lecture notes in complex analysis jyh-haur teh · lecture notes in complex analysis jyh-haur teh...

Lecture notes in complex analysis

Jyh-Haur Teh

Department of Mathematics, National Tsing Hua University of TaiwanEmail address : [email protected]

Contents

Chapter 1. Main structure 11.1. A map of theorems 1

Chapter 2. Complex numbers 32.1. Complex numbers 32.2. Cubic equations 4

Chapter 3. Analytic polynomials 73.1. Analytic polynomials 73.2. FAQ 9

Chapter 4. Power series 114.1. Power series 11

Chapter 5. Analytic functions 155.1. Analytic functions 15

Chapter 6. Complex line integrals 216.1. Complex line integrals 216.2. FAQ 246.3. Fundamental theorem of line integrals 246.4. Rectangle theorem 256.5. Integral theorem 286.6. Closed curve theorem 29

Chapter 7. Cauchy integral formula 31

Chapter 8. Maximum principle and open mapping theorem 39

Chapter 9. Morera’s theorem 49

Chapter 10. Simply connected regions 53

Chapter 11. Singularities 57

Chapter 12. Laurent expansions 61

Chapter 13. Residues 67

Chapter 14. Applications in integration: I 75

Chapter 15. Log and the argument principle 81

v

Chapter 16. Applications in integration: II 89

Chapter 17. The Riemann mapping theorem 95

Chapter 18. The Riemann zeta function 101

Glossary of mathematical terms 105

vi

CHAPTER 1

Main structure

1.1. A map of theorems

ML-formula Fundamental theorem of complex line integrals

Rectangle theorem

Integral theorem

Closed curve theorem

Cauchy integral formula

Liouville’s theorem Fundamental theorem of algebra

Extended Liouville’s theorem

Mean value theorem

Maximum principle

Minimum principle

Open mapping theorem Schwarz’s lemma

Power series expansion

Morera’s theorem

Laurent series

Residue theorem Argument principle Rouche’s theorem

Generalized Cauchy’s integral formula

Riemann mapping theorem Hurwitz’s theorem

Classification of isolated singularities

Analytic extension theorem

Casorati-Weierstrass theorem

1

CHAPTER 2

Complex numbers

2.1. Complex numbers

We recall the definition of a field.

Definition 2.1. A field (F,+, ·) is a set with two operations

+ : F × F → F

· : F × F → F

such that

(1) (F,+) is an abelian group;(2) (F − {0}, ·) is an abelian group where 0 is the additive identity of (F,+);(3) a · (b+ c) = a · b+ a · c for all a, b, c ∈ F .

Proposition 2.2. On R2, define

(x1, y1) + (x2, y2) := (x1 + x2, y1 + y2) (x1, y1) · (x2, y2) := (x1x2 − y1y2, x1y2 + x2y1)

then (R2,+, ·) is a field with additive identity (0, 0) and multiplicative identity (1, 0).

Proof. (R2,+) is clearly an abelian group and the distribution law is easy to check.Here we only find the multiplicative inverse of a nonzero element (x1, y1). Suppose that(x1, y1) · (x2, y2) = (1, 0) where (x1, y1) 6= (0, 0), then we have a system of equations{

x1x2 = 1 + y1y2

x1y2 = −x2y1

For the case x1 = 0, y1 6= 0, we have x2 = 0, y2 = − 1y1

; for the case x1 6= 0, we have

x2 =x1

x21 + y2

1

, y2 = − y1

x21 + y2

2

�

Definition 2.3. The field of complex numbers is

C := (R2,+, ·)Let i := (0, 1). Then a complex number (x, y) is uniquely expressed as x+ iy. If z = x+ iy,the real part of z is Rez := x, the imaginary part is Imz := y, the conjugate of z is z := x−iy,the norm of z is

|z| :=√x2 + y2 =

√zz

The distance between z1, z2 ∈ C is

d(z1, z2) := |z1 − z2|3

The argument for z 6= 0 is a real number Argz ∈ (−π, π] such that

cos(Argz) =Rez

|z|, sin(Argz) =

Imz

|z|Example 2.4.

Arg(i) =π

2, Arg(−1) = π, Arg(

√2− i) = −π

6

Example 2.5. Draw pictures of the following sets:

(1) A = {z : −π2< Argz < 2π

3}.

(2) B = {z : |z − (1 + 2i)| < 3}.

Remark 2.6. (1) As topological spaces, C and R2 are the same. In particular, C isa complete metric space.

(2) There are two natural operations

(x1, y1) + (x2, y2) := (x1 + x2, y1 + y2)

(x1, y1)× (x2, y2) := (x1x2, y1y2)

which makes R2 a commutative ring, but obviously not a field. So the algebraicstructures of R2 and C are totally different.

2.2. Cubic equations

For a general cubic equation

z3 + az2 + bz + c = 0

where a, b, c ∈ R, we may take z = x− a3, then

(x− a

3)3 + a(x− a

3)2 + b(x− a

3) + c = 0

Expanding each term, we get

x3 − 3x2(a

3) + 3x(

a

3)2 − a3

27+ ax2 − 2x(

a

3) + (

a

3)2 + bx− ab

3+ c = 0

the degree 2 terms are cancelled out, therefore we reduce the equation to the form

x3 + px+ q = 0

Letx = y − p

3yThen

(y − p

3y)3 + p(y − p

3y) + q = 0

Expanding each term, we get

y3 − py +p2

3y− p3

27y3+ py − p2

3y+ q = 0

Therefore

y3 + q − p3

27y3= 0

4

and

y6 + qy3 − p3

27= 0

Consider y3 as a variable, by the formula of the solutions of quadratic equations, we have

y3 =−q ±

√q2 + 4

27p3

2= −q

2±√

∆

where

∆ =q2

4+p3

27

Let

w+ = 3

√−q

2+√

∆, w− = 3

√−q

2−√

∆

Note that

w+w− =3

√q2

4− (

q2

4+p3

27) = −p

3

Let

ξ =−1 +

√3i

2

which is a solution of the equation z3 = 1. Then we see that

x0 := w+ − p

3w+= w+ + w−

x1 := w+ξ − p

3w+ξ= w+ξ + w−ξ2

x2 := w+ξ2 − p

3w+ξ2= w+ξ2 + w−ξ

are all the solutions of x.We have the following theorem.

Theorem 2.7. (General solution of cubic equations) If p, q ∈ R are not both zeros, thesolutions of the cubic equation

x3 + px+ q = 0

are

xk = ξk 3

√−q

2+√

∆ + ξ2k 3

√−q

2−√

∆

where k = 0, 1, 2 and ∆ = q2

4+ p3

27.

Remark 2.8. Three Greek problems of antiquity: Using only compass and straightedge:

(1) circle squaring,(2) cube duplication,(3) angle trisection.

5

Let θ := 20◦ and x = cos θ. By the formula

cos 3θ = 4 cos3 θ − 3 cos θ

we have1

2= 4x3 − 3x

and hence

x3 − 3

4x− 1

8= 0

Then

∆ = − 3

256,√

∆ =

√3

16i

The solutions of the cubic equation are

xk = ξk3

√1

16+

√3

16i+ ξ2k 3

√1

16− 3

16i

for k = 0, 1, 2. But cos θ is also a root. So why there are 4 roots for a cubic equation? Laterin the class, we will make it clear the meaning of

3√i

and then easy to show one of xk’s is cos θ.

6

CHAPTER 3

Analytic polynomials

3.1. Analytic polynomials

Definition 3.1. A complex-valued polynomial of two variables of degree n is a functionp : R2 → C defined by

p(x, y) =n∑k=0

cn,kxn−kyk +

n−1∑k=0

cn−1,kxn−1−kyk + ...+ (c1,0x+ c0,1y) + c0,0

where all cj,k ∈ C are some constants. The term∑`

k=0 c`,kx`−kyk is called the term of

degree ` of p. The polynomial p(x, y) will be called an analytic polynomial if there exitsα0, α1, ..., αn ∈ C such that

p(x, y) = α0 + α1(x+ iy) + α2(x+ iy)2 + ...+ αn(x+ iy)n

We will then say that p is a polynomial in z and write it as

p(z) = α0 + α1z + αnzn

Example 3.2.

p(x, y) = x2 − y2 + 2ixy = (x+ iy)2

is analytic.

Example 3.3.

p(x, y) = x2 − y2 − 2ixy

is not analytic.

Proof. Assume that p is analytic. Then x2 − y2 − 2ixy ≡∑2

k=0 αk(x + iy)k for someαk ∈ C. Since the left hand side has only degree 2 terms, we have α0 = α1 = 0. Then theequality x2 − y2 − 2ixy = α2(x+ iy)2 forces α2 to be 1 and −1 which is impossible since α2

is a constant. �

Definition 3.4. Suppose that f : U → C is a function where U ⊂ R2 is an open set.Write f(x, y) = u(x, y) + iv(x, y) where u and v are real-valued functions. We denote

fx := ux + ivx,

fy := uy + ivy.

Proposition 3.5. A complex-valued polynomial p(x, y) is analytic if and only if

py = ipx

7

Proof. Let p(x, y) be a polynomial of degree n. If p(x, y) is analytic, we have

p(x, y) = α0 + α1(x+ iy) + α2(x+ iy)2 + ...+ αn(x+ iy)n

for some α0, · · · , αn ∈ C. Then

px(x, y) = α1 + 2α2(x+ iy) + ...+ nαn(x+ iy)n−1

andpy(x, y) = α1(i) + 2α2(x+ iy)(i) + ...+ nαn(x+ iy)n−1(i) = ipx(x, y)

For the converse, suppose that py = ipx. Let

Q(x, y) = c0xm + c1x

m−1y + c2xm−2y2 + ...+ cmy

m

be the degree m term of p where c0, c1, · · · , cn ∈ C. Since Qx and Qy are the degree (m− 1)terms of px and py respectively, from the equality py = ipx, we have Qy = iQx. Then

c1xm−1 + 2c2x

m−2y + ...+mcmym−1 = i[mc0x

m−1 + (m− 1)c1xx−2y + ...+ cm−1y

m−1]

Comparing coefficients of both sides, we have

C1 = imC0 = i

(m1

)C0

C2 = i(m− 1)

2C1 = i2

m(m− 1)

2C0 = i2

(m2

)C0

C3 = i(m− 2)

3C2 = i3

m(m− 1)(n− 2)

3 · 2C0 = i3

(m3

)C0

and by induction,

Ck = i1

kCk−1 = ik

(mk

)C0

Therefore

Q(x, y) =m∑k=0

Ckxm−kyk = C0

m∑k=0

(mk

)xm−k(iy)k = C0(x+ iy)m

is analytic. Then each degree term of p is analytic, so is p. �

Definition 3.6. (Cauchy-Riemann equations) Let f(x, y) = u(x, y)+ iv(x, y) where u, vbe real-valued functions. The equation fy = ifx is equivalent to uy + ivy = iux − vx, i.e.,{

ux = vyuy = −vx

These equations are called the Cauchy-Riemann equations.

Example 3.7. Show that

p(x, y) = x5 + y5 − 3ix3y2 + 4x2y3

is not analytic.

Proof. Letu(x, y) = x5 + y5 + 4x2y3, v(x, y) = −3x3y2

We see that ux = 5x4 + 8xy3 and vy = −6x3y 6= ux. Therefore p is not analytic. �

8

Definition 3.8. A complex-valued function f(z), defined in a neighborhood of z0, issaid to be complex differentiable at z0 if

limz→z0

f(z)− f(z0)

z − z0

exists, which means that there is L ∈ C such that for any ε > 0, there is δ > 0 such that if|z − z0| < δ, then

|f(z)− f(z0)

z − z0

− L| < ε

We denote this limit by f ′(z0).

Since the definition of a differentiable function is formally same as the definition of adifferentiable function in calculus, results such as product rule, quotient rule and chain ruleare also true for complex differentiable functions. In particular, we get the following familiarformula for the derivatives of polynomials in z.

Proposition 3.9. Letp(z) = α0 + α1z + ...+ αnz

n

then p is complex differentiable at all z ∈ C and

p′(z) = α1 + ...+ nαnzn−1

3.2. FAQ

What is the difference between complex differentiation and differentiation of real vector-valued functions of two variables?

Recall 3.10. Let U be an open subset of R2. A function g : U → R2 is said to bedifferentiable at X0 = (x0, y0) ∈ U if there is a linear transformation Dg(X0) : R2 → R2 suchthat

limX→X0

||g(X)− g(X0)−Dg(X0)(X− X0)||||X− X0||

= 0

Proposition 3.11. Let U ⊂ C be an open set and z0 ∈ U . If a complex-valued functionf : U → C is differentiable at z0 then f considered as a two real variables function f(x, y) isdifferentiable at (x0, y0) where x0 + iy0 = z0.

Proof. Let f ′(z0) = a0 + ib0. Let X0 = (x0, y0),X = (x, y). Then

limz→z0

f(z)− f(z0)

z − z0

= a0 + ib0

implies that

limz→z0

f(z)− f(z0)− (a0 + ib0)(z − z0)

z − z0

= 0.

Note that

(a0 + ib0)(z − z0) = (a0Re(z − z0)− b0Im(z − z0)) + i(a0Im(z − z0) + b0Re(z − z0)),

Re(z − z0) = x− x0 = first component of X− X0

andIm(z − z0) = y − y0 = second component of Y− Y0.

9

So this complex number corresponds to the point A(X− X0) ∈ R2 where

A =

(a0 −b0

b0 a0

)By the continuity of the norm || · ||, we have

| limz→z0

f(z)− f(z0)− (a0 + ib0)(z − z0)

z − z0

| = limz→z0

|f(z)− f(z0)− (a0 + ib0)(z − z0)||z − z0|

= limX→X0

||f(X)− f(X0)− A(X− X0)||||X− X0||

= 0

This implies that f is differentiable at X0. �

Remark 3.12. The converse is not true.

Example 3.13. Show that f(z) = z is not differentiable anywhere, but f(x, y) = (x,−y)is differentiable everywhere.

Proof. By definition,

limh→0

f(z + h)− f(z)

h= lim

h→0

z + h− zh

= limh→0

h

h=

{1, along the x−axis−1, along the y−axis

Therefore f is not differentiable at z. For f(x, y) = (x,−y), each partial derivatives of f arecontinuous functions, hence f(x, y) is differentiable at any point of R2. �

10

CHAPTER 4

Power series

4.1. Power series

Definition 4.1. A power series in z centered at a ∈ C is a series of the form

∞∑k=0

ck(z − a)k

where ck ∈ C.

Definition 4.2. We say that a sequence {pn}∞n=1 in C converges to p ∈ C if for anyε > 0, there is N ∈ N such that for all n ≥ N,

|pn − p| < ε

and a series∑∞

k=1 ak converges to L ∈ C if the sequence of partial sums {∑n

k=1 ak}∞n=1

converges.

Definition 4.3. Let {ak}∞k=1 be a sequence in R. Define

lim supn→∞

an := limn→∞

supk≥n{ak}

The proof of the following root test is similar to the one in calculus.

Recall 4.4. (Root test) Given a power series∑∞

k=0 ck(z − a)k. Let

L = limn→∞

(supk≥n{|ck|

1k })

The root tests tell us that

(1) If L = 0, the power series converges for all z ∈ C.(2) If L =∞, the power series converges for z = a only.(3) If 0 < L <∞, set R = 1

L. The power series converges for |z − a| < R and diverges

for |z − a| > R.

Definition 4.5. The number

R =

∞, if L = 0;1L, if 0 < L <∞;

0, if L =∞.

is called the radius of convergence of the power series∑∞

k=0 ck(z − a)k.

The following result is from the ratio test. It is useful in computation.

11

Proposition 4.6. Give a power series∑∞

k=0 ck(z− a)k with radius of convergence R. If

the limit limk→∞

|ck+1||ck|

exists or equal to infinity, then

R = limk→∞

|ck||ck+1|

The proof of the following is similar to its counterpart in calculus.

Theorem 4.7. Suppose that the radius of convergence of∑∞

n=0 cn(z − a)n is R > 0. Letf(z) :=

∑∞n=0 cn(z−a)n. Then f ′(z) exists, the radius of convergence of

∑∞n=1 ncn(z−a)n−1

is R and

f ′(z) =∞∑n=1

ncn(z − a)n−1

for |z − a| < R.

Definition 4.8. If the radius of convergence of the power series∑∞

k=0 ck(z − a)k is Rand R > 0, then the open ball

BR(a) := {z ∈ C : |z − a| < R}

is called the domain of convergence of the power series.

Corollary 4.9. Power series are infinitely complex differentiable within their domainof convergence.

Proof. Let f(z) =∑∞

n=0 cn(z − a)n. Since

f ′(z) =∞∑n=1

ncn(z − a)n−1

has the same radius of convergence of f hence f ′ is complex differentiable on BR(a) Theresults follows by induction. �

Corollary 4.10. If f(z) =∑∞

n=0 cn(z − a)n has a nonzero radius of convergence, then

cn =f (n)(a)

n!

Proof. f(0) = c0, f ′(z) =∑∞

n=1 ncnzn−1, f ′(0) = C1. The result follows from induction.

�

Lemma 4.11. Given a power series∑∞

n=0 cnzn. Suppose that {zk}∞k=1 is a nonzero se-

quence which converges to 0. If∑∞

n=0 cnznk = 0 for all k ∈ N, then the power series

∑∞n=0 cnz

n

is identically zero, i.e.,∞∑n=0

cnzn = 0

for all z ∈ C.

Proof. Let f(z) =∑∞

n=0 cnzn and R be the radius of convergence of f . Since f is

defined on each zk 6= 0, each zk is in the domain of convergence of f and hence R > 0. A

12

power series is continuous on its domain of convergence and hence f is continuous at 0. Byassumption,

f(zk) = 0

for all k ∈ N. Then

c0 = f(0) = f( limk→∞

zk) = limk→∞

f(zk) = limk→∞

0 = 0.

Assume that cj = 0 for j = 0, 1, ..., n. Let

g(z) =f(z)

zn+1= cn+1 + cn+2z + · · ·

By the ratio test, we see that g has the same radius of convergence of f and hence iscontinuous at z = 0. Therefore

cn+1 = g(0) = g( limk→∞

zk) = limk→∞

g(zk) = limk→∞

f(zk)

zn+1= lim

k→∞0 = 0.

�

Recall 4.12. Let (X, d) be a metric space and Y ⊂ X a subset. For p ∈ X, let

Br(p) := {q ∈ X|d(p, q) < r}be the open ball of radius r at p. A point p ∈ X is called an accumulation point of Y if

(Br(p)− {p}) ∩ Y 6= ∅for any r > 0.

Proposition 4.13. If∑∞

n=0 an(z−a)n and∑∞

n=0 bn(z−a)n converge and agree on a setof points with an accumulation point at a, then an = bn for all n.

Proof. Let {zk}∞k=1 be a sequence such that zk → a and∑∞

n=0 an(zk−a)n =∑∞

n=0 bn(zk−a)n. Since (zk − a)→ 0 and

∑∞n=0(an− bn)(zk − a)n = 0, by Lemma above, the power series∑∞

n=0(an − bn)zn ≡ 0, so an = bn for n ≥ 0. �

13

CHAPTER 5

Analytic functions

5.1. Analytic functions

Definition 5.1. Let U be an open set of C. A function f : U → C is analytic at z0 ∈ Uif f is complex differentiable at some neighborhood of z0. The function f is analytic on U iff is analytic at every point of U . An entire function is a function which is analytic on C.

