Lecture Notes for Math 185 (Follow Stein & Shakarchi's Book) Rui Wang

Lecture Notes for Math 185 (Follow Stein & Shakarchi’s book)

Rui Wang

Contents

Chapter 1. Preliminaries to Complex Analysis 5 1. Complex numbers and the complex plane 5 1.1. Sequences and convergence in ℂ 6 1.2. Open sets in ℂ 6 2. Functions on ℂ 7 2.1. Continuous functions 7 2.2. Holomorphic functions 8 2.3. Complex-valued functions as mappings 8 3. Power series 10 4. Integration along curves 13

Chapter 2. Cauchy’s Theorem and Its Applications 19 1. Cauchy’s theorem and Goursat’s theorem 19 2. Application of Cauchy’s theorem 22 3. Cauchy’s integral formulas 25 3.1. Cauchy’s integral formula and Cauchy’s inequality 25 3.2. Holomorphic functions are analytic 29 4. Important Applications of Cauchy’s thoerem 31 4.1. Morera’s theorem 31 4.2. Sequence of holomorphic functions 32 4.3. Holomorphic functions deﬁned in terms of integrals 33 4.4. Schwarz reﬂection principle 34 4.5. Runge’s approximation theorem 35

Chapter 3. Meromorphic functions and the logarithm 39 1. Zeros and Poles 39 2. The residue formula 42 3. Singularities and meromorphic functions 44 4. The argument principle and applications 48 5. The complex logarithm 51 5.1. Homotopy of paths 51 5.2. The complex logarithm 53

Chapter 4. Conformal mappings 57 1. Conformal equivalence and examples 57 1.1. The unit disk and upper half plane 58 1.2. More examples 59 2. Automorphism of D and ℍ 59

3 4 CONTENTS

2.1. Automorphism of D 59 2.2. Automorphism of ℍ 62 3. Riemann mapping theorem 64 3.1. Normal family and Montel’s theorem 65 3.2. The proof of the Riemann mapping theorem 67 CHAPTER 1

Preliminaries to Complex Analysis

1. Complex numbers and the complex plane

Set of complex numbers is the same as ℝ2 and is denoted by

ℂ = {z = x + iyðx, y ∈ ℝ}. In z = x + it, the x is called the real part and the y is called the imaginary part of z, and write as

x = Re(z), y = Im(z).

One can extend addition + and multiplication ⋅ over ℝ to ℂ (as real part) via

z + w = (Re(z) + Re(w)) + i(Im(z) + Im(w)), and z ⋅ w = (Re(z)Re(w) − Im(z)Im(w)) + i(Re(z)Im(w) + Im(z)Re(w)). In particular, i2 ∶= i ⋅ i = −1. It is easy to check (ℂ, +, ⋅) is a ﬁeld as a ﬁeld extension of ℝ. It is much easier to express the multiplication using polar coordinates. For each r ≥ 0, ∈ ℝ, rei ∶= (r cos ) + i(r sin ) ∈ ℂ.

Conversely, every nonzero complex number z ∈ ℂ∗, there is a unique r > 0 and [] ∈ ℝ∕2 so that z = rei. Such (r, ) is called a polar coordinate of z. i Assume zk = rke k , k = 1, 2. Then

i(1+2) z1 ⋅ z2 = (r1r2)e . The geometric meaning of r is the norm of z, i.e., the distance between the origin and z, as √ ðzð ∶= Re(z)2 + Im(z)2. The geometric meaning of is the argument of z, i.e., the angle between the real axis and the radical line ⃗oz, as y tan = . x Another operation over ℂ is the conjugation, which is the reﬂection about the real axis, i.e.,

̄z = Re(z) − iIm(z).

It is easy to check 2 1 ̄z ðzð = z ̄z = ̄zz, = , z ðzð2 and z + ̄z z − ̄z Re(z) = , Im(z) = . 2 2i The norm ð ⋅ ð introduces a natural distance

d(z, w) = ðz − wð 5 6 1. PRELIMINARIES TO COMPLEX ANALYSIS over ℂ. As a metric space, (ℂ, d) behaves exactly the same as ℝ2. We now review some basic properties for this metric space.

1.1. Sequences and convergence in ℂ. A sequence {zn} is called convergent in ℂ, if there is some w ∈ ℂ so that

lim zn − w = 0. n→∞ ð ð Such w, if exists, must be unique, and we denote such convergence as limn→∞ zn = w, or

zn → w, as n → ∞.

LEMMA 1.1. limn→∞ zn = w if and only if

lim Re(zn) = Re(w), lim Im(zn) = Im(w). n→∞ n→∞ PROOF. Details are left to you.

A sequence {zn} is called a Cauchy sequence, if for any > 0, there exists some N > 0 so that any m, n > N, ðzn − zmð < . Clearly, every convergent sequence is a Cauchy sequence. The converse is also true for ℂ. In another word, we have

THEOREM 1.2. ℂ is complete.

PROOF. {zn} is a Cauchy sequence if and only if {Re(zn)}, {Im(zn)} are both Cauchy sequences. Then the completeness of ℂ follows from completeness of ℝ. 1.2. Open sets in ℂ. Denote by

Dr(z0) ∶= {z ∈ ℂððz − z0ð < r}, r > 0 the open disk of radius r centered at z0. (All open disks form a topology base of ℂ. ) A closed disk is deﬁned as

Dr(z0) ∶= {z ∈ ℂððz − z0ð ≤ r}, r > 0. We use D to denote the unit (open) disk center at the origin. ◦ For a subset Ω ⊂ ℂ, a point z ∈ Ω is called an interior point, if there is some Dr(z) ⊂ Ω. We use Ω to denote the set of interior points of Ω and it the interior of Ω. The set Ω is open if and only if Ω = Ω◦. A set Ω is closed if Ωc is open. A set is closed if and only if it contains all limit points. For any set Ω, its closure Ω is deﬁned as the union of itself with its limit points. It is closed. The boundary )Ω is deﬁned as )Ω ∶= Ω ⧵ Ω◦.

The closed disk Dr(z0) is the closure of the open disk Dr(z0), and

)Dr(z0) = )Dr(z0) = Cr(z0) ∶= {z ∈ ℂððz − z0ð = r0} the circle of radius r centered at z0. A subset Ω in ℂ is called bounded, if there exists some M > 0 so that ðzð ≤ M for any z ∈ Ω. For a bounded set Ω, deﬁne its diameter as

diam(Ω) ∶= sup ðz − wð, z,w∈Ω which is a ﬁnite number. Then the following statements are equivalent for a subset Ω of ℂ: ∙Ω is compact; 2. FUNCTIONS ON ℂ 7

∙Ω is both closed and bounded; ∙Ω is sequentially compact, i.e., every sequence in Ω has a convergent subsequence. The following proposition is useful in proving the Goursat’s theorem later.

PROPOSITION 1.3. Assume Ω1 ⊃ Ω2 ⊃ ⋯ is a sequence of nested subsets of ℂ, with each one nonempty, compact and

lim diam(Ωn) = 0. n→∞

Then their intersection contains a unique point z0 ∈ ℂ.

PROOF. We ﬁrst prove that ∩nΩn is not empty, i.e., it contains some point z0 ∈ ℂ. For this, take zn ∈ Ωn for each n = 1, 2, ⋯ and we obtain a Cauchy sequence {zn} in ℂ since limn→∞ diam(Ωn) = 0.

By the completeness of ℂ, the sequence {zn} converges to some point z0 ∈ ℂ. The limit z0 lives in each

Ωn since each Ωn is (sequentially) compact. This proves z0 ∈ ∩nΩn. z z¨ Next we show 0 is the only point in ∩nΩn. Assume there is another point 0 ∈ ∩nΩn. Then their distance

¨ ðz0 − z0ð ≤ diam(Ωn), n = 1, 2, ⋯ .

n z z¨ z¨ z Take → ∞, there must be ð 0 − 0ð = 0 and then 0 = 0.

An open subset Ω of ℂ is called connected, if there is no way to write Ω as a union of two disjoint nonempty open sets in ℂ. This is equivalent to say Ω is path-connected, i.e., for any two points z0, z1 ∈

Ω, there is a continuous path in Ω which connects z0 and z1. We call Ω a region in ℂ if it is both open and connected.

2. Functions on ℂ

2.1. Continuous functions. Assume Ω is a subset of ℂ and z0 ∈ Ω. A function f ∶ Ω → ℂ is called continuous at z0, if for any > 0, there exists some > 0 so that

ðf(z) − f(z0)ð < , for any ðz − z0ð < , z ∈ Ω.

The following equivalence is left as a homework problem: f is continuous at z0 if and only if

f(zn) → f(z0) for any sequence zn → z0 as n → ∞.

Clearly, f is continuous at z0 if and only if both Re(f) and Im(f) are continuous at z0. 0 We write f ∈ C (Ω) if f is continuous on every z0 ∈ Ω. The definition of continuity and be generalized to functions defined over any metric spaces, for example, over ℂ × ℂ. Then it is easy to check addition, subtraction, multiplication, division, norm are all continuous functions on the corresponding natural domains. The following property for continuous function defined over a compact domain is very useful.

THEOREM 2.1. If Ω is a compact subset in ℂ and f ∈ C0(Ω), then f is bounded and can obtain its maximum and minimum on Ω. 8 1. PRELIMINARIES TO COMPLEX ANALYSIS

2.2. Holomorphic functions. Now there comes the key concept in complex analysis.

DEFINITION 2.2. Assume Ω is an open subset in ℂ. A function f ∶ Ω → ℂ is called holomorphic at z0 ∈ Ω, if the function f(z0 + ℎ) − f(z0) ℎ is convergent as ℎ → 0 (with z0 + ℎ ∈ Ω). The limit is called the derivative of f at z0 and is denoted by ¨ f (z0).

¨ If f (z0) exists for every z0 ∈ Ω, then we say the function f is a holomorphic function over Ω.A holomorphic function f deﬁnes a derivative function f ¨ ∶ Ω → ℂ. We are going to prove it is also holomorphic over Ω in later sections. A holomorphic function over ℂ is called an entire function.

EXAMPLE 2.3. (1) f(z) = z is an entire function with f ¨(z) = 1. n (2) Any polynomial p(z) = a0 + a1z + ⋯ + anz , a0, a1, ⋯ , an ∈ ℂ, an ≠ 0 is an entire function with ¨ n−1 p (z) = a1 + 2a2z + ⋯ + nanz . (3) f(z) = 1 is holomorphic over ℂ∗ with z 1 f ¨(z) = − . z2 (4) f(z) = ̄z is NOT a holomorphic function. ̄ Consider f(z0+ℎ)−f(z0) = ℎ . It has limit 1 if ℎ converges to 0 along the real axis and has ℎ ℎ limit −1 if ℎ converges to 0 along the imaginary axis. Hence f(z0+ℎ)−f(z0) has no limit as ℎ → 0. ℎ

Assume f is holomorphic at z0 ∈ Ω, then we can write ¨ f(z0 + ℎ) − f(z0) = ℎf (z0) + ℎ (ℎ), with (ℎ) → 0 as ℎ → 0. Then it follows that f must be continuous at z0. Similar to real functions, derivatives of complex functions has the following properties.

PROPOSITION 2.4. (1) Assume f, g are holomorphic functions over Ω. Then f ± g, f ⋅ g, f∕g at g(z) ≠ 0 are all holomorphic with (a) (f + g)¨ = f ¨ + g¨; (b) (f ⋅ g)¨ = f ¨g + fg¨; ¨ ¨ (c) (f∕g)¨ = f g−fg . g2 (2) Assume f is holomorphic over Ω and g is holomorphic over U with U ⊃ f(Ω). Then g◦f is holomorphic over Ω and

(g◦f)¨(z) = g¨(f(z)) ⋅ f ¨(z), z ∈ Ω.

2.3. Complex-valued functions as mappings. Any complex valued function

f ∶ Ω → ℂ is equivalent to a pair of real valued functions

u ∶ Ω → ℝ, v ∶ Ω → ℝ with u = Re(f), v = Im(f), i.e., f = u + iv. 2. FUNCTIONS ON ℂ 9

Equivalently, we can regard u, v as real-valued functions deﬁned over the open subset Ω in ℝ2, and consider the mapping F = (u, v) ∶ ℝ2 ⊃ Ω → ℝ2. We can ask what properties F has if f is a holomorphic function.

THEOREM 2.5. Assume Ω is an open subset of ℂ. The function f is holomorphic over Ω if and only if F is differentiable over Ω and the partial derivatives )u )u )v )v , , , )x )y )x )y satisfy the Cauchy–Riemann equations (C–R equations) )u )v )u )v = , = − . )x )y )y )x Recall from real analysis (e.g., Rudin Deﬁnition 9.11) that, the function F is differentiable over 2 Ω ⊂ ℝ if for every (x0, y0) ∈ Ω, there is a 2 × 2 matrix JF so that L M ℎ1 ðF (x0 + ℎ1, y0 + ℎ2) − F (x0, y0) − J ð ℎ2 lim = 0. ⃗ ℎ ℎ=(ℎ1,ℎ2)→(0,0) ð ð

It is not hard to check that the matrix JF must be of the form L )u )u M )x )y JF ∶= )v )v )x )y if it exists. (Why?) It is called the Jacobian matrix of the map F . 2 2 The Jacobian matrix JF ∶ ℝ → ℝ is the linearization of the map F (assuming F is differentiable) and the Cauchy–Riemann equations force the determinant of the Jacobian matrix is nonnegative for holomorphic function since )u )u det J = ( )2 + ( )2 0. F )x )y ≥

EXAMPLE 2.6. We know the function f(z) = ̄z is not holomorphic, and its Jacobian matrix is L M 1 0 J ∶= F 0 −1 with determinant −1.

Now let’s prove Theorem 2.5.

PROOF. (1)" ⇒". Take any (x0, y0) ∈ Ω and denote by z0 = x0 + iy0. Since f is holomorphic

at z0, the limit f(z + ℎ) − f(z ) (u(z + ℎ) − u(z )) + i(v(z + ℎ) − v(z )) 0 0 = 0 0 0 0 ℎ ℎ exists as ℎ → 0. In particular, take ℎ ∈ ℝ, and look at the real part only, we obtain that u(z + ℎ) − u(z ) lim 0 0 ℎ→0,ℎ∈ℝ ℎ exists and this shows )u (x , y ) exists. Similarly, the other three partial derivatives also exist. )x 0 0 At the same time, above calculation also shows that )u )v 1 )u )v f ¨ = + i = ( + i ). )x )x i )y )y 10 1. PRELIMINARIES TO COMPLEX ANALYSIS

Hence it follows )u )v )u )v Re(f ¨) = = , Im(f ¨) = = − , )x )y )y )x which are the Cauchy–Riemann equations.

To see F is differentiable at z0, we rewrite L M ℎ1 ðF (x0 + ℎ1, y0 + ℎ2) − F (x0, y0) − J(x0, y0) ð ℎ2 ðℎð f(z + ℎ) − f(z ) − f ¨(z )ℎ (2.1) = ð 0 0 0 ð ðℎð using the Cauchy–Riemann equations, which converges to zero as ℎ → 0 and we are done. "⇐". This follows from (2.1) under the assumption of differentiability. We remark that the assumption Ω is open in ℂ is crucial. (See HW1 problem 12 from the book. )

EXAMPLE 2.7. Assume f = u + iv is holomorphic over Ω (with u, v smooth). Then u, v must be harmonic function, i.e., Δu = 0 = Δv. This follows from the C–R equations and the equality of mixed partial derivatives: )2u )2u ) )v ) )v Δu = + = − = 0. )x2 )y2 )x )y )y )x We can introduce some useful notations. Deﬁne ) 1 ) ) ) 1 ) ) = ( − i ), = ( + i ). )z 2 )x )y ) ̄z 2 )x )y Then the C–R equations are equivalent to )f = 0, ) ̄z and when f is holomorphic, )f f ¨ = . )z

3. Power series

We have seen that any polynomial

2 n p(z) = a0 + a1z + a2z + ⋯ + anz , a0, a1, a2, ⋯ , an ∈ ℂ, an ≠ 0 is an entire function, i.e., a holomorphic function over ℂ. A power series can be considered as a generalization of a polynomial by allowing inﬁnitely many nonzero terms, i.e., ∞ n Σn=0anz , an ∈ ℂ, n = 0, 1, 2, ⋯ . Then we have to answer the question that when a power series is convergent, and further, if it is holomorphic over its convergent domain.

EXAMPLE 3.1. (1) Consider the power series zn z2 z3 Σ∞ = 1 + z + + + ⋯ . n=0 n! 2 3! 3. POWER SERIES 11

The same as the real variable case, this series is convergent everywhere and this deﬁnes the exponential function zn ez ∶= Σ∞ , z ∈ ℂ. n=0 n! We will see later it is holomorphic over ℂ. (2) Consider the following power series which is called the geometric series

∞ n 2 3 Σn=0z = 1 + z + z + z + ⋯ . It can be explicitly calculated out as follows: Look at partial sums

N n 2 3 N SN (z) = Σn=0z = 1 + z + z + z + ⋯ + z . By multiplying z, we get

N n 2 3 N+1 zSN (z) = zΣn=0z = z + z + z + ⋯ + z , and so 1 − zN+1 zS (z) − S (z) = zN+1 − 1,S = N N N 1 − z whenever z ≠ 1. (Obviously, the series is divergent at z = 1.) From the explicit expression of the partial sum, it becomes apparent that this series is convergent (and also holomorphic) if ðzð < 1 and it is divergent if ðzð > 1. Now we give a complete answer for the convergence question which works for any power series. First, deﬁne 1 R ∶= . √n lim supn→∞ ðanð , ∞ a zn It is a real number in [0 +∞] and is called the convergence radius of the power series Σn=0 n . zn EXAMPLE 3.2. (1) The exponential function Σ∞ has convergent radius R = +∞. n=0 n! ∞ zn R (2) The geometric series Σn=0 has convergent radius = 1.

The open disk DR is called the convergent disk and the reason is the following theorem.

HEOREM The power series ∞ a zn T 3.3. Σn=0 n (1) is absolutely convergent if ðzð < R; (2) It is divergent if ðzð > R.

PROOF. R = 0 or +∞ cases are left to you and we assume here 0 < R < +∞.

(1) Take any z0 with ðz0ð < R, we prove now the series is convergent at z0. √n First, ðz0ð < R implies that ðz0ð lim supn→∞ ðanð < 1. We insert some r so that

√n ðz0ð lim sup ðanð < r < 1, n→∞ and further some small > 0 so that

√n ðz0ð(lim sup ðanð + ) < r < 1, n→∞ Then there exists some N (may depend on ) so that every n > N,

√n √n ðanð < lim sup ðanð + . n→∞ It follows √n √n ðz0ð ðanð < ðz0ð(lim sup ðanð + ) < r < 1 n→∞ 12 1. PRELIMINARIES TO COMPLEX ANALYSIS

and n n ðanððz0ð < r . rn < r < a zn The series Σ is convergent since 0 1. Then the comparison theorem implies that Σ n 0 must be absolutely convergent. (2) This is assigned as a homework problem.

