1′ z = z + ,2

so z z ′ − 1

2 w = = . ′1 + z z + 1

2

So |w| < 1 is equivalent to ′ Rez > 0,

or equivalently 1

Rez > − . 2

Problem 9 on p.41

Write z n 1

= . 1 + z2n zn + z−n

Write an ∼ bn if � an � � � −→ c = 06 . � bn

�

1

Then if |z| > 1,

1

∼ z −n , zn + z−n

and if |z| < 1, 1

∼ z n . zn + z−n

So in both cases we have convergence. If z = e it, we have

1 1 = ,

−n 2 cos nt zn + z

so the terms do not tend to 0, so we have divergence.

2

Lecture 5: Exponentials and Trigonometric

Functions

(Text 42-47)

Remarks on Lecture 5

Since cos z is even, arccos z can just as well defined as √

arccos z = −i log(z + z2 − 1).

This in fact more appropriate because then the derivative is

1 −√ ,

1 − z2

which is better because then the derivative is < 0 for z = 0.

Note that in any case cos (arccos z) = z,

√ √since z + z2 − 1 and z − z2 − 1 are reciprocals.

1

Lecture 7: Linear Transformations

(Text 80-89)

Remarks on Lecture 6

Concerning Definition 13 p. 81, formula (11) shows that the definition does not depend on the choice of z1, z2, z3.

Exercise 2 on page 88 requires a minor correction. For example w = −z is hyperbolic according to definition on page 86, yet when written in the form

az + b

cz + d

with ad − bc = 1,

we must take a = −d = i,

so a + d = 0.

The transformation w = z causes other ambiguities.

Thus we modify the definition a bit.

Definition 1

• S is parabolic if either it is the identity or has exactly one fixed point. • S is strictly hyperbolic if k > 0 in (12) but k = 1. • S is elliptic if |k| = 1 in (12) p. 86 but S is not identity.

Then the statement of Exercise 2 holds with hyperbolic replaced by strictly hyperbolic.

1

(i) the condition for exactly one fixed point for

αz + β Sz =

γz + δ

is (α − δ)2 = −4βγ (wrong sign in text).

With the normalization αδ − βγ = 1

this amounts to (α + δ)2 = 4

as desired.

(ii) Assume two fixed points are a and b, so

w − a z − a = k ,

w − b z − b

which we write as αz + β

w = Tz = . γz + δ

Put

α β A =

γ δ

and define (TraceA)2

Tr2(T ) = . det A

By linear algebra, Trace(BAB−1) = Trace(A)

and det (BAB−1) = det A.

Define z − a

z1 = Sz = , z − b

w − a w1 = Sw = .

w − b

Then Tr2(T ) = Tr2(STS−1).

Now w1 = STz = STS

−1 z1,

2

so w1 = kz1.

Then

Tr2(T ) = Tr2(STS−1) = k +1

+ 2. k

If T is strictly hyperbolic, we have

k > 0, k 1,

so Tr2(T ) > 4,

which under the assumption αδ − βγ = 1

amounts to (α + δ)2 > 4

as stated.

Conversely, if (α + δ)2 > 4,

then k > 0. So the transformation

w1 = kz1

maps each line through 0 and ∞ into itself. So T maps each circle C1 into itself with k > 0. Thus T is strictly hyperbolic.

(iii) If |k| = 1, then iθ w1 = e z1

and we find

Tr2(T ) = 2 cos θ

2

< 4 2

since the possibility θ = 0 is excluded.

Conversely, if −2 < α + δ < 2, αδ − βγ = 1

we have

Tr2(T ) = (α + δ)2 = k +1

+ 2 < 4. k

Writing k = re iθ (r > 0)

3

6=

( )

this implies 1 1

r + cos θ + i r − sin θ < 2, r r

which implies r = 1 or θ = 0 or θ = π.

If r = 1, then |k| = 1, so T is elliptic. Since r + 1

r ≥ 2, the possibility θ = 0 is ruled out.

Finally if θ = π, then k = −r, so

(α + δ)2 = −r − 1

+ 2. r

But r ≥ 0, so since α+δ is real, this implies r = 1, so k = −1, and T is thus elliptic.

4

( ( ))

Lecture 9: Cauchy-Goursat Theorem

(Text 109-115)

Remarks on Lecture 9

Property (ii) (page 116) of the winding number follows since

a −→ n(γ, a)

is a continuous function on C \ γ and the image of each component is a connected set of integers, hence constant.

1

Lecture 10: The Special Cauchy’s Formula and Applications

(Text 118-126)

Remarks on Lecture 10

Exercise 6 on page 108

The values of f(z) lie in the disk |w − 1| < 1 which is contained in the slit plane where Logw is defined. thus Logf(z) is well-defined and holomorphic in Ω and has derivative

f(

1

z) f ′ (z).

Thus 1

f ′ (z)dz = 0

γ f(z)

by the Primitive theorem.

Exercise 2 on page 120

By using the substitution w = ϕ(z) = −z we have

1 1

dw −dz = .

w2 + 1 z2 + 1 ϕ(γ) γ

Since ϕ(γ) = γ (including the orientation). Thus the integral is 0.

Also 1 1 1

= − z2 + 1 z − i z + i

1

� �

� �

� �

and n(γ, i) = n(γ,−i),

so again the total integral is 0.

Exercise 3 on page 120

On |z| = ρ, we can write z = ρeiθ, thus

dz ρeiθ= i,

dθ

so dz

= i dθ, z

and dz

|dz| = ρdθ = −iρ . z

Thus 1 1

|dz| dz = −iρ

|z|=ρ |z − a|2

|z|=ρ z(z − a)(ρ

z

2 − ā)

1 1

1 dz ā dz = −iρ + .

z − ρ2 − |a|2 azρ2 − |a|2 |z|=ρ a |z|=ρ ρ2 − ¯

If |a| > ρ, the first term is 0, the other term is

1

1 dz 1 ρ2

= −2πi , ā ¯|z|=ρ − z a ā

so the result is 2πρ

. |a|2 − ρ2

If |a| < ρ, then the second is 0 and the other is

1 2πρ −iρ 2πi = .

ρ2 − |a|2 ρ2 − |a|2

Thus in both cases the result is

� 2πρ � .

ρ2 − |a|2

2

◮ The Taylor’s Theorem (with remainder) proved in pp.125-126 should be stated as follows:

Theorem 1 (Taylor’s Theorem) If f(z) is analytic in a region Ω containing a, one has

f ′ (a) fn−1(a) a)n−1f(z) = f(a) + (z − a) + · · · + (z − + fn(z)(z − a)

n ,1! (n − 1)!

where fn(z) is analytic in Ω. Moreover, if C is the boundary of a closed disk contained in Ω with center a, then fn(z) has the representation

1

1 f(ζ) dζ fn(z) = (z inside C).

2πi C (ζ − a)n(ζ − z)

3

�

Lecture 11: Isolated Singularities

(Text 126-130)

Remarks on Lecture 11

Singularities: Let f(z) be holomorphic in a disk 0 < |z − a| < δ with the center a removed.

(i) If

lim f(z)

z a→

exist or if just lim f(z)(z − a) = 0, z a→

then a is a removable singularity and f extends to a holomorphic function on the whole disk < δ.|z − a|

(ii) If

lim f(z) = ∞,

z a→

a is said to be a pole. In this case

f(z) = (z − a)−hfh(z),

where h is a positive integer and fh(z) is holomorphic at a and fh(a) = 0. We also have the polar development

f(z) = Bh(z − a)−h + + B1(z − a)−1 + ϕ(z),· · ·

where ϕ(z) is holomorphic at a.