Corollary 5.2. All polynomials in z are entire functions.

Proposition 5.3. If f = u+ iv is complex differentiable at z, then fx and fy exist thereand satisfy the Cauchy-Riemann equation

fy = ifx

at z.

Proof. Let z = x+ iy. Since f is complex differentiable at z, along the x-axis, we have

f ′(z) = limh→0,h∈R

f(z + h)− f(z)

h= lim

h→0,h∈R

f(x+ h, y)− f(x, y)

h= fx(x, y)

On the other hand, along the y-axis, we have

f ′(z) = limh→0,h∈R

f(z + ih)− f(z)

ih= lim

h→0,h∈R

f(x, y + h)− f(x, y)

ih=fy(x, y)

i

which implies that

fy = ifx

�

Example 5.4. The following example shows that the converse of the above result is nottrue. Let

f(x, y) =

{xy(x+iy)x2+y2

, if z 6= 0;

0, if z = 0

Then f(x, 0) = f(0, y) = 0 for any x, y ∈ R. By definition of partial derivatives, a directcalculation give us

fx(0, 0) = fy(0, 0) = 0

Therefore f satisfies the Cauchy-Riemann equation at (0, 0), but f is not differentiable atz = 0 since the limit

limh→0,h∈R

f(0 + h+ ih)− f(0)

h+ ih= lim

h→0,h∈R

h2(h+ih)h2+h2

h+ ih=

1

26= 0 = fx(0, 0).

15

Proposition 5.5. Suppose that fx and fy exist in a neighborhood of z. If fx and fy arecontinuous at z and satisfy the Cauchy-Riemann equation

fy = ifx

at z, then f is complex differentiable at z.

Proof. Recall that for a differentiable function g on [α, β], the Mean Value Theoremfrom calculus tells us that

g(α)− g(β) = g′(α + θ(β − α))(α− β)

for some θ ∈ (0, 1). Let f = u+ iv, h = a+ bi. Then

u(z + h)− u(z)

h=u(x+ a, y + b)− u(x, y)

a+ bi=u(x+ a, y + b)− u(x, y + b) + u(x, y + b)− u(x, y)

a+ bi

=a

a+ biux(x+ θ1a, y + b) +

b

a+ biuy(x, y + θ2b)

Similarly, we have

v(z + h)− v(z)

h=

a

a+ bivx(x+ θ3a, y + b) +

b

a+ bivy(x, y + θ4b)

where 0 < θj < 1 for j = 1, 2, 3, 4. This implies that

f(z + h)− f(z)

h=

a

a+ bi[ux(z1) + ivx(z2)] +

b

a+ bi[uy(z3) + ivy(z4)]

where |zk − z| → 0 as h→ 0 for k = 1, 2, 3, 4. From the condition fy(z) = ifx(z), we get

a

a+ bifx(z) +

b

a+ bify(z) =

a

a+ bifx(z) +

ib

a+ bifx(z) = fx(z)

So

f(z + h)− f(z)

h− fx(z) =

a

a+ bi[ux(z1) + ivx(z2)− fx(z)] +

b

a+ bi[uy(z3) + ivy(z4)− fy(z)]

But | aa+bi|, | b

a+bi| ≤ 1 and fx, fy are continuous at z, we get

limh→0

f(z + h)− f(z)

h= fx(z)

�

Recall that a topological space X is connected if it can not be written as a disjoint unionof two nonempty open subsets.

Definition 5.6. A region is an open and connected subset of C.

Note that by the topology of open sets in R2, an open set U ⊂ C is connected if and onlyif any two points can be joined by a piecewise linear curve.

Proposition 5.7. If f = u + iv is analytic in a region D, and u is constant, then f isconstant.

Proof. Because u is constant, ux = uy = 0. By the Cauchy-Riemann equations, vx =vy = 0. Therefore v is also a constant. �

16

Proposition 5.8. If f = u+ iv is analytic on a region and |f | is constant, then f is aconstant.

Proof. If |f | = 0, f = 0. We may assume |f | 6= 0. Therefore u2 + v2 ≡ c 6= 0. Takepartial derivatives {

2uux + 2vvx ≡ 02uuy + 2vvy ≡ 0

By the Cauchy-Riemann equations, we have{uux − vuy ≡ 0uuy + vux ≡ 0

and {u2ux − uvuy ≡ 0uvuy + v2ux ≡ 0

Therefore the sum of two equations gives (u2 + v2)ux ≡ 0 which implies that ux ≡ 0. On theother hand, we also have {

uvux − v2uy ≡ 0u2uy + uvux ≡ 0

This implies that (u2 + v2)uy ≡ 0 and hence uy ≡ 0. Then u is a constant function. Byprevious result, we see that f is a constant. �

Recall 5.9. The exponential function exp : R → R − {0} defined by exp(x) = ex is agroup homomorphism which is also a smooth function. In the following, we want to extendexp to a function

exp : C→ C− {0}which is a group homomorphism and also an analytic function.

Definition 5.10. Defineez := ex cos y + iex sin y

where z = x+ iy.

Proposition 5.11. The function ez is an entire function and also a group homomor-phism from C to C− {0}. The restriction of the function ez to R is exp(x).

Proof. Since

(ez)x = ex cos y + iex sin y

(ez)y = −ex sin y + iex cos y = i(ez)x

which satisfy the Cauchy-Riemann equations and (ez)x, (ez)y are continuous everywhere, by

Theorem, ez is entire. By elementary trigonometry identities, we see that

ez1+z2 = e(x1+x2)+i(y1+y2) = ex1+x2 [cos(y1 + y2) + i sin(y1 + y2)]

= ex1ex2 [cos y1 cos y2 − sin y1 sin y2 + i sin y1 cos y2 + i cos y1 sin y2]

= ex1ex2 [(cos y1 + i sin y1) cos y2 + i(i sin y1 + cos y1) sin y2]

= ex1 [cos y1 + i sin y1]ex2 [cos y2 + i sin y2] = ez1ez2

for any z1 = x1 + iy1, z2 = x2 + iy2 ∈ C and hence a group homomorphism from C toC− {0}. �

17

Theorem 5.12. (Uniqueness of exponential function) Suppose that f is an entire functionwhich satisfies {

f(z1 + z2) = f(z1)f(z2), for all z1, z2 ∈ Cf(x) = ex, for all x in R

Then f(z) = ez for z ∈ C.

Proof. Note that f(z) = f(x+ iy) = f(x)f(iy) = exf(iy). Let

f(iy) = A(y) + iB(y)

Thenf(z) = exA(y) + iexB(y)

and

fx(z) = exA(y) + iexB(y)

fy(z) = exA′(y) + iexB′(y)

Since f is entire, f has to satisfy the Cauch-Riemann equations fy = ifx everywhere, whichis equivalent to {

exA′(y) = −exB(y)exB′(y) = exA(y)

This implies that A′′ = −A. Solving this differential equation, we get

A(y) = α cos y + β sin y

where α, β in R andB(y) = −A′(y) = α sin y − β cos y.

Since f extends exp, f(0) = 1, which gives us the condition{A(0) = 1B(0) = 0

.

Therefore α = 1, β = 0 and

f(z) = ex cos y + iex sin y = ez

�

The following properties are clear.

Proposition 5.13. (1) eiθ = cos θ + i sin θ for θ ∈ R.(2) |ez| = ex for any z = x+ iy in C.(3) Given α in C, α 6= 0, the equation ez = α has infinitely many solutions.

In the following, we are trying to extend the function sin and cos to entire functions.Note that {

eiy = cos y + i sin ye−iy = cos y − sin y

Therefore

cos y =eiy + e−iy

2,

sin y =eiy − e−iy

2i.

18

Definition 5.14. We define

cos z =eiz + e−iz

2

sin z =eiz − e−iz

2i

for z in C. Then cos z, sin z are entire functions which extends the usual cosine and sinefunctions.


sin 2z = 2 sin z cos z

for z ∈ C.

Proof.

2 sin z cos z = 2(eiz − e−iz

2i)(eiz + e−iz

2) =

ei(2z) − e−i(2z)

2i= sin 2z

�


cos 3x = 4 cos3 x− 3 cosx

for x ∈ R.

Proof. Note that

e3ix = cos 3x+ i sin 3x = (eix)3 = (cosx+ i sinx)3

= cos3 x+ 3(cos2 x)(i sinx) + 3(cos x)(i sinx)2 + (i sinx)3

= cos3 x− 3(cosx)(1− cos2 x) + i(3 cos2 x sinx− sin3 x)

= 4 cos3 x− 3 cosx+ i(3 cos2 x sinx− sin3 x)

Comparing the real parts of both sides, we get the result. �


ez =∞∑n=0

zn

n!

Proof. Let

f(z) =∞∑n=0

zn

n!

The radius of convergence of f is ∞, therefore f is an entire function. Since

f(x) =∞∑n=0

xn

n!= ex

19

for all x in R,

f(z)f(w) = (∞∑n=0

zn

n!)(∞∑n=0

wn

n!)

= 1 + (z

1!+w

1!) + (

z2

2!+ zw +

w2

2!) + ...

=∞∑n=0

(z + w)n

n!= f(z + w)

Therefore by Theorem,f(z) = ez

�

20

CHAPTER 6

Complex line integrals

6.1. Complex line integrals

Definition 6.1. Let f(t) = u(t) + iv(t) be a continuous complex-valued functions,t ∈ [a, b] ⊂ R. Define ∫ b

a

f(t)dt :=

∫ b

a

u(t)dt+ i

∫ b

a

v(t)dt

Recall 6.2. Recall that a function α : [a, b] → R is C1 if α is continuous on [a, b],differentiable on (a, b) and α′ is continuous on (a, b).

Definition 6.3. Let z(t) = x(t) + iy(t), t ∈ [a, b] be a curve on the complex plane.The curve is said to be piecewise C1 if x, y are continuous on [a, b] and there exist somet1 < t2 < · · · < tn in (a, b) such that the restrictions of x, y to intervals [a, t1], [t1, t2], · · · , [tn, b]are C1.

Example 6.4. (1) The curve

z(t) =

{t+ it, t ∈ [0, 1]t− it, t ∈ [−1, 0]

is a piecewise C1 curve.(2) The curve

z(t) = t+ imax{cos t, sin t}for t ∈ [−π, π] is a piecewise C1 curve.

Recall 6.5. In calculus, for a continuous function f : [a, b]→ R, the Riemann integralis defined by ∫ b

a

f(x)dx := limn→∞

n∑k=0

f(xk)(xk+1 − xk)

where a = t0 < t1 < t2 < ... < tn = b is a partition of [a, b].Now for a piecewise C1 curve which lies in the domain of a complex valued function g,

mimic the definition of Riemann integral, we may define∫C

f(z)dz := limn→∞

n∑k=0

g(γ(tk))(γ(tk+1)− γ(tk))

By the mean value theorem, this equals to

limn→∞

n∑k=0

g(γ(tk))γ′(ck)(tk+1 − tk) =

∫ b

a

g(γ(t))γ′(t)dt.

21

If g is a real-valued function and the curve is γ(t) = t+ i0 for t ∈ [a, b]. Then∫[a,b]

g(z)dz =

∫ b

a

g(γ(t))γ′(t)dt =

∫ b

a

g(t)dt.

So this definition of integration extends usual Riemann integration.

Definition 6.6. Let C be a piecewise C1 curve given by z : [a, b] → U where U ⊂ Cis open. Let f : U → C be a function which is continuous at every point of C. Then thecomplex integral of f along C is defined to be∫

C

f(z)dz :=

∫ b

a

f(z(t))z′(t)dt

Example 6.7. Let f(z) = z2, z ∈ C, the curve C : z(t) = cos t+ i sin t, t ∈ [0, 2π]. Then∫C

f(z)dz =

∫ 2π

0

(cos t+ i sin t)2(− sin t+ i cos t)dt = 0

Example 6.8. Let f(z) = 1z, z 6= 0. Let C : γ(t) = cos t+ i sin t, 0 ≤ t ≤ 2π,∫

C

f(z)dz =

∫ 2π

0

1

cos t+ i sin t(− sin t+ i cos t)dt =

−1

i

∫ 2π

0

(−i)(− sin t+ i cos t)

cos t+ i sin tdt

=

∫ 2π

0

idt = 2πi.

Proposition 6.9. (Change of variables formula) Suppose that f : [a, b]→ C is continu-ous and λ : [c, d]→ [a, b] is C1 such that λ(c) = a, λ(d) = b. Then∫ d

c

f(λ(t))λ′(t)dt =

∫ b

a

f(s)ds

Proof. Let f = u+iv and s = λ(t). Then by the change of variables formula in calculus,we have ∫ d

c

f(λ(t))λ′(t)dt =

∫ d

c

u(λ(t))λ′(t)dt+ i

∫ d

c

v(λ(t))λ′(t)dt

=

∫ b

a

u(s)ds+ i

∫ b

a

v(s)ds =

∫ b

a

f(s)ds

�

Definition 6.10. Suppose that C is given by z(t), a ≤ t ≤ b. Then −C is defined by

z(b+ a− t), a ≤ t ≤ b.

Proposition 6.11. Suppose that f : U → C is continuous and C is a curve in U . Then∫−C

f(z)dz = −∫C

f(z)dz.

Proof. By definition,∫−C

f(z)dz = −∫ b

a

f(z(b+ a− t))z′(b+ a− t)dt

22

Let w = b+ a− t, dw = −dt. When t = a, w = b, when t = b, w = a, then∫ b

a

f(z(b+a−t))z′(b+a−t)dt =

∫ a

b

f(z(w))z′(w)(−dw) =

∫ b

a

f(z(w))z′(w)dw =

∫C

f(z)dz.

�

The following simple facts tells us that line integrals are linear.

Proposition 6.12. Let C be a piecewise C1 curve, f and g be continuous functions onC and α be any complex number. Then

(1) ∫C

[f(z) + g(z)]dz =

∫C

f(z)dz +

∫C

g(z)dz

(2) ∫C

αf(z) = α

∫C

f(z)dz

Lemma 6.13. Suppose that G : [a, b]→ C is continuous, then

|∫ b

a

G(t)dt| ≤∫ b

a

|G(t)|dt

Proof. Let ∫ b

a

G(t)dt = Reiθ

for some R ≥ 0. Then

|∫ b

a

G(t)dt| = R =

∫ b

a

e−iθG(t)dt

=

∫ b

a

Re(e−iθG(t))dt+ i

∫ b

a

Im(e−iθG(t))dt

So the imaginary part is 0 and

|∫ b

a

G(t)dt| =∫ b

a

Re(e−iθG(t))dt ≤∫ b

a

|Re(e−iθG(t))|dt

≤∫ b

a

|e−iθG(t)|dt =

∫ b

a

|G(t)|dt

�

Recall 6.14. Let C be a piecewise C1-curve given by γ(t), t ∈ [a, b]. The length of C is

L =

∫ b

a

|γ′(t)|dt.

Theorem 6.15. (ML formula) Suppose that C is a piecewise C1 curve of length L, f iscontinuous on C, and |f | ≤M on C. Then

|∫C

f(z)dz| ≤ML

23

Proof. Suppose that C is given by γ(t), a ≤ t ≤ b. Then

|∫C

f(z)dz| = |∫ b

a

f(γ(t))γ′(t)dt| ≤∫ b

a

|f(γ(t))||γ′(t)|dt ≤M

∫ b

a

|γ′(t)|dt

= ML

�

6.2. FAQ

What is the difference between line integrals and complex line integrals?

Recall 6.16. Suppose that F : U → R2 is a continuous on some open set U ⊂ R2 andr : [a, b]→ U is a smooth curve. The line integral of F along C is defined to be∫

C

F (r) • dr :=

∫ b

a

F (r(u)) • r′(u)du

Its physical meaning is the work done by F from r(a) to r(b). Not that the symbol • inF (r(u)) • r′(u) is the inner product, but in complex line integral, f(γ(t))γ′(t) is just thecomplex multiplication, they are very different.

6.3. Fundamental theorem of line integrals

Theorem 6.17. (Fundamental theorem of complex line integrals) Let C : z(t), a ≤ t ≤ bbe a piecewise C1-curve lying in some open set U ⊂ C. Suppose that F : U → C is ananalytic function and f(z) = F ′(z) for z ∈ U . If f is a continuous function, then∫

C

f(z)dz = F (z(b))− F (z(a))

Proof. Let γ(t) := F (z(t)) where t ∈ [a, b]. Write

γ(t) = γ1(t) + iγ2(t)

where γ1, γ2 are real-valued functions. Then

γ′(t) = F ′(z(t))z′(t)

Therefore ∫C

f(z)dz =

∫ b

a

f(z(t))z′(t)dt =

∫ b

a

F ′(z(t))z′(t)dt =

∫ b

a

γ′(t)dt

=

∫ b

a

γ′1(t)dt+ i

∫ b

a

γ′2(t)dt

By the Fundamental Theorem of Calculus, this integral equals to

γ′1(b)− γ′1(a) + i(γ′2(b)− γ′2(a)) = γ1(b) + iγ2(b)− (γ1(a) + iγ2(a)) = γ(b)− γ(a)

= F (z(b))− F (z(a))

�

24

Definition 6.18. A curve C : γ(t), a ≤ t ≤ b, is a closed curve if

γ(a) = γ(b)

The curve C is a simple curve if γ(t1) = γ(t2) for some t1 < t2 implies t1 = a, t2 = b.

This means that a simple curve has no self-intersection except at the end point.