Over the boundary ðzð = R, the series can be either convergent or divergent.

EXAMPLE 3.4. (Homework problem). Consider the following there series with convergent radius R = 1 n (1) Σnz . It is divergent on ðzð = 1. n (2) Σ z . It is convergent on z = 1. n2 ð ð n (3) Σ z . It is convergent on z = 1 except at z = 1. n ð ð Next, we answer the question about if the power series is holomorphic. ∞ a zn R Assume the power series Σn=0 n has convergent radius , and denote by ∞ n f(z) = Σn=0anz , z ∈ DR. ∞ a nzn−1 By differentiate it term by term, we obtain another power series Σn=1 n . It is easy to check this power series also has R as its convergent radius, hence deﬁnes a function

∞ n−1 g(z) = Σn=1annz , z ∈ DR. We have the following theorem.

¨ THEOREM 3.5. The function f is holomorphic over DR with f = g.

PROOF. We assume 0 < R < ∞. The cases R = 0 and R = +∞ are left you to handle. Denote by N n ∞ n SN (z) ∶= Σn=0anz ,EN (z) ∶= Σn=N+1anz ; and ̃ N n−1 ̃ ∞ n−1 SN (z) ∶= Σn=1annz , EN (z) ∶= Σn=N+1annz . Obviously from deﬁnition, ̃ ̃ ¨ ̃ f = SN + EN , g = SN + EN ,SN = SN . ¨ We show now for any z0 ∈ DR, f (z0) exists and it is g(z0).

For this, we ﬁrst pick some ﬁx some 0 < r < R so that z0, z0 + ℎ ∈ Dr and write f(z + ℎ) − f(z ) 0 0 − g(z ) ℎ 0 S (z + ℎ) − S (z ) (3.1) = N 0 N 0 − S̃ ℎ N E (z + ℎ) − E (z ) (3.2) + N 0 N 0 ℎ ̃ (3.3) + (−EN ).

. For each ﬁxed N, the term ð(3.1)ð is small as ℎ is close to 0; the term ð(3.3)ð is small for large N. We focus on estimating ð(3.2)ð. 4. INTEGRATION ALONG CURVES 13

For it, we look at an estimate of the general term as n n n−1 n−2 n−2 n−1 an(z0 + ℎ) − anz an(z0 + ℎ − z0)((z0 + ℎ) + (z0 + ℎ) z0 + ⋯ + (z0 + ℎ)z + z ) 0 = n 0 ð ℎ ð ð ℎ ð n−1 n−2 n−2 n−1 an(z0 + ℎ − z0)((z0 + ℎ) + (z0 + ℎ) z0 + ⋯ + (z0 + ℎ)z + z ) = n 0 ð ℎ ð n−1 ≤ ðanðnr . n−1 Then since the series Σðanðnr is convergent, the term ð(3.2)ð converges to zero as N → ∞. Now we wrap up the proof using − arguments: For any > 0, there we can pick N large enough so that (3.2) < , (3.3) < , ð ð 3 ð ð 3 and then for such N, take > 0 small so that (3.1) < for any 0 < ℎ < . ð ð 3 ð ð By this way, for any 0 < ℎ < , f(z0+ℎ)−f(z0) − g(z ) , and this shows ð ð ð ℎ 0 ð ≤

f(z0 + ℎ) − f(z0) lim = g(z0). ℎ→0 ℎ An immediate corollary from this theorem as

OROLLARY Assume the power series ∞ a zn has R as its convergence radius. Then it is C 3.6. Σn=0 n inﬁnitely order complex differentiable in the convergence disk DR, and the complex derivatives can be calculated term by term.

The power series can be generalized to ones with any z0 ∈ ℂ as center. To be more precise, a power series with z0 ∈ ℂ as center is deﬁned as ∞ n Σn=0an(z − z0) .

Then the above results can be stated for this case by shifting the convergence disk from DR to DR(z0). We end this section by introducing the following deﬁnition.

DEFINITION 3.7. Assume Ω is an open subset in ℂ. A function f ∶ Ω → ℂ is called analytic at z0 ∈ Ω, if there exists an open disk DR(z0) ⊂ Ω so that f can be written as a power series centered at z0 over DR(z0). If f is analytic at every point z0 ∈ Ω, we call f is analytic over Ω.

We just proved that if f is analytic over Ω, then f must be holomorphic over Ω. In fact, the reverse statement is also true which will be proved next chapter.

4. Integration along curves

By a parametrized curve, we mean a map

z ∶ [a, b] → ℂ.

We need the following terminology for parametrized curves: A parametrized curve z ∶ [a, b] → ℂ ∙ is called smooth, if (1) derivative z¨ exists and continuous on [a, b], i.e., z ∈ C1([a, b]); 14 1. PRELIMINARIES TO COMPLEX ANALYSIS

(2) z¨(t) is nowhere vanishing; ∙ is called piecewise smooth, if there exists a splitting

a = a0 < a1 < a2 < ⋯ < an = b so that z ∶ [a , a ] → ℂ is smooth, i = 0, 1, ⋯ , n − 1; ð[ai,ai+1] i i+1 ∙ is called simple, if it is injective over (a, b); ∙ is called a loop, if z(a) = a(b). Two parametrized smooth curves

zk ∶ [ak, bk] → ℂ, k = 1, 2, 1 are called equivalent, if there exists a C bijective map ∶ [a1, b1] → [a2, b2] so that ¨ z2◦ = z1, and (t) > 0. It is an equivalence relation on the set of smooth parametrized curves. A smooth curve is an equivalence class of smooth parametrized curves. For a smooth curve, its length is deﬁned as b ¨ length( ) ∶= ðz (t)ðdt, Êa where z ∶ [a, b] → ℂ is a smooth parametrization. Notice length( ) is independent of choices of parametrizations.

EXAMPLE 4.1. The circle CR(z0) can be parametrized by it z ∶ [0, 2] → ℂ, z(t) = z0 + Re and can be also parametrized by

− −it z ∶ [0, 2] → ℂ, z(t) = z0 + Re . These two parametrized curves are NOT equivalent. The ﬁrst one has positive orientation and the sec- + ond one has negative orientation. We use CR(z0) to describe the equivalence class of positive oriented − smooth parametrizations and use CR(z0) to describe the equivalence class of negative oriented smooth parametrizations.

Now assume ⊂ ℂ is a curve with a smooth parametrization z ∶ [a, b] → ℂ. We say f as a map from the image of to ℂ is a continuous function deﬁned over , if there exists a smooth parametrization (so for all equivalent smooth parametrizations) z ∶ [a, b] → ℂ so that

f◦z ∶ [a, b] → ℂ is continuous. Then the integration of f over is deﬁned as b f(z)dz ∶= f(z(t)) ⋅ z¨(t)dt. Ê Êa

LEMMA 4.2. The integration ∫ f(z)dz is independent of smooth parametrization.

PROOF. This immediately follows from substitution rule for deﬁnition integrations.

2 it 2 it EXAMPLE 4.3. (1) ∫ + dz = ∫ d(Re ) = ∫ Rie dt = 0; CR 0 0 1 2 1 it 2 (2) ∫ + dz = ∫ it d(Re ) = ∫ idt = 2i. CR z 0 Re 0 The integration of a continuous function along smooth curves has the following properties: 4. INTEGRATION ALONG CURVES 15

LEMMA 4.4. Assume is a smooth curve, and f and g are two continuous functions over . Then (1) For any a, b ∈ ℂ,

(af(z) + bg(z))dz = a f(z)dz + b g(z)dz. Ê Ê Ê (2) f(z)dz = − f(z)dz, Ê − Ê where − the curve with reverse orientation. (3)

ð f(z)dzð ≤ sup ðf(z)ð ⋅ length( ). Ê z∈

PROOF. (1) This is left to you as an exercise. (2) Assume z ∶ [a, b] → ℂ is a smooth parametrization of , and deﬁne

z−(t) ∶= z(a + b − t)

which is a smooth parametrization of −. Then the equality follows from calculation

b f(z)dz = f(z(a + b − t))z¨(a + b − t)(−dt) Ê − Êa a = f(z(s))z¨(s)ds Êb b = − f(z(s))z¨(s)ds Êa = − f(z)dz. Ê (3) By taking a smooth parametrization z ∶ [a, b] → ℂ, we calculate

b ¨ ð f(z)dzð = ð f(z(t))z (t)dtð Ê Êa b ¨ ≤ ðf(z(t))ððz (t)ðdt Êa b ¨ ≤ sup ðf(z)ð ⋅ ðz (t)ðdt z∈ Êa = sup ðf(z)ð ⋅ length( ). z∈

The above statements can all be generalized to piecewise-smooth curves. An integration of a continuous function over a piecewise-smooth curve is deﬁned as the sum of integrations of the function over each piece of smooth curve. Clearly from this deﬁnition, the integration is independent of choices of smooth parametrization of each smooth piece. Moreover, it is also independent of choices of divisions to make it into smooth pieces.

PROPOSITION 4.5. Assume is a piece-wise smooth curve, and f and g are two continuous functions over . Then all three results in Lemma 4.4 hold. 16 1. PRELIMINARIES TO COMPLEX ANALYSIS

Next we discuss some ways for calculation of integrations. In real analysis, we know a deﬁnite integration can be calculated from fundamental theorem of calculus using antiderivatives. In complex analysis, we have a similar result.

DEFINITION 4.6. Assume f is a function over an open set Ω in ℂ. A primitive of f is a holomorphic function F over Ω whose derivative is f, i.e., F ¨(z) = f(z).

1 EXAMPLE 4.7. (1) The function f(z) = z has a primitive F (z) = z2 over ℂ. 2 (2) We will see later the function f(z) = 1 has no primitive over ℂ∗. z

PROPOSITION 4.8. Assume a continuous function f has a primitive F over Ω, where Ω is an open subset of ℂ, and is a piece-wise smooth curve in Ω starting at z0 and ending at z1. Then

f(z)dz = F (z1) − F (z0). Ê

In particular, when is a loop, ∫ f(z)dz = 0.

PROOF. The key ingredient in the proof is the fundamental theorem of calculus for real functions. We ﬁrst deal with the case when is a smooth curve. Take a smooth parametrization

z ∶ [a, b] → ℂ and calculate b f(z)dz = F ¨(z(t))z¨(t)dt Ê Êa b d = F (z(t))dt Êa dt = F (z(b)) − F (z(a))

= F (z1) − F (z0).

Here the last second equality is the fundamental theorem of calculus. Next assume is piece-wise smooth, with a division

a0 = a < a1 < ⋯ < an−1 < an = b for [a, b] so that there is a parametrization z ∶ [a, b] → ℂ which is continuous on [a, b] and smooth on each piece. Denote by the curve from parametrization z , i = 0, 1, ⋯ , n − 1. Then from the i ð[ai,ai+1] smooth case,

n−1 n−1 f(z)dz = Σi=0 f(z)dz = Σi=0 (F (z(ai+1)) − F (z(ai))) = F (z(b)) − F (z(a)) = F (z1) − F (z0). Ê Ê i

EXAMPLE 4.9. (1) A polynomial in z of the form

n f(z) = a0 + a1z + ⋯ + anz , an ≠ 0,

has a primitive as 1 1 F (z) = a z + a z2 + ⋯ a zn+1. 0 2 1 n + 1 n 4. INTEGRATION ALONG CURVES 17

(2) From Proposition 4.8, we see the function f(z) = 1 has no primitive since z 1 dz = 2i ≠ 0. ÊC+ z 1 COROLLARY 4.10. Assume Ω is a region, i.e., Ω is open and connected in ℂ, and f ¨ = 0, then f must be constant over Ω.

PROOF. Take z0 ∈ Ω and consider the set

A ∶= {z ∈ Ωðf(z) = f(z0)}. The set A is closed since f is continuous. The set A is also open, because Ω is open so any z ∈ A we can take a small disk of radius > 0

D(z) ⊂ Ω.

Any open w ∈ D(z) can be connected to z via a radical line starting at z and ending at w, which must be smooth. Then apply Proposition 4.8 and using f ¨ = 0, we have f(w) = f(z). This proves the open disk D(z) lives in A, and thus A is open. Now because Ω is assumed to be connected, the open and closed subset A can only be Ω since it is never empty. This proves any z ∈ Ω, f(z) = f(z0), i.e., a constant. (Remark: Because we only discuss integration over piecewise-smooth curves but a priori, a path connecting two points may be quite pathetical in the sense that it may be only continuous but not differentiable anywhere. To avoid this subtle issue, we avoid the proof from the book using path-connectedness property of Ω and use connectedness directly.)

CHAPTER 2

Cauchy’s Theorem and Its Applications

1. Cauchy’s theorem and Goursat’s theorem

We ﬁrst see two introductory examples.

EXAMPLE 1.1. Assume f is a holomorphic function over ℂ and has continuous partial derivatives (In fact, we are going to prove any holomorphic function has continuous partial derivatives, but here we need to assume it.). Write f(z) = u(z) + iv(z), u, v ∶ ℂ → ℝ. Denote by D the unit disk centered at origin, and by C ∶= )D its boundary. Notice C has a canonical orientation induced from D as the counter-clockwise one. We calculate ∫C f(z)dz by considering real and imaginary parts separately. First, we calculate

f(z(t))z¨(t) = (u(z(t)) + iv(z(t)))z¨(t) = (u(z(t)) + iv(z(t)))(x¨(t) + iy¨(t)) = (u(z(t))x¨(t) − v(z(t))y¨(t)) + i(u(z(t))y¨(t) + v(z(t))x¨(t)).

Hence b f(z)dz = f(z(t))z¨(t) ÊC Êa b b = (u(z(t))x¨(t) − v(z(t))y¨(t))dt + i (u(z(t))y¨(t) + v(z(t))x¨(t))dt Êa Êa = udx − vdy + i udy − vdx ÊC Ê )v )u )u )v = ( − )dxdy + i ( + )dxdy, ËD )x )y ËD )x )y and this is zero due to the Cauchy–Riemann equations. We have seen many times that 1 is not zero but 2i. ∫C z Such baby examples indicate a reasonable conclusion: Integration of any holomorphic function over interior of a simple closed loop vanishes. After adding some technically assumptions, this is what Cauchy’s theorem states.

THEOREM 1.2. [Cauchy’s theorem for a disk] Assume f is a holomorphic function over an open disk D. Then for any closed piece-wise smooth curve inside D, there is ∫ f(z)dz = 0.

We will prove this theorem by showing the following proposition, and then Theorem 1.2 immediately follows by Proposition 4.8.

19 20 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

PROPOSITION 1.3. Any holomorphic function over an open disk D has a primitive.

The proof of this proposition needs the Goursart’s theorem whose proof is postponed after we prove this propositon.

THEOREM 1.4 (Goursat’s theorem). Assume Ω is an open subset of ℂ and T is a triangle whose interior is contained in Ω. Then for any holomorphic function f over Ω, there is

f(z)dz = 0. ÊT

PROOFOF PROPOSITION 1.3. For this, we try to deﬁne a function F ∶ D → ℂ and check it is a primitive of f. Denote by z0 the center of the disk D. Take z ∈ D, denote by z the radical line starting at z0 and ending at z. It has a natural smooth parametrization as

w ∶ [0, 1] → ℂ, w(t) = (1 − t)z0 + tz.

Deﬁne F (z) ∶= f(w)dw, z ∈ D. Ê z We now check F is holomorphic over D with F ¨ = f. For this, calculate

1 1 (F (z + ℎ) − F (z)) = ( f(w)dw − f(w)dw) ℎ ℎ Ê z+ℎ Ê z 1 1 = f(w)dw − ( f(w)dw + f(w)dw + f(w)dw), ℎ ℎ − Ê ℎ Ê z+ℎ Ê z Ê ℎ z z ℎ − where ℎ denotes the line path starting at and ending at + . The three paths z+ℎ ∪ z ∪ ℎ form a closed piecewise-smooth curve which is actually a triangle. Then it follows from the Goursart’s theorem 1.4 that the integration of f over it is zero. We only need to deal with 1 f(w)dw now. First, write ℎ ∫ℎ 1 1 f(w)dw = ( f(w)dw − f(z)) + f(z) ℎ ℎ Êℎ Êℎ 1 = (f(w) − f(z))dw + f(z). ℎ Êℎ Next, we estimate 1 1 (f(w) − f(z))dw sup f(w) − f(z) ⋅ length( ) ðℎ ð ≤ ℎ ð ð ℎ Êℎ ð ð w∈ℎ 1 = sup f(w) − f(z) ⋅ ℎ ℎ ð ð ð ð ð ð w∈ℎ = sup ðf(w) − f(z)ð, w∈ℎ which converges to zero as ℎ → 0 since f is continuous at z. This proves 1 F ¨(z) = lim (F (z + ℎ) − F (z)) = f(z), for any z ∈ D. ℎ→0 ℎ

Now we prove the Goursat’s theorem. 1. CAUCHY’S THEOREM AND GOURSAT’S THEOREM 21

PROOFOF GOURSAT’STHEOREM. Assume T is a triangle with three vertices as A, B, C counter- clockwise, and then

f(z)dz = f(z)dz + f(z)dz + f(z)dz. ÊT ÊAB⃗ ÊBC⃗ ÊCA⃗ By taking middle point of each side, we obtain four small triangles whose boundaries are assigned with canonical orientations and are denoted as T1,T2,T3,T4. Notice

f(z)dz = f(z)dz + f(z)dz + f(z)dz + f(z)dz. ÊT ÊT1 ÊT2 ÊT3 ÊT4 Assume T 1 is one of these four so that

ð f(z)dzð ∶= max {ð f(z)dzð}, 1 i=1,2,3,4 ÊT ÊTi and denote by T̂ 1 the closure of the interior of T 1. It follows that

ð f(z)dzð ≤ 4ð f(z)dzð. ÊT ÊT 1 Further, by taking middle point of each side of T 1, we obtain another four small triangles and again, 2 ̂ 2 2 denote the one largest ð ∫⋅ f(z)dzð as T . Denote by T the closure of the interior of T . By this way, we obtain a sequence of compact sets

T̂ 1 ⊃ T̂ 2 ⊃ ⋯ ⊃ T̂ n ⊃ ⋯ , with length(T n) = 2−nlength(T ), diam(T n) = 2−ndiam(T ).