If neither (i) nor (ii) holds, a is said to be an essential singularity.

1

Theorem 9 A holomorphic function comes arbitrarily close to any complex value in every neighborhood of an essential singularity.

Simplified Proof: Suppose statement false. Then ∃A ∈ C and δ > 0 and � > 0 such that

|f(z) − A| > δ for |z − a| < �. Then

lim(z − a)−1(f(z) − A) = ∞. z a→

So (z − a)−1(f(z) − A)

has a pole at z = a. Thus

f(z) − A = (z − a)(z − a)−h g(z),

where h ∈ Z+ and g(z) is holomorphic at z = a. If h = 1, f(z) has a removable singularity at z = a. If h > 1, f(z) − A has a

pole at z = a and so does f(z). Both possibilities are excluded by assumption, so the proof is complete. Q.E.D.

Exercise 4 on p.130.

Suppose f is meromorphic in C ∪ {∞}. We shall prove f is a rational function. If ∞ is a pole, we work with g = 1/f , so we may assume ∞ is not a pole. It is not an essential singularity, so ∞ is a removable singularity. Thus for some R > 0, f(z) is bounded for |z| ≥ R. Since the poles of f(z) are isolated, there are just finitely many poles in the disk |z| ≤ R. (Poles of f(z) are zeroes of 1/f(z).) At a pole a, use the polar development near a

f(z) = Bh(z − a)−h + + B1(z − a)−1 + ϕ(z).· · ·

The equation shows that ϕ extends to a meromorphic function on C ∪∞ with one less pole than f(z). We can then do this argument with ϕ(z) and after iteration we obtain

n � �� 1 f(z) = Pi + g(z),

z − aii=1 where Pi are polynomials and g is holomorphic in C. The formula shows that g is bounded for |z| > geR and being analytic on |z| ≤ R, it thus must be bounded on C. By Liouville’s theorem, it is constant. So f is a rational function.

2

� �

Lecture 12: The Local Mapping. Schwarz’s Lemma

and non-Euclidean interpretation

(Text 130-136)

Remarks on Lecture 12

Proof of Theorem 11, p.131.

The last formula on p.131 reads

n(γ, zj (a)) = n(γ, zj (b)) (1) j j

provided a and b belong to the same component of the set C − Γ. In (1) we take for γ the circle

z(t) = z0 + �eit (0 ≤ t ≤ 2π),

where � > 0 is so small that z0 is the only zero of

f(z) − w0 = 0

inside γ. (The zeroes of f(z) − w0 = 0 are isolated.) As before let Γ = f(γ) and let Ω0 denote the component of C − Γ containing w0.

z0

Ω0

γ

Γ

w0

1

This model satisfies all Euclid’s axioms except the famous Parallel Axiom:

Given a point p outside a line l, there is exactly one line through the point which does not intersect l.

This axiom clearly fails in the above model; thereby solving the 2000 year old problem of proving the Parallel Axiom on the basis of the other axioms. It cannot be done!

Since γ is a circle,

n(γ, zj (a)) = 0 or 1

depending on whether zj (a) is outside γ or inside γ (Since a /∈ Γ, zj (a) ∈/ γ). Thus the left hand side of (1) equals the number of points inside γ where f takes the value a.

In (1) we now take a arbitrary in Ω0 and take b = w0. By the choice of γ he right hand side of (1) equals n. Equation (1) therefore shows that each value in Ω0 is taken n times by f inside γ (multiple roots counted according to their multiplicity). In particular, this holds for any disk |w−w0| < δ inside Ω0 with center w0. Q.E.D.

Remark: In dealing with winding numbers n(δ, z), we have to pay attention to the parametrization. Thus if

nγ(t) = e it (0 ≤ t ≤ 2π), f(z) = z

and Γ = f(γ), we have n(γ, 0) = 1, n(Γ, 0) = n

although γ and Γ are represented geometrically by the same point set.

Non-Euclidean Plane

This is the unit disk |z| < 1 with the following convention: a) Non-Euclidean point = Point in disk;

b) Non-Euclidean line = Arc in disk perpendicular to the boundary.

We can now introduce distance in the non-Euclidean plane D.

2

���� ����

� ���� ���� �� �

z3

z 1

z2v

u

Given z1, z2 D, mark the∈points u, v of intersection with |z| = 1, in the order indicated. We put

1 d(z1, z2) = log (z1, z2, v, u)

2

= 1

log ( z1 − v

: z2 − v

). 2 z1 − u z2 − u

The cross ratio is real (page 79) and the geometry shows easily that the cross ratio is 1.≥So

d(z1, z2) ≥ 0,

d(z1, z2) = d(z2, z1).

It is also easy to show from the formula that

d(z1, z3) = d(z1, z2) + d(z3, z2).

Consider a fractional linear transformation

z −→ e iϕ 1 z −−

z̄

z

2

2

z

mapping

z2 −→ 0 and z1 −→ z1 − z2 1 − z̄2z1

.

Then by the order of the points v, z2, z1, u we see that

v −→ −1 and u −→ 1.

The invariance of the cross ratio gives

1

d(z2, z1) = log 0,

2

z2 − z1 1 − z̄2z1

, −1, 1

= 1 2

log |1 − z̄2z1| + |z1 − z2||1 − z̄2z1| − |z1 − z2|

(Exercise 7).

� �Thus

d(z, z + Δz) =

1 2

log 1 + 2|Δz|

|1 − z̄(z + Δz)| − |Δz| .

3

So since log(1 + x)

lim = 1 x 0 x→

we deduce d(z, z + Δz) 1

lim = . Δz 0 Δz z 2 → | | 1 − | |

This suggests defining a non-Euclidean length of a curve

γ : z(t) (α ≤ t ≤ β)

by � L(γ) =

|dz| .

zγ 1 − | |2

Exercises 1 and 6 now give the following geometric interpretation of Schwarz’s Lemma:

Ex 1. Formula (30) implies by division, letting z z0,→

1 −|f|

�

f

(z

(

)

z

|)|2

≤ 1 −

1 |z|2

.

Ex 6. Let f : D D be holomorphic and f(γ) the image curve →

w(t) = f(z(t)) α ≤ t ≤ β

of the curve γ above. Then

L(f(γ)) = � β |w�(t)|

dt α 1 − w(t) 2� β f �(z|(t))z|�(t)

= | |

dt �αβ 1 − |f(z(t))|2 ≤ |z

�(

z

t

(

)

t

|)

(by Ex 1.) α 1 − | |2

= L(γ).

Thus L(f(γ)) ≤ L(γ)

as stated.

4

�

Lecture 13: The General Cauchy Theorem

(Replacing Text 137-148)

Here we shall give a brief proof of the general form of Cauchy’s Theorem. (cf: John D. Dixon, A brief proof of Cauchy’s integral theorem, Proc. Amer. Math. Soc. 29, (1971) 625-626.)

Definition 1 A closed curve ζ in an open set � is homologous to 0 (written ζ � 0) with respect to � if

n(ζ, a) = 0 for all a /∩ �.

Definition 2 A region is simply connected if its complement with respect to the extended plane is connected.

Remark: If � is simply connected and ζ → � a closed curve, then ζ � 0 with respect to �. In fact, n(ζ, z) is constant in each component of C − ζ, hence constant in C − � and is 0 for z sufficiently large.

Theorem 1 (Cauchy’s Theorem) If f is analytic in an open set �, then ⎩

f(z) dz = 0

for every closed curve ζ → � such that ζ � 0. ⎨

In particular, if � is simply connected then � f(z) dz = 0 for every closed ζ → �.