6.4. Rectangle theorem

Definition 6.19. A rectangle in C is a set of the form

R = {x+ iy|a ≤ x ≤ b, c ≤ y ≤ d}for some a, b, c, d ∈ R. The boundary of R is the set⋃

x=a,b

{x+ iy|y ∈ [c, d]}⋃ ⋃

y=c,d

{x+ iy|x ∈ [a, b]}

Lemma 6.20. Suppose that U ⊂ C is an open set and f : U → C is defined by

f(z) = α + βz

Let Γ be the boundary of a rectangle R ⊂ U . Then∫Γ

f(z)dz = 0

Proof. Suppose that Γ is given by γ(t), a ≤ t ≤ b. Since the boundary of a rectangleis a simple closed curve, we have γ(a) = γ(b). Let

F (z) = αz +β

2z2

Then F ′(z) = f(z). By the Fundamental theorem of line integrals∫Γ

f(z)dz = F (γ(b))− F (γ(a)) = 0

�

Theorem 6.21. (Rectangle Theorem) Let U ⊂ C be an open set. Suppose that f : U → Cis analytic and Γ is the boundary of a rectangle R ⊂ U . Then∫

Γ

f(z)dz = 0

Proof. Suppose that Γ is in counterclockwise direction. Let I =∫

Γf(z)dz. Split R

into 4 equal subrectangles. Let Γj, j = 1, 2, 3, 4 be the boundaries of these 4 subrectanglesall with counterclockwise directions. Note that the common sides of these subrectangles arein opposite directions, therefore the integral along them cancel, and hence∫

Γ

f(z)dz =4∑j=1

∫Γj

f(z)dz

Let Γ(1) be one of the boundaries such that

|∫

Γ(1)

f(z)dz| = maxj=1,2,3,4

{|∫

Γj

f(z)dz|}

25

Then

|I| = |∫

Γ

f(z)dz| ≤4∑j=1

|∫

Γj

f(z)dz| ≤ 4|∫

Γ(1)

f(z)dz|

and

|∫

Γ(1)

f(z)dz| ≥ |I|4

Let R(1) be the rectangle bounded by Γ(1). Repeat this process to the rectangle R(1), we getR(2) and continuous in this manner, we get a sequence of rectangles

R(1) ⊃ R(2) ⊃ R(3) ⊃ · · ·and their boundaries

Γ(1),Γ(2),Γ(3), · · ·and we have

|∫

Γ(k)

f(z)dz| ≥ |I|4k

Since each R(k) is compact and nonempty, by the nested property of compact sets, thereexists

z0 ∈∞⋂k=1

R(k)

By the analyticity of f at z0,

limz→z0

f(z)− f(z0)

z − z0

= f ′(z0)

Thereforef(z) = f(z0) + f ′(z0)(z − z0) + ε(z)(z − z0)

where

ε(z) =f(z)− f(z0)

z − z0

− f ′(z0)→ 0

as z → z0. Then∫Γ(n)

f(z)dz =

∫Γ(n)

f(z0) + f ′(z0)(z − z0)dz +

∫Γ(n)

ε(z)(z − z0)dz =

∫Γ(n)

ε(z)(z − z0)dz

which follows from the Lemma above. Let S be the length of the largest side of Γ. Then thelength of the largest side of Γ(n) is S

2n, and the length of Γ(n) ≤ 4S

2n. Hence the length of the

diagonal of Γ(n) is ≤√

2S2n

. We have

|z − z0| ≤√

2S

2n

for all z in R(n). Given δ > 0, take N such that if |z − z0| ≤√

2S2N

,

|ε(z)| < δ

By the ML-formula,

|∫

Γ(N)

f(z)dz| = |∫

Γ(N)

ε(z)(z − z0)dz| ≤ (δ ·√

2S

2N) · 4S

2N= δ

4√

2S2

4N.

26

From the inequality|I|4N≤ |∫

Γ(N)

f(z)dz|

we have|I| ≤ δ(4

√2S2)

Let δ → 0, we get I = 0. �

Theorem 6.22. (Rectangle Theorem II) Suppose that f : U → C is analytic where U ⊂ Cis open. For a ∈ U , let

g(z) :=

{f(z)−f(a)

z−a , if z ∈ U − {a}f ′(a), if z = a

Then ∫Γ

g(z)dz = 0

where Γ is the boundary of a rectangle R ⊂ U .

Proof. case 1: a is not in R: Note that 1z−a is analytic throughout R and hence

g is analytic throughout R. Therefore by the rectangle theorem,∫Γ

g(z)dz = 0

case 2: a is in Γ: Since f is analytic at a, g is continuous on R. By the compactnessof R and the continuity of |g|, |g| attains a maximum M on R. For any ε > 0, divideR into 6 subrectangles R1, · · · , R6 such that a is in the boundary of R1 only and thelength of R1 is < ε. Let Γ1, · · · ,Γ6 be the boundaries of R1, · · · , R6 respectively.Then by the ML-formula,

|∫

Γ1

g(z)dz| ≤Mε

and by the rectangle theorem ∫Γk

g = 0

for k 6= 1. Therefore

|∫

Γ

g(z)dz| = |6∑

k=1

∫Γk

g(z)dz| ≤Mε

which implies that ∫Γ

g(z)dz = 0

case 3: a is in R−Γ: For any ε > 0, divide R into 9 subrectangles R1, ..., R9 such thata lies in the interior of R1 and the length of R1 is < ε. Then as previous argument,we see that ∫

Γ

g(z)dz = 0

�

27

6.5. Integral theorem

Definition 6.23. We denote by ∫ z+h

z

f(ξ)dξ

for the line integral along the straight lines from z to z+Re(h) and from z+Re(h) to z+h.

Theorem 6.24. (Integral theorem) Suppose that r ∈ (0,∞]. If f : Br(z0) → C isanalytic, then there exists an analytic function F on Br(z0) such that

F ′(z) = f(z)

for all z ∈ Br(z0).

Proof. Consider z0 = 0 first. Let

F (z) =

∫ z

0

f(ξ)dξ

Then by the rectangle theorem,∫ Re(z+h)

Re(z)

f(ξ)dξ +

∫ z+Re(h)

Re(z+h)

f(ξ)dξ −∫ z+Re(h)

z

f(ξ)dξ −∫ z

Re(z)

f(ξ)dξ = 0

Therefore ∫ z+Re(h)

Re(z)

f(ξ)dξ =

∫ z

Re(z)

f(ξ)dξ +

∫ z+Re(h)

z

f(ξ)dξ

and

F (z + h)− F (z) =

∫ z+h

0

f(ξ)dξ −∫ z

0

f(ξ)dξ

=

∫ Re(z)

0

f(ξ)dξ +

∫ Re(z+h)

Re(z)

f(ξ)dξ +

∫ z+h

Re(z+h)

f(ξ)dξ −∫ Re(z)

0

f(ξ)dξ −∫ z

Re(z)

f(ξ)dξ

=

∫ Re(z+h)

Re(z)

f(ξ)dξ +

∫ z+Re(h)

Re(z+h)

f(ξ)dξ +

∫ z+h

z+Re(h)

f(ξ)dξ −∫ z

Re(z)

f(ξ)dξ

=

∫ z

Re(z)

f(ξ)dξ +

∫ z+Re(h)

z

f(ξ)dξ +

∫ z+h

z+Re(h)

f(ξ)dξ −∫ z

Re(z)

f(ξ)dξ

=

∫ z+Re(h)

z

f(ξ)dξ +

∫ z+h

z+Re(h)

f(ξ)dξ =

∫ z+h

z

f(ξ)dξ

Since f is continuous, for any ε > 0, there is δ > 0 such that if |h| < δ, then

|f(z + h)− f(z)| < ε

Combine with the fact that ∫ z+h

z

1dξ = (z + h)− z = h

28

we have

|F (z + h)− F (z)

h− f(z)| = |1

h

∫ z+h

z

f(ξ)dξ − 1

h

∫ z+h

z

f(z)dξ| = |1h

∫ z+h

z

f(ξ)− f(z)dξ|

≤ 1

|h|ε|h| = ε

Therefore F ′(z) = f(z). Now for any z0 ∈ C and f : Br(z0) → C analytic, we make atranslation of f by defining

g(z) := f(z + z0)

for z ∈ Br(0). Then g is analytic and for any rectangle R ⊂ Br(0),∫Γ

g(z)dz = 0

where Γ is the boundary of R. By the case before, there is an analytic function G on Br(0)such that G′ = g on Br(0). Let

F (w) = G(w − z0)

Then F is analytic on Br(z0) and

F ′(w) = G′(w − z0) = g(w − z0) = f(w)

�

Analysis the proof of integral theorem above, if a continuous function f satisfies∫Γ

f(z)dz = 0

for Γ of the boundary of rectangles R ⊂ Br(z0), the Integral Theorem is then a directconsequence. Therefore we have the following result.

Theorem 6.25. (Integral theorem II) Suppose that r ∈ (0,∞]. If f : Br(z0) → C isanalytic and

g(z) :=

{f(z)−f(a)

z−a , if z ∈ Br(z0)− {a}f ′(a), if z = a

then there exists an analytic function F on Br(z0) such that

F ′(z) = g(z)

for all z ∈ Br(z0).

6.6. Closed curve theorem

Theorem 6.26. (Closed curve theorem) Suppose that r ∈ (0,∞]. If f : Br(z0) → C isanalytic, then for any closed piecewise C1-curve C in Br(z0),∫

C

f(z)dz = 0

29

Proof. Suppose that C is given by z(t) where t ∈ [a, b]. Since C is closed, z(a) = z(b).Furthermore, since f is analytic on Br(z0), by the integral theorem, F ′ = f for some analyticF on Br(z0). By the fundamental theorem of complex line integrals,∫

C

f(z)dz = F (z(b))− F (z(a)) = 0

�

By integral theorem II, we also have the following result.

Theorem 6.27. (Closed curve theorem II) Suppose that r ∈ (0,∞]. If f : Br(z0) → Cis analytic and

g(z) :=

{f(z)−f(a)

z−a , if z ∈ Br(z0)− {a}f ′(a), if z = a

then for any closed piecewise C1-curve C in Br(z0),∫C

g(z)dz = 0

30

CHAPTER 7

Cauchy integral formula

Definition 7.1. (Uniformly convergence) Given a sequence {fn}∞n=1 of complex-valuedfunctions defined on a set E ⊂ C. We say that {fn}∞n=1 converges uniformly on E if there isa function f : E → C such that for any ε > 0, there exists N such that if n ≥ N , then

|fn(x)− f(x)| < ε

for all x in E. We say that a series∑∞

i=1 fi converges uniformly on E if the sequence of thepartial sums

{n∑i=1

fi}∞n=1

converges uniformly on E. We say that the series∑∞

i=1 fi is uniformly absolutely-convergentif the series

∞∑n=1

|fi|

converges uniformly.

We recall the following extremely useful result.

Theorem 7.2. (Weierstrass M-test) Let E ⊂ C and fn : E → C be a function wheren ∈ N. Suppose that for each n ∈ N,

supx∈E|fn(x)| ≤Mn

and∞∑n=1

Mn <∞

then∞∑i=1

fi

converges uniformly on E.

Theorem 7.3. If the power series∑∞

n=0 an(z − z0)n has radius of convergence R > 0,

then for any r < R, the power series is uniformly absolutely-convergent on Br(z0).

Proof. Let t = r+R2

. Then r < t < R and hence the power series converges at z = z0 +t.Therefore {antn}∞n=0 is a bounded sequence. Let

M := sup{|antn|}∞n=0 <∞31

For |z − z0| ≤ r, |z−z0|t≤ r

t< 1,

|an(z − z0)n| = |antn||z − z0

t|n ≤M |z − z0

t|n

and the geometric series∞∑n=0

M |z − z0

t|n

is convergent, therefore by the Weierstrass M-test, the power series∑∞

n=0 |an(z − z0)n| con-

verges uniformly on Br(z0). �

We recall the following uniform limit theorem from advanced calculus. It says that theuniform limit of any sequence of continuous functions is continuous.

Theorem 7.4. (Uniform limit theorem) Let X be a topological space and Y be a metricspace. If fn : X → Y is continuous and fn → f uniformly for some function f : X → Y ,then f is continuous.

Proposition 7.5. Let C : γ(t), a ≤ t ≤ b be a piecewise C1-curve. Suppose that{fn : C → C} is a sequence of continuous functions and fn → f uniformly on C, then

limn→∞

∫C

fn(z)dz =

∫C

f(z)dz.

Proof. Let L > 0 be the arc length of C. Since fn → f uniformly on C, for all ε > 0,there exists N ∈ N such that if n ≥ N , then

|fn(z)− f(z)| < 1

Lε

for all z in C. Then

|∫C

fn(z)dz −∫C

f(z)dz| = |∫C

(fn(z)− f(z))dz| ≤∫C

|fn(z)− f(z)|dz ≤ (1

Lε)(L) = ε

Therefore

limn→∞

∫C

fn(z) =

∫C

f(z)dz

�

Corollary 7.6. Let C : γ(t), a ≤ t ≤ b be a piecewise C1-curve. Suppose that {fn :C → C} is a sequence of continuous functions and

∑∞n=1 fn converges uniformly on C, then∫

C

∞∑n=1

fn(z)dz =∞∑n=1

∫C

fn(z)dz

Proof. Let gm :=∑m

n=1 fn. Since {gm}∞m=1 converges uniformly to some g : C → C, wehave for each x ∈ C,

g(x) = limm→∞

gm(x) =∞∑n=1

fn(x)

We get the desire result by applying result above. �

32

Lemma 7.7. Let Cr be the circle centered at z0 with radius r, and |a− z0| < r. Then∫Cr

1

z − adz = 2πi.

Proof. Let Cr : z0 + reiθ, 0 ≤ θ ≤ 2π. Then∫Cr

1

z − z0

dz =

∫ 2π

0

1

reiθ(rieiθ)dθ = 2πi.

and ∫Cr

1

(z − z0)k+1dz =

∫ 2π

0

1

rk+1e(k+1)iθrieiθdθ =

i

rk

∫ 2π

0

e−kiθdθ = 0

(7.1)

Note that1

z − a=

1

(z − z0)− (a− z0)=

1

(z − z0)[1− a−z0z−z0 ]

=1

z − z0

· 1

1− wwhere

w =a− z0

z − z0

and |w| = |a−z0|r

< 1. Recall that the geometric series

1 + w + w2 + · · ·converges uniformly on Br(0) for r < 1 to 1

1−w and hence we may interchange the integraland summation to get∫

Cr

1

z − adz =

∫Cr

1

z − z0

1

1− wdz =

∫Cr

1

z − z0

∞∑k=0

wkdz =

∫Cr

1

z − z0

+∞∑k=1

(a− z0

z − z0

)kdz

=

∫Cr

1

z − z0

dz +∞∑k=1

∫Cr

(a− z0)k

(z − z0)k+1dz =

∫Cr

1

z − z0

dz = 2πi.

�

Theorem 7.8. (Cauchy integral formula) Suppose that f is analytic on Br(z0) for somer ∈ (0,∞] and a ∈ Br(z0). Let C be the circle C : z0 + Reiθ where 0 ≤ θ ≤ 2π andr > R > |a− z0|. Then

f(a) =1

2πi

∫C

f(z)

z − adz.

Proof. By the closed curve theorem II,∫C

f(z)− f(a)

z − adz = 0

Therefore ∫C

f(z)

z − adz = f(a)

∫C

1

z − adz = 2πif(a)

�

33

Theorem 7.9. (Power series expansion) If f is analytic on Br(z0) for some r ∈ (0,∞],then

f(z) =∞∑k=0

f (k)(z0)

k!(z − z0)k

for all z in Br(z0).

Proof. Let r > R > 0 and C : z0 + Reiθ, 0 ≤ θ ≤ 2π be the circle centered at z0 withradius R. Suppose that |z − z0| < R. By the Cauchy Integral formula,

f(z) =1

2πi

∫C

f(w)

w − zdw

By the geometric series, we have

1

w − z=

1

(w − z0)− (z − z0)=

1

(w − z0)(1− z−z0w−z0 )

=1

w − z0

(1 +z − z0

w − w0

+(z − z0)2

(w − w0)2+ · · · )

=1

w − z0

+z − z0

(w − z0)2+

(z − z0)2

(w − z0)3+ · · ·

Fix z and consider z−z0w−z0 as a function in w ∈ C, since

| z − z0

w − z0

| = |z − z0|R

< 1

the convergence is uniformly on the circle C, therefore we may interchange the summationand integral to get

f(z) =1

2πi

∫C

f(w)

w − zdw =

1

2πi

∫C

f(w)∞∑k=0

(z − z0)k

(w − z0)k+1dw

=∞∑k=0

(1

2πi

∫C

f(w)

(w − z0)k+1dw)(z − z0)k =

∞∑k=0

ck(R)(z − z0)k

where

ck(R) =1

2πi

∫C

f(w)

(w − z0)k+1dw

Therefore f has a power series expansion and by the uniqueness of power series, we have

ck(R) =f (k)(z0)

k!

which is independent of R. �

Corollary 7.10. If f : U → C is analytic where U is open in C, then f is infinitelycomplex differentiable.

Proof. For p ∈ U , take an open ball Br(p) ⊂ U . Then f is analytic on Br(p), hence bytheorem, f is equal to a power series on Br(p) and hence infinitely complex differentiable atp. �

34

Proposition 7.11. Let U ⊂ C be an open set and a ∈ U . If f : U → C is analytic andif

g(z) =

f(z)− f(a)

z − a, if z ∈ U − {a}

f ′(a), if z = a

then g is analytic on U .

Proof. The function g is clearly analytic on U − {a}. We show that g is analytic at a.Expand f into a power series

f(z) = f(a) + f (1)(a)(z − a) + f (2)(a)(z − a)2 + · · ·

in a small neighborhood of a. Then for z 6= a,

g(z) =f(z)− f(a)

z − a= f (1)(a) + f (2)(a)(z − a) + · · ·

and at the point a, g(a) = f ′(a) which implies that g equals to the power series

∞∑k=1

f (k)(a)

k!(z − a)k−1

therefore complex differentiable at a and hence analytic on U . �

Corollary 7.12. Suppose that f is analytic on Br(z0) for some r ∈ (0,∞] and a1, a2, ..., aN ∈Br(z0) are some distinct zeros of f . If g is defined by

g(z) =

f(z)

(z − a1)(z − a2)...(z − aN), if z ∈ Br(z0)− {a1, ..., aN}

limz→ak

f(z)

(z − a1)...(z − aN), if z ∈ {a1, ..., aN}.

then g is analytic on Br(z0).

Proof. We prove it by induction on the number of zeros. If N = 1,

limz→a1

f(z)

z − a1

= lima→a1

f(z)− f(z1)

z − a1

= f ′(a1)

Therefore

g(z) :=

f(z)− f(a1)

z − a1

, if z 6= a1

f ′(a1), if z = a1

which is analytic by Proposition before. Suppose that the result is true for distinct N zerosof f . Let a1, ...., aN+1 be some distinct N + 1 zeros of f . Let

h(z) =

f(z)

(z − a1) · · · (z − aN), if z ∈ Br(z0)− {a1, ..., aN}

limz→ak

f(z)

(z − a1)...(z − aN), if z ∈ {a1, ..., aN}

Then by induction hypothesis, h is an analytic function.

35

Let

g(z) :=

h(z)

z − aN+1

, if z ∈ Br(z0)− {aN+1}

limz→ak

h(z)

z − aN+1

, if z = aN+1

Since aN+1 is a zero of h, by the case N = 1, we see that g is analytic.�

Theorem 7.13. (Liouville’s Theorem) A bounded entire function is constant.

Proof. Let a, b in C and R > max{|a|, |b|}. Let C : Reiθ, 0 ≤ θ ≤ 2π be the circle withradius R. By the Cauchy integral formula,

f(b)− f(a) =1

2πi

∫C

f(z)

z − bdz − 1

2πi

∫C

f(z)

z − adz =

1

2πi

∫C

f(z)(z − a)− f(z)(z − b)(z − a)(z − b)

dz

=1

2πi

∫C

f(z)(b− a)

(z − a)(z − b)dz.

Note that a is enclosed inside the circle, the shortest distance from a to the circle isR− |a|. Therefore for z ∈ C,

|(z − a)(z − b)| ≥ (R− |a|)(R− |b|)Let M = maxz∈C{|f(z)|}. Then

|f(b)− f(a)| ≤ 1

2π

∫C

| f(z)(b− a)

(z − a)(z − b)|dz ≤ 1

2π

M |b− a|2πR(R− |a|)(R− |b|)

→ 0

as R→∞. Therefore f(b) = f(a) which implies that f is a constant. �

Theorem 7.14. (The Extended Liouville’s Theorem) If f is entire and if for someinteger k ≥ 0, there exist constants A, B such that

|f(z)| ≤ A+B|z|k

for all z ∈ C, then f is a polynomial of degree at most k.