∞ T̂ n z By Proposition 1.3, ∩n=1 is a set contains only one point, which is denote by 0. Because f is holomorphic at z0, we can write

¨ f(z) = f(z0) + f (z0)(z − z0) + (z), (z) → 0 as z → z0,

n for z ∈ Ω so that ðz − z0ð small enough. Then for diam(T ) small, it follows

n ð f(z)dzð ≤ 4 ð f(z)dzð ÊT ÊT n

n ¨ = 4 ð (f(z0) + f (z0)(z − z0) + (z)(z − z0))dzð ÊT n

n ¨ = 4 ðf(z0) dz + f (z0) (z − z0)dz + (z)(z − z0)dzð ÊT n ÊT n ÊT n n = 4 ð0 + 0 + (z)(z − z0)dzð ÊT n n ≤ 4 ð (z)(z − z0)dzð ÊT n n n n ≤ 4 length(T )diam(T ) sup ð (z)ð z∈T n = length(T )diam(T ) sup ð (z)ð, z∈T n which converges to zero as n → ∞, and this proves that ∫T f(z)dz = 0. 22 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

2. Application of Cauchy’s theorem

In this sections, we see some calculations using the Cauchy’s theorem.

z 2 EXAMPLE 2.1. Calculate the integration of the function f(z) = e + z + sin z over )D3(i).

SOLUTION. Because f is an entire function, it follows f(z)dz = 0. ∫)D3(i) n EXAMPLE 2.2. Calculate the integration of the function f(z) = z over )D3(i).

SOLUTION. We need to discuss different situations of n ∈ ℤ. (1) If n = 0, 1, 2, ⋯, the function zn is an entire function, and so zndz = 0. ∫)D3(i) n ∗ (2) If n = −2, −3, −4, ⋯, the function z is holomorphic and has a primitive over ℂ . Since )D3(i) doesn’t go through the origin, zndz = 0. ∫)D3(i) n C+ (3) If = −1, we may construct a circle 1 (0). Using Cauchy’s theorem, we can prove that

z−1dz = z−1dz = 2i. Ê)D3(i) Ê)D1(0)

1 EXAMPLE 2.3. Calculate the integration of the function f(z) = over )D (i). z2−3z 3 1 1 1 1 SOLUTION. Write = ( − ). Hence z2−3z 3 z−3 z 1 1 1 1 f(z)dz = dz − dz 3 z − 3 3 z Ê)D3(i) Ê)D3(i) Ê)D3(i) 2i 2i = 0 − = − . 3 3

EXAMPLE 2.4. Prove that for any ∈ ℝ, +∞ 2 2 e− = e−x e−2ixdx. Ê−∞ +∞ −x2 PROOF. (1) If = 0, we prove ∫−∞ e dx = 1. To see it, consider the double integration e−(x2+y2)dxdy. ∬DR(0) Change the Cartesian coordinate to the polar coordinate, we have 2 R 2 2 2 e−(x +y )dxdy = e−r rdrd ËDR(0) Ê0 Ê0 2 R 1 2 = e−r d(r2)d 2 Ê0 Ê0 2 = 1 − e−R .

Taking R → ∞, we obtain

2 2 e−(x +y )dxdy = 1. Ëℝ2 On the other hand, the Fubini theorem allows us to write +∞ 2 2 2 e−(x +y )dxdy = ( e−x dx)2. Ëℝ2 Ê−∞ +∞ −x2 Hence, ∫−∞ e dx = 1. 2. APPLICATION OF CAUCHY’S THEOREM 23

(2) If > 0, we consider an entire function f(z) = e−z2 and a closed piecewise-smooth loop consisting of four paths as

= R ∪ 1 ∪ ∪ 2,

∙ R is the path z ∶ [−R, R] → ℂ, z(t) = t;

∙ 1 is the path z ∶ [0, ] → ℂ, z(t) = R + it; ∙ is the path z ∶ [−R, R] → ℂ, z(t) = −t + i;

∙ 2 is the path z ∶ [0, ] → ℂ, z(t) = −R + i( − t) Since f is entire function, by Cauchy’s theorem we have

f(z)dz = f(z)dz + f(z)dz + f(z)dz + f(z)dz = 0, Ê Ê R Ê 1 Ê 2 Ê for any R > 0. We calculate the four integrations now. (a) lim f(z)dz = +∞ e−t2 dt = 1. R→+∞ ∫ R ∫−∞ (b) −(R2+2iRt−t2) ð f(z)dzð = ð e idtð Ê 1 Ê0 −(2iRt−t2) −R2 = ð e dtð ⋅ e Ê0 −(2iRt−t2) −R2 ≤ ðe ðdt ⋅ e Ê0 t2 −R2 = ðe ðdt ⋅ e Ê0 2 −R2 ≤ ⋅ e ⋅ e , which converges to 0 as R → +∞. (c) R 2 f(z)dz = − e−(t+i) dt Ê Ê−R R 2 2 = −e e−t e−2itdt. Ê−R Taking R → +∞, this is ∞ 2 2 −e e−x e−2ixdx. Ê−∞ (d) This term converges to 0 as R → +∞ for the same reason as (b). Above all, we obtain

0 = lim ( f(z)dz + f(z)dz + f(z)dz + f(z)dz) R→+∞ Ê R Ê 1 Ê 2 Ê ∞ 2 2 = 1 + 0 + 0 − e e−x e−2ixdx, Ê−∞ i.e., ∞ 2 2 e− = e−x e−2ixdx. Ê−∞ (3) If < 0, the equality can be shown by a similar way to the case > 0. Details are left to you. 24 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

EXAMPLE 2.5. Prove that +∞ 1 − cos t dt = . 2 Ê0 t 2

1−eiz PROOF. We consider the function f(z) = and a closed piecewise-smooth loop (an indented z2 semicircle) consisting of four paths as

= ∪ + ∪ R ∪ −, 0 < < R.

i(−t) ∙ is the path z ∶ [0, ] → ℂ, z(t) = e ;

∙ + is the path z ∶ [, R] → ℂ, z(t) = t; it ∙ R is the path z ∶ [0, ] → ℂ, z(t) = Re ;

∙ − is the path z ∶ [−R, −] → ℂ, z(t) = t. Since f is a holomorphic function over ℂ∗, by Cauchy’s theorem we have

f(z)dz = f(z)dz + f(z)dz + f(z)dz + f(z)dz = 0, Ê Ê Ê + Ê R Ê − for any 0 < < R. We calculate the four integrations now. (1) 1 − eiz f(z)dz = dz 2 Ê Ê z 1 − (1 + iz + 1 (iz)2 + ⋯) = 2 dz 2 Ê z 1 = −i dz − g(z)dz z Ê Ê = − − g(z)dz, Ê

where g(z) = 1 i2 + 1 i2z + ⋯ which is convergent over ℂ, and uniformly bounded over a small 2 3! enough disk. Hence from

lim − g(z)dz sup g(z) = 0, →0+ ð ð ≤ ð ð Ê z∈C we obtain lim f(z)dz = −. →0+ Ê (2)

f(z)dz + f(z)dz Ê + Ê − R 1 − eit − 1 − eit = dt + dt 2 2 Ê t Ê−R t R 1 − cos t − 1 − cos t R sin t − sin t = ( dt + dt) − i( dt + dt) 2 2 2 2 Ê t Ê−R t Ê t Ê−R t R 1 − cos t = 2 dt, 2 Ê t 3. CAUCHY’S INTEGRAL FORMULAS 25

where the last equality is due to cos t is an even function and sin t is an odd function. As → 0+ and R → +∞, +∞ 1 − cos t lim ( f(z)dz + f(z)dz) = 2 dt. →0+,R→+∞ t2 Ê + Ê − Ê0 which converges to 0 as R → +∞. (3) 1 − eiz f(z)dz = dz. 2 Ê R Ê R z Over R, we have the estimate 1 − eiz 1 − eiz 1 + eiz = ð ð ð ð ð z2 ð R2 ≤ R2 1 + e−Imz 2 = . R2 ≤ R2 Hence 2 2 f(z)dz ⋅ R = ð ð ≤ 2 R Ê R R which converges to 0 as R → +∞. Above all, we obtain

0 = lim ( ∪ + ∪ R ∪ −) →0+,R→+∞ +∞ 1 − cos t = − + 2 dt. 2 Ê−∞ t Thus, +∞ 1−cos t dt = . ∫−∞ t2 2

3. Cauchy’s integral formulas

3.1. Cauchy’s integral formula and Cauchy’s inequality.

EXAMPLE 3.1. Assume f(z) is an analytic function

∞ n f(z) = Σn=0an(z − z0) , with z0 ∈ D. We calculate ∞ n f(z) Σ an(z − z0) dz = n=0 dz = Ê)D z − z0 Ê)D z − z0 a0 ∞ n−1 = dz + Σn=1an(z − z0) dz Ê)D z − z0 Ê)D = 2ia0 + 0 = 2if(z0), and obtain the following formula 1 f(z) f(z0) = dz. 2i Ê)D z − z0 This is a baby example of Cauchy’s integral formula.

THEOREM 3.2 (Cauchy’s integral formula (n = 0 case)). Assume Ω is an open subset of ℂ and f is holomorphic over Ω. Assume D is disk with D ⊂ Ω. Then for any z0 ∈ D, 1 f(z) f(z0) = dz. 2i Ê)D z − z0 26 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

PROOF. We write 1 f(z) 1 f(z ) + (f(z) − f(z )) dz = 0 0 dz 2i Ê)D z − z0 2i Ê)D z − z0 1 f(z ) 1 f(z) − f(z ) = 0 dz + 0 dz 2i Ê)D z − z0 2i Ê)D z − z0 1 f(z) − f(z0) = f(z0) + dz. 2i Ê)D z − z0

1 f(z)−f(z0) We now calculate ∫)D dz. By taking small enough > 0 so that D(z0) ⊂ D and write 2i z−z0 ¨ f(z) − f(z0) = f (z0)(z − z0) + (z)(z − z0), lim sup (z) = 0, →0+ ð ð z∈)D(z0) we have 1 f(z) − f(z ) 0 dz 2i Ê)D z − z0 1 f(z) − f(z ) = 0 dz 2i z − z Ê)D(z0) 0 1 = (f ¨(z ) + (z))dz. 2i 0 Ê)D(z0) The norm 1 (f ¨(z ) + (z))dz ð2i 0 ð Ê)D(z0) 1 f ¨(z ) + (z) dz ≤ 2 ð 0 ð Ê)D(z0) ¨ ≤ ⋅ (ðf (z0)ð + sup ð(z)ð), z∈)D(z0) which converges to zero as → 0+. This proves 1 f(z) − f(z ) 0 dz = 0 2i Ê)D z − z0 and we are done.

ez+z2+3 EXAMPLE 3.3. Calculate . ∫)D3(0) z−1

SOLUTION. Since 1 ∈ D3(0), by the Cauchy’s integral formula, 1 ez + z2 + 3 = (ez + z2 + 3) = e + 4. 2i z − 1 ðz=1 Ê)D3(0)

Next, we improve the Cauchy’s integral formula to arbitrary order derivatives.

THEOREM 3.4 (Cauchy’s integral formula). Assume Ω is an open subset of ℂ and f is holomorphic over Ω. Then f has inﬁnitely many complex derivatives over Ω. Moreover, for any disk D with D ⊂ Ω and any z0 ∈ D, the n-th derivative at z0 is calculated as n! f(z) f (n)(z ) = dz. 0 n+1 2i Ê)D (z − z0) 3. CAUCHY’S INTEGRAL FORMULAS 27

PROOF. From The Cauchy’s integral formula for n = 0 case, we can write

0! f(z) f (0)(z ) = dz. 0 0+1 2i Ê)D (z − z0)

Now assume for any n = k,

k! f(z) (3.1) f (k)(z ) = dz, 0 k+1 2i Ê)D (z − z0)

(k+1) we calculate f (z0). By deﬁnition,

(k) (k) (k+1) f (z0 + ℎ) − f (z0) f (z0) = lim . ℎ→0 ℎ Using (3.1), we can write

(k) (k) f (z0 + ℎ) − f (z0) ℎ k! 1 1 = f(z) − )dz k+1 k+1 2iℎ Ê)D (z − (z0 + ℎ)) (z − z0) k! (z − z )k+1 − (z − (z + ℎ))k+1 = f(z) 0 0 dz k+1 k+1 2iℎ Ê)D (z − (z0 + ℎ)) (z − z0) k! ℎ((z − z )k + (z − z )k−1(z − (z + ℎ)) + ⋯ + (z − z )(z − (z + ℎ))k−1 + (z − (z + ℎ))k) = f(z) 0 0 0 0 0 0 dz k+1 k+1 2iℎ Ê)D (z − (z0 + ℎ)) (z − z0) k! (z − z )k + (z − z )k−1(z − (z + ℎ)) + ⋯ + (z − z )(z − (z + ℎ))k−1 + (z − (z + ℎ))k = f(z) 0 0 0 0 0 0 dz. k+1 k+1 2i Ê)D (z − (z0 + ℎ)) (z − z0)

Taking ℎ → 0, it converges to

k! (k + 1) (k + 1)! f(z) f(z) dz = dz, k+2 k+2 2i Ê)D (z − z0) 2i Ê)D (z − z0) and we are done with k + 1 case. Such mathematical induction argument shows that

n! f(z) f (n)(z ) = dz 0 n+1 2i Ê)D (z − z0) for all n = 0, 1, 2, ⋯.

A corollary from Cauchy’s integral formula is the following Cauchy’s inequality.

THEOREM 3.5 (Cauchy’s inequality). Assume Ω is an open subset of ℂ and f is holomorphic over

Ω. Then for each disk center at z0 ∈ Ω with DR(z0) ⊂ Ω, there is the estimate

n!‖f‖)D (z ) f (n)(z ) R 0 . ð 0 ð ≤ Rn

Here f ∶= sup f(z) . ‖ ‖)DR(z0) z∈)DR(z0) ð ð 28 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

PROOF. By Cauchy’s integral formula, n! f(z) f (n)(z ) = dz ð 0 ð ð2i z z n+1 ð Ê)DR(z0) ( − 0) n! f ⋅ length()D (z )) ≤ ‖ n+1 ‖)DR(z0) R 0 2 (z − z0) n! f(z) = sup ⋅ 2R ð n+1 ð 2 z∈)DR(z0) R n!‖f‖)D (z ) = R 0 . Rn A corollary is the following Liouville’s theorem.

THEOREM 3.6 (Liouville’s theorem). A bounded entire function can only be constant.

PROOF. Assume f is an entire function with ðf(z)ð ≤ M for any z ∈ ℂ, where M > 0 is a constant. Then from the Cauchy’s inequality, for any z0 ∈ ℂ,

‖f‖)D (z ) f ¨(z ) R 0 , ð 0 ð ≤ R ¨ for any R ∈ ℝ. Take R → +∞, we see that f (z0) must be zero. ¨ This argument shows that f ≡ 0 over ℂ. By the connectedness of ℂ, from Corollary 4.10, this implies f is a constant. We use the Liouville’s theorem to give a proof of the fundamental theorem of algebra now.

n n−1 THEOREM 3.7. For any polynomial p(z) = anz + an−1z + ⋯ a1z + a0 with

an, an−1, ⋯ , a1, a0 ∈ ℂ, an ≠ 0 and n ≥ 1, the equation p(z) = 0 has n roots in ℂ.

PROOF. (1) We ﬁrst prove that there must exist one root for p(z) = 0. Assume this in not true, then p(z) ≠ 0 for any z ∈ ℂ and so the function 1 f(z) ∶= p(z) is an entire function over ℂ. We write n p(z) = anz (1 + g(z)), with a zn−1 + ⋯ a z + a a a a a g(z) = n−1 1 0 = n−1 + n−2 + ⋯ + 1 + 0 . n 2 n−1 n anz anz anz anz anz Notice lim g(z) = 0, there must exist some M > 0 so that whenever z > M, g(z) < 1 . z→0 ð ð ð ð 2 It follows 1 1 + g(z) > , z > M. ð ð 2 ð ð Then 1 1 1 2 f z , ð ( )ð = ð ð = ð n ðð ð ≤ n p(z) anz 1 + g(z) ðanððzð 3. CAUCHY’S INTEGRAL FORMULAS 29

and so lim f(z) = 0. Then we can take some M¨ > 0 so that f(z) < 1. Further ðzð→+∞ ð ð ð ð since f(z) is continuous in z, we can take C ∶= max{sup{ðf(z)ðz ∈ DM¨ (0)}, 1}, which is a positive real number, so that any z ∈ ℂ

ðf(z)ð ≤ C. This shows f is a bounded entire function. By the Liouville’s theorem, such f must be constant, but then it contracts with the assumption that the degree n of f is at least 1. Hence p(z) = 0 has at least one solution in ℂ. (2) Next we show p(z) = 0 has n (possibly equal) solutions.

Assume z1 is a solution of p(z) = 0 from (1). We can factor p(z) and write

p(z) = (z − z1)p1(z),

with p1 is a polynomial of degree n − 1.

If n − 1 = 0, then p(z) has only one root z1 and we are done. Otherwise by (1) again, we

ﬁnd z2 as a solution of p1(z) = 0. Write

p1(z) = (z − z2)p2(z), p(z) = (z − z1)(z − z2)p2(z).

If n − 2 = 0, then p(z) has two roots z1, z2 and we are done. Otherwise repeat the procedure above again. Since n is ﬁnite, after n steps, we obtain

p(z) = an(z − z1)(z − z2) ⋯ (z − zn). This proves the theorem. (We remark that Step (2) actually works for any UFD (unique factorization integral domain). ) 3.2. Holomorphic functions are analytic. We have seen before that every analytic function is holomorphic. Now, we use Cauchy’s integral formula to prove that in fact every holomorphic function is analytic.

THEOREM 3.8. Assume Ω is an open subset of ℂ and f is holomorphic over Ω. Then for any open disk D(z0) with D(z0) ⊂ Ω, f can be written as a convergent power series centered at z0 as f (n)(z ) f(z) = Σ∞ a (z − z )n, z ∈ D(z ), a = 0 . n=0 n 0 0 n n!

PROOF. Take any open disk D(z0) with D(z0) ⊂ Ω. For any z ∈ D(z0), by Cauchy’s integral formula, we have 1 f(w) f(z) = dw. 2i Ê)D w − z z0 The term 1 for z ∈ D(z ), w ∈ )D(z ) can be written as w−z 0 0 1 1 1 1 = = . z−z0 w − z w − z0 − (z − z0) w − z0 1 − w−z0 Notice that for such z, there exists some 0 < r < 1 so that z−z0 < r, and then we can further write ð w−z0 ð 1 z − z = Σ∞ ( 0 )n, z−z0 n=0 1 − w − z0 w−z0 30 2. CAUCHY’S THEOREM AND ITS APPLICATIONS and then

1 1 ∞ z − z0 n = Σn=0( ) w − z w − z0 w − z0 (z − z )n = Σ∞ 0 . n=0 n+1 (w − z0) Plug it into the integration, we have 1 f(w) f(z) = dw 2i Ê)D w − z z0 1 (z − z )n = f(w)Σ∞ 0 dw n=0 n+1 2i Ê)D (w − z ) z0 0 1 f(w) = Σ∞ ( dw)(z − z )n n=0 n+1 0 2i Ê)D (w − z ) z0 0 ∞ n =∶ Σn=0an(z − z0) .