We shall first prove

Theorem 2 (Cauchy’s Integral Formula) Let f be holomorphic in an open set �. Then

⎩ 1 f(�)

n(ζ, z)f(z) = d� (1)2�i � � − z

where ζ � 0 with respect to �.

1

�

�

Proof: The prove is based on the following three claims.

Define g(z, �) on � × � by

f(�)−f(z) for z =∪ �, � �−z

g(z, �) = f �(z) for z = �.

Claim 1: g is continuous on � × � and holomorphic in each variable and g(z, �) = g(�, z).

Clearly g is continuous outside the diagonal in � × �. Let (z0, z0) be a point on the diagonal and D → � a disk with center z0. Let z ∪= � in D. Then by Theorem 8

1 g(z, �) − g(z0, z0) = f

�(�) + f2(z)(z − �) − f �(z0).

2

So the continuous at (z0, z0) is obvious.

For the holomorphy statement, it is clear that for each �0 ∩ � the function

z ⊂� g(z, �0)

is holomorphic on � − �0. Since

lim g(z, �0)(z − �0) = 0 z��0

the point �0 is a removable singularity (Theorem 7, p.124), so

z ⊂� g(z, �0)

is indeed holomorphic on �. This proves Claim 1.

2

�

Let �� = {z ∩ C − (ζ) : n(ζ, z) = 0}.

Define function h on C by

⎩ 1

h(z) = g(z, �) d�, z ∩ �; (2)2�i �

⎩ h(z) =

1 f(�) d�, z ∩ �� . (3)

2�i � � − z

Since both expression agree on � ≡ �� and since � ∼ �� = C, this is a valid definition.

Claim 2: h is holomorphic.

This is obvious on the open sets �� and � − ζ. To show holomorphy at z0 ∩ ζ, consider a disk D → � with center z0. Let � be any closed curve in D. Then

� ⎧⎩ ⎩ ⎩

1 h(z) dz = g(z, �) d� dz

� 2�i � � � ⎧

⎩ ⎩ 1

= g(z, �) dz d�. 2�i � �

For each �, z ⊂� g(z, �)

is holomorphic on D (even �). So by the Cauchy’s theorem for disks, ⎩

g(z, �) dz = 0.

Now the Morera’s Theorem implies h is holomorphic.

Now we can prove:

Claim 3: h � 0, so (1) holds.

We have z ∩ �� for |z| sufficiently large. So by (3),

lim h(z) = 0. z��

By Liouville’s Theorem, h � 0. Q.E.D.

3

Proof of Theorem 1: To derive Cauchy’s theorem, let z0 ∩ � − ζ and put

F (z) = (z − z0)f(z).

By (1),

⎩ ⎩ 1 1 F (z)

f(z) dz = dz 2�i � 2�i � z − z0

= n(ζ, z0)F (z0)

= 0.

Q.E.D.

Note finally that Corollary 2 on p.142 is an immediately consequence of Cauchy’s Theorem.

4

�

Lecture 14: The Residue Theorem and Application

(Replacing Text 148-154)

Let Ω be a region and a ∈ Ω. Let f(z) be holomorphic in Ω� = Ω − a

Definition 1 The residue is defined as

1 R = Resz=af(z) � f(z) dz,

2πi C where C is any circle contained in Ω with center a.

If C � is another circle with

C

C `

Fig. 14-1

center a and

C � ⊂ Ω,

then Cauchy’s Theorem for the annulus shows that

Resz=af(z)

is independence of the choice of C.

While the definition can be shown to be equivalent to Definition 3 on p.149 in the text, we shall not need this.

In place of Theorem 17 (Text p.150) we shall prove the following version:

Theorem 17 � Let f be analytic except for isolated singularities aj in a region Ω. Let γ be a simple closed curve which has interior contained in Ω and aj ∈/ γ (all j). Then �

1 � f(z) dz = Resz=aj f(z). 2πi γ j

where the sum ranges over all aj inside γ.

1

Proof:

Ω

γ

Fig. 14-2By compactness of γ and its interior, the sum above is finite. For simplicity let

a1, a2 be the singularities inside γ.

γ

a₁a₂

Fig 14-3

The outside of γ is connected and if we take two disks D1, D2 around a1 and a2 and connect their boundaries to γ with “bridges” as in Fig. 14-3, the piece remaining in the interior of γ is simply connected (the complement is connected). Thus the integral over the boundary of this region is 0.

Letting the widths of the bridges tend to 0, the theorem follows. Q.E.D.

2

� �

�

� �

Calculation of residues.

1. If lim f(z)(z − a) z a→

exists and is finite, then it equals Resz=af(z).

In fact a is then a pole of f(z), so

f(z) = Bh(z − a)−h + · · · + B1(z − a)−1 + ϕ(z), Bh �= 0.

Then � 1

f(z) dz = B12πi C

and since the singular part above equals

(z − a)−h(Bh + Bh−1(z − a) + + B1(z − a)h−1)· · ·

the finiteness of the limit implies h ≤ 1.

2. If f(z) = hg((zz)) where g(a) = 0 � and h(z) has a simple zero at z = a, then

g(a)Resz=af(z) = .

h�(a)

In fact 1 g(a)

zlim

a f(a)(z − a) =

zlim

a g(z)

h(z)−h(a) = h�(a) .

→ →z−a

3. If f has a pole of order h, then

1 dh−1 Resz=af(z) = .

(h − 1)! dzh−1 (z − a)hf(z)

z=a

In fact f(z) = (z − a)−h g(z),

where g is holomorphic at a. So

g(h−1)(a) = (h − 1)! 2

1 πi (z

g

− (z

a

)

)h dz = (h − 1)!Resz=af(z).

C

Example: (from text p.151.)

ez d f(z) = = Resz=af(z) = e

z = e a . (z − a)2

⇒dz z=a

3

�

Application: The Argument Principle.

Theorem 18 � Let f(z) be meromorphic in Ω, γ ⊂ Ω a simple closed curve with interior inside Ω. Assume γ passes through no zeros nor poles of f . Then

1 f �(z)dz = N − P,

2πi γ f(z)

where N is the number of zeros, P the number of poles inside γ, all counted with multiplicity.

Proof: By theorem 17�, the integral is the sum of the residues of f �(z)/f(z).

At a zero a of order h, we have

f(z) = (z − a)hfh(z), fh(a) = 0 �

and f �(z) h fh

� (z) f(z)

= z − a

+ fh(z)

= ⇒ Residue h,

At a pole b of order k, we have similarly

f

f

�

(

(

z

z

)

) = z −− k

b + f

fh

h

� (

(

z

z

)

) = ⇒ Residue − k.

Now the result follows from Theorem 17�. Q.E.D.

Corollary 1 (Rouche’s Theorem) Let f and g be holomorLet γ be a simple closed curve in Ω with interior ⊂ Ω. Assume

|f(z) − g(z)| < f(z) on γ.

Then f and g have the same number of zeros inside γ, say Nf

phic in a

and Ng.

region Ω.

Proof: (The text does not take into account the case when f and g have common zeros). The inequality implies that f and g are zero-free on γ. Put

g(z)ψ(z) = ,

f(z)

then |ψ(z) − 1| < 1

4

� �

� � � �

on γ, so the curve Γ = ψ(γ) lies in the disk |ζ − 1| < 1. Hence

1 ψ�(z) dζ dz = = n(Γ, 0) = 0

2πi γ ψ(z) Γ ζ

(book p.116). Now

1 g�(z)Ng = dz

2πi γ g(z)

1 ψ�f + ψf � = dz

2πi γ ψf

1 ψ�(z) 1 f �(z) = dz + dz

2πi γ ψ(z) 2πi γ f(z)

= Nf .

This proves the result. Q.E.D.