Proof. When k = 0, it is the Liouville’s Theorem. We do induction on k. Let

g(z) =

{f(z)− f(0)

z, z 6= 0

f ′(0), z = 0

By proposition, g is entire. By the continuity of g, for |z| ≤ 1, |g(z)| ≤M for some M > 0.For |z| > 1,

|g(z)| = |f(z)− f(0)||z|

≤ A+B|z|k + |f(0)||z|

≤ A+ |f(0)|+B|z|k−1.

Therefore|g(z)| ≤ D + E|z|k−1

for some constants D,E > 0. By induction hypothesis, g(z) is a polynomial of degree atmost k − 1. Then

f(z) = f(0) + g(z)z

is a polynomial of degree at most k. �

36

Definition 7.15. Given a sequence {zk} in C, we say that

zk →∞if

|zk| → ∞Suppose that U = C − B where B is some compact subset of C. Let f : U → C be afunction. We say that

f(z)→∞ as z →∞denoted as

limz→∞

f(z) =∞

if for any M > 0, there is R > 0 such that

|f(z)| > M

for |z| > R.

Recall that|a− b| ≥ ||a| − |b||

for any a, b ∈ C.

Lemma 7.16. Suppose that p(z) is a polynomial of degree k > 0. Then

limz→∞

p(z) =∞

Proof. Given M > 0. For a 6= 0 and |z| ≥ M+|b||a| ,

|az + b| ≥ ||a||z| − |b|| ≥ |a|(M + |b||a|

)− |b| ≥M

Suppose that the result is true for k − 1. Let

p(z) = akzk + · · ·+ a0

be a polynomial of degree k. By the induction hypothesis, there exists R > 0 such that if|z| > R, then

|akzk−1 + · · ·+ a1| > 1

Then for |z| > |a0|+M +R,

|p(z)| ≥ ||z||akzk−1 + · · ·+ a1| − |a0|| ≥ |a0|+M +R− |a0| > M

This completes the induction. �

Theorem 7.17. (The Fundamental Theorem of Algebra) Every non-constant polynomialhas a zero in C.

Proof. Letp(z) = akz

k + ...+ a1z + a0

be a polynomial of degree k > 0. Then ak 6= 0. If p(z) 6= 0, for all z in C, then

f(z) :=1

p(z)

37

is an entire function. Since limz→∞ p(z) = ∞, there exists R > 0 such that if |z| > R,|p(z)| > 1. Therefore

|f(z)| = 1

|p(z)|< 1

if |z| > R. By the continuity of f , f is bounded on the ball BR(0) which implies that f is abounded entire function. By Liouville’s Theorem, f is a constant and hence p is a constantwhich is a contradiction. Therefore there exists z in C such that p(z) = 0. �

We recall the following well known results of complex polynomials.

Theorem 7.18. (Euclidean division of polynomials) Given two polynomials f(z) andg(z) with complex coefficients, there exist unique polynomials q(z) and R(z) with complexcoefficients such that

f(z) = q(z)g(z) +R(z)

where R(z) = 0 or deg(R) < deg(g).

A direct consequence of above result is the following factor theorem.

Corollary 7.19. (Factor theorem) The polynomial z − a is a factor of a polynomialz − a if and only if

f(a) = 0

Corollary 7.20. Every degree n ∈ N polynomial has n zeros in C counting multiplici-ties.

Proof. Let p(z) be a degree n ∈ N polynomial. By the Fundamental Theorem ofAlgebra, there is a1 ∈ C such that p(a1) = 0. By the factor theorem, we may write

p(z) = (z − a1)q(z)

where q(z) is a polynomial of degree n− 1. The result follows from induction. �

38

CHAPTER 8

Maximum principle and open mapping theorem

Example 8.1. Let f(z) = 1z−1

, z 6= 1. Then f is analytic on C− {1}. The power seriesexpansion of f at 2 can be obtained as following:

1

z − 1=

1

1 + z − 2=∞∑k=0

(−1)k(z − 2)k,

where |z − 2| < 1, and the power series of f at 3 is

1

z − 1=

1

2 + (z − 3)=

1

2

1

1 + 12(z − 3)

=1

2

∞∑k=0

(−1)k(1

2)k(z − 3)k

Note that

limk→∞|(1

2)k+1(z − 3)k+1

(12)k(z − 3)k

| < 1

so |z − 3| < 2. Therefore the two power series of f(z) have different radius of convergence.There does not exist a power series which equals to f everywhere.

Remark 8.2.

{entire functions} = { power series with infinite radius of convergence }{analytic functions} = {functions that can be expressed by some power series locally}

Recall that an open set U ⊂ C is connected if U can not be written as a disjoint unionof two nonempty open subsets of U and a region is an open connected subset of C.

Theorem 8.3. (Uniqueness theorem) Suppose that f is analytic in a region D and that

f(zn) = 0

for some sequence of distinct points {zn} such that zn → z0 in D. Then

f ≡ 0

in D.

Proof. Since f has a power series expansion at z0, by the uniqueness result of powerseries, f ≡ 0 in a disc of z0. Let

A = {z ∈ D|z is the limit of zeros of f}B = D − A

Then

A ∪B = D and A ∩B = φ

39

If z in A, then f = 0 in a disc of z which implies that A is open. For z in B, if for everyr > 0, there is a zero of f in Br(z), then z is the limit of zeros of f which is a contradiction.Hence there exists r > 0 such that f(w) 6= 0, for all w in Br(z)− {z}. Therefore Br(z) ⊂ Band hence B is open. Since z0 ∈ A, A is nonempty. A and B are open and disjoint, but Dis connected, therefore B = ∅ and hence A = D. So f ≡ 0 in D. �

Corollary 8.4. If two analytic functions f, g in a region D agree on a set of pointswith an accumulation point in D, then

f ≡ g

in D.

Proof. By taking h = f − g, this result follows from previous theorem directly. �

The following is an obvious consequence of the result above.

Corollary 8.5. If f, g are analytic in a region D, and f |U = g|U for some nonemptyopen set U ⊂ D, then

f ≡ g

in D.

Theorem 8.6. An entire function f is a polynomial if and only if

f(z)→∞

as z →∞.

Proof. By Lemma, if f is a polynomial, then f(z)→∞ as z →∞. The hard work ison another direction. So let f be an entire function such that

f(z)→∞ as z →∞

Then there exists R > 0 such that

|f(z)| > 1

for |z| > R. Therefore the zeros of f are contained in the disc BR(0). If f has infinitely

many zeros, since BR(0) is compact, the zero set of f has an accumulation points. Then byCorollary,

f ≡ 0

which is a contradiction. Therefore f has at most finitely many zeros, namely,

p1, ..., pn

in BR(0). For each j ∈ {1, ..., n}, let kj be the largest positive integer such that

f(pj) = f (1)(pj) = · · · = f (kj−1)(pj) = 0, f (kj)(pj) 6= 0

We note that kj <∞ otherwise the m-th coefficient of the power series expansion of f at pjis

f (m)(pj)

m!= 0

40

and hence f ≡ 0 which is a contradiction.Let

g(z) :=

f(z)

(z − p1)k1(z − p2)k2 · · · (z − pn)knif z /∈ {p1, ..., pn};

limw→pj

f(w)

(w − p1)k1(w − p2)k2 · · · (w − pn)knif z = pj, j ∈ {1, ..., n}

By repeatedly applying Corollary 7.12, g is entire. Since g(z) 6= 0, for all z in C, the function

h(z) :=1

g(z)

is entire. Let N = k1 + k2 + · · ·+ kn and

M = max|z|6R+1

{|h(z)|}

For |z| > R + 1, we have |f(z)| > 1 and

|z − pi|ki 6 (|z|+ |pi|)ki 6 (|z|+ |z|)ki = 2ki |z|ki ,Therefore

|h(z)| ≤ |z − p1|k1 ...|z − pn|kn|f(z)|

6 2N |z|N

Then

|h(z)| 6M + 2N |z|N

for all z in C. By the extended Liouville’s theorem h(z) is a polynomial, but since h(z) 6= 0,for all z in C, the fundamental theorem of algebra says that

h ≡ c 6= 0

for some constant c. Therefore

g(z) =1

cand

f(z) =1

c(z − p1)k1 ...(z − pn)kn

�

Theorem 8.7. (The Mean Value Theorem) If f is analytic in a region D, then

f(a) =1

2π

∫ 2π

0

f(a+ reiθ)dθ

for any r such that Br(a) ⊂ D.

Proof. By the Cauchy integral formula, f(a) = 12πi

∫Cr

f(z)z−adz where Cr : a + reiθ,

0 ≤ θ ≤ 2π is a circle in D. Let z = a+ reiθ, then

f(a) =1

2πi

∫ 2π

0

f(a+ reiθ)

reiθrieiθdθ =

1

2π

∫ 2π

0

f(a+ reiθ)dθ.

�

41

Recall 8.8. If g : [0, 2π] → R is a nonnegative continuous function and g(t) > 0 forsome t ∈ [0, 2π], then ∫ 2π

0

g(t)dt > 0

Theorem 8.9. (The maximum principle) Suppose that f is a nonconstant analytic func-tion in a region D. For all z in D and δ > 0, there exists w in Bδ(z) ∩D such that

|f(w)| > |f(z)|

In particular, f has no maximum in D.

Proof. Fix z ∈ D and δ > 0. Assume that there does not exist w in Bδ(z) ∩ D suchthat

|f(w)| > |f(z)|

For r < δ such that Br(z) ⊂ D, we have

|f(z + reiθ)| ≤ |f(z)|

and by the mean valued theorem,

|f(z)| = | 1

2π

∫ 2π

0

f(z + reiθ)dz| ≤ 1

2π

∫ 2π

0

|f(z + reiθ)|dz ≤ 1

2π

∫ 2π

0

|f(z)|dz = |f(z)|

which implies ∫ 2π

0

|f(z + reiθ)|dz ≤∫ 2π

0

|f(z)|dz

and

0 =1

2π

∫ 2π

0

|f(z)| − |f(z + reiθ)|dθ.

But

|f(z)| − |f(z + reiθ)| ≥ 0

and is a continuous function in θ, by the result from calculus that we recall before, we have

|f(z)| = |f(z + reiθ)|

for all r < δ such that Br(z) ⊂ D. Therefore

|f(z)| = |f(z)|

for z ∈ Br(z). By theorem f is a constant on Br(z). By the uniqueness theorem, f is aconstant in D which is a contradiction. �

Corollary 8.10. Suppose that f is analytic in a bounded region D and continuous onD, then |f | has a maximum in the boundary ∂D of D.

Proof. Since |f | is a continuous on D and D is compact, by Theorem, |f | has a maxi-mum at some point p in D. By the maximum principle, p /∈ D, therefore p ∈ ∂D.

�

42

Theorem 8.11. (The minimum principle theorem) If f is a non-constant analytic func-tion in a region D, and

|f(z)| ≤ |f(w)|for all w in D, then

f(z) = 0

Proof. Suppose that f(z) 6= 0. Then

g(w) =1

f(w)

is analytic on D. But then

|g(z)| ≥ |g(w)|for all w in D which contradicts to the maximum principle theorem, therefore

f(z) = 0

�

Theorem 8.12. (The open mapping theorem) The image of a region under a nonconstantanalytic mapping is an open set.

Proof. Let U be a region and g : U → C be a nonconstant analytic mapping. Fixa ∈ U . Let

f(z) = g(z)− g(a)

then

f(a) = 0

Suppose that Bδ(a) ⊂ U for some δ > 0. There exists a circle C of radius r < δ around asuch that f(z) 6= 0 for z in C otherwise a would be the accumulation point of zeros of f andby the uniqueness theorem, f would be a constant. Let

ε =1

2minz∈C|f(z)|

We claim that

Bε(0) ⊆ f(Br(a))

Let w in Bε(0). For z ∈ C,

|f(z)− w| ≥ |f(z)| − |w| ≥ 2ε− ε = ε

and when z = a,

|f(a)− w| = | − w| < ε

Therefore the minimum value of |f(z) − w| occurs at some z0 in Br(a). By the minimumprinciple, f(z0) = w. Therefore we have

Bε(0) ⊆ f(Br(a))

In term of g, we have

Bε(g(a)) = Bε(0) + g(a) ⊆ f(Br(a)) + g(a) = g(Br(a)) ⊂ g(U)

Therefore g is an open mapping. �

43

Theorem 8.13. (Schwarz’s Lemma) Suppose that f : B1(0)→ C is analytic. If

|f(z)| ≤ 1

for all |z| < 1 andf(0) = 0

then

(1) |f(z)| ≤ |z| for all z ∈ B1(0) and(2) |f ′(0)| ≤ 1.

Furthermore,f(z) ≡ eiθz

for some θ if and only if |f ′(0)| = 1 or there is some 0 6= z0 ∈ B1(0) such that

|f(z0)| = |z0|

Proof. Let

g(z) =

{f(z)

z, 0 < |z| < 1

f ′(0), z = 0

By theorem, g is analytic on B1(0). Let

Cr : reiθ, 0 ≤ θ ≤ 2π, 0 < r < 1

On Cr,

|g(z)| = |f(z)||z|

≤ 1

r

Let r → 1−, by the maximum principle,

|g(z)| ≤ 1

for |z| < 1 which proves (1), (2).If

f(z) ≡ eiθz

for some θ, then|f(z)| = 1

andf ′(z) = eiθ

for all z ∈ B1(0). Therefore|f ′(0)| = 1

Now suppose that |f ′(0)| = 1 or there is some 0 6= z0 ∈ B1(0) such that

|f(z0)| = |z0|That means

|g(0)| = 1 or |g(z0)| = 1

By the maximum principle, g is a constant with norm 1. Therefore

g(z) ≡ eiθ

for some θ. Hence for z 6= 0,f(z) = eiθz

44

But f(0) = 0 which means thatf(z) = eiθz

on B1(0). �

Proposition 8.14. Suppose that 0 < |α| < 1. Define

Aα(z) =z − α1− αz

Then

(1) Aα is analytic on B 1|α|

(0); in particular, Aα is analytic on the unit disc.

(2) For |z| = 1,|Aα(z)| = 1

and henceAα(B1(0)) ⊂ B1(0)

Proof. (1) Note that for |z| < 1|α| ,

|αz| = |α||z| < |α|( 1

|α|) = 1

therefore 1− αz 6= 0. This implies that Aα(z) is analytic on B 1|α|

(0).

(2) On |z| = 1,

|Aα|2 = AαAα = (z − α1− αz

)z − α1− αz

=zz − zα− αz + αα

1− αz − αz + ααzz=

1− αz − αz + |α|2

1− αz − αz + |α|2= 1.

By the maximum principle, Aα maps the unit disc to the unit disc.�

Example 8.15. Suppose that f is analytic and bounded by 1 in the unit disc and thatf(1

2) = 0. Estimate |f(3

4)| and |f ′(1

2)|.

Solution 8.16. Let

g(z) =

f(z)− f(1

2)

A 12(z)

=f(z)

z − 12

(1− 12z), z 6= 1

2

3

4f ′(1

2), z = 1

2

Since

limz→ 1

2

f(z)

z − 12

(1− 1

2z) =

3

4f ′(

1

2)

by theorem, g is analytic on B1(0). Since

|g(z)| ≤|1− 1

2z|

|z − 12|

=1

|A 12(z)|→ 1

as z → 1. Therefore by the maximum principle,

|g(z)| ≤ 1

on B1(0) and hence|f(z)| ≤ |A 1

2(z)|

45

for |z| < 1. Then

|f(3

4)| ≤ |

34− 1

2

1− 12(3

4)| =

1458

=2

5

Furthermore,

|34f ′(

1

2)| ≤ 1

implies

f ′(1

2) ≤ 4

3

Proposition 8.17. If f is entire and

|f(z)| ≤ 1

|Im(z)|for all Im(z) 6= 0, then

f ≡ 0

Proof. Let R > 0 andg(z) = (z2 −R2)f(z)

For z such that|z| = R,Re(z) ≥ 0, Im(z) 6= 0

we have

|(z −R)f(z)| ≤ |z −R||Im(z)|

= sec θ

where 0 ≤ θ ≤ π4. Since

1 ≤ sec θ ≤√

2

we have|(z −R)f(z)| ≤

√2

The continuity of (z −R)f(z) implies

|(z −R)f(z)| ≤√

2

for all |z| = R,Re(z) ≥ 0. Similarly, for |z| = R and Re(z) ≤ 0,

|(z +R)f(z)| ≤√

2

For |z| = R,|z −R| ≤ |z|+R = 2R and |z +R| ≤ 2R

then|g(z)| = |z +R||(z −R)f(z)| ≤ (2R)

√2 ≤ 3R

By the maximum principle,|g(z)| < 3R

for all |z| < R. Then

|f(z)| < 3R

|z2 −R2|for all |z| < R and any R > 0. Fix z and let R→∞, we see that

|f(z)| → 0

46

which implies thatf(z) ≡ 0

�

47

CHAPTER 9

Morera’s theorem

Theorem 9.1. (Morera’s Theorem) Let f be a continuous function on an open set D. If∫Γ

f(z)dz = 0

whenever Γ is the boundary of a rectangle R ⊂ D, then f is analytic on D.

Proof. Fix z0 ∈ D. Take r > 0 such that Br(z0) ⊂ D. Define

F (z) =

∫ z

z0

f(z)dz

for z ∈ Br(z0). Then by the assumption, as done in the integral theorem, we have

F (z0 + h)− F (z0)

h=

1

h

∫ z0+h

z0

f(ξ)dξ

Since f is continuous,

|F (z0 + h)− F (z0)

h− f(z0)| = |1

h

∫ z0+h

z0

f(ξ)− f(z0)dξ|

≤ 1

|h|max

|w−z0|≤|h||f(w)− f(z0)|(2|h|)→ 0

as h→ 0. Therefore F is complex differentiable at z0 and

F ′(z0) = f(z0)

Since z0 is arbitrary, F is analytic on D and

F ′(z) = f(z)

for z ∈ D. By theorem, the derivative of an analytic function is analytic, therefore F ′ isanalytic and so is f . �

Definition 9.2. Suppose that fn, f : D → C are functions. We say that fn convergesto f uniformly on compacta if

fn → f

uniformly on every compact subset K ⊂ D.

Theorem 9.3. Suppose that fn, f : D → C are functions,each fn is analytic, and fn → funiformly on compacta. Then f is analytic on D.

49

Proof. For any z in D, take δ > 0 such that Bδ(z) ⊂ D. Since fn → f uniformly on

Bδ(z), by the uniform limit theorem, f is continuous on Bδ(z). Since z is arbitrary, f iscontinuous on D. For any rectangle R ⊂ D with boundary Γ, since fn → f uniformly on R,∫

Γ

f =

∫Γ

limn→∞

fn = limn→∞

∫Γ

fn = limn→∞

0 = 0

By Morera’s theorem, f is analytic on D. �


f(z) =

∫ ∞0

e−zt

t+ 1dt

defined onD = {z ∈ C|Re(z) > 0}

is analytic.