Moreover, by Cauchy’s integral formula, the coefﬁcients an can be written as 1 f(w) f (n)(z ) a = dw = 0 . n n+1 2i Ê)D (w − z ) n! z0 0

Notice, in the proof, we never use the radius of the disk D(z0) but only use D(z0) ⊂ Ω. Hence, every entire function can be written as a power series (with any point as a center)

∞ n f(z) = Σn=0an(z − z0) . Now we are ready to prove the unique analytic continuation property of holomorphic functions.

THEOREM 3.9. Assume Ω is an open and connected subset of ℂ and f is holomorphic over Ω. If there exists a convergent sequence of distinct points {zn} in Ω so that f(zn) = 0, then f must vanish on Ω.

PROOF. Assume the sequence zn → z0 in Ω. Since any holomorphic function is continuous, f(zn) =

0, n = 1, 2, ⋯, imply f(z0) = 0.

We take a small disk D(z0) so that D(z0) ⊂ Ω. From Theorem 3.8, we can write ∞ n f(z) = Σn=0an(z − z0) , over D(z0). If there exists some coefﬁcient ak ≠ 0, WLOG, we can assume such k is the coefﬁcient of smallest degree term, we can write 2 k ak+1(z − z0) ak+2(z − z0) f(z) = ak(z − z0) (1 + g(z)), g(z) = + + ⋯ , z ∈ D(z0). ak ak

Since g is holomorphic so continuous, limn→∞(1 + g(zn)) = 1 + g(z0) = 1, which is away from 0. It follows there exists some N so that any n > N, 1 + g(zn) ≠ 0. On the other hand, f(zn) = 0 for all n. Hence we must have zn = z0 for all n > N, which contradicts with the assumption that {zn} are distinct.

This proves that the function f vanishes over the disk D(z0). Now we construct a subset of Ω as

Z ∶= {z ∈ Ωðf(z) = 0}, 4. IMPORTANT APPLICATIONS OF CAUCHY’S THOEREM 31

◦ and U = Z , the interior of Z. We have proved that U ≠ since D(z0) ⊂ U. U is open by deﬁnition. We now show U is also closed. To see this, take any sequence {wn} of distinct points in U with wn → w0, we show that w0 ∈ U. First w0 ∈ Z because f is continuous and f(wn) = 0 for all n. Moreover, w0 is an interior point of Z follows from the same proof for z0 before. Above all, we ﬁnd a nonempty open and closed subset of Ω. At last, by the connectedness of Ω, U must be the same as Ω and we are done. DEFINITION 3.10. (1) Assume U ⊂ V are two sets. If there are two functions

f ∶ U → ℂ, f̃ ∶ V → ℂ ̃ ̃ with fðU = f, then we call f an extension of f. (2) If U and V are both open subsets of ℂ, f, f̃ are both analytic (holomorphic), then we call f̃ an analytic extension or analytic continuiation of f from U to V . From the above theorem, we obtain the following unique continuation property for analytic functions.

COROLLARY 3.11 (Unique Continuation Theorem). Assume Ω ⊂ ℂ is an open and connected subset, and U is an open subset in Ω. Then there is at most one analytic continuations of any analytic function f ∶ U → ℂ from U to Ω. ̃ ̃ PROOF. Assume f1, f2 are two analytic continuations of f from U to Ω. Then consider ̃ ̃ g ∶= f1 − f2 ̃ ̃ over Ω, which is analytic and vanishes on U. Apply Theorem 3.9, g must be zero on Ω, i.e., f1, f2 must be the same on Ω. We make two remarks for the continuation theorem: (1) It is possible that some analytic function over an open set doesn’t have any analytic extension. E.g., f(z) = 1 , z ∈ ℂ∗. z (2) If we change the terms from ‘holomorphic (analytic)’ to ‘smooth’, then there always exist inﬁnitely many different extensions. This is another big difference between smooth functions and holomorphic functions.

4. Important Applications of Cauchy’s thoerem

4.1. Morera’s theorem. Moreta’s theorem is a reserve of Goursat’s theorem 1.4.

THEOREM 4.1. Assume f is a continuous function over an open disk D in ℂ so that for triangle T with T ⊂ D, f(z)dz = 0. Ê)T Then f is holomorphic over D.

PROOF. Recall the proof of Cauchy theorem 1.2, the assumption ∫)T f(z)dz = 0 implies that f has a primitive F so that F ¨(z) = f(z), z ∈ D. Then the regularity of F (Cauchy’s integral formula for F ) implies F has second order complex derivative, i.e., f is holomorphic. Morera’s theorem helps us to understand behavior of sequence of holomorphic functions. 32 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

4.2. Sequence of holomorphic functions.

THEOREM 4.2. Assume Ω is an open subset of ℂ and {fn} is a sequence of holomorphic functions over Ω that uniformly converges to f over each compact subset K in Ω. Then f is holomorphic over Ω. ¨ ¨ Moreover, {fn} uniformly converges to f over each compact subset K in Ω.

PROOF. (1) We ﬁrst prove f is holomorphic using the Morera’s theorem 4.1.

For each point z0 ∈ Ω, we take a small disk D(z0) ⊂ Ω. Then for each triangle T with

T ⊂ D(z0), the uniform convergence of {fn} to f implies

lim fn(z)dz = lim fn(z)dz = f(z)dz = 0, n→∞ Ê)T Ê)T n→∞ Ê)T

since )T is compact. Then apply the Morera’s theorem, f is holomorphic over D(z0) and so

at z0. By this way, we prove that f is holomorphic over Ω. ¨ ¨ (2) Take any compact subset K in Ω. We prove now {fn} uniformly converges to f over K. For each z ∈ K, take > 0 so that D (z) ⊂ Ω. Then {D (z) z ∈ K} form an open z 2z z ð cover of K. Since K is compact, we can pick a ﬁnite subcover {D (zi) i = 1, ⋯ ,L} from zi ð it. Denote by r ∶= inf{ i, j = 1, 2, ⋯ ,L}. Since there are only ﬁnitely many such disks, r zi ð must be positive.

Now for any z ∈ K, there must exist some i so that z ∈ D (zi). Any point w ∈ Dr(z), zi we have

ðw − zið ≤ ðw − zð + ðz − zið ≤ r + i ≤ 2i.

Hence w ∈ D (zi) ⊂ D (zi)Ω. 2 zi 2 zi ̃ ̃ Construct K = ∪z D (zi). Clearly, K contains K and is compact. The above argument i 2 zi shows that ̃ Dr(z) ⊂ K, for any z ∈ K. ̃ Now for any > 0, since {fn} uniformly converges to f over K, there exists some N > 0 so that any n > N, for any w ∈ K̃ ,

ðfn(w) − f(w)ð < .

Now for any z ∈ K, consider Dr(z) and by the Cauchy’s inequality for the holomorphic func-

tions fn − f,

supw∈)D (z) ðfn(w) − f(w)ð sup ̃ f (w) − f(w) f ¨(z) − f ¨(z) r w∈K ð n ð . ð n ð ≤ r ≤ r ≤ r ¨ ¨ ¨ This proves {fn − f } uniformly converges to zero over K, i.e., {fn} uniformly converges to f ¨ over K. (k) (k) As an immediate corollary, {fn } uniformly converges to f over each compact subset K in Ω for any k = 1, 2, ⋯. We remark that this theorem is another difference of holomorphic functions from smooth functions. In real analysis we know any continuous function can be approached by polynomials uniformly over compact sets; however, on the other hand, there does exist some continuous but nowhere differentiable functions. 4. IMPORTANT APPLICATIONS OF CAUCHY’S THOEREM 33

4.3. Holomorphic functions deﬁned in terms of integrals.

THEOREM 4.3. Assume Ω is an open subset of ℂ, and F is complex valued function deﬁned on Ω × [0, 1], such that (1) F is continuous on Ω × [0, 1]; (2) For each s ∈ [0, 1], F (⋅, s) ∶ Ω → ℂ is holomorphic. 1 Then the function f(z) ∶= ∫0 F (z, s)ds is holomorphic over Ω.

PROOF. Since F is continuous in s, the Riemann integral is well-deﬁned and we can use a sequence of Riemann sums to approach it: For each n = 1, 2, ⋯, deﬁne

1 k f (z) ∶= Σn F (z, ), n n k=1 n which is holomorphic over Ω.

Now take any z ∈ Ω, we use {fn} to prove that f is holomorphic at z. For this, take some ball D so that D ⊂ Ω. We prove {fn} converges to f uniformly on D. To see this, we ﬁrst notice that F is continuous implies F is uniformly continuous over D × [0, 1] since D × [0, 1] is compact. Then for any

> 0, there exists some > 0 so that any ðs1 − s2ð < ,

sup ðF (w, s1) − F (w, s2)ð < . w∈D

Then we consider n’s large enough so that 1 < . If follows for any w ∈ D, n

k∕n n k ðfn(w) − f(w)ð = ðΣk=1 (F (w, ) − F (w, s))dsð Ê(k−1)∕n n k∕n n k ≤ Σk=1 ðF (w, ) − F (w, s)ðds Ê(k−1)∕n n Σn = , ≤ k=1 n and this proves {fn} uniformly converges to f over D. Then for each compact subset K in D, {fn} uniformly converges to f over K. By applying Theorem 4.2 for D, we obtain that f is holomorphic over D and in particular, holomorphic at z.

Another way to prove this theorem is to use Morera’s theorem: For any triangle T ⊂ T ⊂ Ω,

1 f(z)dz = F (z, s)dsdz Ê)T Ê)T Ê0 1 = F (z, s)dzds Ê0 Ê)T 1 = 0ds = 0. Ê0 Here the last second equality if the Fubini’s theorem, which can be proved by the Fubini’s theorem for real functions. The last equality if the Goursat’s theorem. 34 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

4.4. Schwarz reﬂection principle. A set Ω is called symmetric about real axis, if a point z ∈ Ω if and only if ̄z ∈ Ω. For a set, we denote by

+ − Ω ∶= {z ∈ ΩðIm(z) > 0}, Ω ∶= {z ∈ ΩðIm(z) < 0}, and I ∶= {z ∈ ΩðIm(z) = 0}. Clearly, Ω = Ω+ ∪ Ω− ∪ I. If Ω is open in ℂ, then Ω+, Ω− are open in ℂ. The Schwarz reﬂection principle answers a question that: given a holomorphic function f over Ω+, when can we extend it holomorphically to Ω? From the unique continuation theorem 3.11, we know such extension if exists, must be unique.

THEOREM 4.4 (Schwarz Reﬂection Principle). Assume f is holomorphic over Ω+ and can be continuously extended to I, which means there exists a continuous function

f̃ ∶ Ω+ ∪ I → ℂ ̃ ̃ so that fðΩ+ = f. If fðI is a real function, then f has a holomorphic extension to Ω.

PROOF. We construct a function F deﬁned on Ω as ⎧ f(z), z ∈ Ω+ F z ⎪ ̃ ( ) = ⎨ f(z), z ∈ I ⎪ f( ̄z), z ∈ Ω− ⎩ Now we prove F is holomorphic over Ω. For this, we take any point z ∈ Ω. + + ∙ If z ∈ Ω , then take a disk D(z) centered at z with D(z) ⊂ Ω . Since F = f , ðD(z) ðD(z) + f is holomorphic over Ω implies F is holomorphic over D(z), so holomorphic at z. − − ∙ If z ∈ Ω , then take a disk D(z) centered at z with D(z) ⊂ Ω . If follows

D( ̄z) is a disk centered at ̄z in Ω+. Since f is holomorphic over Ω+, we can write

∞ n f(w) = Σn=0an(w − ̄z) , w ∈ D( ̄z).

If follows for any w ∈ D(z), w̄ ∈ D( ̄z) and hence F w f w̄ ∞ a w̄ ̄z n ∞ a w z n, ( ) = ( ) = Σn=0 n( − ) = Σn=0 n( − )

which is convergent. Hence F is holomorphic over D(z), and in particular holomorphic at z. ∙ If z ∈ I, again we take some open disk D centered at z in Ω, and prove that F is holomorphic over D. For this, consider any triangle T with T ⊂ D. For each > 0 and sufﬁciently small, we can divide T into three parts as:

+ − T ∶= {z ∈ T ðIm(z) > },T ∶= {z ∈ T ðIm(z) < −},T ∶= {z ∈ T ð − ≤ Im(z) ≤ }. It follows

F (z)dz = F (z)dz + F (z)dz + F (z)dz. + − Ê)T Ê)T Ê)T Ê)T 4. IMPORTANT APPLICATIONS OF CAUCHY’S THOEREM 35

Since F is holomorphic over Ω+, Ω− as we have shown, the integrals

F (z)dz = 0 = F (z)dz. + − Ê)T Ê)T Let’s show lim F (z)dz = 0. →0 ∫)T Typically, )T consists of four paths, which we denote by L1, the horizontal one with

orientation from left to right; L2, the one across the real axis from down to up; L3, another

horizontal one with orientation from right to left; L4, the one across the real axis from up to down. Notice that f̃ is continuous on Ω+ ∪ I and real on I implies F is continuous on Ω. Hence the integrals

ð F (z)dzð ≤ sup ðF (z)ð ⋅ length(Ti) → 0 ÊLi z∈Li as → 0 for i = 2, 4. For F (z)dz + F (z)dz, The uniform continuity of F over T implies it converges to ∫L1 ∫L3 ̃ ̃ ∫I f(z)dz + ∫I− f(z)dz which is constant zero.

This proves that ∫)T F (z)dz = 0 for any such triangle T . By Morera’s theorem, F is holomorphic over D, in particular, holomorphic at z. The proof of Schwarz reﬂection principle can be generalized to the following symmetry principle.

THEOREM 4.5 (Symmetry Principle). Assume f ± are holomorphic over Ω± and can be extended continuously to I respectively (still denote by f ±), and

f +(z) = f −(z), z ∈ I.

Then the function T f +(z), z ∈ Ω+ ∪ I F (z) = f −(z), z ∈ Ω− is holomorphic over Ω.

4.5. Runge’s approximation theorem.

EXAMPLE 4.6. (1) Holomorphic function f over D. (2) Function f(z) = 1 over C. z

THEOREM 4.7 (Runge’s Appoximation Theorem). For any compact subset K of ℂ and a holomorphic function f deﬁned over an open neighborhood Ω of K, there exists a sequence of rational functions c {rn} with singularities living in K which uniformly converge to f on K. Further more, if Kc is connected, then such rational functions can be taken as polynomials.

PROOF. (1) The proof is constructive using Riemann sums. First since K is a compact subset in Ω, the number (possible inﬁnity when Ω = ℂ)

d ∶= inf d(z1, z2) > 0. z1∈K,z2∈)Ω Again because K is compact, we enclose it into a big enough square I × I, and then divided it into N × N equal squares using grids so that the diameter of each small square is smaller than d∕100. By this way, we can use K to color the small squares whose closures have nonempty 36 2. CAUCHY’S THEOREM AND ITS APPLICATIONS

intersection with K. Denote by S the union of such colored small squares. The boundary )S is a piecewise-smooth curve, which has no intersection with K by construction. By Cauchy’s integral formula (and some careful deformation for the integral domain), for any z ∈ K, we can write 1 f(w) f(z) = dw. 2i Ê)S w − z Further, over each smooth piece of the piecewise-smooth curve )S, we can write f(w) 1 f(w(s))w¨(s) dw = ds, Ê)S w − z Ê0 w(s) − z where w ∶ [0, 1] → ℂ is a smooth parametrization of a smooth piece of )S. ¨ Deﬁne F (z, s) ∶= f(w(s))w (s) , with (z, s) ∈ Ω × [0, 1]. Notice F is not deﬁned on w(s)−z {(z, s)ðw(s) = z}, however, it is continuous on K × [0, 1] and for each s ∈ [0, 1], F (⋅, s) is holomorphic over S◦. ¨ We denote by f (z) the Riemann sum from the Riemann integral 1 f(w(s))w (s) ds by diving n ∫0 w(s)−z [0, 1] into n equal pieces, for each z ∈ K, and by rn(z) the one after taking sums for all smooth

pieces of )S. Just as the proof in Theorem 4.3, rn uniformly converges to f on K.

By noticing each fn is a rational function with (at most n) singularities living on )S which is in Kc. We are done with the ﬁrst part. c (2) Next we prove that in the case K is connected, each rational function fn in that sequence can be uniformly approximated over K by a sequence of polynomials {pn,kðk = 1, 2, ⋯}. After this is done, for f , take k so that f − p < 1; For f , take k > k and f − p < 1 ; 1 1 ‖ 1 1,k1 ‖K 2 2 1 ‖ 2 2,k2 ‖K 2 ... Keep doing this, we obtain a sequence of polynomials {p n = 1, 2, ⋯} so that Notice that n,kn ð 1 p − f p − f + f − f + f − f . ‖ n,kn ‖K ≤ ‖ n,kn n‖K ‖ n ‖K ≤ n ‖ n ‖K Hence polynomial sequence {p n = 1, 2, ⋯} can uniformly approximate f over K. n,kn ð

LEMMA 4.8. Assuming Kc is connected. Then any rational function r with singularities in c K can be uniformly approximated over K by a sequence of polynomials {pn}.

c PROOF. First, notice that it is enough to prove that for any z0 ∈ K , one can uniformly approximate 1 over K by a sequence of polynomials {p }. z−z n 0 c If there exists a disk D so that K ⊂ D and z0 ∈ D , then 1 1 1 zn = − = Σ∞ , z n=1 n+1 z − z0 z0 1 − −z z0 0 1 N zn which shows can be uniformly approximated by polynomials pN (z) = Σ n+1 . z−z0 n=1 −z c 0 Otherwise, we can enclose K by a large enough disk and take z1 ∈ D and do the above approximation for 1 . z−z1 c Then since K is connected, we can connect z0 and z1 via a continuous path with (0) = c z0, (1) = z1, and ⊂ K . we cover via ﬁnite open disks

D(z0),D(w1), ⋯ ,D(wl),D(z1), z0 =∶ w0, w1, w2, ⋯ , wl, wl+2 ∶= z1 ∈ ,

which are disjoint from K and wi ∈ wi−1 for all i = 1, 2, ⋯ , l + 1. Using them, one can

approach z1 from z0 through these ﬁnite points w1, w2, ⋯ , wl. 4. IMPORTANT APPLICATIONS OF CAUCHY’S THOEREM 37

By writing 1 1 1 (z − w )n = = Σ∞ 0 1 , z0−w1 n=0 n+1 z − z0 z − w1 1 − (z − w1) z−w1 n 1 N (z0−w1) one can uniformly approximate by rational functions Σ n+1 , N = 1, 2, ⋯. Then z−z0 n=0 (z−w1) each 1 can be uniformly approximated by rational functions from summations and multi- z−w1 plications of 1 ... Using the same argument at the beginning of (2), we see that 1 can be z−w2 z−z0 uniformly approximated by rational functions from summations and multiplications of 1 . z−w2 Keep doing this, after ﬁnite steps, this proves that 1 can be uniformly approximated by z−z0 rational functions from summations and multiplications of 1 . Then from the polynomial z−z1 approximation for 1 we have constructed, the proof is done. z−z1

CHAPTER 3

Meromorphic functions and the logarithm

1. Zeros and Poles

By a point singularity or isolated singularity of a function f, we mean a point z0 ∈ ℂ where f is not ∗ defined, but there exists an open disk D centered at z0 so that f is defined on D ⧵ {z0}. We use D to denote D ⧵ {z0} and called it a punctured disk with puncture z0. There are three types of isolated singularities, which are ∙ Removable singularities: if lim f(z) exists. For example, f(z) = z, z ∈ ℂ∗. Then 0 is a z→z0 removable singularity. ∙ Poles (main object of this section). We will define it precisely later. For example, f(z) = 1 has z 0 as a pole. ∙ Essential singularities: a singularity which is not removable and not a pole. For example, f(z) = e1∕z has 0 as an essential singularity.