Exercise 2 p.154

We use Rouche’s theorem twice, first on γ : |z| = 2 and then on γ : |z| = 1. For γ : |z| = 2, take f(z) = z4, g(z) = z4 − 6z + 3. For γ : |z| = 1, take f(z) = −6z, g(z) = z4 − 6z + 3.

5

Lecture 15: Contour Integration and Applications

(Text 154-161)

Remarks on Lecture 15

In parts 4 and 5 (p. 154-160) some clarification of the use of the logarithm are called for.

Example 4 p.159

The relation 2πiα 2α(−z)2α = e z

which is crucial for proof deserves explanation.

We consider the function l logθ z = log |z| + iargθz

in the region C − lθ (the plane with the ray lθ removed) where the angle is fixed by

Fig. 15-1 θ < argθz < θ + 2π.

In the problem of computing

∞

x αR(x) dx 0

we consider log−π (z)

2

1

in the plane C with the negative imaginary axis removed and us the Residue theorem on the contour in Fig. 4.13. As in the text we arrive at the integral

∞ ∞ (

z 2α+1R(z 2) dz = z 2α+1 + (−z)2α+1)

R(z 2) dz. −∞ 0

On the right z belongs to (0, ∞) and

π π −π 3π( )

log−π

2

(z) = log |z| + − i, + < arg−π 2

z < 2

,

2 2 2

π 3π log−π

2

(−z) = log |z| + − + i 2 2

log−π 2

(z) + iπ, z > 0.=

Thus for z > 0,

= e

(2α+1) log

− π 2

(−z)(−z)2α+1

(2α+1)(log |z|+iπ)= e2αiπ 2α+1 = −e z ,

so the last integrals combine to ∞

2αiπ) 2α+1(1 − e z R(z 2) dz. 0

For z > 0 we have from the above

log−π 2

(z) = log |z|,

so ∞ ∞1 2α+1 2)

1 2αiπ) 2α+1 2)z R(z dz = (1 − e x R(x dx 2πi −∞ 2πi

0 (1) ∞ = −

1 e απi sin πα x 2α+1R(x 2) dx.

π 0

The left hand side of (1) is the sum of the residues of

z 2α+1R(z 2) = f(z)

in the upper half plane. If g(z)

R(z 2) = , h(z)

where g and h are holomorphic, g(a) 0, and h has a simple zero at = a, then

(2α+1)(a)g(a)

Resz=af(z) = z , (2) h ′(a)

where 2α+1 (2α+1) log

− π 2

(z) .z = e

2

Example: Exercise 3(g) p.161

To calculate ∞ dx

,21 + x

dz 5

1

3x 0

2we use x = t and arrive at ∞

3z 41 + z−∞

in (1). The poles in the upper half plane are

π

4 π

2

π

4

i( )i +and z = e z = e .

We use (2) to calculate the residues:

1 153

5

3

π

4

( )

iRes π 4i

= zz e π 44 i )31 +z=e 4(ez

π

4i 153 (e )log− π 2= e π

4i )34(e 15

3

π

4)(i= e π 4i )34(e

1 −i= e π

3,

4

and

1 1( )53

5

3

π

4i3

eRes π 4i3

z=e z = z

π

44(e i341 + z )3

1 π

4

π

4

5

4(e i3

1 −iπ = e . 4

Thus (1) gives

5

3

π

2(i(− )) log +π 2−= e )3

∞1 1 1 π dx π 3

1 5

−i −iπ πi sin = −+ 3 3e e e x ,

4 4 π 3 1 + x0 so

∞ dx π = √

5

3x . 0 1 + x

4 2 3

3

4

∫

∫

∫

∫

)

)

(

(

Example 5 p.160

The last four lines on the page are a bit misleading because the specific logarithm has already been chosen. So here is a completion of the proof after the equation

π

Log(−2ieix sin x) dx = 0. 0

We know (Lecture 2) that

Log(z1z2) = Logz1 + Logz2, if − π < Argz1 + Argz2 < π. (3)

Using this for z = 2 sin x we get

π π

log(2 sinx) dx + Log(−ieix) dx = 0. (4) 0 0

But πi ix Log(−i) = − , Loge = ix (0 < x < π),2

so since −π 2 + x is in (−π, π), (3) implies

Log(−ieix) = − πi

+ ix. 2

Now (4) implies the result

π

log sin θ dθ = −π log 2. 0

4

∫

∫ ∫

∫

� �

Lecture 16: Harmonic Functions

(Replacing Text 162-170)

While integrals like ϕ f(z) dz and ϕ M dx+N dy have been defined in the text

(p.101), differential forms like dx, dy and dz = dx + i dy have not been defined (and the definition is more subtle), we shall develop the theory of harmonic functions (p.162-170) without differential forms.

Definition 1 A real-valued function u(z) = u(x, y) in a region � is harmonic if it is C2 and satisfying the equation

�2u �2u + = 0.

�x2 �y2

The Cauchy-Riemann equations for a holomorphic function imply quickly that the real and imaginary parts of a holomorphic function are harmonic. The converse holds if � is simply connected:

Theorem 1 If � is simply connected and u harmonic in �, there exists a holomorphic function f(z) such that

u(z) = Ref(z).

Remark: Note the condition � is simply connected can not be removed, for example u(z) = log |z| is harmonic in the punctured plane C − {0}, but it cannot be written as real part of a holomorphic function.

Proof: Put �u �u

g(z) = �x

− i �y

= u1 + iv1.

Then �u1 �

2u �2u �v1 �x

= �x2

= − �y2

= �y

,

�u1 �2 u �v1

�y =

�x�y = −

�x .

1

�

� �

So by the Cauchy-Riemann equation, g is holomorphic. By p.142, since � is simply connected,

g(z) = f �(z)

for some holomorphic function f . Writing

f(z) = U(x, y) + iV (x, y),

we have by the Cauchy-Riemann equation

�U �U g(z) = f �(z) =

�x − i

�y ,

so u(x, y) = U(x, y) + constant.

Thus u(z) = Ref(z) + constant.

Q.E.D.

Corollary 1 (cf. (34) p.134) If u is harmonic in �, then if the disk |z − z0| � r lies in �,

1 �

2�

u(z0) = u(z0 + re i�) dα.

2ϕ 0

More generally, if the annulus r1 � |z − z0| � r2 belongs to a region �, we have

Theorem 20 If u is harmonic in �, and {z : r1 � |z − z0| � r2} ≥ �, then

1 2� u(z0 + re

i�) dα = � log r + ∂, r1 � r � r2, (1)2ϕ 0

where � and ∂ are constants.

Proof: The function z √� u(z0 + z) is harmonic, so writing the Laplacian in polar coordinates,

�2 1 � 1 �2 � = + + .

�r2 r �r r2 �α2

Denote the left hand side of (1) by V (r), then

�2V 1 �V + = 0.

�r2 r �r

Writing this as � �V

r = 0, �r �r

the theorem follows. Q.E.D.