Proof. Let

fn(z) =

∫ n

0

e−zt

t+ 1dt

where z ∈ D. Let Γ be the boundary of a rectangle R ⊂ D. Note that for each fixed t > 0,e−zt is entire. Using Fubini’s theorem to interchange the order of integrals, we have∫

Γ

fn(z)dz =

∫Γ

∫ n

0

e−zt

t+ 1dtdz =

∫ n

0

∫Γ

e−zt

t+ 1dzdt =

∫ n

0

0dt = 0

By Morera’s theorem, fn is analytic on D. Furthermore, for n ∈ N,

|fn(z)− f(z)| = |∫ ∞n

e−zt

t+ 1dt| ≤

∫ ∞n

|e−zt|dt =

∫ ∞n

e−Re(z)tdt =e−Re(z)t

Re(z)

∣∣∣∣∣∞

n

=e−Re(z)n

Re(z)

For K ⊂ D a compact set, there is a > 0 such that

a ≤ Re(z)

for z ∈ K. For any ε > 0, there is N ∈ N such that

e−an

a< ε

for n ≥ N . Then for n ≥ N and z ∈ K, we have

−Re(z)n ≤ −anand

|fn(z)− f(z)| ≤ e−Re(z)n

Re(z)≤ e−an

a< ε

Hence fn → f uniformly on K. By theorem, f is analytic on D. �

Theorem 9.5. Let D be a region. Suppose that f is continuous on D and analytic onD − L where L is a line segment. Then f is analytic on D.

Proof. By making the transformation z → Az + B for some constants A,B, we mayassume that L is contained in the real line.

Let Γ be the boundary of a rectangle R ⊂ D.

50

Case 1: : R ∩ L = ∅. Since f is analytic on R.∫Γ

f(z)dz = 0

Case 2: : L ∩ R ⊂ Γ. We consider R = [a, b] × [0, d] in the upper half-plane. Thecase R in the lower half-plane is similar. Define g : [0, d]→ C by

g(y) :=

∫ b

a

f(x+ iy)dx

For any given ε > 0, since f is uniformly continuous on R, there is δ > 0 such thatif |z1 − z2| < δ,

|f(z1)− f(z2)| < ε

b− aFix y0 ∈ [0, d]. If |y − y0| < δ,

|g(y)− g(y0)| ≤∫ b

a

|f(x+ iy)− f(x+ iy0)|dx < ε

b− a(b− a) = ε

Therefore g is a continuous function. Let Γε be the boundary of the rectangle

Rε = [a, b]× [ε, d]

shifted the bottom side of R up by ε. Since

limy→0+

g(y) = g(0)

we have

limε→0+

∫ b

a

f(x+ iε)dx =

∫ b

a

f(x)dx

Therefore

limε→0+

∫Γε

f(z)dz =

∫Γ

f(z)dz

Since f is analytic on Rε, ∫Γε

f(z)dz = 0

Therefore ∫Γ

f(z)dz = 0

By Morera’s theorem, f is analytic on D.Case 3: : R ∩ L * Γ. Split

R = R1 ∪R2

where R1, R2 are rectangles such that

R1 ∩R2 = Γ1 ∩ Γ2 and L ∩R ⊂ Γ1 ∩ Γ2

where Γ1 = ∂R1 and Γ2 = ∂R2. Then use Case 2.

�

51

CHAPTER 10

Simply connected regions

Definition 10.1. We say that a continuous function Γ : [a, b] × [c, d] → C is piecewiseC1 if for any s0 ∈ [a, b], Γs0 : [c, d]→ C defined by

Γs0(t) = Γ(s0, t)

is a piecewise C1-curve and for any t0 ∈ [c, d], Γt0 : [a, b]→ C defined by

Γt0(s) = Γ(s, t0)

is a piecewise C1-curve.

Definition 10.2. If γ : [0, 1]→ X is a closed continuous curve where X is a topologicalspace, then the point

γ(0) = γ(1)

is called the based point of γ.

Definition 10.3. Let γ0, γ1 : [0, 1]→ D be two closed piecewise C1-curves in a region D.We say that γ0 is homotopic to γ1 if there is a piecewise C1 function H : [0, 1]× [0, 1]→ Dsuch that {

H(s, 0) = γ0(s), H(s, 1) = γ1(s), where 0 ≤ s ≤ 1;H(0, t) = H(1, t), where 0 ≤ t ≤ 1.

Such H is called a homotopy between γ0 and γ1.

Definition 10.4. Let U ⊂ C be an open set. A piecewise C1-curve γ : [0, 1]→ U is saidto be a constant curve if γ is a constant function. The set U is said to be simply connectedif U is connected and every closed piecewise C1-curve in U is homotopic to a constant curve.

Proposition 10.5. For any z0 ∈ C and any r ∈ (0,∞], any two closed piecewise C1-curves in Br(z0) with the same based point are homotopic. In particular, the open ball Br(z0)is simply connected.

Proof. Let γ1, γ2 : [0, 1] → Br(z0) be two closed piecewise C1-curves with the samebased point z. Define H : [0, 1]× [0, 1]→ Br(z0) by

H(s, t) := (1− t)γ1(s) + tγ2(s)

Then H is a homotopy between γ1 and γ2. �

Theorem 10.6. (Homotopy theorem) If γ0 and γ1 are two homotopic closed piecewiseC1-curves in a region D, then ∫

γ0

f(z)dz =

∫γ1

f(z)dz

for any analytic function on D.

53

Proof. Let I = [0, 1] and H be a homotopy between γ0 and γ1. Since H is continuousand I2 is compact, H is uniformly continuous on I2 and H(I2) is a compact set. Let

r := distance between H(I2) and C−D:= inf{|z − w||z ∈ H(I2), w ∈ C−D}

From advanced calculus, we know that r > 0. Hence there is n ∈ N such that if(s, t), (s′, t′) ∈ I2 and

√(s− s′)2 + (t− t′)2 < 2

n,

|H(s, t)−H(s′, t′)| < r

Let

zj,k := H(j

n,k

n), 0 ≤ j, k ≤ n

and

Ij,k := [j

n,j + 1

n]× [

k

n,k + 1

n]

for 0 ≤ j, k ≤ n− 1. Since the diameter of Ij,k is√

2

n<

2

n

we have

H(Ij,k) ⊂ Br(zj,k)

Let Pj,k be the closed curve H|∂(Ij,k). Then the image of Pj,k is contained in Br(zj,k). By theclosed curve theorem of an analytic function in a ball, we have∫

Pj,k

f(z)dz = 0

Let Qk be the curve H|I×{ kn}. Since H(0, k

n) = H(1, k

n), this curve is closed. Denote

[z0,k, z0,k+1] = H|{0}×[ kn, k+1n

]

Sincen−1∑j=0

∫Pj,k

f(z)dz = 0

we have ∫Qk

f(z)dz +

∫[zn,k,zn,k+1]

f(z)dz +

∫−Qk+1

f(z)dz +

∫[z0,k+1,z0,k]

f(z)dz = 0

Furthermore, since

z0,k = H(0,k

n) = H(1,

k

n) = zn,k

we have

[z0,k+1, z0,k] = −[z0,k, z0,k+1] = −[zn,k, zn,k+1]

and hence ∫[z0,k+1,z0,k]

f(z)dz +

∫[zn,k,zn,k+1]

f(z)dz = 0

54

This implies that ∫Qk

f(z)dz =

∫Qk+1

f(z)dz

Therefore ∫γ0

f(z)dz =

∫Q0

f(z)dz =

∫Q1

f(z)dz = · · · =∫Qn

f(z)dz =

∫γ1

f(z)dz

�

Corollary 10.7. (General integral theorem) Let D be a simply connected region andf : D → C be an analytic function. There exists an analytic function F : D → C such that

F ′ = f

Proof. Fix z0 ∈ D. For z ∈ D, let γ : [0, 1] → D be a piecewise C1-curve such thatγ(0) = z0, γ(1) = z. Define

F (z) :=

∫γ

f(w)dw

We need to show that F is well defined. Let γ1, γ2 be two piecewise C1-curves joining z0 toz. Since D is simply connected, the closed curve γ1 − γ2 is homotopic to a constant curveγ0. Therefore ∫

γ1

f(w)dw −∫γ2

f(w)dw =

∫γ1−γ2

f(w)dw =

∫γ0

f(w)dw = 0

So ∫γ1

f(w)dw =

∫γ2

f(w)dw

which means that F is well defined. Take r > 0 such that Br(z) ⊂ D. Then for anyu ∈ Br(z),

F (u) +

∫ z

u

f(w)dw = F (z)

Thus

|F (u)− F (z)

u− z− f(z)| = | 1

u− z

∫ z

u

f(w)− f(z)dw| ≤ supu∈Br(z)

{|f(u)− f(z)|}

Since f is a continuous function, as r → 0+, the supremum goes to 0. This implies that Fis differentiable at z and

F ′(z) = f(z)

�

Corollary 10.8. (General closed curve theorem) Suppose that f is analytic in a simplyconnected region D and C is a closed piecewise C1-curve in D. Then∫

C

f(w)dw = 0

55

Proof. Suppose that C is given by γ : [0, 1]→ C. Since C is closed, we have γ(0) = γ(1).By the general integral theorem, there is an analytic function F : D → C such that

F ′ = f

Then∫C

f(w)dw =

∫ 1

0

f(γ(t))γ′(t)dt =

∫ 1

0

F ′(γ(t))γ′(t)dt =

∫ 1

0

d

dt(F◦γ)dt = F (γ(1))−F (γ(0)) = 0

�

56

CHAPTER 11

Singularities

Definition 11.1. A deleted neighborhood of z0 is a set of the form

B′r(z0) := Br(z0)− {z0}A function f has an isolated singularity at z if there is r > 0 such that f is analytic on B′r(z0)but f is not defined or not continuous at z0. The point z0 is called a removable singularityif there is r > 0 and an analytic function g : Br(z0)→ C such that

g(z) = f(z)

for z ∈ B′r(z0).

Example 11.2. From the power series expansion of sin z at 0, we see that the function

g(z) =

{sin z

z, if z 6= 0

1, if z = 0

is entire and equal tosin z

zon C−{0}. Therefore the function sin z

zhas a removable singularity at 0. It is clear that the

singularity of 1z

is not removable.

Theorem 11.3. Suppose that f has an isolated singularity at z0. Then z0 is a removablesingularity if and only if

limz→z0

(z − z0)f(z) = 0

Proof. If z0 is a removable singularity of f , there is r > 0 and analytic function gdefined on Br(z0) such that

g(z) = f(z)

for z ∈ B′r(z0). Thenlimz→z0

(z − z0)f(z) = limz→z0

(z − z0)g(z) = 0

For the converse, let

h(z) =

{(z − z0)f(z), z 6= z0

0, z = z0

By hypothesis, h is continuous at z0. Since h is analytic in a deleted neighborhood of z0, bytheorem, h is analytic at z0. Let

g(z) =

h(z)

z − z0

, z 6= z0

h′(z0), z = z0

57

By theorem, g is analytic. Sinceg(z) = f(z)

for z 6= z0, by definition, z0 is a removable singularity of f . �

Corollary 11.4. If f is bounded in a deleted neighborhood of an isolated singularity,the singularity is removable.

Proof. Suppose that f is bounded on B′r(z0). Let

M = max{|f(z)||z ∈ B′r(z0)}Then

|z − z0||f(z)| ≤ |z − z0|M → 0

as z → z0. Thereforelimz→z0|z − z0||f(z)| = 0

By theorem above, z0 is removable. �

Definition 11.5. Suppose that h is an analytic function in Br(z0). We say that z0 is azero of h of order k ∈ N if

h(z0) = h(1)(z0) = · · · = h(k−1)(z0) = 0

buth(k)(z0) 6= 0

Definition 11.6. Suppose that f is analytic in B′r(z0). If f can be written in the form

f(z) =g(z)

h(z)

in B′r(z0) where g, h are analytic in Br(z0), g(z0) 6= 0 and h(z0) = 0, then we say that f hasa pole at z0. If h has a zero of order k at z0, we say that f has a pole of order k. If f hasneither a removable singularity nor a pole at z0, we say that f has an essential singularityat z0.

Theorem 11.7. Suppose that f is analytic in B′r(z0). Then f has a pole of order k atz0 if and only if

limz→z0

(z − z0)kf(z) 6= 0

butlimz→z0

(z − z0)k+1f(z) = 0

Proof. Suppose that f has a pole of order k at z0. By definition,

f(z) =g(z)

h(z)

on B′r(z0) where g, h are analytic functions in Br(z0). Since z0 is a zero of order k of h, thepower series expansion of h at z0 is

h(z) =∞∑n=0

h(n)(z0)

n!(z − z0)n =

∞∑n=k

h(n)(z0)

n!(z − z0)n = (z − z0)kp(z)

58

where

p(z) =∞∑n=k

h(n)(z0)

n!(z − z0)n−k

is analytic in Br(z0) and p(z0) 6= 0. Therefore

limz→z0

(z − z0)kf(z) = limz→z0

(z − z0)kg(z)

(z − z0)kp(z)=g(z0)

p(z0)6= 0

and

limz→z0

(z − z0)k+1f(z) = limz→z0

(z − z0)k+1 g(z)

(z − z0)kp(z)= lim

z→z0(z − z0)

g(z)

p(z)= 0.

For the converse, let

g(z) =

{(z − z0)k+1f(z), z 6= z0

0, z = z0

Then g is continuous on Br(z0) and by theorem, analytic in B′r(z0). Let

h(z) :=

g(z)

z − z0

, z 6= z0

g′(z0), z = z0

Then h is analytic in a neighborhood of z0. Because

h(z) = (z − z0)kf(z)

in B′r(z0) and

h(z0) = limz→z0

(z − z0)kf(z) 6= 0

Therefore

f(z) =h(z)

(z − z0)k

for z 6= z0 and hence z0 is a pole of order k of f . �

Example 11.8. (1) The function 1zn

is clearly has a pole of order n at 0.

(2) The function e1z has an essential singularity at 0 since

limz→0

zne1z = lim

z→0zn

∞∑k=0

1

k!z−k

is never 0 for any n ∈ N.

Remark 11.9. We may view a removable singularity as a pole of order 0, and an essentialsingularity as a pole of order ∞. The theorem above gives a unified method to determinewhat kind of isolated singularity of a function at a point is.

Theorem 11.10. (Casorati-Weierstrass theorem) If f has an essential singularity at z0

and f is analytic in B′r(z0), then the range of f

range(f) = {f(z)|z ∈ B′r(z0)}is dense in the complex plane.

59

Proof. If range(f) is not dense in C, there exists w in C and δ > 0 such that

Bδ(w) ∩ range(f) = φ

Therefore |f(z)− w| > δ and1

|f(z)− w|<

1

δ

for all z in B′r(z0). By theorem, this boundedness implies that

1

f(z)− whas a removable singularity at z0. Then there is a function g(z) analytic in Br(z0) such that

g(z) =1

f(z)− wfor z ∈ B′r(z0). Then

f(z) = w +1

g(z)

and hence f has a pole at z0 if g(z0) = 0 and has a removable singularity at z0 if g(z0) 6= 0which contradicts to the given hypothesis that z0 is an essential singularity of f . �

60

CHAPTER 12

Laurent expansions

Definition 12.1. The radius of convergence of the series

p(z) =∞∑k=1

a−k(z − z0)−k

is

R = inf{r ∈ [0,∞] : p(z) converges, for all z where |z − z0| > r}

Proposition 12.2. The radius of convergence of the series

p(z) =∞∑k=1

a−k(z − z0)−k

is

R = limk→∞

sup |a−k|1k

Proof. By the root test, p(z) converges if

limk→∞

sup |a−k(z − z0)−k|1k < 1

and p(z) diverges if

limk→∞

sup |a−k(z − z0)−k|1k > 1

This implies that the series converges if

|z − z0| > limk→∞

sup |a−k|1k

and the series diverges if

|z − z0| < limk→∞

sup |a−k|1k

This completes the proof. �

Remark 12.3. If the limit

limk→∞|a−(k+1)

a−k|

exists, then

R = limk→∞|a−(k+1)

a−k|

61

Definition 12.4. We say that the series∞∑

k=−∞

µk = L

if∞∑k=0

µk and∞∑k=1

µ−k

converge and∞∑k=0

µk +∞∑k=1

µ−k = L

Proposition 12.5. Let

f(z) =∞∑

k=−∞

ak(z − z0)k

Then f is analytic in the region

D = {z : R1 < |z − z0| < R2}where

R1 = radius of convergence of∞∑k=1

a−k(z − z0)−k

R2 = radius of convergence of∞∑k=0

ak(z − z0)k

Proof. Let

g1(z) =∞∑k=1

a−k(z − z0)−k

g2(z) =∞∑k=0

ak(z − z0)k

Since g1(z) is convergent for |z − z0| > R1,

h(z) := g1(1

z − z0

+ z0) =∞∑k=1

a−k(z − z0)k

is convergent for |z − z0| < 1R1

. By theorem, h is analytic in the region |z − z0| < 1R1

.Therefore

g1(z) = h(1

z − z0

+ z0)

is analytic if |z − z0| > R1. By definition, g2(z) converges for |z − z0| < R2, hence

f(z) = g1(z) + g2(z)

is analytic in the intersection region R1 < |z − z0| < R2. �

Example 12.6. Find the radius of convergence of

62

(1)∞∑k=1

1

zk

(2)∞∑k=1

2k1

zk

Solution 12.7. (1) If

limk→∞

| 1zk+1 |

1|zk|

=1

|z|< 1

we have |z| > 1 and hence R = 1.(2) If

limk→∞

|2k+1 1zk+1 |

|2k 1zk|

=2

|z|< 1

we have |z| > 2 and hence R = 2.

Definition 12.8. An annulus centered at z0 with radius R1 and R2 is a set of the form

A(R1, R2; z0) := {z|R1 < |z − z0| < R2}We may just write A(R1, R2) if the center is the origin.

If

f(z) =∞∑

k=−∞

ak(z − z0)k

in some annulus A(R1, R2; z0), the series∞∑

k=−∞

ak(z − z0)k

is called the Laurent series of f centered at z0.

Theorem 12.9. If f is analytic in the annulus A(R1, R2; z0), then f has a Laurentexpansion

f(z) =∞∑

k=−∞

ak(z − z0)k

in A(R1, R2; z0).