EXAMPLE 1.1. Consider the function f(z) = e1∕z over ℂ∗. There are the following limiting behavior near z0 = 0: 1∕z ∙ limz→0,z∈ℝ+ e = +∞; 1∕z ∙ limz→0,z∈ℝ− e = 0; 1∕z 1∕z ∙ The function e oscillates rapidly as z → 0, z ∈ iℝ, so the limit limz→0,z∈iℝ e does not exist. This is a typical example of essential singularity.

Assume f is a holomorphic function deﬁned over an open subset Ω in ℂ. A point z0 ∈ Ω is called a zero of f if f(z0) = 0. By the Analytic Continuation Theorem 3.9, any nontrivial (i.e., not constant zero) holomorphic function has only isolated zeros, i.e., its zero points must be isolated points in Ω. By extending the proof of Theorem 3.9, we will obtain the following result.

THEOREM 1.2. Assume Ω is an open subset in ℂ and f is holomorphic over Ω with a zero but not constant zero. Then for each zero z0 ∈ Ω, there exists some open disk D centered at z0, a nowhere vanishing holomorphic function g over D, an integer n ∈ ℤ+ so that

n f(z) = (z − z0) g(z), z ∈ D.

PROOF. By the Analytic Continuation Theorem 3.9, there exists some open disk D centered at z0 so that z0 is the unique zero of f on D, and one can write f as a power series centered at z0 as

∞ k f(z) = Σk=0ak(z − z0) , z ∈ D.

+ Since f(z0) = 0, so a0 = 0. Since f is not indetifcally zero over D, there must be some n ∈ ℤ so that

a1 = ⋯ = an−1 = 0, an ≠ 0. 39 40 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

Hence we can write

n ∞ k f(z) = (z − z0) g(z), with g(z) ∶= Σk=0an+k(z − z0) .

Notice that g is holomorphic over D since it is analytic. Moreover, g(z0) ≠ 0, and it is not vanishing on D ⧵ {z0} either since f doesn’t vanish on D ⧵ {z0}. We are done with the proof. Notice the integer n and the function g are uniquely determined by the function f. To see this, assume there exists some m ∈ ℤ+ and ℎ so that

n m f(z) = (z − z0) g(z) = (z − z0) ℎ(z). Assuming that n < m, then we can write

m−n g(z) = (z − z0) ℎ(z), and then g(z0) = 0 which contradicts with our requirement. Similarly, n can not be bigger than m, and so they must be the same, which then implies g = ℎ over D.

Such uniquely determined n is called the multiplicity of the zero z0 or the order of the zero z0. If n = 1, then such zero point is called a simple zero.

Next we introduce the definition of a pole. We say a function f has a pole at z0, if f is defined over ∗ 1 1 D (z0) and after define = 0, is holomorphic and has z0 as a zero. For example, z0 = 0 is a f(z0) f(z0) pole of f(z) = 1 . z

THEOREM 1.3. If f has a pole at z0, then there exists some open disk D(z0) and a nonwhere vanishing holomorphic function ℎ and a unique integer n ∈ ℤ+ so that ℎ(z) f z . ( ) = n (z − z0) 1 PROOF. Since f has a pole at z , there exists some open disk D(z ) so that the function is holo- 0 0 f morphic over D(z0) and has z0 as its unique zero. Using Theorem 1.2, there exists a unique n ∈ ℤ+ so that 1 = (z − z )ng(z), f(z) 0 with g nowhere vanishing over D(z0). It follows that ℎ(z) 1 f z , ℎ z . ( ) = n ( ) ∶= (z − z0) g(z) The uniqueness of n follows from the uniqueness of n in Theorem 1.2. Again, the integer n is called the order or multiplicity of the pole. If n = 1, then we say the pole is simple.

THEOREM 1.4. If f has a pole of order n at z0, then over some small disk D(z0) centered at z0, f can be written as a a−(n−1) a f(z) = −n + + ⋯ + −1 + g(z) n n−1 (z − z0) (z − z0) z − z0 where g is holomorphic over D(z0). 1. ZEROS AND POLES 41

PROOF. By writing ℎ(z) f z ( ) = n (z − z0) over some small disk centered at z0, we can further shrink the disk to some D(z0) so that ℎ(z) has a power series expansion as ∞ k ℎ(z) = Σk=0ck(z − z0) . It follows ∞ k ℎ(z) Σ ck(z − z0) c c c f(z) = = k=0 = 0 + 1 + ⋯ + n−1 + Σ∞ c (z − z )k−n. n n n n−1 k=n k 0 (z − z0) (z − z0) (z − z0) (z − z0) z − z0 By deﬁning ∞ k−n a−k = cn−k, k = 1, 2, ⋯ , n, g(z) = Σk=nck(z − z0) , we are done with the proof. The sum of n terms a a−(n−1) a P (z) ∶= −n + + ⋯ + −1 f n n−1 (z − z0) (z − z0) z − z0 is called the principal part of f at the pole z0. The coefﬁcient a−1 is called the residue of f at the pole, and denoted as res f ∶= a . z0 −1 Notice 1 1 res f = P (z)dz = f(z)dz. z0 2i f 2i Ê)D(z0) Ê)D(z0) This is the simplest case of residue formula. (We leave more discussion on this to next section. ) The following theorem is helpful in calculation of residue.

THEOREM 1.5. If f has a pole of order n at z0, then 1 dn−1 res f = lim ((z − z )nf(z)). z0 n−1 0 z→z0 (n − 1)! dz

a−1 PROOF. ∙ If n = 1, we can write f(z) = + g(z) over some open disk D(z0). If follows z−z0

(z − z0)f(z) = a−1 + (z − z0)g(z), which converges to a = res f as z → z . −1 z0 0 ∙ Higher order cases are left to you as exercise. We see some examples about calculation residues.

e−z EXAMPLE 1.6. (1) Consider the function f(z) = . Find singularities in ℂ, determine their 1+z2 types and calculate the residue at poles.

SOLUTION: The function has singularities at 1 + z2 = 0, i.e., z = ±i. Since 1∕f(z) = ez(1 + z2) is holomorphic over ℂ, both singularities are poles and their orders are both 1. −z −z −i −z i Write f(z) = e . The residue res (f) = e = e and res (f) = e = − e . (z−i)(z+i) i z+i ðz=i 2i −i z−i ðz=−i 2i 2 2 (2) Consider the function f(z) = (z −1) . Find singularities in ℂ, determine their types and z2(z−2)(2z−1) calculate the residue at poles. 42 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

1 1 SOLUTION: The function has 0, 2, as singularities. z = 0 is an order 2 pole, z = 2, are 2 2 two order 1 pole. The residue (z2 − 1)2 3 (z2 − 1)2 3 res2(f) = lim = , res 1 (f) = lim = − . z→2 z2(2z − 1) 4 2 z→2 2z2(z − 2) 4 The residue d (z2 − 1)2 5 res0(f) = lim = . z→0 dz (z − 2)(2z − 1) 4

Another way to calculate res0(f) is to use definition. Write z2 2 1 f(z) = − + . (z − 2)(2z − 1) (z − 2)(2z − 1) z2(z − 2)(2z − 1) Near z = 0, the first two parts are holomorphic so has no contribution to the residue at 0. It follows 1 res (f) = res . 0 0 z2(z − 2)(2z − 1) To calculate res 1 we can either use the formula or write it as 0 z2(z−2)(2z−1) 1 1 z z = (1 + + ( )2 + ⋯)(1 + (2z) + (2z)2 + ⋯). z2(z − 2)(2z − 1) 2z2 2 2 The residue is the coefficient of order 1 term in 1 (1 + z + ( z )2 + ⋯)(1 + (2z) + (2z)2 + ⋯) 2 2 2 which is 1 (2 + 1 ) = 5 . 2 2 4

2. The residue formula

THEOREM 2.1. Assume Ω is an open subset in ℂ and f is holomorphic over Ω except at ﬁnite poles z1, z2, ⋯ , zk ∈ Ω. Then for any disk D including z1, z2, ⋯ , zk with D(z0) ⊂ Ω, there is

f(z)dz = 2iΣk res (f). j=1 zj Ê)D

PROOF. By the ﬁniteness of poles and openness of D, there exists some > 0 so that each D(zj) ⊂ D and all these disks are disjoint from each other. Using Cauchy’s theorem,

k f(z)dz = Σj=1 f(z)dz, Ê)D Ê)D(zj ) and each f(z)dz = 2iΣk res (f) by Theorem 1.4, we are done with the proof. ∫)D(zj ) j=1 zj

We see some applications of residue formula.

+∞ 1 EXAMPLE 2.2. (1) From calculus, we know dx = arctan x +∞ = − (− ) = . ∫−∞ 1+x2 ð−∞ 2 2 We now use complex analysis method to calculate it. For this, consider the function f(z) = 1 , which has two pole singularity of order 1 are ±i. Take D+ as the half disk centered at 1+z2 R zero with radius R > 0. The path integral

f(z)dz = 2iresi(f) = . + Ê)DR 2. THE RESIDUE FORMULA 43

On the other hand, R 1 f(z)dz = dx + f(z)dz, + 1 + x2 Ê)DR Ê−R Ê R

where R denotes the path from R to −R along the half circle CR. We estimate 1 2 f(z)dz R ⋅ sup R → 0 ð ð ≤ ð 2 ð ≤ 2 Ê R R 1 + z R as R → +∞. Hence by taking R → +∞, we obtain +∞ 1 dx = . 2 Ê−∞ 1 + x (2) For each n ∈ ℤ+, calculate +∞ 1 dx. ∫−∞ (1+x2)n+1 Similar as previous one, we consider the function f(z) = 1 , which has two pole (1+z2)n+1 + singularity of order n + 1 are ±i. Take the same loop DR, the integral

f(z)dz = 2iresi(f), + Ê)DR and we calculate 1 dn 1 resi(f) = lim n! z→i dzn (1 + x2)n+1 1 (−1)n(n + 1)(n + 2) ⋯ (2n) = n! (2i)2n+1 1 (2n − 1)!! = . 2i (2n)!! Use the same way we prove

ð f(z)dzð → 0 Ê R as R → +∞. Hence +∞ 1 (2n − 1)!! dx = . 2 n+1 Ê−∞ (1 + x ) (2n)!!

+∞ eax +∞ ex∕2 EXAMPLE 2.3. Prove the formula dx = , 0 < a < 1. (e.g., dx = .) ∫−∞ 1+ex sin a ∫−∞ 1+ex PROOF. Consider the complex-valued function eaz f(z) = 1 + ez over the rectangle region

RecR ∶= {z ∈ ℂð − R ≤ Re(z) ≤ R, 0 ≤ Im(z) ≤ 2}.

It has an order 1 pole at z = i in RecR, the residue is calculated as eaz(z − i) resi(f) = lim (z − i)f(z) = lim z→i z→i 1 + ez eaz(z − i) = lim z→i −((z − i) + 1 (z − i)2 + ⋯) 2 = −eai.

Hence ai f(z)dz = 2iresi(f) = −2ie . Ê)RecR 44 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

On the other hand, f(z)dz consists of four parts, two horizontal parts and two vertical parts A ,A . ∫)RecR l r The two vertical part can be bounded as 2 ea(R+it) f(z)dz dt Ce(a−1)R ð ð ≤ ð eR+it ð ≤ ÊAr Ê0 1 + which converges to zero since a < 1. The two horizontal integrations then give the identity we want.

3. Singularities and meromorphic functions

THEOREM 3.1 (Riemann’s theorem on removable singularities). Assume Ω is an open subset of ℂ and z0 ∈ Ω. Assume f is holomorphic over Ω ⧵ {z0} and bounded. Then f can be extended holomorphically to Ω.

PROOF. It is enough to work locally for some disk with D(z0) ⊂ ℂ.

We deﬁne a function over D(z0) as 1 f(w) g(z) ∶= dw. 2i w − z Ê)D(z0)

It is well-deﬁned and holomorphic since f is holomorphic on Ω ⧵ {z0}. We now show that g(z) = f(z) ∗ on D (z0). Then it follows the unique analytic continuity that f can be extended to z0 as g(z0). ∗ ∗ To see g(z) = f(z) on D (z0), for each z ∈ D (z0), pick some small > 0 so that D(z0) ∩ D(z) = ç. Then we write 1 f(w) 1 f(w) 1 f(w) dw = dw + dw. 2i w − z 2i w − z 2i w − z Ê)D(z0) Ê)D(z0) Ê)D(z) By Cauchy’s integral formula, 1 f(w) dw = f(z). 2i w − z Ê)D(z) The ﬁrst term, sup f 1 f(w) D(z0) ð ð dw ð ð ≤ 1 2i Ê)D (z ) w − z z − z 0 2 ð 0ð which converges to zero as → 0. Here we use f is bounded so that sup f is a number. D(z0) ð ð ∗ Hence this proves g(z) = f(z) on D (z0). A powerful consequence from Riemann’s theorem on removable singularities is the following corollary, which gives an easy criterion of determining the type of an isolated singularity.

COROLLARY 3.2. Assume Ω is an open subset of ℂ and z0 ∈ Ω. Assume f is holomorphic over Ω ⧵ {z } with z as an isolated singularity. Then z is a pole if and only if lim f(z) = ∞. 0 0 0 z→z0 ð ð g(z) PROOF. Assume z0 is a pole, then near z0, f can be written as f(z) = n , for some g(z) nowhere (z−z0) vanishing near z0. Then it follows g(z) lim f(z) = lim = ∞. z z ð ð z z ð n ð → 0 → 0 (z − z0) Conversely, assume lim f(z) = ∞. We consider g(z) ∶= 1 near z . It is holomorphic on the z→z0 ð ð f(z) 0 punctured disk and bounded and so z0 is a removable singularity of g. This shows that z0 is a pole of f.

COROLLARY 3.3. For an isolated singularity z0 of f, it is

(1) removable, if f is bounded near z0; 3. SINGULARITIES AND MEROMORPHIC FUNCTIONS 45

(2) a pole, if lim f(z) = ∞; z→z0 ð ð (3) essential, otherwise.

∗ THEOREM 3.4 (Casorati–Weierstrass theorem). Assume f is holomorphic on D (z0) with z0 an ∗ essential singularity. Then the image f(D (z0)) is dense in ℂ.

Actually, under the same assumption, there is a much stronger result called great Picard’s theorem.

∗ THEOREM 3.5 (Great Picard’s theorem). Assume f is holomorphic on D (z0) with z0 an essential singularity. Then f takes every point in ℂ inﬁnitely many times with at most one exception.

∗ PROOFOF CASORATI–WEIERSTRASSTHEOREM. Assume the image f(D (z0)) is not dense in ℂ, then there exist some w0 ∈ ℂ and a small disk D(w0) in ℂ so that

∗ D(w0) ∩ f(D (z0)) = ç.

1 ∗ ∗ Consider the function g(z) ∶= , z ∈ D (z0). It is well-deﬁned and holomorphic over D (z0). f(z)−w0 Moreover, it is bounded, since 1 1 ðg(z)ð = ≤ . ðf(z) − w0ð ∗ Then by Riemann’s theorem on removable singularities, g can be holomorphically extended to D (z0). If for the extended g, g(z0) ≠ 0, then z0 is a removable singularity of f; If for the extended g, g(z0) = 0, then z0 is a pole singularity of f. Both cases contradict with the assumption that z0 is an essential singularity of f.

EXAMPLE 3.6. In Great Picard’s theorem, the conclusion on possible one exception can not be removed. For example, the function f(z) = e1∕z has 0 as an essential singularity. The image of f is ℂ∗, which doesn’t contain 0.

DEFINITION 3.7. A function f deﬁned on an open subset Ω of ℂ is called a meromorphic function, if there exists a sequence of points

{z0, z1, z2, ⋯} which have no limit points in Ω and such that f is holomorphic over Ω ⧵ {z0, z1, z2, ⋯} and has poles at

{z0, z1, z2, ⋯}.

We can extend this deﬁnition by including ∞ and consider the extended complex plane ℂ.

For a function f defined outside some large disk DR, we say it has a pole/removable/essential singularity at infinity ∞, if the function 1 F (z) ∶= f( ) z D∗ defined on 1∕R has a pole/removable/essential singularity at 0. A meromorphic function f over the extended complex plane ℂ is a meromorphic function over ℂ which has ∞ as a removable singularity or a pole. The following theorem completely characterizes meromorphic functions over ℂ, which are in fact rational functions.