2

�

The Poisson Formula

Let u be harmonic on |z| � 1. Then

u = Re(f)

where f is holomorphic on |z| � 1. Consider

z + a S(z) =

1 + āz , (|a| < 1)

which maps the unit disk onto itself. Then f ∞S is holomorphic and u∞S is harmonic (the real part of f ∞ S). Use the corollary on it with z0 = 0, then

1 �

2�

u(a) = u(S(0)) = u(S(e i�)) d�. 2ϕ 0

But i�S(e i�) =

ei� + a = e ,

1 + āei�

so i�

i� e − a e = . 1 − ¯ i�ae

Hence

iei� d�

= iei� − |a|2iei�

,dα (1 − ¯ i�)2ae

or d� iei� a 2iei� 1 ae i�

= − | |

i�)2 1

i�

− ¯dα (1 − ¯

· · e − a

(2) ae i a

=1 i�

− | |2

2 .

|e − a|

This gives

Poisson’s Formula ((63) in text)

i�)u(a) = 1

� 2�

u(e d�

dα =1 1 − |a|2

u(z) dα. 2ϕ 0 dα 2ϕ |z − a|2 |z|=1

3

�

� �

�

Schwarz� Theorem

Theorem 2 (Schwarz� Theorem) Let U be a real piecewise continuous function on z = 1 and define the Poisson integral u(z) = PU (z) by| |

u(a) = 2

1 ϕ 0

2�

a 1 −

− e

|ai�

|2

2 U(e i�) d�, |a| < 1. (3)

| |

Then u is harmonic, and lim u(z) = U(e i�0 )

z�ei�0

if U is continuous at e i�0 .

Proof: We may assume �0 = 0. Since

1 − 2 �

i� + z �

|z|= Re

e,

|z − ei�|2 ei� − z

u is the real part of a holomorphic function, hence harmonic.

Because of (2) formula (3) can be written

1 �

2�

u(S(0)) = U(S(e i�)) d�. 2ϕ 0

Taking a = tanh t we obtain as t � →

1 �

2� ei� + tanh t u(tanh t) = U d�

2ϕ 0 tanh tei� + 1

1 2� −�

2ϕ 0 U(1) d�

= U(1).

Q.E.D.

4

�

� � �

� �

� � � �

Exercise 5, p.171

Since log 1 + z is harmonic in z < 1 we have by the mean-value theorem | | | |

1 �

2ϕ log |1 + re i� | dα = log 1 = 0 (4)

−�

for r < 1. We shall now show that

�log 1 + re i� | |

is bounded by an integrable function g(α). So by the dominated convergence theorem we can let r � 1 under the integral sign, giving the desired result

−� log |1 + e i� | dα = 0. (5)

Since the integrand log 1 + e i� changes sign on the circle, we split the circle 2� 2�

| | 4�into the two arcs (− , ) and (2� , ), where we have

3 3 3 3

|1 + e i� | � 1

and |1 + e i� | � 1

respectively. In the first interval we have cos α � − 1 so 2

≤3 i� i� α 2ϕ 1

2 � |1 + re | � |1 + e | = 2 cos

2, |α| �

3 , and r �

2 . (6)

In the second interval we put α = ϕ +� and we see from the geometry, since |�| � � ,3

that

α 2ϕ 4ϕ 1 � |1 + re i� | = |1 − re i� | � 1 − cos � = 2 cos2

2,

3 � α �

3 . (7)

α i�Since log �

�cos 2 �� is integrable, the estimates (6) and (7) show that | log |1 + re || is

bounded by an integrable function g(α), so (5) is established.

5

� �

� � � �

� �

� �

�

Lecture 17: Mittag-Leffer’s Theorem

(Text 187-190)

Theorem 1 (Mittag-Leffer’s Theorem) Let {bν} be a sequence in C such that

lim bν = ∞, ν→∞

and Pν(ζ) polynomials without constant term. Then there exist functions f meromorphic in C with poles at just the points bν and corresponding singular parts

1 Pν .

z − bν

The most general f(z) of this kind can be written

1 f(z) = g(z) + Pν − pν(z) (1)

z − bνν

where g is holomorphic in z and the pν are polynomials.

1 Proof: We may assume all bν = 0. Consider the Taylor series for Pν

z − bν around z = 0. It is analytic for |z| < |bν |. Let pν(z) be the partial sum up to z

nν

(nν to be determined later). Consider the finite Taylor series of

1 ϕ(z) = Pν

z − bν

in a disk D with center 0. By (29) on p.126,

1 ϕ(ζ)ϕn(z) = dζ.

2πi C ζn(ζ − z)

1

� �

� � � �

� �

� � � �

� � � �

� �

|bν |Taking C as the circle with center 0 and radius and n = nν + 1 we deduce

2

1 |bν | Mν |bν ||ϕnν+1(z)| ≤ 2π for |z| ≤ ,2π 2 (1 |bν |)nν+1 ·

|bν | 4 2 4

where

1 Mν = max Pν . z∈C z − bν

Thus by Theorem 8 on p.125,

nν+1 1 2|z| |bν | Pν − pν(z) ≤ 2Mν for |z| ≤ . (2) z − bν |bν | 4

We now select nν large enough so that

2nν ≥ Mν2ν .

Then 2|z|

nν+1 |bν |≤ 2−ν2Mν for |z| ≤ .

|bν | 4

We claim now that the sum (1) converges uniformly in each disk |z| ≤ R (except at the poles) and thus represents a meromorphic function h(z). To see this we split the sum in (1):

1 1 h(z) =

|bν | 4 ≤R

Pν z − bν

− pν(z) + |bν | 4 >R

Pν z − bν

− pν(z) . (3)

Because of (2), the second sum is holomorphic for |z| ≤ R since R ≤ |bν | 4 . The

first sum is finite and has 1

Pν z − bν

as the singular part at the pole bν .

This proves the existence. If f is any other meromorphic function with these properties, then f(z)−h(z) is holomorphic. Q.E.D.

2

( )

)( )

)

) )))

)

(

((

(

((

(

∑ ∑

�

Exercise 3 on p.178

Here we need some preparation on series of the form

∞

anvn n=1

and use on an = (−1)

n , vn = (1 + n)−s , s = σ + it.

We have if An = a0 + · · · + an,

then N N−1

A0v0 + (An − An−1)vn − An(vn − vn+1) = ANvN . n=1 n=0

Lemma 1 If (An) is bounded, vn → 0, and

∞

|vn − vn+1| < ∞, n=1

∞ L

then anvn converges. n=0

This is obvious from the identity above.

In our example,

vn = |(1 + n)−s| =

1 ,

(1 + n)σ

so vn → 0 even uniformly on compact subsets of Res > 0. For vn − vn+1 we have

1 1 n+2 vn − vn+1 = − = s x

−s−1 dx, (n + 1)s (n + 2)s n+1

so 1

|vn − vn+1| ≤ |s| . (n + 1)σ+1

Thus ∞

(−1)n−1 1

ns n=1

converges, and actually uniformly on compact sets in the region σ > 0 because this L

is the case with vn → 0 and |vn − vn+1|.

3

∑

∑∑

∑

∑

∫

Lecture 18: Infinite Products

(Text 191-200)

Remarks on Lecture 18

Problem 1 on p.197: Suppose that an → ∞ (all different, a condition missing in text) and An arbitrary complex numbers. Show that there

exists an entire function f(z) which satisfies f(an) = An.

Proof: (A simpler alternative to the hint in text). Let g(z) be an analytic function with simple zeros at the an. By the Mittag-Leffler theorem, there exists a meromorphic function h on C with poles exactly at the points an with the corresponding singular part

An/bn

z − an , g(z) = (z − an)k(z) , k(an) = bn �= 0 .

Then f(z) = g(z)h(z)

has the desired property.

Q.E.D.

Remarks on the formula for π cotπz (line 8 p.197)

Since the product formula for sin πz has infinitely many factors taking the logarithmic derivative requires justification. Generally, write

∞ N I I

f(z) = fn(z) = lim fn(z) = lim gN(z)N→∞ N→∞

1 1

the convergence being uniform on compacts. By Theorem 1,

f ′ (z) = lim g ′ (z)NN→∞

so ′f ′(z) gN(z) = lim .

f(z) N→∞ gN(z)

1

�

� �

� �

� �

�

Here g ′ (z)/gN (z) is given by the rule for differentiating a product. NThis remark justifies to proof of (27) as well.