Proof. Fix z ∈ A(R1, R2; z0). Let C1, C2 be two circle centered at z0 with radius r1 andr2 respectively where

R1 < r1 < |z − z0| < r2 < R2

For w ∈ A(R1, R2; z0), let

g(w) =

f(w)− f(z)

w − z, w 6= z

f ′(z), w = z

63

Then g is analytic in A(R1, R2; z0). Join C1 and C2 by two line segments L1 and L2. Then

A(r1, r2; z0) is separated into two closed subsets D1 and D2. For each Dj, take an open setUj which is simply connected and contains Dj for j = 1, 2. By the closed curve theorem,∫

∂D1

g(w)dw =

∫∂D2

g(w)dw = 0

Since in the curve∂D1 ∪ ∂D2

L1 and L2 are crossed twice in different direction, we have∫C2−C1

g(w)dw =

∫∂D1

g(w)dw +

∫∂D2

g(w)dw = 0

From the definition of g, we have∫C2−C1

f(w)

w − zdw =

∫C2−C1

f(z)

w − zdw = f(z)[

∫C2

1

w − zdw −

∫C1

1

w − zdw] = (2πi)f(z)

Hence

f(z) =1

2πi

∫C2

f(w)

w − zdw − 1

2πi

∫C1

f(w)

w − zdw

For w in C1, |w − z0| < |z − z0|, we have

1

w − z=

1

(w − z0)− (z − z0)=−1

z − z0

1

1− w−z0z−z0

= − −1

z − z0

∞∑n=0

(w − zz − z0

)n

For w in C2, |w − z0| > |z − z0|, we have

1

w − z=

1

(w − z0)− (z − z0)=

1

w − z0

1

(1− z−z0w−z0 )

=1

w − z0

∞∑n=0

(z − z0

w − z0

)n

Hence

f(z) =1

2πi

∫C2

(∞∑k=0

f(w)(z − z0)k

(w − z0)k+1)dw +

1

2πi

∫C1

∞∑k=1

f(w)(w − z0)k−1

(z − z0)kdw

=∞∑k=0

(1

2πi

∫C2

f(w)

(w − z0)k+1dw)(z − z0)k +

−∞∑k=−1

(1

2πi

∫C1

f(w)

(w − z0)k+1dw)(z − z0)k

Note thatf(w)

(w − z0)k+1

is analytic in A(R1, R2; z0), by the closed curved theorem,∫C

f(w)

(w − z0)k+1dw =

∫C1

f(w)

(w − z0)k+1dw =

∫C2

f(w)

(w − z0)k+1dw

for any circle C in A(R1, R2; z0).Therefore

f(z) =∞∑

k=−∞

ak(z − z0)k

64

where

ak =1

2πi

∫C

f(w)

(w − z0)k+1dw

for any circle C in A(R1, R2; z0). �

Theorem 12.10. The Laurent expansion of an analytic function in A = A(R1, R2; z0) isunique.

Proof. Let

f(z) =∞∑

n=−∞

an(z − z0)n

be the Laurent expansion of f . Since the series∞∑

n=−∞

an(z − z0)n

converges uniformly on C where C is a circle in A centered at z0,∫C

f(z)

(z − z0)k+1dz =

∞∑n=−∞

∫C

an(z − z0)n−k−1dz

Moreover, from the calculation∫C

(z − z0)pdz =

{2πi, p = −10, otherwise

we have ∫C

f(z)

(z − z0)k+1dz =

∫C

ak(z − z0)k−k−1dz = 2πiak

which shows that

ak =1

2πi

∫C

f(z)

(z − z0)k+1dz

Since by the closed curve theorem, this integral is independent of the choice of the circle C,ak is uniquely determined by f . �

Example 12.11. Find the Laurent expansion of

f(z) =1

z2(1− z)

in some maximal annulus centered at

(1) z = 0(2) z = 1

Solution 12.12. (1)

f(z) =1

z2(1− z)=

1

z2(∞∑k=0

zk) =1

z2+

1

z+ 1 + z + z2 + · · ·

for 0 < |z| < 1. Therefore the maximal annulus for this expansion is A(0, 1; 0).

65

(2)

f(z) =1

z2(1− z)=

−1

(1− (1− z))2(z − 1)= − 1

z − 1(∞∑k=0

(1− z)k)2

=−1

z − 1(1− 2(z − 1) + 3(z − 1)2 − 4(z − 1)3 + 5(z − 1)4 + · · · )

= − 1

z − 1+ 2− 3(z − 1) + 4(z − 1)2 − 5(z − 1)3 + · · ·

in A(0, 1; 1).

Corollary 12.13. If f : B∗r (z0) → C is an analytic function, then f has a Laurentseries expansion

f(z) =∞∑n=0

an(z − z0)n +∞∑m=1

bm(z − z0)−m

and the principal part

P (1

z − z0

) :=∞∑m=1

bm(z − z0)−m

is analytic in C− {z0} which converges uniformly on compacta on C− {z0}.

Proof. By Theorem, f has a unique Laurent series expansion on 0 < |z − z0| < r.Therefore the principal part P of f converges and is analytic on |z − z0| > 0. This gives usthe desire result. �

66

CHAPTER 13

Residues

Definition 13.1. If

f(z) =∞∑

k=−∞

ak(z − z0)k

is the Laurent expansion of f about an isolated singularity z0, the series−1∑

k=−∞

ak(z − z0)k

is called the principal part of f at z0;∑∞

k=0 ak(z − z0)k is called the analytic part.

Theorem 13.2. (Classification of isolated singularities by Laurent series) Suppose thatf has an isolated singularity at z0 and

f(z) =∞∑

k=−∞

ak(z − z0)k

is the Laurent series of f in A(R1, R2; z0) for some 0 < R1 < R2.

(1) f has a removable singularity at z0 if and only if

a−k = 0

for all k = 1, 2, ....(2) f has a pole of order k if and only if a−k 6= 0 and

a−(k+1) = a−(k+2) = · · · = 0

(3) f has an essential singularity at z0 if and only if for any N > 0, there exists k > Nsuch that

a−k 6= 0

Proof. (1) Suppose that f has a removable singularity at z0. Let g be an analyticfunction in Br(z0) such that

g(z) = f(z)

for z in B∗r (z0). By the uniqueness of Laurent series, the power series expansion ofg at z0 is the Laurent series of f centered at z0. Therefore all

a−k = 0

for k = 1, 2, .... On the other hand, if all a−k = 0, then

f(z) =∞∑k=0

ak(z − z0)k

67

in B∗r (z0) is a power series. But

∞∑k=0

ak(z − z0)k

is analytic in Br(z0) and hence z0 is a removable singularity of f .(2) Suppose that f has a pole of order k at z0. Then we may write

f(z) =Q(z)

(z − z0)k

where Q(z) is analytic in a neighborhood of z0 and Q(z0) 6= 0. By the power seriesexpansion of Q, we have

Q(z) =∞∑n=0

bn(z − z0)n

Since Q(z0) 6= 0, b0 6= 0. Then the Laurent series of f at z0 is

f(z) =∞∑n=0

bn(z − z0)n−k =∞∑

n=−k

dn(z − z0)n

where dn = bn+k, d−k = b0 6= 0 which proves the result. For the converse, supposethat a−k 6= 0, and a−(k+1) = a−(k+2) = · · · = 0, then

f(z) =∞∑

n=−k

an(z − z0)n =

∑∞n=−k an(z − z0)n+k

(z − z0)k=

∑∞n=0 an−k(z − z0)n

(z − z0)k

Since∞∑n=0

an−k(z − z0)n

is analytic in a neighborhood of z0 and 6= 0 when z = z0, by definition, f has a poleof order k at z0.

(3) From the above results, we see that z0 is an essential singularity if and only if theprincipal part has infinitely many terms.

�

Definition 13.3. If

f(z) =∞∑

k=−∞

ck(z − z0)k

in some deleted neighborhood of z0, c−1 is called the residue of f at z0. We write

Res(f ; z0) := c−1

Proposition 13.4. Suppose that f has a pole at z0 of order k, then

Res(f ; z0) =1

(k − 1)!limz→z0

dk−1

dzk−1[(z − z0)kf(z)]

68

Proof. Let

f(z) =c−k(z − z0)−k + c−(k−1)(z − z0)−(k−1) + · · ·+ c−1(z − z0)−1 + c0 + c1(z − z0) + · · ·g(z) =(z − z0)kf(z) = c−k + c−(k−1)(z − z0)1 + c−1(z − z0)k−1 + c0(z − z0)k + c1(z − z0)k+1 + · · · .

Therefore

dk−1

dzk−1g(z) = (k − 1)!c−1 +

k!

1!c0(z − z0) +

(k + 1)!

2!c1(z − z0)2 + · · ·

and hence

limz→z0

dk−1g

dzk−1= (k − 1)!c−1

�

Example 13.5. Find the residues:

(1) Res( 1z100−1

; 1).

(2) Res( 1z100−1

; i).

(3) Res( 1z100−1

; 2i).

(4) Res( 1(z−1)100

; 1).

(5) Res( 1(z−2)2(z−3)101(z−4)2

; 3).

(6) Res(sin 1z−1

; 1).

Solution 13.6. (1) 1 is a simple pole of 1z100−1

, hence the residue is

limz→1

(z − 1)1

z100 − 1= lim

z→1

1

100z99=

1

100

(2) i is a simple pole of 1z100−1

, hence the residue is

limz→i

(z − i) 1

z100 − i= lim

z→i

1

100z99=

i

100

(3) Since 1z100−1

is analytic at 2i, the residue is 0.

(4) Since 1 is a pole of order 100 of 1(z−1)100

, and the Laurent series of 1(z−1)100

at 1 is

itself, we see that its residue at 1 is 0.(5) Since 3 is a pole of order 101 of

1

(z − 2)2(z − 3)101(z − 4)2

to calculate the residue, by proposition above, we need to find the coefficient of theterm (z − 3)100 of the power series of

1

(z − 2)2(z − 4)2

69

expanded at 3. Write

1

(z − 2)2= (

1

1− (−(z − 3)))2 = (

∞∑n=0

(−1)n(z − 3)n)2 = 1− 2(z − 3) + 3(z − 3)2 − · · ·

1

(z − 4)2= (

−1

1− (z − 3))2 = (

∞∑n=0

(z − 3)n)2 = 1 + 2(z − 3) + 3(z − 3)2 + · · ·

Therefore, the coefficient of the term (z − 3)100 is

100∑k=0

(−1)k(k+1)(101−k) =49∑n=0

(2n+1)(101−2n)−(2n+2)(100−2n)+101 =49∑n=0

(4n−99)+101 = 51

which is the desire residue.(6) The power series expansion of sin at 0 is

sinw = w − w3

3!+w5

5!− · · ·

therefore

sin1

z − 1= (z − 1)−1 − 1

3!(z − 1)−3 +

1

5!(z − 1)−5 − · · ·

and hence the residue is 1.

Definition 13.7. Suppose that γ is a closed piecewise C1 curve and a is not in γ. Then

n(γ; a) :=1

2πi

∫γ

dz

z − a

is called the winding number of γ around a.

We first show that this number coincides with our intuition of how many times a closedcurve going around a point.

Example 13.8. Let

γ(θ) = a+ reiθ, 0 ≤ θ ≤ 2kπ

where k ∈ N. Then

n(γ; a) =1

2πi

∫γ

1

z − adz =

1

2πi

∫ 2kπ

0

1

reθrieiθdθ

=1

2π

∫ 2kπ

0

dθ =2kπ

2π= k

Proposition 13.9. For any closed piecewise C1 curve γ in C and a point a not in γ,n(γ; a) is an integer.

Proof. Suppose that γ is given by z(t), 0 ≤ t ≤ 1. Define F : [0, 1]→ C by

F (s) =

∫ s

0

z′(t)

z(t)− adt

70

Since γ is piecewise C1, z′(t), and hence z′(t)z(t)−a , has at most finitely many discontinuities.

From the Fundamental Theorem of Calculus, we have

F ′(s) =z′(s)

z(s)− aLet

g(s) = (z(s)− a)e−F (s)

Then

g′(s) = z′(s)e−F (s) + (z(s)− a)e−F (s)(−F ′(s)) = z′(s)e−F (s) − (z(s)− a)e−F (s) z′(s)

z(s)− a= 0

except for possibly finitely many points 0 = s0, s1, s2, ..., sn−1, sn = 1. Therefore

g|[sj ,sj+1] ≡ cj

for some constants cj, j = 0, 1, 2, ..., n− 1. But g is continuous, so

c1 = c2 = · · · = cn−1 = c

are the same constant. Then

(z(s)− a)e−F (s) ≡ c

Since F (0) = 0,

c = z(0)− aFurthermore, since γ is a closed curve,

z(0) = z(1)

Then

e−F (1) =g(1)

z(1)− a=z(0)− az(1)− a

= 1

which implies

F (1) = 2πki

for some k in Z. From the above calculation, we have

n(γ; a) =1

2πi

∫γ

1

z − adz

=1

2πi

∫ 1

0

z′(s)

z(s)− ads =

1

2πF (1)

= k ∈ Z

�

Theorem 13.10. (Residue Theorem) Suppose that f is analytic in a simply connecteddomain D except for isolated singularities at z1, z2, ..., zm. Let γ be a closed piecewise C1

curve not intersecting any of the singularities. Then∫γ

f(z)dz = 2πim∑k=1

n(γ; zk)Res(f ; zk)

71

Proof. Let Pk(1

z−zk) be the principal part of the Laurent expansion of f at zk. By

Corollary 12.13

g(z) := f(z)− P1(1

z − z1

)− · · · − Pm(1

z − zm)

is analytic in D. Since D is simply connected,∫γ

g(z)dz = 0

and hence ∫γ

f(z)dz =m∑k=1

∫γ

Pk(1

z − zk)dz

Recall that ∫γ

1

(z − zk)n= 0

for n 6= 1. Since

Pk(1

z − zk) =

c−1

z − zk+

c−2

(z − zk)2+ · · ·

we have ∫γ

Pk(1

z − zk)dz =

∫γ

c−1

z − zkdz +

∫γ

c2

(z − zk)2dz + · · ·

= c−1

∫γ

1

z − zkdz = 2πin(γ; zk)Res(f ; zk)

Therefore ∫γ

f(z)dz = 2πim∑k=1

n(γ; zk)Res(f ; zk)

�

Definition 13.11. We say that f is meromorphic in a region D if f is analytic in Dthere except at isolated poles.

Theorem 13.12. (Argument principle) Let f be a meromorphic function in a simplyconnected region D with poles p1, ..., pm and zeros z1, ..., zn. Let ord(pi) = the order of polepi, and ord(zi) = the order of zero zi. If γ is a closed curve in D and not passing throughp1, ..., pm, z1, ..., zn, then

1

2πi

∫γ

f ′(z)

f(z)dz =

n∑k=1

ord(zk)n(γ; zk)−m∑j=1

ord(pj)n(γ; pj)

Proof. If f has a zero of order k at z = a,

f(z) = (z − a)kg(z)

where g(a) 6= 0, sof ′(z) = k(z − a)k−1g(z) + (z − a)kg′(z)

Hencef ′(z)

f(z)=

k

z − a+g′(z)

g(z)

72

This shows that f ′

fhas a simple pole at a with residue k. Similarly, if f has a pole of order

l at z = b, thenf(z) = (z − b)−lh(z)

where h(b) 6= 0. Thereforef ′(z)

f(z)= − l

z − b+h′(z)

h(z)

Hence f ′

fhas a simple pole at b with residue −l. By the residue theorem,

1

2πi

∫γ

f ′(z)

f(z)dz =

n∑k=1

n(γ; zk)ord(zk)−m∑j=1

n(γ; pj)ord(pj)

�

73

CHAPTER 14

Applications in integration: I

In the following, we denote LR to be the line segment from −R to R,

H+ := {z ∈ C : Imz ≥ 0}H− := {z ∈ C : Imz ≤ 0}

to be the upper half-plane and lower half-plane respectively, and Γ+R,Γ

−R to be the semi-circles

with radius R centered at 0 in H+ and H− respectively. Furthermore, let

C+R := LR ∪ Γ+

R, C−R := Γ−R ∪ (−LR)

be the upper half-circle and lower half-circle oriented counterclockwise. For a polynomialQ(z), we write Z(Q) for the set of all zeros of Q, and

Z+(Q) := Z(Q) ∩H+, Z−(Q) := Z(Q) ∩H−

Proposition 14.1. Suppose that P (x), Q(x) are real polynomials with the conditionsthat Q(x) 6= 0 for all x ∈ R and degQ− degP ≥ 2. Then∫ ∞

−∞

P (x)

Q(x)dx = 2πi

∑z∈Z+(Q)

Res(P

Q; z) = −2πi

∑z∈Z−(Q)

Res(P

Q; z)

Proof. By the Cauchy residue theorem,∫C+R

P (z)

Q(z)dz = 2πi

∑z∈Z+(Q)

Res(P

Q; z)

Then ∫ 0

−R

P (x)

Q(x)dx+

∫ R

0

P (x)

Q(x)dx+

∫Γ+R

P (z)

Q(z)dz = 2πi

∑z∈Z+(Q)

Res(P

Q; z)

Since degQ− degP ≥ 2,

|P (z)

Q(z)| ≤ A

R2

for |z| large and some constant A > 0. By the ML-formula, we have

limR→∞

|∫

Γ+R

P (z)

Q(z)dz| ≤ lim

R→∞πR

A

R2= 0

Therefore∫ ∞−∞

P (x)

Q(x)dx = lim

R→∞

∫ 0

−R

P (x)

Q(x)dx+ lim

R→∞

∫ R

0

P (x)

Q(x)dx+ = 2πi

∑z∈Z+(Q)

Res(P

Q; z)

75

Instead of C+R , if we use C−R ,∫

Γ−R

P (z)

Q(z)dz +

∫ 0

R

P (x)

Q(x)dx+

∫ −R0

P (x)

Q(x)dx = 2πi

∑z∈Z−(Q)

Res(P

Q; z)

Then same calculation gives the desire result. �

Example 14.2. Compute ∫ ∞−∞

1

x4 + 1dx

Solution 14.3. If z4 + 1 = 0, then z4 = −1 = eπi, so we have

z = eπ+2kπ

4i where k = 0, 1, 2, 3

So

z1 = eπ4i =

√2

2+ i

√2

2

z2 = e3π4i = −

√2

2+ i

√2

2

are the two simple poles of 1z4+1

that lie in the upper half-plane. We compute

Res(1

z4 + 1; z1) = lim

z→z1(z − z1)

1

z4 + 1= lim

z→z1

1

4z3=

1

4z31

= −1

4z1 = −

√2

8− i√

2

8

Similarly,

Res(1

z4 + 1; z2) =

√2

8− i√

2

8so ∫ ∞

−∞

1

x4 + 1dx = 2πi(−i

√2

4) =

√2π

2

Proposition 14.4. Suppose that P,Q are polynomials, degQ > degP , and Q(x) 6= 0 forx ∈ R. Then ∫ ∞

−∞

P (x)

Q(x)cosxdx = Re(2πi

∑z∈Z+(Q)

Res(P (z)

Q(z)eiz; z))

and ∫ ∞−∞

P (z)

Q(z)sinxdx = Im(2πi

∑z∈Z+(Q)

Res(P (z)

Q(z)eiz; z))

Proof. First we show that∫Γ+R

P (z)

Q(z)eizdz → 0 as R→∞

Fix h ∈ (0, R). LetA = {z ∈ Γ+

R|Imz ≥ h}and

B = {z ∈ Γ+R|Imz < h}76

Since

|P (z)

Q(z)| ≤ k

R

for R large enough and some k > 0,

|eiz| = e−Imz

we have

|∫A

P (z)

Q(z)eizdz| ≤ k

Re−hπR = πke−h

Let ` = Rθ be the arc length of B in the first quadrant, then h = R sin θ, and

1 ≤ `

h=

θ

sin θ≤ π

2< 2

for θ ∈ [0, π2] which implies that ` < 2h. Then for z ∈ B, |eiz| = e−Imz < e0 = 1, and

|∫B

P (z)

Q(z)eizdz| ≤ k

R· 1 · 4h = 4k

h

R

Take h =√R, then

|∫

Γ+R

P (z)

Q(z)eizdz| ≤ πke−

√R + 4k

√R

R= πke−

√R +

4k√R→ 0 as R→∞

Therefore,

limR→∞

∫C+R

P (z)

Q(z)eizdz =

∫ ∞−∞

P (x)

Q(x)eixdx = 2πi

∑z∈Z+(Q)

Res(P (z)

Q(z)eiz; z)

Then ∫ ∞−∞

P (x)

Q(x)cosxdx = Re

∫ ∞−∞

P (x)

Q(x)eixdx = Re(2πi

∑z∈Z+(Q)

Res(P (z)

Q(z)eiz; z))

and ∫ ∞−∞

P (z)

Q(z)sinxdx = Im

∫ ∞−∞

P (x)

Q(x)eixdx = Im(2πi

∑z∈Z+(Q)

Res(P (z)

Q(z)eiz; z))

�


cosx

x2 + 1dx

Solution 14.6. Since

Res(1

z2 + 1eiz; i) = lim

z→i

z − iz2 + 1

eiz =1

2ie−1

we have∫ ∞−∞

cosx

x2 + 1dx = Re(

∫ ∞−∞

P (x)

Q(x)eixdx) = Re(2πiRes(

1

z2 + 1eiz; i)) = Re(2πi(

1

2ie−1)) = πe−1

77

Remark 14.7. (1) If Q(x) and sinx have a common simple zero x0, to calculate theintegral ∫ ∞

−∞

P (x)

Q(x)sinxdx

we modify the above method by considering∫C+R

P (z)

Q(z)(eiz − 1)dz, if cosx0 = 1;∫

C+R

P (z)

Q(z)(eiz + 1)dz, if cosx0 = −1

Note that if sinx0 = 0, cosx0 = ±1, and

eiz − 1

Q(z)or

eiz + 1

Q(z)

has a removable singularity at x0.(2) If Q(x) and cosx have a common simple zero x0, to calculate the integral∫ ∞

−∞

P (x)

Q(x)cosxdx

we modify the above method by considering∫C+R

P (z)

Q(z)(eiz − i)dz, if sinx0 = 1;∫

C+R

P (z)

Q(z)(eiz + i)dz, if sinx0 = −1


sinx

xdx

Solution 14.9. Since ∫ ∞−∞

sinx

xdx = Im

∫ ∞−∞

eix − 1

xdx

andeiz − 1

zhas a removable singularity at 0, we have∫

C+R

eiz − 1

zdz = 0

Hence ∫ R

−R

eix − 1

xdx =

∫Γ+R

1− eiz

zdz =

∫Γ+R

1

zdz −

∫Γ+R

eiz

zdz = πi−

∫Γ+R

eiz

zdz

As calculated before ∫Γ+R

eiz

zdz → 0 as R→∞

78

we have ∫ ∞−∞

eix − 1

xdx = πi

and hence ∫ ∞−∞

sinx

xdx = π

79

CHAPTER 15

Log and the argument principle

Definition 15.1. Let U be a region. A function f : U → C is said to be a branch oflog z in U if f is continuous in U and

ef(z) = z

for z ∈ U .