THEOREM 3.8. A function f is meromorphic over ℂ if and only if it is a rational function. 46 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

PROOF. For any rational function f, we can write it as p(z) f(z) = , q(z) with both p, q are some polynomial functions. Then f has pole singularity at

{zðq(z) = 0} in ℂ, which is a finite set. The infinity point is removable, if the order of p is smaller than the order of q; and it is a pole, if otherwise. This shows that any rational function is meromorphic over ℂ. Now we prove the reverse direction. Assume f is a meromorphic function over ℂ. Then f can has only finitely many pole singularities which are z1, ⋯ , zn ∈ ℂ and removable or pole singularity ∞. Near each zk, k = 1, 2, ⋯ , n, we write

f(z) = fk(z) + gk(z), with fk the principal part and gk the holomorphic part. For the inﬁnity, outside a large disk we can write f(1∕z) = f∞(z) + g∞(z), with f∞ the principal part at 0 and g∞(z) the holomorphic part. Denote by f∞(z) = f(1∕z). Then we consider the holomorphic function

n ℎ(z) ∶= f(z) − Σk=1fk(z) − f∞(z), z ∈ ℂ ⧵ {z1, ⋯ , zn}. Notice that near each z , k0 ℎ(z) = g (z) − Σ f (z) − f (z) k0 k≠k0 k ∞ which is bounded, and hence ℎ can be extended to ℂ as an entire function. Similarly,

n ℎ(z) = g∞(z) − Σk=1fk(z), which is bounded near ∞. Hence ℎ is bounded over ℂ. By the Liouville’s theorem, ℎ must be a constant, say c ∈ ℂ. It follows n f(z) = Σk=1fk(z) + f∞(z) + c, which is a rational function. Now we introduce a geometric way of understanding the extended complex plane. We extend the complex plane ℂ to the three dimensional euclidean space ℝ3 with the xy-plane taken 1 1 1 as the ℂ-plane. Place a ball with radius at center (0, 0, ), and denote it by B 1 (0, 0, ). The boundary 2 2 2 2 of it is a sphere, which we denote by 1 S ∶= )B 1 (0, 0, ). 2 2 It has a north pole N = (0, 0, 1) and a south pole S = (0, 0, 0). Take a point P from the sphere S, whenever P ≠ N, there is a unique line going through P and N. We denote this line by PN. The line PN intersects with XY -plane at a unique point, which we denote by z(P ) ∈ ℂ. By this way, we construct a map

N ∶ S ⧵ {N} → ℂ,P ↦ z(P ).

It is easy to check that N is a bijection. We can extend the map N to N by deﬁning N (N) = ∞. This sets up a bijection between S and the extended complex plane ℂ. The map N is called a stereographic projection at N. 3. SINGULARITIES AND MEROMORPHIC FUNCTIONS 47

You can check that a stereographic projection maps circles to circles and preserves angles. How- ever, it doesn’t preserve distance. A map which preserves angles is called a conformal mapping. So a stereographic projection is conformal.

Use this model, next we explain why the behavior of a function f(z) near ∞ can be understood via the behavior of f(1∕z) near 0. For it, similarly, we can construct a stereographic projection at S. For it, we place the complex plane at Z = 1, so that the real axis, imaginary axis and Z-axis with reverse orientation satisfy the right-hand rule. Then using the line through the south pole and a point on the sphere, we set up another bijective map

S ∶ S → ℂ, which maps the south pole to inﬁnity. We now explicitly calculate out these two maps. Take P = (X,Y,Z) ∈ S⧵{P,N}. The coordinates must satisfy 1 1 (3.1) X2 + Y 2 + (Z − )2 = ( )2. 2 2 The stereographic projections can be calculated as X Y X Y (X,Y,Z) = + i , (X,Y,Z) = − i . N 1 − Z 1 − Z S Z Z Using (3.1), it is straightforward to check that there multiplication X Y X Y ( + i )( − i ) 1 − Z 1 − Z Z Z ∗ is constant 1, which says for a point z ∈ ℂ as a coordinate of some point P in S ⧵ {P,N} under N , then the point P should have coordinate 1 ∈ ℂ∗ as its coordinate via . Such correspondence also z S matches the formal notation that 1 = ∞ and 1 = 0. 0 ∞ As a summary, the two stereographic projections provide two coordinate charts in ℂ for S and they together cover the whole sphere. The map 1 ' ∶ ℂ∗ → ℂ, z ↦ z 48 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM which maps a coordinate to anther for the overlapping region is called the transition function. Notice, it is injective onto its image and it is holomorphic. In fact, you can check neither the position of the sphere nor the size of the sphere effects this transition 1 , which is an intrinsic property of the sphere S. With these structure, the sphere S is referred as the z Riemann sphere (https://www.youtube.com/watch?v=usCCkgkD_2s) and also denoted by ℂP 1. It is one of Riemann surfaces.

4. The argument principle and applications

We are going to introduce the argument principle and several important applications from it.

THEOREM 4.1 (The argument principle). Assume Ω ⊂ ℂ is open which contains some closed disk D. Assume f is meromorphic over Ω and has no zero, no pole on )D. Then 1 f ¨(z) dz = n0 − np, 2i Ê)D f(z) where n0 is the number (with multiplicity) of zeros of f inside D and np is the number (with multiplicity) of poles of f inside D

PROOF. Since D is compact, there are only ﬁnitely many zeros and poles inside D. We can take

> 0 small enough so that all such D(z) lives inside D and disjoint from each other for all such z a zero or a pole of f.

For each order n zero z0 ∈ D of f, we can write

n f(z) = (z − z0) g(z) with g a holomorphic function which is nowhere vanishing on a small neighborhood including D(z0). Calculate ¨ n−1 n ¨ f (z) = n(z − z0) g(z) + (z − z0) g (z), ∗ and then on a small enough punctured disk D (z0), f ¨(z) n g¨(z) = + . f(z) z − z0 g(z)

Integrate it over )D(z0), we have 1 f ¨(z) dz = n. 2i f(z) Ê)D(z0) ¨ Similarly, for each order n pole z0 ∈ D of f, we can write ℎ(z) f(z) = n¨ (z − z0) with ℎ a holomorphic function which is nowhere vanishing on a small neighborhood including D(z0). Calculate ℎ(z) ℎ¨(z) f ¨(z) = −n¨ + , n¨−1 n¨ (z − z0) (z − z0) ∗ and then on a small enough punctured disk D (z0), f ¨(z) n¨ ℎ¨(z) = − + . f(z) z − z0 ℎ(z) 4. THE ARGUMENT PRINCIPLE AND APPLICATIONS 49

Integrate it over )D(z0), we have 1 f ¨(z) dz = −n¨. 2i f(z) Ê)D(z0) At last, notice that 1 f ¨(z) 1 f ¨(z) 1 f ¨(z) dz = Σ dz + Σ dz, 2i f(z) z0 is a zero 2i f(z) z0 is a pole 2i f(z) Ê)D Ê)D(z0) Ê)D(z0) we obtain the identity that we want to prove. Next, we see several applications. The ﬁrst one is the Rouché’s theorem. It’s proof involves the idea of homotopy which is very com- monly used in all ﬁelds of mathematics.

THEOREM 4.2 (The Rouché’s theorem). Assume Ω ⊂ ℂ is open which contains some closed disk D. For any two holomorphic functions f, g over Ω. If

ðf(z)ð > ðg(z)ð, z ∈ )D, then the number of zeros of f inside D equals the number of zeros of f + g inside D (count with multiplicities).

PROOF. We construct a homotopy from f to f + g using

ft(z) ∶= f(z) + tg(z), t ∈ [0, 1].

Then it follows that f0 = f and f1 = f + g. The assumption that ðf(z)ð > ðg(z)ð over z ∈ )D implies ft never vanish on )D for any t ∈ [0, 1]. Then we can apply the argument principle to ft(z) and obtain ¨ 1 ft (z) dz = nt, 2i Ê)D ft(z) where nt is the number of zeros of ft inside D, so is integer valued.

On this other hand, the LHS is continuous in t ∈ [0, 1]. By the intermediate value theorem, nt must be constant in t, so it proves that n0 = n1 and we are done. We now see two applications of Rouché’s theorem. The ﬁrst one is another proof of the fundamental theorem of algebra, and the second one is the open mapping theorem.

n n−1 THEOREM 4.3 (Fundamental theorem of algebra). For any polynomial p(z) = anz + an−1z +

⋯ a1z + a0 with

an, an−1, ⋯ , a1, a0 ∈ ℂ, an ≠ 0 and n ≥ 1, the equation p(z) = 0 has n roots in ℂ.

n−1 PROOF. WLOG, we can assume an = 1. Denote by g(z) = an−1z + ⋯ a1z + a0. Since the order of g is smaller than n, for large enough R, there is

n n ðg(z)ð < R = ðz ð, ðzð = R.

From Rouché’s theorem, we obtain that the number of zeros of f inside DR is the same as n. 50 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

THEOREM 4.4 (The open mapping theorem). Assume Ω is an open subset of ℂ and f is a noncon- stant holomorphic function over Ω. Then f is open, which means any set U ⊂ Ω which is open in ℂ has its image f(U) open in ℂ.

PROOF. Assume U ⊂ Ω is open in ℂ. We show f(U) must be open. For it, take any point w0 ∈ f(U), we prove it must be an interior point.

Since w0 ∈ f(U), there is some z0 ∈ U so that f(z0) = w0. Since U is open, we take some small disk so that D(z0) ⊂ U. Moreover, since f is not constant, z0 must be an isolated zero of the holomorphic function f(z) − w0, and we can further take such small enough so that z0 is the only zero of f(z) − w0 in D(z0). Then deﬁne 1 = inf ðf(z) − w0ð. 2 )D(z0)

Since ðf(z) − w0ð is a continuous function on the compact set )D(z0), must be positive. Now for any w ∈ D(w0), we have

ðf(z) − w0ð ≥ inf ðf(z) − w0ð = 2 > > ðw − w0ð )D(z0) for any z ∈ )D(z0). Apply Rouché’s theorem, ðf(z) − w0ð should have the same number of zeros inside D(z0) as the function

(f(z) − w0) + (w0 − w) = f(z) − w.

Since f(z0) = w0, this shows that there must be some z ∈ D(z0) so that f(z) = w. In another word, w ∈ f(U).

Then we conclude that the whole disk D(w0) ⊂ f(U), and this proves f(U) is open.

A useful consequence from the open mapping theorem is the following maximum modulus principle.

THEOREM 4.5 (Maximum modulus principle). Assume Ω is an open subset of ℂ and f is holomorphic over Ω. Then the norm function ðfð has no maximum in Ω unless f is constant.

PROOF. Assume this is not true, and ðfð obtain its maximum at z0 ∈ Ω. Because Ω is open, we can take some open disk D(z0) ⊂ Ω. The open mapping theorem states that f(D(z0)) is open in ℂ. ¨ ¨ Then there must be some other point z ∈ D(z0) so that ðf(z )ð > ðf(z0)ð, which contradicts with the assumption that ðfð obtain its maximum at z0.

An immediate corollary from it is the following corollary.

COROLLARY 4.6. Assume Ω is an open subset of ℂ with Ω compact (bounded). Then for any holomorphic function f over Ω which can be continuously extended to Ω, the maximum of ðfð can only obtained on boundary )Ω.

PROOF. Since ðfð is continuous on the compact set Ω, the function ðfð must have maximum. If the maximum is obtain on boundary, we are done. If it is obtained somewhere in the interior, then f must be constant, and the maximum is also obtained on boundary. 5. THE COMPLEX LOGARITHM 51

5. The complex logarithm

5.1. Homotopy of paths.

DEFINITION 5.1. Assume Ω is a subset of ℂ. Two piecewisely smooth paths

0, 1 ∶ [a, b] → Ω with 0(a) = 1(1) =∶ za, 0(b) = 1(b) = zb are called homotopic in Ω, if there is some continuous map ℎ ∶ [a, b] × [0, 1] → Ω so that ℎ(⋅, 0) = 0, ℎ(⋅, 1) = 1 and each ℎ(⋅, s) is piecewisely smooth.

For short, let’s denote by ℎs = ℎ(⋅, s), s ∈ [0, 1]. For each s ∈ [0, 1], ℎs is a path in Ω which has initial point at za and endpoint at zb. The choice of such homotopy ℎ is not unique.

We use the notation 0 ∼Ω 1, if 1 and 2 are homotopic in Ω.

THEOREM 5.2. Assume Ω is an open subset in ℂ and f is a holomorphic function over Ω. For

0 ∼Ω 1, then f(z)dz = f(z)dz. Ê 0 Ê 1

PROOF. Assume ℎ ∶ [a, b] × [0, 1] → Ω is a homotopy between 0 and 1. Let’s consider a map

Φ ∶ [0, 1] → ℂ, s ↦ f(z)dz. Êℎs Because ℎ is continuous and f is holomorphic, the map Φ is continuous. Denote by A ∶= f(z)dz, ∫ 0 and deﬁne K ∶= {s ∈ [0, 1]ðΦ(s) = A}. K is closed since Φ is continuous, nonempty since 0 ∈ K. We now prove it is also open.

Fort this, take some s0 ∈ K, i.e., Φ(s0) = A, we show there exists a small neighborhood of s0 in [0, 1] so that Φ is constant on this neighborhood. First because ℎ is a path with endpoints, its image is a compact subset in Ω. Then we can cover its s0 image with ﬁnite many disks of radius ∕2 for some small > 0 and centers are

ℎ (t ), ℎ (t ), ℎ (t ), ⋯ , ℎ (t ), t = a, t = b, s0 0 s0 1 s0 2 s0 N 0 N with the property that ℎ ([t , t ]) ⊂ D (ℎ (t )). s0 i i+1 ∕2 s0 0 At the same time, notice ℎ is deﬁned on a compact set [a, b] × [0, 1] so it is uniformly continuous, and hence there exists some > 0 so that

ðℎ(t, s1) − ℎ(t, s2)ð < ∕2, for any t ∈ [a, b], ðs1 − s2ð < .

In particular, it follows for any t ∈ [ti, ti+1], ℎ (t) − ℎ (t ) < ∕2, ð s0 s0 i ð and so ℎ (t) − ℎ (t ) ℎ (t) − ℎ (t) + ℎ (t) − ℎ (t ) < ∕2 + ∕2 = ð s s0 i ð ≤ ð s s0 ð ð s s0 i ð for any ðs − s0ð < . This says for any s in the neighborhood of s0, the curve ℎs satisﬁes ℎ ([t , t ]) ⊂ D (ℎ (t )). s i i+1 s0 0 52 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

Then we apply Cauchy’s theorem for the loop consisting of

(1) ℎs(ti) to ℎs(ti+1) along ℎs as t changes; (2) ℎ (t ) to ℎ (t ) along ℎ (t ) as s changes; s i+1 s0 i+1 s i+1 (3) ℎ (t ) to ℎ (t ) along ℎ as t changes; s0 i+1 s0 i s0 (4) ℎ (t ) to ℎ (t ) along ℎ (t ) as s changes s0 i s i s i with the positive orientation inside each disk D (ℎ (t )), and then take summation of all these integra- s0 0 tions. It follows the integrations on (2)(4) cancel out with each other, and the rest gives

f(z)dz = f(z)dz. Êℎ Êℎ s0 s Now we have proved that K is nonempty open and closed subset of the connected set [0, 1], and hence it must be the whole [0, 1], which in particular states that

f(z)dz = f(z)dz, Êℎ0 Êℎ1 i.e., f(z)dz = f(z)dz. Ê 0 Ê 1

DEFINITION 5.3. An subset Ω of ℂ is called simply-connected if any two paths with the same endpoints are homotopic to each other.

EXAMPLE 5.4. (1) A subset Ω is ℂ is called convex if any z0, z1 ∈ Ω, then for any t ∈ [0, 1],

(1 − t)z0 + tz1 ∈ Ω.

Any convex set is simply-connected.

For it, take any two paths 0, 1 with the same endpoints, we can construct the homotopy ℎ as

ℎ ∶ [a, b] × [0, 1] → Ω, ℎ(t, s) = (1 − s) 0(t) + s 1(t). (2) The slit ℂ ⧵ (−∞, 0] is simply-connected. It is not convex, but we can transform it to a convex region using the polar coordinate transformation as ℂ → (0, +∞) × (−, ), z ↦ (r, ).

Then since (0, +∞) × (−, ) is convex, we can construct the homotopy for any two paths

i0 i1 0 = r0e , 1 = r1e

with the same endpoints using the homotopy

i((1−s)0+s1) ℎ(t, s) = ((1 − s)r0 + sr1)e .

COROLLARY 5.5. Assume Ω is an open subset in ℂ and simply-connected, and f is a holomorphic function over Ω. Then the integration of f over any loop inside Ω is zero.

We are going to use these concepts to deﬁne complex logarithm in next section. 5. THE COMPLEX LOGARITHM 53

5.2. The complex logarithm. If we want to solve the equation ez = w, we can write z = x + iy, x, y ∈ ℝ and w = rei. Then the equation is equivalent to

exeiy = rei.

Then x is uniquely determined as x = log r, but y has inﬁnitely many choices as

y = + 2k, k ∈ ℤ.

A geometric view is as follows. We can cut ℂ at (−∞, 0], and then glue a copy of ℂ with another copy of ℂ along (−∞, 0] so that the second quadrant of the ﬁrst copy us connected with the third quadrant of the second copy. By doing it for ℤ-copies of ℂ, we obtain a space, denoted by ℂ, which contains ℤ-copies of ℂ. Each ℂ is called a sheet of this space, and this space contains countable many sheets. Then the exponential map can be considered as a map from ℂ to ℂ. The logarithm can be deﬁned as the inverse function of the exponential function exp(z) ∶= ez over a subset U in ℂ so that exp on it is injective. If we denote the logarithm by log, then

logΩ ∶ Ω → U, Ω ∶= exp(U) with exp ◦ logΩ = idΩ. The following theorem answers the question that what kind of region can make the logarithm be well-deﬁned.

THEOREM 5.6. Assume Ω is an open, connected subset of ℂ and it is simply-connected, with 1 ∈ Ω and 0 ∉ Ω. Then in Ω, there is a branch of the logarithm F (z) = logΩ(z) so that (1) F is holomorphic over Ω; (2) eF (z) = z; (3) F (r) = log r whenever r is real and near 1. Such F is the unique function that satisﬁes these properties.

PROOF. We define a function F and check it satisfies these requirements. Take any point z ∈ Ω, since Ω is open and connected, it must be path-connected. We can find a piecewise smooth path (z) connecting 1 and z. Define 1 F (z) ∶= dw. Ê (z) w This definition makes sense, since 0 is not in Ω, and this integral is independent of choices of paths by the simply-connectedness of Ω. This defines a function

F ∶ Ω → ℂ.

By the similar way as we did in proving Goursat’s theorem (see the proof of Proposition 1.3), we know F is holomorphic over Ω with F ¨(z) = 1 . z Now we check eF (z) = z. For it, consider the function g(z) ∶= e−F (z)z. It is holomorphic since it is composition and multiplicaiton of holomorphic functions. By the product rule and the chain rule,

g¨(z) = −e−F (z)F ¨(z)z + e−F (z) = −e−F (z) + eF (z) = 0.

Then the connected of Ω ensures that g is a constant on Ω, which must be the same as the value taken at 1. Notice g(z) = e−F (1) = 1, −F (z) F (z) we proved that e z ≡ 1 and so e = z on Ω. 54 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

For any r ∈ ℝ in an open disk neighborhood center at 1, we can connect 1 and r using the path ∶ [0, 1] → ℝ, (t) = 1 + (r − 1)t. It follows

1 r − 1 t=1 F (r) = dt = log(1 + (r − 1)t)ðt=0 = log r, Ê0 1 + (r − 1)t where is the real logarithm, i.e., the inverse function of the real exponential function. At last, by the uniqueness of analytic continuation, such F must be unique.