In the text the Gamma function is defined by means of the product formula (29) in §2.4 and the integral formula (42) derived by an interesting residue calculus due to Lindelöf. Here we go a shorter way and derive the product formula from the definition in terms of the integral formula.

The Gamma function can be defined by ∞

tz−1Γ(z) = e −tdt Rez > 0. 0

Writing n

t−z−1fn(z) = e −tdt

0

fn is holomorphic and ∞ ∞

� tz−1 Rez−1|Γ(z) − fn(z)| ≤ e −tdt ≤ t e −tdt

n n

which → 0 uniformly in each half plane Rez > δ (δ > 0). Thus Γ(z) is holomorphic in Rez > 0. Here are some of its properties (i) Γ(z + 1) = zΓ(z). This follows by integration by parts. (ii) Γ(z) extends to a meromorphic function on C with simple poles at z = 0,−1,−2, . . .. The function

Γ(z + 1) H(z) =

z

is meromorphic in Rez > −1 with a pole at z = 0. Since

lim zH(z) 0 z→0

the pole is simple. The residue is Γ(1) = 1. Also H(z) = Γ(z) for Rez > 0. Thus Γ(z) is meromorphic in Rez > −1 with simple pole at z = 0. Statement (ii) follows by repetition. (iii) For x > 0, y > 0

1 Γ(x)Γ(y)tx−1(1 − t)y−1dt = .

0 Γ(x + y)

r ′ Γ(z)Γ(w)Extend to Re z > 0 , Rew > 0 tz−1(1 − t)w−1 dt = 0 Γ/2+w

Proof: ∞ ∞

tx−1Γ(x)Γ(y) = e −tdt sy−1 e −sds 0 0

2

6=

�

�

�

�

�

�

�

�

�

�

�

�

Put s = tv. Since integrands are positive, integrals can be interchanged. We get ∞ ∞

tx−1 y−1ty −tvdv Γ(x)Γ(y) = e −tdt v e 0 0

u tx+y−1 −(v+1)tdt =

∞

vy−1dv ∞

e t = 1 + v0 0

x+y−1 = ∞

vy−1dv ∞

u e −u(1 + v)−x−ydu 0 0

∞ y−1 1 v= Γ(x + y) dv = Γ(x + y) s x−1(1 − s)y−1ds (1)

0 (1 + v)x+y

0

the last expression coming from v = s−1(1 − s). This proves (iii), and it extends to Re z > 0, Re w > 0. (iv) Γ(z)Γ(1 − z) = π

sinπz

From (1) we obtain ∞ −xv

Γ(x)Γ(1 − x) = dv 1 + v0

which evaluates to π/ sin(πx) by the method of Exercise 3(g) p. 161, done in Lecture 15. This proves (iv) by meromorphic continuation.

Since the poles of Γ(z) are canceled by zeros of sin πz, Γ(1 − z) is never 0. By (iii) we have for 0 < h < x

2 z = x + iy

Γ(z − h)Γ(h) 1 = (1 − t)z−h−1th−1 dt

Γ(z) 0 1

= 1

+ [(1 − t)z−h−1 − 1]th−1 dt . h 0

In the integral we use the dominated convergence theorem to justify letting h → 0 under the integral sign. In the interval [1

2, 1] there is no problem bounding the

< x 1integrand uniformly for h 2. On the interval [0,

2] we have (with α = z − h − 1)

h−1 (1 − t)α − 1

((1 − t)α − 1)t ≤ t

and by l‘Hospital’s rule this has limit |α| = |z−h−1| ≤ |z|+2 so again the integrand is bounded. Thus we let h → 0 and obtain

Γ(z − h)Γ(h) 1 1 = + [(1 − t)z−1 − 1]t−1 dt + o(1) as h → 0 .

Γ(z) h 0

The left hand side is using Taylor for h → Γ(z − h) and Laurent for h → Γ(h) both at h = 0

1 1 (Γ(z) − hΓ ′ (z) + · · · ) + A + Bh · · ·

Γ(z) h

3

∫∫

∫∫

∫ ∫

∫∫

∫

∫

∫

∫

∫

∫

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

where { } is the Laurent series for Γ(h) with center h = 0. Equating the constant terms on left hand side and right hand side we get

Γ′(z) 1 = (1 − (1 − t))z−1t−1 dt − A x > 0 .

Γ(z) 0 L

Writing t−1 = ∞ 0

(1 − t)n the expression is

1 L

[(1 − t)n − (1 − t)n+z−1] dt− A , 0

0

and since expression in [ ] equals (1−t)n (1−(1−t)z−1 )

t which is bounded by (1− t)nKt,t

rL

with integral K/(n + 1)(n + 2) we can exchange and n by the dominated convergence theorem. Thus our expression equals

1 L

[(1 − t)n − (1 − t)n+z−1] dt − A 0

0

∞ ∞ L L1 1 1 1 1

= − − A = 1 − + − − A n + 1 n + z z n + 1 n + z

0 1

∞ ∞ L L1 1 1 1 1

= − − + − − A n n + 1 z n + 1 n + z

1 1

∞ L1 1 1

= − + − − A z n n + z

1

so

∞ LΓ′(z) 1 1 1

= + = − − A . Γ(z) z n n + z

1

Having justified taking logarithmic derivative of an infinite product this gives

I

( )1 ∞

zCz − n= ze 1 + e z

C = const . Γ(z) n

1

Putting z = 1 we have

∞ I 1C − 1

n1 = e 1 + e n

1

so

( )

1C −(1+ 1 +··· N )21 = e lim (N + 1)e

N→∞

so

4

∫

∫

∫

Lecture 19: Normal Families

(Replacing Text 219-227)

Theorem 1 Let � → C be a region, F a family of holomorphic functions on � such that for each compact E → �, F is uniformly bounded on E. Then F has a subsequence converging uniformly on each compact subset of �.

First we prove that on each compact subset E → �, the family F is equicontinuous. This means, given π > 0 there exists a � > 0 such that for all f ≤ F ,

|f(z �) − f(z ��)| < π if |z � − z ��| < �, z � , z �� ≤ E. (1)

The distance function x � d(x, C − �) is continuous and has a minimum > 0 on the compact set E. Let d > 0 be such that (D denoting disk) F =

�

D(x, 2d)x�E ¯has closure F → �.

Let z�, z�� ≤ E satisfy |z � − z ��| < d

and let ζ denote the circle ζ : |z − z �| = 2d.

Then ζ → F̄ and z� and z�� are both inside ζ. Also |� − z�| = 2d, |� − z��| � d for

� ≤ ζ.

By Cauchy’s formula for f ≤ F ,

z� − z�� �

f(�)f(z �) − f(z ��) = d�,

2�i � (� − z�)(� − z��)

so if M(F̄ ) is the maximum of f on F̄

M(F̄ )|f(z �) − f(z ��)| � |z � − z ��| .

d

Hence (1) follows.

To conclude the proof of Theorem 1 choose any sequence (zj) which is dense in �. Let fm be any sequence in F . The sequence fm(z1) is bounded so fm has

1

a subsequence fm,1 converging at z1. Form this take a subsequence fm,2 which converges at z2. Continuing we see that the subsequence fm,m converges at each zj .

¯By the first part of the proof, F is equicontinuous on the compact set F . Given ¯π > 0 there exists a � < d such that (1) holds for all z � , z �� ≤ F , f ≤ F . If z ≤ E the

disk D(z, �) contains some zj so D(zj , �) contains z.