Example 15.2. Let

D = {z ∈ C||z − 1| < 1}Find a branch f of log z in D such that

f(1) = 2πi

Solution 15.3. Let

f(z) = ln |z|+ iArgz

and take3

2π < Argz <

5

2π

for z ∈ D. Then f is a branch of log z in D and f(1) = 2πi.

Example 15.4. Let

D = C− {z ∈ R|z ≤ 0}and take

Argz ∈ (π, 3π)

for z ∈ D. Find log i and log(1 + i).

Solution 15.5. We have

log i = ln |i|+ iArgi = ln 1 + i5π

2=

5π

2i

and

log(1 + i) = ln |1 + i|+ iArg(1 + i) = ln√

2 + i(9

4π)

Proposition 15.6. If f, g are two branches of log z in a region U , then

f(z) = g(z) + 2πki

for some k ∈ Z.

81

Proof. Let

h(z) =1

2πi[f(z)− g(z)]

Then h is continuous on U . Write

f(z) = u(z) + iv(z)

For z ∈ U , let

z = Reiθ

for some θ. Since

ef(z) = z

we have

R = eu(z) and θ = v(z) + 2kπ

for some k ∈ Z. Therefore

f(z) = ln |z|+ i(θ − 2kπ)

Similarly,

g(z) = ln |z|+ i(θ − 2k′π)

This implies that

h(z) =1

2πi[i2(k′ − k)π] = k′ − k ∈ Z

Since h is a continuous function, this implies that h(z) ≡ m ∈ Z is a constant function.Therefore

f(z) = g(z) + 2mπi

�

Definition 15.7. Let U = C− {z ∈ R|z ≤ 0} and

f(z) = ln |z|+ iArgz

where −π < Argz < π. Then f is called the principal branch of log z.

Definition 15.8. Fix z0 ∈ C− {0} and fix

θ0 = Argz0 ∈ R

Let

D = C− {tz0|t ∈ [0,∞)}For each z ∈ D, we take Argz ∈ (θ0, θ0 +2π). Then D with the choice θ0 is called the branchcut of the logarithm.

Definition 15.9. (Complex power) Let

D = C− {tz0|t ∈ [0,∞)}

and fix θ0 = Argz0 ∈ [0, 2π) to form a branch cut of the logarithm. For each z, w ∈ D, define

zw := ew(ln |z|+iArgz)

82

Example 15.10. (1) Let

D = C− {z ∈ R|z ≤ 0}and for z ∈ D, take Argz ∈ (−π, π). Then

ii = ei(ln |i|+iArgi) = ei(0+iπ2

) = e−π2

and

iii

= ie−π2 = ee

−π2 (ln 1+iπ2

) = ee−i(π2 )2

(2) LetD = C− {ix|x ≥ 0}

and take Argz ∈ (3π2, 7π

2). Then

ii = ei(ln |i|+iArgi) = ei(i(5π2

)) = e−5π2

and √1 + i = e

12

(ln |1+i|+iArg(1+i)) = e12

(ln√

2+i 9π4

) = 214 ei

9π8

Proposition 15.11. If f is a branch of log z in a region U , then f is analytic and

f ′(z) =1

zin U .

Proof. Recall that if g(f(z)) = z and if g′(f(z)) 6= 0, then by the chain rule, f isdifferentiable at z and

f ′(z) =1

g′(f(z))Since

(ez)′ = ez 6= 0

everywhere andef(z) = z

we see that

f ′(z) =1

ef(z)=

1

zis analytic in U . �

Theorem 15.12. (Existence of branch of log) Suppose that U is simply connected and0 /∈ U . Take z0 ∈ U and fix a value of log z0. Set

f(z) =

∫γ

1

wdw + log z0

where γ is a piecewise C1 curve in U from z0 to z. Then f is a branch of log z in U .

Proof. Since 0 /∈ U , 1z

is analytic on U . The region U is simply connected, so thegeneralized integral theorem tells us that f is an analytic function on U and

f ′(z) =1

zLet

g(z) = ze−f(z)

83

Then

g′(z) = e−f(z) + ze−f(z)(−f ′(z)) = e−f(z) + ze−f(z)(−1

z) = 0

Therefore g is a constant function. Furthermore,

g(z0) = z0e−f(z0) = z0e

− log z0 = 1

we have have g(z) ≡ 1 which implies that

ef(z) ≡ z

So f is a branch of log z in U . �

Definition 15.13. (Inside and outside of a curve) Let γ be a piecewise C1 simple closedcurve in a region D. For a ∈ D that not in γ, a is said to be inside of γ if

n(γ; a) = 1 or − 1

and outside of γ if

n(γ; a) = 0

Definition 15.14. Let γ be a piecewise C1 simple closed curve oriented counterclockwisein a region D, and f be a meromorphic function in D whose zeros and poles do not lie in γ.Let

Zγ(f) = the number of zeros of f inside γ counted according to their order

and

Pγ(f) = the number of poles of f inside γ counted according to their order

Example 15.15. If γ is the unit circle and

f(z) =(z − 1

2)5(z + 2)8

z2(z + 22i)6(z − 3)9

Then

Zγ(f) = 5,Pγ(f) = 8

If γ is the circle |z| = 0.5, then

Zγ(f) = 0, Pγ(f) + 2

Remark 15.16. If γ is positively oriented Jordan curve, then by theorem,

1

2πi

∫γ

f ′(z)

f(z)dz = Zγ(f)− Pγ(f)

According to the argument principle, we have the following result.

Corollary 15.17. If γ is a piecewise C1 simple closed curve oriented counterclockwisein a region D and f is an analytic function on D, then

Zγ(f) =1

2πi

∫γ

f ′(z)

f(z)dz

84

Remark 15.18. Suppose that γ(t), t ∈ [a, b] is a piecewise C1 curve in a region D, andf is an analytic function on D. Let α(t) = f(γ(t)). Then

dα(t) = f ′(γ(t))γ′(t)dt

and

1

2πi

∫ b

a

f ′(γ(t))


1

2πi

∫ b

a

1

α(t)dα(t) =

1

2πi

∫α

1

zdz

= n(α; 0) = n(f ◦ γ; 0)

Lemma 15.19. Suppose that γ is a closed piecewise C1 curve and f is a meromorphicfunction. If f ◦ γ is contained in some region D where a branch of log z exists in D, then∫

γ

f ′(z)

f(z)dz = 0

Proof. Suppose that γ is given by γ(t), t ∈ [0, 1].∫γ

f ′(z)

f(z)dz =

∫ 1

0

f ′(γ(t))


∫ 1

0

1d log(f ◦ γ)(t) = log(f ◦ γ)(t)|10

= log(f ◦ γ)(1)− log(f ◦ γ)(0) = 0

�

Theorem 15.20. (Rouche’s theorem) Suppose that f and g are meromorphic functionson a region D and γ is a piecewise C1 simple closed curve in D oriented counterclockwisewhich does not contain zeros and poles of f and g. If

|f(z)| > |g(z)|

for z ∈ γ, then

Zγ(f + g)− Pγ(f + g) = Zγ(f)− Pγ(f)

Proof. Note that if

h(z) = A(z)B(z)

then

h′(z) = A′(z)B(z) + A(z)B′(z)

thereforeh′(z)

h(z)=A′(z)

A(z)+B′(z)

B(z)

This implies that ∫γ

h′(z)

h(z)dz =

∫γ

A′(z)

A(z)dz +

∫γ

B′(z)

B(z)dz

Consider

f + g = f(1 +g

f)

85

by theorem, we have

Zγ(f + g)− Pγ(f + g) =1

2πi

∫γ

(f + g)′(z)

(f + g)(z)dz

=1

2πi

∫γ

f ′(z)

f(z)dz +

1

2πi

∫γ

(1 + gf)′(z)

(1 + gf)(z)

dz

By assumption

|g(z)| < |f(z)|for z ∈ γ, we see that the curve

(g

f) ◦ γ

lies in the unit disc B1(0) and hence the curve 1 + ( gf) ◦ γ lies in the open ball B1(1). The

open disc B1(1) is a simply connected region, so a branch of log z exists in B1(1). Accordingto Lemma, ∫

γ

(1 + gf)′(z)

(1 + gf)(z)

dz = 0

By Theorem,

1

2πi

∫γ

f ′(z)

f(z)dz = Zγ(f)− Pγ(f)

so we have

Zγ(f + g)− Pγ(f + g) = Zγ(f)− Pγ(f)

�

Example 15.21. (1) Find the number of zeros of

2z10 + 4z2 + 1

inside the unit disc |z| < 1.(2) Show that the equation

z4 + 26z + 2 = 0

has exactly 3 distinct solutions in A(52, 3; 0).

Solution 15.22. (1) Since |4z2| = 4 and

|2z10 + 1| ≤ 2 + 1 = 3

on |z| = 1, we have

|4z2| > |2z10 + 1|on the unit circle C. But

ZC(4z2) = 2

by Rouche’s theorem, 2z10 + 4z2 + 1 has two zeros in the unit disc.

86

(2) On C1 = {z : |z| = 3}, |z4| = 81,

|26z + 2| ≤ 26(3) + 2 = 80

so|z4| ≥ |26z + 2|

on C1. By Rouche’s theorem,

ZC1(z4 + 26z + 2) = ZC1(z

4) = 4

On C2 = {z||z| = 52}, |26z| = 26(5

2) = 65,

|z4 + 2| ≤ |z|4 + 2 =5

2

4

+ 2 < 65

by Rouche’s theorem,

ZC2(z4 + 26z + 2) = ZC2(26z) = 1

This implies that z4 + 26z + 2 = 0 has 3 solutions in A(52, 3; 0). Let

h(z) = z4 + 26z + 2

Thenh′(z) = 4z3 + 26

In A(52, 3; 0),

|h′(z)| = |4z3 + 26| ≥ 4|z|3 − 26 ≥ 4(5

2)3 − 26 > 0

Thus all zeros of h in A(52, 3; 0) are simple.

Corollary 15.23. (The Fundamental Theorem of Algebra) Let

p(z) = zd + a1zd−1 + ...+ ad

be a degree d complex polynomial, then p has d zeros counted according to their order.

Proof. Sincep(z)

zd= 1 +

a1

z+ · · ·+ an

zn→ 1

as z →∞, there exists R large enough such that

|p(z)

zd− 1| < 1

for |z| = R. Therefore|p(z)− zd| < |zd|

on |z| = R. By Rouche’s theorem,

ZCR(p) = ZCR(zd) = d

Since p has at most d zeros, this implies that p has exactly d zeros. �

87

CHAPTER 16

Applications in integration: II

Definition 16.1. A keyhole-shaped contour Kε,R is a curve consisting of the following4 pieces:

(1) the line segment I1,ε from iε to√R2 − ε2 + iε;

(2) the circular arc CR of radius R traced counterclockwise from√R2 − ε2 + iε to√

R2 − ε2 − iε;(3) the horizontal line segment I2,ε from

√R2 − ε2 − iε to −iε;

(4) the semi-circle Cε of radius ε traced clockwise from −iε to iε.

Proposition 16.2. Suppose that P (x), Q(x) are polynomials, degQ − degP ≥ 2 andQ(x) 6= 0 for x ≥ 0. Then ∫ ∞

0

P (x)

Q(x)dx = −

∑a∈Z(Q)

(P (z)

Q(z)log z; a)

Proof. We fix a branch of log z in the region

D = C− {x ∈ R|x ≥ 0}

and take Argz ∈ (0, 2π). By the residue theorem,

limε→0+

R→∞

∫Kε,R

P (z)

Q(z)log zdz = 2πi

∑a∈Z(Q)

Res(P (z)

Q(z)log z; a)

Since P (z)Q(z)

is continuous at 0, there is r > 0 and A > 0 such that

|P (z)

Q(z)| < A

for z ∈ Br(0). Then if ε is small,

|∫Cε

P (z)

Q(z)log zdz| ≤ πεmax

z∈Cε{|P (z)

Q(z)log z|} ≤ πεA(| ln ε|+ 2π)→ 0

as ε→ 0+. For R large,

|∫CR

P (z)

Q(z)log zdz| ≤ 2πR

B

R2(lnR + 2π)→ 0

as R→∞. Note that∫I1,ε

P (z)

Q(z)log zdz =

∫ √R2−ε2

0

P (x+ iε)

Q(x+ iε)(ln√x2 + ε2 + iArg(x+ iε))dx

89

The function inside the integral is continuous, hence uniformly continuous on [δ, R] for smallδ > 0. Then we may interchange the limit and integral to get

limε→0+

∫ √R2−ε2

δ

P (x+ iε)

Q(x+ iε)(ln√x2 + ε2 + iArg(x+ iε))dx =

∫ R

δ

P (x)

Q(x)lnxdx

Note that for any ε ≥ 0,

|∫ δ

0

P (x+ iε)

Q(x+ iε)(ln√x2 + ε2 + iArg(x+ iε))dx| ≤ δA(ln

√δ2 + ε2 + 2π)→ 0

as δ → 0+. So we have ∫I1,ε

P (z)

Q(z)log zdz →

∫ R

0

P (x)

Q(x)lnxdx

as ε→ 0+ and

limε→0+

R→∞

∫I1,ε

P (z)

Q(z)log zdz =

∫ ∞0

P (x)

Q(x)lnxdx

Note that if z ∈ I2,ε,Argz → 2π

as ε→ 0+ and the direction of I2,ε is opposite to the direction of I1,ε, so

limε→0+

R→∞

∫I1,ε

P (z)

Q(z)log zdz = −

∫ ∞0

P (x)

Q(x)(lnx+ 2π)dx

Combining all together, we have

−2πi

∫ ∞0

P (x)

Q(x)dx = 2πi

∑k

(P (z)

Q(z)log z; zk)

and hence ∫ ∞0

P (x)

Q(x)dx = −

∑k

(P (z)

Q(z)log z; zk)

�

Example 16.3. Compute ∫ ∞0

1

1 + x3dx

Solution 16.4. Let

z1 =1

2+

√3

2i, z2 = −1, z3 =

1

2−√

3

2i

We calculate

Res(log z

1 + z3; z1) = lim

z→z1(z − z1)(

log z

1 + z3) = −iπ

9(1

2−√

3

2i)

Res(log z

1 + z3; z2) =

πi

3

Res(log z

1 + z3; z3) = −5πi

9(1

2−√

3

2i)

90

By formula above, we have ∫ ∞0

1

1 + x3dx =

2π

9

√3

Remark 16.5. We are able to make some transformation of some types of integrals toapply result above.

(1) For ∫ ∞α

P (x)

Q(x)dx

where degQ−degp ≥ 2 and Q(x) 6= 0 for x ≥ a, make a change of variable y = x−a,then the integral ∫ ∞

a

P (x)

Q(x)dx =

∫ ∞0

P (y + a)

Q(y + a)dy

and we may apply result above.(2) For ∫ b

a

P (x)

Q(x)dx

where degQ− degP ≥ 2 and Q(x) 6= 0 for x ≥ a, we have∫ b

a

P (x)

Q(x)dx =

∫ ∞a

P (x)

Q(x)dx−

∫ ∞b

P (x)

Q(x)dx

and the apply results above.

Example 16.6. Evaluate ∫ 2

1

x2 + 1

x4 + 1dx

Proposition 16.7. Suppose that degQ > α + 1 > 0, Q(x) 6= 0 for x ≥ 0,∫ ∞0

xα

Q(x)dx =

2πi

1− e2πiα

∑a∈Z(Q)

Res(zα

Q(z); a)

Proof. We use notation as in the proof of Proposition 16.2. Since the keyhole-shapedcontour Kε,R is simply connected and 0 /∈ Kε,R, we may fix a branch of log z in Kε,R andtake Argz ∈ (0, 2π) for z inside Kε,R. The function

zα = eα log z

is analytic in Kε,R. For z ∈ I1,ε, as ε→ 0+,

zα = eα(ln |z|+iArgz) → eα lnx = xα

For z ∈ I2,ε,

zα = eα(ln |z|+iArgz) → eα(lnx+2πi) = xαe2πiα

as ε→ 0+. The integrals along CR and Cε are zero when R→∞ and ε→ 0+. Therefore

limε→0+

R→∞

∫Kε,R

zα

Q(z)dz =

∫ ∞0

xα

Q(x)dx−

∫ ∞0

xαe2πiα

Q(x)dx

91

This implies that

(1− e2πiα)

∫ ∞0

xα

Q(x)dx = lim

ε→0+

R→∞

∫Kε,R

zα

Q(z)dz = 2πi

∑a∈Z(Q)

Res(zα

Q(z); a)

and hence ∫ ∞0

xα

Q(x)dx =

2πi

1− e2πiα

∑a∈Z(Q)

Res(zα

Q(z); a)

�

Example 16.8. Evaluate ∫ ∞0

1

x13 (x2 + 1)

dx

Solution 16.9. ∫ ∞0

1

x13 (x2 + 1)

dx =

∫ ∞0

x−13

x2 + 1dx

Take a branch cut {z|z real ,z ≥ 0}, 0 < Argz < 2π. We calculate

Res(z−

13

z2 + 1; i) =

i−13

2i= −1

2i · e−

13

(iArgi) = − i2e−

π6i = − i

2(

√3

2− i

2) = −

√3

4i− 1

4

Res(z−

13

z2 + 1;−i) =

(−1)−13

−2i=i

2e−

13

(iArg(−i)) =i

2e−

π2i =

i

2(−i) =

1

2.