The so-deﬁned function F (z) is called the branch of logarithm with respect to Ω and we use logΩ to denote it. In particular, when we take Ω = ℂ ⧵ (−∞, 0], the so-deﬁned F is called the principal logarithm, and we use Log to denote it.

i PROPOSITION 5.7. For the principal logarithm, for any z = re ∈ ℂ with r ∈ ℝ and ðð < ,

Log(z) = log r + i.

PROOF. For any z ∈ Ω, we take a piece-wisely smooth path with pieces 0 and 1 deﬁned as

0 ∶ [0, 1] → ℝ, (t) = 1 + (r − 1)t, and it 1 ∶ [0, 1] → ℂ, (t) = re . Then 1 1 F (z) = dw + dw = log r + i. w w Ê 0 Ê 1

Notice that for the principal logarithm, in general

Log(z1z2) ≠ Log(z1) + Log(z2).

i 2 For example, take z1 = z2 = e 3 , then

i 4 −i 2 2 Log(z z ) = Log(e 3 ) = Log(e 3 ) = − i, 1 2 3 but Log(z ) = Log(z ) = i 2 . 1 2 3 The principal logarithm has the series representation as

z2 z3 (−1)n−1zn Log(1 + z) = z − + − ⋯ = Σ∞ , z < 1. 2 3 n=1 n ð ð To check it, notice the RHS series has convergence radius 1 and then we can compare the derivatives of both sides which turn out to be the same. Then the connectedness of the slit ℂ ⧵ (−∞, 0] implies that the difference between LHS and RHS is a constant. Take z = 0, we can see the difference in fact is zero, i.e., LHS=RHS. The logarithm can be used to deﬁne z , for ∈ ℂ. For any Ω ⊂ ℂ which is open, connected and simply-connected with 1 ∈ Ω and 0 ∉ Ω, we can deﬁne a branch of z with respect to Ω as

z Ω ∶= e logΩ z. 5. THE COMPLEX LOGARITHM 55

We can check that for any such Ω, there is 1 = e logΩ 1 = e0 = 1. For any = 1∕n, n ∈ ℤ+,

(z1∕n)n = z1∕n ⋅ z1∕n ⋅ ⋯ ⋅ z1∕n 1 log z 1 log z 1 log z = e n Ω ⋅ e n Ω ⋅ ⋯ ⋅ e n Ω

= elogΩ z = z.

We know every nonzero complex number w can be written as ez for some z ∈ ℂ. To obtain this, we can take z = log ðwð + i arg(w) for example. Notice such z is not unique but can be shifted by adding 2ki, k ∈ ℤ. We can now ask such a question: Given a nowhere vanishing holomorphic function f defined on an open set Ω (This can be considered as an Ω-family of w’s), if it is possible to find a holomorphic function g defined on Ω (This can be considered as an Ω-family of z’s) so that f = eg over Ω. A positive answer is given in the next theorem for simply-connected Ω.

THEOREM 5.8. Assume Ω is an open, connected and simply-connected subset of ℂ and f is a nowhere vanishing holomorphic function over it. Then there exists a holomorphic function g over Ω so that f(z) = eg(z), z ∈ Ω. Such g is not unique but can be shifted by a multiple of 2i. On the other hand, any two such g differ by a multiple of 2i.

c0 PROOF. Take an arbitrary point z0 ∈ Ω and choose some c0 so that e = f(z0), we deﬁne f ¨(w) g(z) = dw + c0, Ê (z) f(w) where (z) is a piecewisely smooth path connecting z0 and z in Ω. Since f is nonwhere vanishing, such deﬁnition makes sense. It is independent of choices of path since Ω is simply-connected. The function g is holomorphic because f is holomorphic over Ω, and f ¨(z) g¨(z) = . f(z) To check f(z) = eg(z), we consider the function ℎ(z) ∶= f(z)e−g(z), which is holomorphic over Ω. The derivative ℎ¨(z) = f ¨(z)e−g(z) − f(z)e−g(z)g¨(z) = 0.

Hence ℎ is constant over Ω since Ω is connected. Such constant must be the same as ℎ(z0) which is

−g(z0) −c0 c0 −c0 ℎ(z0) = f(z0)e = f(z0)e = e e = 1. This proves f(z) = eg(z) over Ω. If g satisﬁes the requirement, then g + 2ki for any k ∈ ℤ also satisﬁes the requirement since 2ki e = 1. Now assume both g1, g2 satisfy the requirements. Then

eg1(z)−g2(z) = 1 for all z ∈ Ω. On the other hand, all complex numbers with ew = 1 must be 2ki, for some k ∈ ℤ, hence

g1(z) − g2(z) ∈ {2kiðk ∈ ℤ} for all z ∈ Ω. The continuity of g1 − g2 and the connectedness of Ω require the image of (g1 − g2)(Ω) is connected in {2kiðk ∈ ℤ}, hence can only be a point. 56 3. MEROMORPHIC FUNCTIONS AND THE LOGARITHM

We can use logΩ f(z) to denote the set of all such g’s. By choosing an initial point z0 so that f(z0) ∈ ℂ ⧵ (−∞, 0] and taking c0 = Logf(z0) we can ﬁnd a such g0 in this set. Then this set can be written as {g0 + 2kiðk ∈ ℤ}. We call each element a branch of the logarithm of f(z) with respect to Ω. CHAPTER 4

Conformal mappings

1. Conformal equivalence and examples

DEFINITION 1.1. Assume U,V are two open sets in ℂ. A function f ∶ U → V is called a conformal map or a biholomorphic map, if it is bijective, holomorphic and its inverse function is also holomorphic.

We ﬁrst prove that the requirement that the inverse function is holomorphic is redundant that in fact every bijective holomorphic function automatically has holomorphic inverse.

PROPOSITION 1.2. Assume U,V are two open sets in ℂ and f ∶ U → V is injective and holomorphic. Then the derivative f ¨ is nowhere vanishing on U.

PROOF. We prove it by contradiction. ¨ Assume that there is some point z0 ∈ U so that f (z0) = 0. The injectivity of f requires z0 is an isolated zero of f ¨, since otherwise f ¨ is constant zero and then f can not be injective. ¨ We take a small enough disk D(z0) ⊂ U so that over D(z0), f has only one zero z0. ¨ By the vanishing of f (z0), over D(z0) we can write k f(z) = f(z0) + a(z − z0) + g(z) with a ≠ 0, k ≥ 2, and g is an analytic function with lower order greater than k. Then we can further shrink the disk to some D(z0) with some > 0 so that k ða(z − z0) ð > ðg(z)ð, z ∈ )D(z0), g(z) since limz→z k = 0. 0 ð a(z−z0) ð Denote by ∶= inf { a(z − z )k − g(z) } [This was where I got confused in lecture today that )D(z0) ð 0 ð ð ð k we should use min instead of max]. It is positive, because )D(z0) is compact and ða(z − z0) ð − ðg(z)ð is continuous and positive everywhere on )D(z0). We claim then for any w with 0 < w < 1 , ð ð 2 k ðg(z) − wð < ða(z − z0) ð, z ∈ )D(z0). This can be proved using triangle inequalities as

k k ðg(z) − wð ≤ ðg(z)ð + ðwð = ða(z − z0) ð + (ðwð − (ða(z − z0) ð − ðg(z)ð)) 1 < a(z − z )k + ( − ( a(z − z )k − g(z) )) ð 0 ð 2 ð 0 ð ð ð 1 < a(z − z )k − ð 0 ð 2 k < ða(z − z0) ð. k Then apply Rouché’s theorem, we can claim the number of zeros of a(z − z0) + g(z) − w inside k D(z0) is the same as the number of zeros of a(z − z0) inside D(z0), which is at least 2. Since k a(z − z0) + g(z) − w = f(z) − f(z0) − w, 57 58 4. CONFORMAL MAPPINGS this shows that f(z) − f(z0) − w has at least two zeros inside D(z0). Moreover, the two or more zeros must be different, because if some zero has order greater than 1, ¨ then the derivative of f will vanish at this point but we assumed z0 is the unique zero of f in D(z0). It follows f not be injective then which leads to a contradiction.

COROLLARY 1.3. Assume U,V are two open sets in ℂ and f ∶ U → V is injective and holomorphic. Then f is biholomorphic onto its image.

PROOF. By the open mapping theorem, f(U) is an open subset of ℂ. We prove now the inverse map −1 g ∶= f ∶ f(U) → U is holomorphic. For it, take any w0 = f(z0) ∈ f(U) with z0 ∈ U, for w = f(z) close to w0, there is g(w) − g(w ) 1 1 0 = = w−w0 f(z)−f(z ) w − w0 0 g(w)−g(w ) 0 z−z0 1 ¨ which converges to ¨ since f (z0) ≠ 0. This shows g is holomorphic everywhere. f (z0)

REMARK 1.4. Similarly result doesn’t hold for real functions. For example, the function f(x) = x3 1 is differential and injective, but it has vanishing derivative at x = 0. Its inverse function g(x) = x 3 is not differentiable at x = 0. (The function f(z) = z3 is not injective near 0.)

DEFINITION 1.5. Two open subsets U,V of ℂ are called conformally equivalent if there exists some biholomorphic map f ∶ U → V .

PROPOSITION 1.6. Conformal equivalence is an equivalence relation.

PROOF. (1) Any set is conformally equivalent to itself via the identity map; (2) If f ∶ U → V is biholomorphic, then f −1 ∶ V → U is also biholomorphic; (3) If f ∶ U → V and g ∶ V → W are biholomorphic, then g◦f ∶ U → W is biholomorphic. 1.1. The unit disk and upper half plane. We denote by

ℍ ∶= {z ∈ ℂðIm(z) > 0} the upper half plane. It is somehow surprising that the upper half plane ℍ is conformally equivalent to the unit disk D ∶= {z ∈ ℂððzð < 1}. We now explicitly construct the map i − z F ∶ ℍ → ℂ,F (z) = . i + z THEOREM 1.7. The map F ∶ ℍ → D is biholomorphic.

PROOF. Clearly F is holomorphic over ℍ. Moreover, i − z Re(z)2 + (1 − Im(z))2 F (z) 2 = 2 = < 1 ð ð ði + zð Re(z)2 + (1 + Im(z))2 if and only if Im(z) > 0. Hence the image of F is D. Direct checking shows that F is injective and it has inverse map 1 − w G ∶ D → ℍ,G(z) = i . 1 + w 2. AUTOMORPHISM OF D AND ℍ 59

Clearly, G is holomorphic. All these together proves that F ∶ ℍ → D is biholomorphic. We if we use the topology induced from ℂ, the closure of ℍ in ℂ is

ℍ = ℍ ∪ ℝ ∪ {∞}.

The map F can be extended continuously to ℍ, with

F (∞) = −1,F (0) = 1,F (±1) = ±i, F (ℝ) = )D ⧵ {−1}.

We call a map of the form az + b z ↦ , a, b, c, d ∈ ℂ cz + d a fractional linear transformation (or a Möbius transformation) if the denominator is not multiple of the H I a b numerator. A fractional linear transformation can give rise to a matrix with nonzero determi- c d nant. In fact, we can require the determinant is 1 by rescaling, and then the ambiguity in writing down the matrix is just ±1 to each coefﬁcients. The set (actually group) of such matrices is written as PSL2(ℂ), where P means projective by modeled out ±1 from the special linear group SL2(ℂ) with ℂ coefﬁcients.

(The group extension 1 → ℤ2 → SL2(ℂ) → PSL2(ℂ) → 1 is a central extension with ℤ2 the center of

SL2(ℂ).) 1.2. More examples.

(1) Translation c(z) = z+c, c ∈ ℂ is a conformal map from ℂ to itself. For ℎ ∈ ℝ, the translation

ℎ(z) = z + ℎ is a conformal map from ℍ to ℍ. ∗ (2) Dilation dc(z) = cz, c ∈ ℂ is a conformal map from ℂ to itself. In particular, if ðcð = 1, this is a rotation. If c ∈ ℝ∗, this is a real dilation.

(3) Take 0 < < 2, then f(z) = z takes ℍ to the sector S = {w ∈ ℂð0 < arg(w) < }. For example, we can take the branch of logarithm by deleting ℝ+ and obtain this conformal map. It has inverse map g(w) = w1∕ , where branch of logarithm is chosen as 0 < arg w < . (4) The map f(z) = 1+z from ℍ conformally to the ﬁrst quadrant. 1−z (5) The map z ↦ log z = logr + i deﬁned by the branch of logarithm by deleting ℝ−i maps ℍ conformally to the strip

{w = u + ivðu ∈ ℝ, 0 < v < }.

2. Automorphism of D and ℍ

2.1. Automorphism of D.

DEFINITION 2.1. An conformal equivalence between an open set Ω and itself is called an automorphism of Ω. We use Aut(Ω) to denote the set of automorphisms of Ω.

It is straightforward to check that

LEMMA 2.2. Aut(Ω) is a group with the binary operation as the composition of maps.

Now we consider the case Ω = D and ﬁgure out what Aut(D) is.

i EXAMPLE 2.3. (1) Any rotation by ∈ ℝ as r(z) = e z ∈ Aut(D). (2) The reﬂection z ↦ ̄z is bijective but not holomorphic so not in Aut(D). 60 4. CONFORMAL MAPPINGS

(3) The map z ↦ z2 is holomorphic but not injective so not in Aut(D). (4) Take ∈ D, and deﬁne a map − z (2.1) ∶ D → ℂ, z ↦ . 1 − ̄ z Checking from direct calculation, there are

(a) (0) = , ( ) = 0;

(b) ◦ = id;

LEMMA 2.4. ∈ Aut(D).

PROOF. Since ð ð < 1, the denominator 1 − ̄ z ≠ 0 for any z ∈ D, hence is holomorphic over D. It can be extended to D continuous. Since ð (z)ð = 1 for any z ∈ )D, by the maximum modulus principle,

max ð ð = 1. D Hence any z ∈ D, ð (z)ð ≤ 1 and if there is some point z ∈ D so that ð (z)ð = 1, then is constant which contradicts with the deﬁnition of . Hence holomorphically maps D to D. Then from the fact ◦ = id, we know it is bijective hence lives in Aut(D). The main result we are going to obtain in the section is the following theorem.

THEOREM 2.5. For any f ∈ Aut(D), there exists some ∈ ℝ and ∈ ℂ so that

i f(z) = e (z).

To prove it, we ﬁrst introduce the Schwarz lemma.

LEMMA 2.6 (Schwarz Lemma). Assume f ∶ D → D is holomorphic and f(0) = 0. Then

(1) ðf(z)ð ≤ ðzð for any z ∈ D; (2) If there is some point z0 ∈ D and z0 ≠ 0 so that ðf(z0)ð = ðz0ð, then f must be a rotation; ¨ (3) The derivative at 0, ðf (0)ð ≤ 1, and equals 1 if and only if f is a rotation.

PROOF. (1) Since f is holomorphic and f(0) = 0, we can write

2 f(z) = a1z + a2z + ⋯ , for some a , a , ⋯ ∈ ℂ. Deﬁne g(z) ∶= f(z) , z ∈ D∗, which is holomorphic over D∗. 1 2 z The singularity 0 is removable since

lim g(z) = a1. z→0 This means we can extend g to D and get a holomorphic function, which we still use g to denote.

For any 0 < r < 1, consider the disk Dr ⊂ D. By the maximum modulus principle, ðf(z)ð 1 max g(z) ≤ max g(z) = max < . ð ð )D ð ð )D r r Dr r r Take r → 1, we obtain max ðg(z)ð ≤ 1, D and this is equivalent to ðf(z)ð ≤ ðzð for any z ∈ D. 2. AUTOMORPHISM OF D AND ℍ 61

(2) Assume at z0 ≠ 0 in D, ðf(z0)ð = ðz0ð, then it means ðg(z)ð can obtain its maxima 1 in the interior point z0 since z0 ≠ 0. It follows from the maximum modulus principle that g(z) must be constant of norm 1. Denote this constant by ei for some ∈ ℝ, then

f(z) = eiz,

which is a rotation. (3) We calculate

¨ f(z) − f(0) f(z) f (0) = lim = lim ≤ 1. ð ð ð z→0 z − 0 ð z→0 ð z ð ¨ ¨ Notice on the other hand, g(0) = f (0). If ðf (0)ð = 1, then g will obtain its maximal modulus at interior point 0, which requires g is constant ei. Then f is a rotation.

With the help of Schwarz Lemma, we are ready to prove Theorem 2.5.

PROOFOF THEOREM 2.5. Because f ∈ Aut(D), there must be a unique ∈ D so that f( ) = 0. Deﬁne a function

g ∶= f◦ .

Since f, ∈ Aut(D), it follows g ∈ Aut(D).

Notice g(0) = f( (0)) = f( ) = 0, by the Schwarz Lemma,

ðg(z)ð ≤ ðzð, z ∈ D.

At the same time, g−1 ∈ Aut(D) with g−1(0) = 0, apply the Schwarz Lemma to g−1, we get

−1 ðg (z)ð ≤ ðzð, z ∈ D.

Connect these two inequality,

−1 ðg(z)ð ≤ ðzð = ðg (g(z))ð ≤ ðg(z)ð, we get ðg(z)ð = ðzð for all z ∈ D. Then by the Schwarz Lemma again, g is a rotation. Writing g(z) = eiz, ∈ ℝ, it follows

i f = g◦ = e .

From this theorem, the automorphic group Aut(D) has three dimensions of freedoms. To be more precise, it is a Lie group of dimension 3. If we add a two dimension constraint as follows, there is only one dimension, which is rotation, left.

COROLLARY 2.7. If f ∈ Aut(D) has a ﬁxed point 0, then f must be a rotation.

i PROOF. By Theorem 2.5, we can write f = e ⋅ for some ∈ ℝ, ∈ D. Since f(0) = 0, we obtain i i e (0) = e = 0, i and hence must be 0. Then f(z) = e z, i.e., a rotation. 62 4. CONFORMAL MAPPINGS

2.2. Automorphism of ℍ. Recall from Section 1.1, there is a conformal equivalence i − z F ∶ ℍ → ℂ,F (z) = . i + z between the upper half plane and the unit disk. It relates the automorphism group of ℍ and the automorphism group of D as the following result.

LEMMA 2.8. The map F induces a group isomorphism

F ∶ Aut(D) → Aut(ℍ) as F(f) = F −1◦f◦F. H I a b Recall that given a 2 × 2 complex matrix M = , there is a fractional linear transformation c d deﬁned as az + b f ∶ z ↦ , a, b, c, d ∈ ℂ. M cz + d Moreover, there is the following lemma.