By the compactness of E,

p

E → �

D(zi, �) i=1

for some z1, · · · , zp. Thus given z ≤ E there exists a zi = zi(z) such that |z −zi(z)| < ¯ ¯�. Then zi(z) ≤ F . Thus by (1) for F ,

|f(z) − f(zi(z))| < π. f ≤ F . (2)

There exists N > 0 such that

|fr,r(zi) − fs,s(zi)| < π 1 � i � p, r, s > N. (3)

Given z ≤ E we have with zi = zi(z)

|fr,r(z) − fs,s(z)| � |fr,r(z) − fr,r(zi)| + |fr,r(zi) − fs,s(zi)| + |fs,s(zi) − fs,s(z)|

� 3π by (2) and (3).

The proves the stated uniform convergence on E.

Remark: In the text, p. 223, it is erroneously assumed (and used) that �k ≤ E. This error occurs in many other texts.

2

Lecture 20: The Riemann Mapping Theorem

(Text 229-231)

Remark: The first step is to show that there exists an analytic and univalent function f , mapping Ω into the unit disk D satisfying f(z0) = 0 and f �(z0) > 0. In order to expand f(Ω) we take f �(z0) as large as possible. This indeed accomplishes f(Ω) = D.

The proof concludes with the contradiction

|G�(z0)| > B = |f �(z0)|.

After the proof it is indicated in the text that this contradiction |f �(z0)| < |G�(z0)|is a consequence of Schwarz’s lemma by writing

f(z) = H(W ), W = G(z).

However, H is then only defined on G(Ω), so until we know that G(Ω) = D Schwarz’s lemma (p.135) does not apply.

Thus the phrase “a consequence of Schwarz’s lemma” should properly read “consistent with Schwarz’s lemma”, which probably was the author’s intention.

Koebe’s proof from 1915 is outlined in Pólya and Szegö, Vol. II, §2. This is based on considering the distance of the origin to the boundary of the image region f(Ω). The square root map is then used to increase this distance. By iteration this leads to an explicit sequence converging to the desired limit map.

1

Lecture 21 and 22: The Prime Number Theorem

(New lecture, not in Text)

The location of prime numbers is a central question in number theory. Around1808, Legendre offered experimental evidence that the number π(x) of primes < xbehaves like x/ log x for large x. Tchebychev proved (1848) the partial result thatthe ratio of π(x) to x/ log x for large x lies between 7/8 and 9/8. In 1896 Hadamardand de la Valle Poussin independently proved the Prime Number Theorem that thelimit of this ratio is exactly 1. Many distinguished mathematicians (particularlyNorbert Wiener) have contributed to a simplification of the proof and now (by animportant device by D.J. Newman and an exposition by D. Zagier) a very short andeasy proof is available.

These lectures follows Zagier’s account of Newman’s short proof on the primenumber theorem. cf:

(1) D.J.Newman, Simple Analytic Proof of the Prime Number Theorem, Amer.Math. Monthly 87 (1980), 693-697.

(2) D.Zagier, Newman’s short proof of the Prime Number Theorem. Amer.Math. Monthly 104 (1997), 705-708.

The prime number theorem states that the number π(x) of primes which arex

less than x is asymptotically like :log x

π(x)−→ 1 as x → ∞.

x/ log x

Through Euler’s product formula (I) below (text p.213) and especially throughRiemann’s work, π(x) is intimately connected to the Riemann zeta function

∑

∞1

ζ(s) = ,ns

n=1

which by the convergence of the series in Res > 1 is holomorphic there.

1

The prime number theorem is approached by use of the functions

log pΦ(s) =

∑

, V(x) = log p.ps

p prime p≤x

∑

prime

1Simple properties of Φ will be used to show ζ(s) = 0 and Φ(s)− holomorphic

s − 1for Res ≥ 1. Deeper properties result from writing Φ(s) as an integral on whichCauchy’s theorem for contour integration can be used. This will result in the relationV(x) ∼ x from which the prime number theorem follows easily.

∞1

I =∏

(1 − p−sn

) for Res > 1.ζ(s)

1

∞

Proof: For each n (1 − p−sn )−1 =

∑

p−msn . Putting this into the finite productm=0

∏

N N ∞

(1 − p−s1 )−1 we obtain

∏

(1 − p−s 1n )− = n

1 1

∑

−sk . Now let N → ∞.

k=1

1II ζ(s) − extends to a holomorphic function in Res > 0.

s − 1

Proof: In fact for Res > 1,

∞1

ζ(s) − =s − 1

∑ 1 ∞−

ns

∫

dx

1 xs

n=1∞

=∑

n=1

∫ n+1

n

(

1 1−

ns xs

)

dx

But∣

∣ 1 1 x dy s s∣ −

∣

∣

∣ =∣

∣∫

∣

∣

∣ s ∣ ≤ max ≤ ,ns xs ∣ ∣ ys+1

∣

∣

∣ ∣

∣

∣ys+1 R≤x

∣

sn≤

∣

n e +1yn

so the sum above converges uniformly in each half-plan

∣

∣

e Res ≥ δ (δ > 0).

III V(x) = O(x) (Sharper form proved later).

Proof: Since the p in the interval n < p ≤ 2n divides (2n)! but not n! we have

2n

22n = (1 + 1)2n =∑

k=0

(

2n

k

)

≥

(

2n

n

)

≥n<

∏

p = eV(2n)−V(n),p≤2n

6

2

ThusV(2n) − V(n) ≤ 2n log 2. (1)

xIf x is arbitrary, select n with n < ≤ n + 1, then

2

V(x) ≤ V(2n + 2) ≤ V(n + 1) + (2n + 2) log 2 (by (1))

= V(x

+ 1)

+ (x + 2) log 22

= V(x)

+ log(x

+ 12 2

)

+ (x + 2) log 2.

Thus if C > log 2,

V(x) − V(x)

≤ Cx for x ≥ x0 = x0(C). (2)2

Consider the points

x x x x x x0 rr+1 r-1

2 2 2 2

Use (2) for the points right of x0,

V(x

2

)

− V( x

22

) x≤ C ,

2...

xV

(

2r

)

− V( x

2r+1

) x≤ C .

2r

Summing, we get

xV(x) − V(x0) ≤ V(x) − V

(

2r+1x

)

≤ Cx + · · ·+ C ,2r

soV(x) ≤ 2C(x) + O(1).

3

x x0

x

2r+1

x

2r-1

x

2r

x

2

61

IV ζ(s) = 0 and Φ(s) − is holomorphic for Res ≥ 1.s − 1

Proof: If Res > 1, part I shows that ζ(s) = 0 and

ζ ′(s) log p− =

∑ log p= Φ(s) + .

ζ(s) ps − 1

∑

(3)ps(ps − 1)

p p

1The last sum converges for Res > , so by II, Φ(s) extends meromorphically to

21

Res > with poles only at s = 1 and at the zeros of ζ(s). Note that2

ζ(s) = 0 =⇒ ζ(s̄) = 0.

Let α ∈ R. If s0 = 1 + iα is a zero of ζ(s) of order µ ≥ 0, then

ζ ′(s) µ− = − + function holomorphic near s0.

ζ(s) s − s0

Solim ǫΦ(1 + ǫ + iα) = −µ.ǫ→0

We exploit the positivity of each term in

lΦ(1 + ǫ) =

∑ og p

p1+ǫp

for ǫ > 0. It implies∑ log p i

p1+ǫ

(

p+α

2 + p−iα

2

p

)2

≥ 0,

soΦ(1 + ǫ + iα) + Φ(1 + ǫ − iα) + 2Φ(1 + ǫ) ≥ 0. (4)

By II, s = 1 is a simple pole of ζ(s) with residue +1, so

lim ǫΦ(1 + ǫ) = 1.ǫց0

Thus (4) implies−2µ + 2 ≥ 0,

soµ ≤ 1.