Therefore∫ ∞0

1

x13 (x2 + 1)

dx =2πi

1− e− 2π3i(1

4−√

3

4i) =

(12−√

32i)πi

1 + 12

+√

32i

=(√

32

+ 12i)π(3

2−√

32i)

(32

+√

32i)(3

2−√

32i)

=(3√

34− 3

4i+ 3

4i+

√3

4)π

94

+ 34

=

√3π

3

Proposition 16.10.∫ 2π

0

R(cos θ, sin θ)dθ =

∫|z|=1

R(z + 1

z

2,z − 1

z

2i)dz

iz

where

R(x, y) =P (x, y)

Q(x, y)

and P,Q are polynomials.

Proof. For |z| = 1,

z = eiθ = cos θ + i sin θ

and1

z= e−iθ = cos θ − i sin θ

Therefore

cos θ =z + 1

z

292

and

sin θ =z − 1

z

2iFurthermore,

dz = ieiθdθ

and hence

dθ = −ie−iθdz = − izdz

Therefore∫|z|=1

R(z + 1

z

2,z − 1

z

2i)dz

iz=

∫ 2π

0

R(cos θ, sin θ)ieiθdθ

ieiθ=

∫ 2π

0

R(cos θ, sin θ)dθ

�

Example 16.11. Evaluate ∫ 2π

0

1

3 + sin θdθ

Solution 16.12.∫ 2π

0

1

3 + sin θdθ =

∫|z|=1

1

3 +z− 1

z

2i

dz

iz=

∫|z|=1

2

6i+ z − 1z

dz

z=

∫|z|=1

2

z2 + 6iz − 1dz

If z2 + 6iz − 1 = 0,

z =−6i±

√−36 + 4

2=−6i± 4

√2i

2= −3i± 2

√2i = (−3± 2

√2)i

Note that | − 3− 2√

2| > 1 and | − 3 + 2√

2| < 1. Therefore∫|z|=1

2

z2 + 6iz − 1dz = 2πiRes(

2

z2 + 6iz − 1; (−3+2

√2)i) = 2πi

2

2(−3 + 2√

2)i+ 6i=

2π

3 + 2√

2

Example 16.13. Let γ(t) = 1 + it, −∞ < t <∞. Evaluate∫γ

ez

(z + 2)3dz

Solution 16.14. By the Cauchy residue theorem, we have∫ 1+iR

1−iR

ez

(z + 2)3dz +

∫CR

ezdz

(z + 2)3= 2πiRes(

ez

(z + 2)3;−2)

Since |ez| = eRez, for z in the left half-plane of the vertical line x = 1, Rez ≤ 1. Then

minz∈CR

|z − (−2)| = (R− 1)− 2 = R− 3

|∫CR

ezdz

(R + 2)3| ≤ e

(R− 3)3πR =

πeR

(R− 3)3→ 0

as R→∞. Therefore ∫γ

ez

(z + 2)3dz = 2πiRes(

ez

(z + 2)3;−2)

93

To calculate the residue, we note that

ez = e−2 · ez+2 = e−2(1 + (z + 2) +1

2(z + 2)2 + · · · )

Therefore

Res(ez

(z + 2)3;−2) =

1

2e−2

and we have ∫γ

ez

(z + 2)3dz =

πi

e2

94

CHAPTER 17

The Riemann mapping theorem

Definition 17.1. We say that a C1 curve C : z(t) = x(t) + iy(t) is regular at t0 if

z′(t0) 6= 0

Suppose that C1 : z1(t) = x1(t) + iy1(t) is regular at t1, C2 : z2(t) = x2(t) + iy2(t) is regularat t2, and

z1(t1) = z2(t2)

The tangent vector of C1 at z0 is (x′1(t1), y′1(t1)) and the tangent vector of C2 at z0 is(x′2(t2), y′2(t2)). The angle

∠(C1, C2)(t1,t2)

of the intersection of C1 and C2 at z0 is defined to be the unique θ ∈ [0, π] such that

cos θ =(x′1(t1), y′1(t1)) · (x′2(t2), y′2(t2))

|z′1(t1)||z′2(t2)|where

(x′1(t1), y′1(t1)) · (x′2(t2), y′2(t2)) = x′1(t1)x′2(t2) + y′1(t1)y′2(t2)

is the dot product of R2.

Example 17.2. Given C1 : eiθ, θ ∈ [0, 2π], C2 : 2 +√

2eiθ, θ ∈ [0, 2π]. The equations forthe image of C1 and C2 are

x2 + y2 = 1 and (x− 2)2 + y2 = 2

respectively. Solve these equations, we get

x =3

4, y = ±

√7

16= ±√

7

4

Let

z0 =3

4+ i

√7

4

Suppose that eiθ1 = z0 = 2 +√

2eiθ2 , then√

2eiθ2 = −54

+ i√

74

. We have

z′1(θ1) = ieiθ1 = −√

7

4+

3

4i,

z′2(θ2) =√

2ieiθ2 = −√

7

4− 5

4i.

The dot product of the tangent vectors is

(−√

7

4,3

4) · (−

√7

4,−5

4) = −1

295

The norms are|z′1(θ1)| = |ieiθ1 | = 1, |z′2(θ2)| = |

√2ieiθ2 | =

√2

So the angle of intersection between C1 and C2 is θ ∈ [0, π] which satisfies

cos θ =−1

2√2

= −√

2

4

Definition 17.3. Suppose that f is a function defined in a neighborhood of z0. We saythat f is conformal at z0 if f preserves angles there, that is, for any curves C1 : z1(t) andC2 : z2(t) which are regular at t1 and t2 respectively and

z1(t1) = z2(t2) = z0

we have that f(C1) and f(C2) are regular at t1 and t2 respectively, and

∠(C1, C2)(t1,t2) = ∠(f(C1), f(C2))(t1,t2)

We say that f is conformal in a region D if f is conformal at all points of D.

Definition 17.4. Suppose that D is a region and f : D → C is analytic. For z0 ∈ D,we say that f is locally one-to-one at z0 if there exists some δ > 0 such that for any z1 6= z2

in Bδ(z0),f(z1) 6= f(z2)

Theorem 17.5. Let D1, D2 be two regions. Suppose that f : D1 → D2 is analytic at z0

and f ′(z0) 6= 0. Then f is conformal and locally one-to-one at z0.

Proof. For z ∈ C∗, take Argz ∈ [0, 2π). Let C1 : z1(t), C2 : z2(t) be two curves regularat t1 and t2 respectively and

z1(t1) = z2(t2) = z0

Letw1(t) = f(z1(t)), w2(t) = f(z2(t))

By the chain rule, we have

w′1(t1) = f ′(z0)z′1(t1)

w′2(t2) = f ′(z0)z′2(t2)

Since f ′(z0) 6= 0, we have

Arg(w′1(t1)) ≡ Arg(f ′(z0)) + Arg(z′1(t1)) mod 2π

Arg(w′2(t2)) ≡ Arg(f ′(z0)) + Arg(z′2(t2)) mod 2π

Therefore

Arg(w′1(t1))− Arg(w′2(t2)) ≡ Arg(z′1(t1))− Arg(z′2(t2)) mod 2π

andcos(Arg(w′1(t1))− Arg(w′2(t2))) = cos(Arg(z′1(t1))− Arg(z′2(t2)))

which implies that∠(f(C1), f(C2))(t1,t2) = ∠(C1, C2)(t1,t2)

Next we claim that f is one-to-one in a neighborhood of z0. Let

f(z0) = α

96

and take δ > 0 small enough so that f(z) − α has no other zeros in Bδ(z0). If such δ doesnot exist, then z0 is an accumulation point of zeros of f hence f ≡ 0 which contradicts tothe assumption that f ′(z0) 6= 0. Let C = C(z0; δ) be the circle centered at z0 with radiusδ, and Γ = f(C) be the image of C under f . Observe that since winding number is locallyconstant, by the argument principle,

1 =1

2πi

∫C

f ′(z)

f(z)− αdz =

1

2πi

∫Γ

dw

w − α=

1

2πi

∫Γ

dw

w − βfor all β in Bε(α) for some ε small enough.

Take δ ≤ δ such that

Bδ(z0) ⊂ f−1(Bε(α))

Then for any z1 in Bδ(z0),

1 =1

2πi

∫Γ

dw

w − f(z1)=

1

2πi

∫C

(f(z)− f(z1))′

f(z)− f(z1)dz

Therefore by the argument principle again, the value of f(z1) is assumed once inside C so fis one-to-one on Bδ(z0). �

Corollary 17.6. Let f : D1 → D2 be an analytic function between two regions. Thenf is conformal if and only if f ′(z) 6= 0 for all z ∈ D1.

Proof. Suppose that f is conformal. Assume that there is z0 ∈ D1 such that f ′(z0) = 0.Let C1 : γ1(t), t ∈ [0, 1] be a line such that

γ1(t1) = z0

By the chain rule,

(f ◦ γ1)′(t1) = f ′(γ1(t1))γ′(t1) = f ′(z0)γ′(t1) = 0

Therefore

f(C1)

is not regular at t1. By the definition of conformal function, f is not conformal at z0. Theproof of the other direction follows from the theorem above. �

Example 17.7. (1) Let

f(z) = ez

Since

f ′(z) = ez 6= 0

for any z ∈ C, the function f is conformal and locally one-to-one at all points butf is not one-to-one.

(2) Let

f(z) = z2

Since

f ′(z) = 2z 6= 0

for z 6= 0, f is conformal and locally one-to-one on C∗.97

Definition 17.8. Suppose that D1, D2 are two regions and f : D1 → D2 is analytic.Let k be a positive integer. We say that f is a k-to-1 mapping of D1 onto D2 if for all α inD2, the equation

f(z) = α

has k roots(counting multiplicity) in D1.

Example 17.9. Let p(z) be a polynomial of degree k ∈ N. By the fundamental theoremof algebra, p is a k-to-1 mapping of D1 onto D2.

Lemma 17.10. Let

f(z) = zk

where k in N. For any δ > 0, f is a k-to-1 mapping of Bδ(0) onto the disc Bδk(0).

Proof. If f(z) = α, α 6= 0, then zk = α has distinct k roots in the circle |z| = |α| 1k . Ifα = 0, 0 is the root with multiplicity k. �

Theorem 17.11. Suppose f is analytic at z0 with

f ′(z0) = 0

and f is not a constant. Then in some sufficiently small open set containing z0, f is a k-to-1mapping where k is the least positive integer such that

f (k)(z0) 6= 0

Proof. By making the translation f → f − f(z0), we may assume f(z0) = 0. By theassumption, we have

f(z) = ak(z − z0)k + ak+1(z − z0)k+1 + ak+2(z − z0)k+2 + · · ·= (z − z0)k[ak + ak+1(z − z0) + ak+2(z − z0)2 + · · · ]

where

ak =f (k)(z0)

k!6= 0

Let

g(z) = ak + ak+1(z − z0) + ak+2(z − z0)2 + · · ·Since g(z0) 6= 0, we may take a branch of log z in a small neighborhood of g(z0). Then

[g(z)]1k can be defined in some Bδ(z0) and is analytic.

Let

h(z) = (z − z0)[g(z)]1k

for z in Bδ(z0). Then

f(z) = [h(z)]k

Note that h(z0) = 0 and

h′(z0) = [g(z0)]1k 6= 0

Therefore in a small neighborhood of z0, h is one-to-one and conformal. Since zk a k-to-1map, so does f . �

We get the following result immediately.

98

Corollary 17.12. If f : D → C is analytic and one-to-one, then

f ′(z) 6= 0

for z ∈ D.

Remark 17.13. This result in general is not true for real-valued functions. For example,the function f : R→ R defined by

f(x) = x3

is bijective, but f ′(0) = 0.

Theorem 17.14. Suppose that f : D → C is an one-to-one analytic function where D isa region. Then f and the inverse function f−1 : f(D)→ D are conformal.

Proof. Since f is one-to-one, by Corollary, f ′(z) 6= 0 for z ∈ D. By the chain rule

(f−1)′(f(z)) = (f ′(z))−1 6= 0

for z ∈ D, hence f and f−1 are conformal in D and f(D) respectively. �

Definition 17.15. Let f : D1 → D2 be an analytic function between two regions. Wesay that f is a biholomorphism if f is bijective and in this case, we say that D1, D2 arebiholomorphic.

Remark 17.16. The complex plane C is not biholomorphic to any bounded region D.If ϕ : C→ D is a biholomorphism, then ϕ is entire and bounded, thus by Liouville theorem,ϕ is constant which is a contradiction.

Theorem 17.17. (The Riemann mapping theorem) If D is a simply connected region,D 6= C, and z0 in D, there is a unique biholomorphism

ϕ : D → B1(0)

such thatϕ(z0) = 0 and ϕ′(z0) > 0

99

CHAPTER 18

The Riemann zeta function

Proposition 18.1. The series∞∑n=1

n−z

is analytic for Rez > 1.

Proof. Since

|n−z| = |e−z lnn| = e−(Rez) lnn = n−Rez

for ε > 0 and Rez ≥ 1 + ε,

|n−z| ≤ n−(1+ε)

By the p-series test, we know that∞∑n=1

n−(1+ε)

converges, and then by the Weierstrass M -test,

∞∑n=1

n−z

converges uniformly and absolutely on

{z : Rez ≥ 1 + ε}

For any compact set C in the right-half plane, we may take an ε such that

C ⊂ {z : Rez ≥ 1 + ε}

Hence∑∞

n=1 n−z converges uniformly on compacta on the right-half plane. By theorem,

∞∑n=1

n−z

is analytic for Rez > 1. �

Definition 18.2. For Rez > 1, the Riemann zeta function ξ is defined by

ξ(z) :=∞∑n=1

n−z

Proposition 18.3. Let Rez > 1. Then

101

(1)

ξ2(z) =∞∑n=1

d(n)

nz

where d(n) is the number of divisors of n.(2)

ξ(z)ξ(z − 1) =∞∑n=1

σ(n)

nz

where σ(n) is the sum of the divisors of n.(3)

1

ξ(z)=∞∑n=1

µ(n)

nz

where µ is the Mobius function which is defined by

µ(n) =

1, if n = 1(−1)m, if n = p1 · · · pm where pi are distinct primes0, otherwise.

(4)

ξ(z − 1)

ξ(z)=∞∑n=1

φ(n)

nz

where φ is the Euler φ-function and φ(n) is the number of integers less than n andrelatively prime to n.

Proof. (1)

ξ2(z) = (∞∑n=1

1

nz)(∞∑n=1

1

nz) =

∞∑n=1

∑d|n 1 · 1nz

=∞∑n=1

d(n)

nz

(2)

ξ(z)ξ(z − 1) = (∞∑n=1

1

nz)(∞∑n=1

n

nz) =

∞∑n=1

∑d|n 1 · (n

d)

nz

=∞∑n=1

∑d|n d

nz=∞∑n=1

σ(n)

nz

(3) Let

1

ξ(z)=∞∑n=1

annz

102

Then

ξ(z) · 1

ξ(z)= (

∞∑n=1

1

nz)(∞∑n=1

annz

)

=∞∑n=1

∑d|n ad

nz= 1

which implies that a1 = 1 and ∑d|n

ad = 0

From these two properties, we have

an = µ(n)

(4)

ξ(z − 1)

ξ(z)= (

∞∑n=1

n

nz)(∞∑n=1

µ(n)

nz) =

∞∑n=1

∑d|n d · µ(n

d)

nz

=∞∑n=1

φ(n)

nz

�

Definition 18.4. The Gamma function is defined to be

Γ(z) =eγz

z

∞∏n=1

(1 +z

n)−1e

zn

where γ is the Euler constant

γ = limn→∞

[(1 +1

2+ · · ·+ 1

n)− log n]

Remark 18.5. It is a fact that the Gamma function is meromorphic on C with simplepoles at z = 0,−1,−2, · · · . For Rez > 0, the Gamma function has a more familiar form

Γ(z) =

∫ ∞0

e−ttz−1dt

Theorem 18.6. (The Riemann functional equation) For z 6= 1, there is an equation

ξ(z) = 2(2π)z−1Γ(1− z)ξ(1− z) sin(1

2πz)

Theorem 18.7. The Riemann zeta function can be extended to a meromorphic functionin C with only a simple pole at z = 1 and

Res(ξ; 1) = 1

Definition 18.8. The strip{z : 0 ≤ Rez ≤ 1}

is called the critical strip.

103

Conjecture 18.9. (The Riemann Hypothesis) If z is a zero of the Riemann zeta functionin the critical strip then

Rez =1

2

104

Glossary of mathematical terms

k-to-1 mapping, 98-C, 22

analytic extension theorem, 50analytic function, 15analytic polynomial, 7angle of intersection, 95annulus, 63argument principle, 72

based point, 53biholomorphism, 99branch, 81branch cut, 82

Cauchy integral formula, 33Cauchy-Riemann equations, 8change of variables formula, 22closed curve theorem, 29closed curve theorem II, 30complex differentiable function, 9complex line integral, 22complex numbers, 3complex power, 82conformal, 96constant curve, 53convergent sequence, 11converges uniformly, 31cos z, 19critical strip, 103

domain of convergence, 12

entire function, 15Euclidean division of polynomials, 38exponential function, 17extended Liouville’s theorem, 36

factor theorem, 38

field, 3Fundamental theorem of complex line

integrals, 24

Gamma function, 103general closed curve theorem, 55general integral theorem, 55General solution of cubic equations, 5

homotopic, 53homotopy, 53homotopy theorem, 53

integral of complex-valued functions, 21integral theorem, 28integral theorem II, 29

Laurent series, 63limsup, 11Liouville’s theorem, 36locally one-to-one, 96

maximum principle, 42mean value theorem, 41meromorphic, 72minimum principle, 43ML-formula, 23Morera’s theorem, 49

open mapping theorem, 43order of a zero, 58

piecewise C1, 53piecewise C1-curve, 21power series, 11power series expansion, 34principal branch, 82

quotient zeros, 35

105

radius of convergence, 11

rectangle theorem, 25

rectangle theorem II, 27

region, 16

regular, 95

residue, 68

residue theorem, 71

Riemann functional equation, 103

Riemann hypothesis, 104

Riemann mapping theorem, 99

Riemann zeta function, 101

Rouche’s theorem, 85

Schwarz’s lemma, 44simply connected, 53sin z, 19

the fundamental theorem of algebra, 37

uniform limit theorem, 32uniformly absolutely-convergent, 31uniformly convergence, 31uniqueness of exponential function, 18uniqueness theorem, 39

Weierstrass M-test, 31winding number, 70

106

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