LEMMA 2.9. For any M,M¨, there is

fM ◦fM¨ = fMM¨ .

All 2 × 2 invertible complex matrices forms the general linear group of rank 2 over ℂ and is denoted by GL2(ℂ). It has a normal subgroup consisting of determinant 1 matrices, which is called the special linear group of rank 2 over ℂ and denoted by SL2(ℂ). If a, b, c, d ∈ ℝ, this is a 2 × 2 real matrix.

All invertible real matrices forms the general linear group of rank 2 over ℝ and is denoted by GL2(ℝ).

Similarly as the complex case, GL2(ℝ) has a normal subgroup consisting of determinant 1 matrices, which is called the special linear group of rank 2 over ℝ and denoted by SL2(ℝ).

THEOREM 2.10. There is a 2-to 1 correspondence between SL2(ℝ) and Aut(ℍ) given by M ↦ fM .

PROOF. Deﬁne

Φ ∶ SL2(ℝ) → Aut(ℍ),M ↦ fM .

We first check this map is well-defined, i.e., fM ∈ Aut(ℍ), then we check this map is surjective and 2-to-1. (1) Well-definedness. H I a b For any M = ∈ SL (ℝ), we check f ∈ Aut(ℍ). For it, we take any z ∈ ℍ, c d 2 M

and then calculate the imaginary part of fM (z) as az + b (ad − bc)Im(z) Im(z) Im(fM (z)) = Im( ) = = > 0. cz + d ðcz + dð2 ðcz + dð2

Hence fM maps ℍ to ℍ.

Since fM ◦fM−1 = fI = idℍ = fM−1 ◦fM , we have fM ∈ Aut(ℍ).

(2) Now take any f ∈ Aut(ℍ), assume that f(z0) = i for some z0 ∈ ℍ. We ﬁrst construct some

N ∈ SL2(ℝ) so that

fN (i) = z0. 2. AUTOMORPHISM OF D AND ℍ 63

Assuming H I a b N = , c d we can ﬁgure out N by solving the equation ai + b (ai + b)(−ci + d) (ac + bd) + (ad − bc)i (ac + bd) + i = = = = z . ci + d c2 + d2 c2 + d2 c2 + d2 0 For example, by taking d = 0, we get √ √ 1 c = Im(z0), a = Re(z0)c = Re(z0) Im(z0), b = −√ , d = 0 Im(z0) as a solution. Then the composition

f◦fN ∈ Aut(ℍ) and maps i to i. If we use F−1 to map it to Aut(D), then it must be a rotation. This shows that −1 i F ◦f◦fN ◦F = e , ∈ ℝ. It follows H I cos sin f◦f = F −1◦ei◦F = f ,M ∶= 2 2 ∈ SL (ℝ). N M − sin cos 2 2 2 Hence −1 f = f ◦(f ) = f ◦f −1 = f −1 , M N M N MN −1 with MN ∈ SL2(ℝ). This proves that the map Φ is surjective.

(3) Now assume fM = fM¨ = f ∈ Aut(ℍ), with H I H I a b a¨ b¨ M = ∈ SL (ℝ),M¨ = ∈ SL (ℝ). c d 2 c¨ d¨ 2

This means that az + b a¨z + b¨ = , cz + d c¨z + d¨ for all z ∈ ℍ. Hence it follows the polynomial

g(z) = (az + b)(c¨z + d¨) − (cz + d)(a¨z + b¨)

is constant 0. By looking at the coefﬁcients, there must be M = ±M¨.

The group structure of SL2(ℝ) preserves after being modeled out ±1. The resulting group is called projective special linear group of rank 2 over ℝ and denoted by PSL2(ℝ). The above results show that

Aut(D) ≅ Aut(ℍ) ≅ PSL2(ℝ).

REMARK 2.11. As the complex counterpart, the automorphism group of the Riemann sphere Aut(S) is PSL2(ℂ). 64 4. CONFORMAL MAPPINGS

REMARK 2.12. On the upper half plane, there is a hyperbolic metric deﬁned as ðdzð ds 2 = . ℍ Imz Under the map F , the metric is pushed forward to D and becomes 2ðdwð dsD2 = . 1 − ðwð2 An interesting result is that the orientation-preserving isometry group with respect to this metric is exactly the same as the automorphism group. This indicates the deep relation between complex analysis and hyperbolic geometry.

3. Riemann mapping theorem

The famous Riemann mapping theorem answers the question that what kind of nonempty subset Ω in ℂ is conformally equivalent to the unit disk D. First by the open mapping theorem, Ω must be open. Since the unit disk is connected and simply- connected, the subset Ω must be also connected and simply-connected. Moreover, if Ω is ℂ, but the Liouville’s theorem, there is no biholomorphic map between ℂ and a bounded domain, hence Ω can not be the whole ℂ. The Riemann mapping theorem states that these necessary conditions are in fact sufﬁcient conditions.

THEOREM 3.1 (Riemann mapping theorem). Assume Ω ⊂ ℂ is nonempty, open, connected, simply- connected and not the whole ℂ. Then there exists some conformal equivalence F between Ω and D.

Moreover, by taking a point z0 ∈ Ω and require that ¨ F (z0) = 0,F (z0) > 0, then such conformal map is unique.

We ﬁrst proof the uniqueness part, which is much easier than the existence part.

PROOFOFTHE RIEMANNMAPPINGTHEOREM, THE UNIQUENESS PART. Assume there are two conformal equivalence F,G ∶ Ω → D, satisfying the requirements

¨ ¨ F (z0) = G(z0) = 0,F (z0) > 0,G (z0) > 0. Then we construct the map H ∶= F ◦G−1 ∶ D → D. It is biholomorphic and H(0) = 0. From Corollary 2.7, we know H must be some rotation

H(z) = eiz, for some ∈ ℝ. Further more, ℍ¨(0) = ei = F ¨(G−1(0))(G−1)¨(0) which is positive since 1 F ¨(G−1(0)) = F ¨(z ) > 0, (G−1)¨(0) = > 0. 0 ¨ G (z0) 3. RIEMANN MAPPING THEOREM 65

i Then e = 1 and this shows that H is the identity map. Hence there must be F = G. An immediate corollary from the Riemann mapping theorem is the following statement. A subset of ℂ is called proper if it is not empty and not the ℂ.

COROLLARY 3.2. Any two open, connected and simply-connected proper subsets of ℂ are conformally equivalent.

PROOF. Assume Ω1, Ω2 are both open, connected and simply-connected proper subsets of ℂ. Then from the Riemann mapping theorem, there are conformal maps

F1 ∶ Ω1 → D,F2 ∶ Ω2 → D. We can construct the map −1 G ∶= F2 ◦F1 ∶ Ω1 → Ω2.

Then G is a conformal equivalence between Ω1 and Ω2. In this section, we give a proof of the Riemann mapping theorem. The idea of the proof is to seek for the conformal map from the set

 ∶= {f ∶ Ω → Dðf is injective and holomorphic , f(z0) = 0}. To be more concrete, this map will be constructed as a limit of certain sequence. We ﬁrst do some preparation on introducing the Montel’s theorem on normal family, which guarantees the existence of convergent subsequences. Not only in proving the Riemann mapping theorem, the concepts of normal family and Montel’s theorem themselves are also very useful.

3.1. Normal family and Montel’s theorem.

DEFINITION 3.3. Assume Ω is an open subset of ℂ. A family  of holomorphic functions over Ω is called a normal family, if any sequence {fn} has a pointwise convergent subsequence which is uniformly convergent over any compact subset of Ω.

It follows from Chapter 2 Theorem 4.2, the subsequence limits must be holomorphic over Ω. A family of functions  deﬁned over Ω is called uniformly bounded if there is some M > 0 so that any f ∈  and any z ∈ Ω, there is ðf(z)ð < M. A family of functions  deﬁned over Ω is called equicontinuous if for any > 0 there is some > 0 so that any f ∈  and any z1, z2 ∈ Ω with ðz1 − z2ð < , there is ðf(z1) − f(z2)ð < . The following lemma is a fundamental result of analysis, and we skip its proof in our class.

LEMMA 3.4 (Arzela–Ascoli Lemma). Any uniformly bounded and equicontinuous sequence of functions deﬁned over a compact subset K of ℂ has a uniformly convergent subsequence.

COROLLARY 3.5. Assume Ω is an open subset of ℂ and  is a family of functions deﬁned over Ω. If for any compact subset K ⊂ Ω, the family of functions restricted to K is uniformly bounded and equicontinuous, then it is a normal family.

PROOF. To achieve this, we only need to notice that the open set Ω can be exhausted by a sequence of nested compact sets, i.e., there is a sequence of compact subsets {Kn, n = 1, 2, ⋯} of Ω so that

(1) each Kn lives in the interior of Kn+1; ∞ K (2) Ω = ∪n=1 n. 66 4. CONFORMAL MAPPINGS

Then for each Kn, we apply the Arzela–Ascoli Lemma and then by the diagonal argument, we prove that  is a normal family. We remark that so far, the arguments don’t need the family is holomorphic. Now we prove the following Montel’s theorem.

THEOREM 3.6. Assume  is a family of holomorphic functions over Ω, where Ω is an open subset in ℂ. If  is uniformly bounded over any compact subset K ⊂ Ω, then  is a normal family.

PROOF. By Corollary 3.5, we only need to prove  is equicontinuous over any compact subset K ⊂ Ω. For it, take any compact subset K ⊂ Ω. We can ﬁnd some open set U so that

K ⊂ U ⊂ Ω.

Moreover, we can always ﬁnd a positive number r so that

D2r(z) ⊂ U, for any z ∈ K.

Then for any z1, z2 ∈ K satisfying ðz1 − z2ð < r, the disk

z1, z2 ∈ D2r(z2) ⊂ D2r(z2) ⊂ U.

Using Cauchy’s integral formula over )D2r(z2), 1 1 1 f(z ) − f(z ) = f(w)( − )dw. 1 2 2i w − z w − z Ê)D2r(z2) 1 2 We estimate 1 1 z − z z − z − = ð 1 2ð < ð 1 2ð, w ∈ )D (z ). ð ð 2 2r 2 w − z1 w − z2 ðw − z1ððw − z2ð r Hence 1 2rB z − z B z − z f(z ) − f(z ) ð 1 2ð = ð 1 2ð, ð 1 2 ð ≤ 2 r2 r where B = supz∈K¨ ðf(z)ð < ∞ by the assumption of the uniformly boundedness of  over the compact set K¨ with ¨ K ∶= ∪z∈K D2r(z) which is a compact subset inside Ω by the choice of r. This proves that  is equicontinuous over K. In the proof of the Riemann mapping theorem, we also the following proposition.

PROPOSITION 3.7. Let Ω be an open and connected subset of ℂ and {fn} is a sequence of injective holomorphic functions over Ω. Assume f is a holomorphic function over Ω so that over any compact subset K ⊂ Ω, {fn} uniformly converges to f. then it follows that f is either a constant function or an injective function.

PROOF. Assume f is not injective, we prove it must be a constant.

Take z1 ≠ z2 from Ω so that f(z1) = f(z2). We construct a sequence {gn} by deﬁning

gn(z) = fn(z) − fn(z1).

Clearly, gn(z1) = 0. Since each fn is injective, then gn doesn’t have any zero points other than z1. 3. RIEMANN MAPPING THEOREM 67

At the same time, over any compact subset K ⊂ Ω, it follows {gn} uniformly converges to the holomorphic function g(z) ∶= f(z) − f(z1).

The function g has zeros at z1, z2. If g is constant zero, then f is constant. Otherwise, z1, z2 are isolated zeros of g. We take a small enough disk D centered at z2 so that z2 is the only one zero inside D. Then apply the argument principle, we have 1 g¨(w) dw = 1. 2i Ê)D g(w)

On the other hand, apply the argument principle to each gn, there is ¨ 1 g (w) n dw = 0. 2i Ê)D gn(w) g¨ (w) ¨ However, { n } uniformly converges to g (w) , which leads to contradiction 0 = 1. gn(w) g(w) Hence this proves that f can only be constant if we assume f is not injective. 3.2. The proof of the Riemann mapping theorem. We will give the proof using three steps. Step 1. We show that for any Ω satisfying the assumptions in Riemann mapping theorem, one can construct a conformal map so that it maps Ω into the unit disk and maps z0 to 0. This proves the set

 ∶= {f ∶ Ω → Dðf is injective and holomorphic , f(z0) = 0} ≠ ç. First since Ω ≠ ℂ, we can take some ∉ Ω. Then z − is never zero for any z ∈ Ω. Because Ω is assumed to be connected and simply-connected, the function g(z) ∶= Log(z − ) is well-deﬁned over Ω and it is injective and holomorphic. For the point z0 ∈ Ω, we claim that any z ∈ Ω,

g(z) ≠ g(z0) + 2i.

That is because if there is some z ∈ Ω so that g(z) = g(z0) + 2i, take exponential to both sides, we obtain 2i z = z0 ⋅ e = z0.

Then it follows g(z) = g(z0) which contradicts with the injectivity of g.

Moreover, in fact, there is a small disk D(g(z0) + 2i) so that

D(g(z0) + 2i) ∩ g(Ω) = ç.

That is because if this is not true, there is a sequence {zn} from Ω so that

g(zn) → g(z0) + 2i. 2i Since the exponential function is continuous, it follows zn → z0 ⋅ e = z0, and then it implies g(zn) → g(z0), which contradicts with the uniqueness of limit. Now using the disk, we deﬁne 1 f(z) ∶= . g(z) − (g(z0) + 2i) The function f is injective and holomorphic over Ω. Also, it is bounded by 1 . So f(z) − f(z ) < ð 0 ð 2 + f(z ) =∶ M. ð 0 ð Then we can deﬁne 1 F ∶ Ω → ℂ,F (z) = (f(z) − f(z )). M 0 It is an injective holomorphic map into D (which means F (Ω) ⊂ D), and it maps z0 to 0, i.e., F ∈ . 68 4. CONFORMAL MAPPINGS

Step 2. Our goal is to prove there is an element in  which is also a surjective map. We prove in ¨ this step 2 that if there is an element F ∈  so that ðF (z0)ð achieves the maxima among all elements in  then such F must be surjective. Notice from Step 1., since conformal equivalence is an equivalence relation, we can restrict ourself to the simpler case by assuming Ω ⊂ D and z0 = 0. Then the set  is

 = {f ∶ Ω → Dðf is injective and holomorphic , f(0) = 0}.

Deﬁne a function Φ ∶  → ℝ as ¨ Φ ∶  → ℝ, f ↦ ðf (0)ð.

Assume there is some F ∈  so that

Φ(F ) ≥ Φ(f), for all f ∈ , we prove now such F must be surjective. Assume F is not surjective. Then there is some point ∈ D so that F (z) ≠ for any z ∈ Ω. Consider the map

∶ D → D deﬁned in the same way as in (2.1). In particular, ∈ Aut(D) and maps to 0 and 0 to .

Then we can construct a map by connecting F and

◦F ∶ Ω → D → D. The image of this map doesn’t contain 0. Moreover, the image is connected and simply-connected, so 1 we can deﬁne (⋅) 2 over its image and consider the function

1 ( ◦F ) 2 ∶ Ω → D. On the other hand, denote by g(z) = z2 as a map from D to itself. Notice that

−1 −1 1 g 1∕2 F 2 F. ( ◦ ◦ 1∕2 )◦( ( ◦ ) ) = By denoting 1 −1 −1 F 1∕2 F 2 ,G g , ∶= ( ◦ ) ∶= ◦ ◦ 1∕2 the map F is a injective holomorphic map from Ω to D with F(0) = 0, and the map G is also a holomorphic map from D to D with G(0) = 0 but it is not injective since g is not injective. ¨ The Schwarz’s lemma says that ðG (0)ð ≤ 1 and it is 1 if and only if it is a rotation. Here since it is ¨ not injective, ðG (0)ð must be strictly smaller than 1. On the other hand, by the chain rule, ¨ ¨ ¨ ¨ ¨ ¨ F (0) = G (0)F (0), then ðF (0)ð = ðG (0)ððF (0)ð. ¨ ¨ Hence, we must have ðF (0)ð > ðF (0)ð which contradicts with our assumption for F . Then we prove that such F must be surjective map. By modifying this F by a rotation, we can also state that F maps 0 to 0 and F ¨(0) > 0. Step 3. The last step is to show that Φ can achieve its maxima. This guarantees that we can ﬁnd a F for Step 2., and then we are done with the proof of Riemann mapping theorem. First, we prove the map ¨ Φ ∶  → ℝ, Φ(f) = ðf (0)ð is uniformly bounded and continuous by Cauchy’s inequality. 3. RIEMANN MAPPING THEOREM 69

For it, only need to notice that the Cauchy’s inequality for each f ∈  gives supD (R) ðfð Φ(f) = f ¨(0) 0 ,D ⊂ Ω. ð ð ≤ R R ¨ Take some R > 0 and ﬁx it, this in particular gives Φ(f) = ðf (0)ð ≤ ‖f‖C0(Ω)∕R ≤ 1∕R, which is independent of f ∈ . This proves Φ is uniformly bounded. Also, apply the Cauchy’s inequality to f1 − f2, supD (R) ðf1 − f2ð Φ(f ) − Φ = f ¨(0) − f ¨(0) (f ¨ − f ¨)(0) 0 ,D ⊂ Ω. ð 1 2ð ðð 1 ð ð 2 ðð ≤ ð 1 2 ð ≤ R R Again take some R > 0 and ﬁx it, this in particular gives

ðΦ(f1) − Φ(f2)ð ≤ ‖f1 − f2‖C0(Ω)∕R, which proves that Φ is (uniformly) continuous. Now denote by s = sup Φ(f). The uniformly boundedness of Φ guarantees that s is a ﬁnite f∈ number. Take a sequence {fn} from  so that Φ(fn) → s. The sequence is uniformly bounded (and in fact equicontinuous from the argument ) over any compact subset K of Ω. By the Montel’s theorem, there exists a subsequence which we still denote by {fn} converging to some holomorphic function f over Ω, and the convergence is uniform over any compact subset K ⊂ Ω. It follows from Proposition 3.7, the limit f is either constant or injective. By the continuity of Φ, Φ(f) = s. Notice that the inclusion map Ω → D has value 1 under Φ, so f is not a constant map and then must be injective. To show f ∈  we only need to check then f maps Ω to D. By the continuity of Φ, ðf(z)ð ≤ 1 for any z ∈ Ω. By the maximal modulus principle, ðfð must be strictly smaller than 1 since it is not constant. Together with f(0) = limn→∞ fn(0) = 0, we ﬁnd an element f ∈  so that Φ(f) = s. This says Φ achieves its maxima at f. By the above three steps, we are done with the proof of the Riemann mapping theorem.