6

4

This is not good enough, so we try

∑ log p ( + iα − iα 4p 2 + p 2p1+ǫ

p

)

≥ 0.

Puttinglim ǫΦ(1 + ǫ ± 2iα) = −ν,ǫց0

where ν ≥ 0 is the order of 1 ± 2iα as a zero of ζ(s), the same computation gives

6 − 8µ − 2ν ≥ 0,

1which implies µ = 0 since µ, ν ≥ 0. Now II and (3) imply Φ(s)− holomorphic

s − 1for Res ≥ 1.

∫ ∞ V(x) − xV

2dx is convergent.

1 x

Proof: The function V(x) is increasing with jumps log p at the points x = p. Thus

Φ(s) =∑ log p

psp

= s

∫ ∞ V(x)dx

1 xs+1

∞ pIn fact, writing

∫

as∑

i

∫

i+1

pithis integral becomes

1

∑∞ 1 1 −1i=1 V(pi) ps −

ips

si+1

which by V(pi+1) − V(pi) = log pi+1 reduces to Φ(s). Using the subs

(

titution x

)

= et

we obtain∞

Φ(s) = s

∫

e−stV(et) dt Res > 1.0

Consider now the functions

f(t) = V(et)e−t − 1,

Φ(z + 1) 1g(z) = − .

z + 1 z

f(t) is bounded by III and we have

∫ eT V(x) − x Tdx =

21 x

∫

f(t) dt . (5)0

5

Also, by IV,1

Φ(z + 1) = + h(z),z

where h is holomorphic in Rez ≥ 0, so

Φ(z + 1) 1 h(z) − 1g(z) = − =

z + 1 z z + 1

is holomorphic in Rez ≥ 0.

For Rez > 0 we have

g(z) =

∫ ∞ ∞

e−zt(f(t) + 1) −0

∫

e−zt dt0

∞

=

∫

e−ztf(t) dt.0

Now we need the following theorem:

Theorem 1 (Analytic Theorem) Let f(t) (t ≥ 0) be bounded and locally inte-grable and assume the function

g(z) =

∫ ∞

e−ztf(t) dt Re(z) > 00

extends to a holomorphic function on Re(z) ≥ 0, then

T

lim

∫

f(t) dtT→∞ 0

exists and equals g(0).

This will imply Part V by (5). Proof of Analytic Theorem will be given later.

VI V(x) ∼ x.

Proof: Assume that for some λ > 1 we have V(x) ≥ λx for arbitrary large x. SinceV(x) is increasing we have for such x

∫ λx V(t) − tdt ≥

∫ λx λx − t − sdt =

∫ λ λds = δ(λ) > 0.

t2 x t2

1 s2

x

On the other hand, V implies that to each ǫ > 0, ∃K such that

∣

∣

∫ K2 V(x) − xdx

K1 x2

∣

∣

< ǫ for K1, K2 > K.∣∣

∣

∣

6

Thus the λ cannot exist.

Similarly if for some λ < 1, V(x) ≤ λx for arbitrary large x, then for t ≤ x,

V(t) ≤ V(x) ≤ λx,

so∫ x V(t) − t λ 1 −

≤

∫ x x − t∫

λ s= ds = δ(λ) < 0.

2λx t λx t

2λ s

2

Again this is impossible for the same reason. Thus both

V(x)β = lim sup > 1

x→∞ x

andV(x)

α = lim inf < 1x→∞ x

are impossible. Thus they must agree, i.e. V(x) ∼ x.

Proof of Prime Number Theorem:

We haveV(x) =

∑

log p ≤∑

log x = π(x) log x,p≤x p≤x

soπ(x) log x V(x)

lim inf ≥ lim inf = 1.x→∞ x x→∞ x

Secondly if 0 < ǫ < 1,

V(x) ≥x1−

∑

log pǫ≤p≤x

≥ (1 − ǫ)x1−

∑

log xǫ≤p≤x

= (1 − ǫ) log x(

π(x) + O(x1−ǫ))

thusπ(x) log x 1 V(x)

lim sup ≤ lim supx→∞ x 1 − ǫ x→∞ x

for each ǫ. Thusπ(x) log x

lim = 1.x→∞ x

Q.E.D.

7

Proof of Analytic Theorem: (Newman).

Put

gT (z) =

∫ T

e−ztf(t) dt,0

which is holomorphic in C. We only need to show

lim gT (0) = g(0).T→∞

Fix R and then take δ > 0 small enough so thatg(z) is holomorphic on C and its interior.

|z| = R

δ 0

C

By Cauchy’s formula

g(0) − gT (0) =

1 z2 dz(g(z) − gT (z)) e

zT 1 + . (6)2πi

∫

C

(

R2

)

z

On semicircle

C+ : C ∩ (Rez > 0)

2Bintegrand is bounded by , where

R2

B = sup |f(t)|.t≥0

In fact for Rez > 0,

∞

|g(z) − gT (z)| =

∣

∣

∫

∣ f(t)e−zt dtT

∣

∣

∞

≤

∣

∣

B

∣

∫

|e−zt| dtT

Be−RezT=

Rez

and∣

∣

ezT(

z21 +

R2

)

1

z

∣

∣

∣

R z∣ = e ezT

2Re· (z = Reiθ).

R2∣ ∣

8

BSo the contribution to the integral (6) over C+ is bounded by , namely

R

Be−RezT T z B· eRez2Re 1

· · πR = .Rez R2 2π R

Next consider the integral over

C_ = 0.

Look at g(z) and gT (z) separately. For gT (z) which is entire, this contour can bereplaced by

C'_ =0.

BAgain the integral is bounded by because

R

T

|gT (z)| =

∣

∣

∫

∣ f(t)e−ztdt∣

0

∣

∣

∣

≤ B

∣

∫ T

|e−zt| dt0T

≤ B

∫

|e−zt| dt−∞

Be−RezT=

|Rez|

9

and∣

∣

(

z21 +

R2

)

1

z

∣

∣

on C ′ has the same estimate as bef

∣

∣

ore.

∣

∣

There remains∫

1ezT g(z) d .

C

(

z21 + z

R2

)

z︸

indep︷

.︷

of T︸

C_ =δOn the contour, |ezT | ≤ 1 and

lim |ezT | → 0 for Rez < 0.T→∞

By dominated convergence, the integral → 0 as T →+∞, δ is fixed. It follows that

2Blim sup |g(0) − gT (0)| ≤ .

T→∞ R

Since R is arbitrary, this proves the theorem.

Q.E.D.

Remarks: Riemann proved an explicit formula relating the zeros ρ of ζ(s) in 0 <Res < 1 to the prime numbers. The improved version by von Mangoldt reads

V(x) =∑

log pp≤x

= x −∑ xρ x−2

+ρ

∑n

− log 2π.2n

{ρ} n≥1

1He conjectured that Reρ = for all ρ. This is the famous Riemann Hypothesis.

2

10

C_ =δ

Lecture 23: The Extension of ζ(s) to the whole plane and the Functional Equation

(Text 214-217)

See Riemann’s Collected Works p. 146. In Theorem 10 we consider the function z → (−z)s−1 with z outside the positive real axis R+ . Angles are measured from the positive real axis from −π to +π. Consider contour C.

If z is on the upper part of the cut R+ , −z is below the negative real axis so

s−1 −(s−1)πi arg(−z) = −π so (−z)s−1 = x e .

If z is on the lower par of the cut R+ then −z is above the negative real axis so s−1 (s−1)iπ arg(−z) = +π so (−z)s−1 = x e .

1

page18122syllabuslecture20