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Part A: Analysis – Complex Analysis

2010 Michaelmas Term

Mathematical Institute, Oxford

October 20, 2010

ii

iv CONTENTS

Introduction

These are (actual I hope)) lecture notes for the Part A Analysis, 2010, based on the course and lecture notes designed by previous lecturers, in particular Glenys Luke, whose revised notes are posted on the course web page for additional reading. In Mods Analysis, we have studied convergence of sequences and series, continuity and dif- ferentiability of a real function of one variable, and we have studied the theory of Riemann’s integration for functions of one real variable. In Part A Option Multi-Calculus, we will study continuity and differentiability of vector valued functions of several (real) variables. In this part of Analysis, we will reveal the meaning of the differentiability of a complex function of one complex variable, and prove some surprising results. Here is one example of such results: Suppose f is a complex function in the unit disk D = {z : |z| < 1 } which has derivative f ′(z) at any point in z ∈ D, that is,

f ′(z) = lim h→ 0

f (z + h) − f (z) h

exists for any z ∈ D, then f ′^ is continuous in D. Moreover all high order derivatives f (n)^ exist as well. You might attempt to test your ability and try to prove it, since concepts involved such as limits and differentiability are already familiar to you. We will develop sufficient machinery in order to prove this seemingly simple (and easy you would think) theorem. In contrast with the 1st year analysis, we will prove the continuity and differentiability by means of integration (while, integrals along curves in the complex plane). Some of you may wonder why it is worthy of your effort through 24 hour lectures in order to prove this beautiful statement which seems no use in practice. What I am going to explain to you is that: a theorem like I stated above is not only beautiful, but, in order to prove it, we need to develop a large array of mathematics which prove quite useful in both sides of the St Giles. They are powerful tools in the hands of applied mathematicians, and they are inspiration for pure guys. About notations. I will use the standard notions and notations in current literature. R and C denote the set of real numbers (real line) and the field of complex numbers. A complex number z = x + iy, where x, y are real numbers, x = Rez is called the real part of z, y = Imz is called the imaginary part of z. (x, y) is called the real coordinates of z, and z is called the complex coordinate of the point (x, y). ¯z = x − iy is the conjugate of z, and |z| =

x^2 + y^2 is called the modulus which measures the distance between points z and 0. We note the relations that x = 12 (z + ¯z) and y = (^21) i (z − ¯z).

Chapter 1

Topology of Euclidean spaces

To study the derivative of a complex function f at a point z ∈ C

f ′(z) = lim h→ 0

f (z + h) − f (z) h

where h → 0 happens in the complex plane, we need to know some features of subsets of C which are significantly more complicated than those in the real line. We begin with some common terminologies used by all mathematicians, such as open sets, closed sets, limiting points etc. If A ⊂ R, a point x ∈ A is an interior point of A if (x−ε, x+ε) ⊂ A for some ε > 0. A subset U is open if all points of U are interior points of U. An interval (x − ε, x + ε) can be described as {y ∈ R : d(x, y) < ε}, where d(x, y) = |x − y| is the distance between x and y. d is called a distance function (or called a metric) on R. Thus, in order to define the concept of open sets, for example in a complex plane C, we only need a distance function on C, nothing else such as the field or the vector space structure is relevant. For the benefit of other options as well, let us study the topology of Euclidean spaces Rn.

1.1 Topology on Rn

If n is a natural number, Rn^ is the set of all ordered n-tuples (x 1 , · · · , xn) where all coordinates xi are real numbers. Rn^ is an example of vector spaces in Linear Algebra, but we will not use its vector structure, we will only use the fact that each point x = (x 1 , · · · , xn) in Rn, xi are reals so that we can define a distance function d : Rn^ × Rn^ → [0, ∞) as the following:

d(x, y) =

(x 1 − y 1 )^2 + · · · + (xn − yn)^2. (1.1.1)

for x = (x 1 , · · · , xn) and y = (y 1 , · · · , yn), called the distance between two points x and y. In terms of summation sign

d(x, y) =

∑n

i=

(xi − yi)^2. (1.1.2)

Example. As sets, C may be identified with the two dimensional Euclidean space R^2 by z = x + iy, i.e. z ←→ (x, y). If z 1 = x 1 + iy 1 and z 2 = x 2 + iy 2 then

d(z 1 , z 2 ) =

(x 1 − x 2 )^2 + (y 1 − y 2 )^2 = |z 1 − z 2 |

1.1. TOPOLOGY ON RN^3

Remark 1.1.5 There are many other distance functions on Rn^ one can use. Here are some examples.

1. d∞(x, y) = ||x − y||∞ where ||x||∞ = max{|x 1 |, · · · , |xn|} is the L∞-norm.
2. d 1 (x, y) = ||x − y|| 1 where ||x|| 1 = |x 1 | + · · · + |xn| is the L 1 -norm.
3. For p ≥ 1 , dp(x, y) = ||x − y||p where ||x||p = p

|x 1 |p^ + · · · + |xn|p^ called the Lp-norm.

We can now define the topology of Rn: the collection of open subsets in Rn. Let r > 0 and a ∈ Rn. The open ball Br(a) centered at a with radius r is the subset {x ∈ Rn^ : d(x, a) < r}.

Definition 1.1.6 A subset U ⊂ Rn^ is open (in Rn) if for every a ∈ U there is ε > 0 such that Bε(a) ⊂ U.

Example 1.1.7 1) Any ball Br(a) ⊂ Rn^ is open. Indeed, if x ∈ Br(a) then d(x, a) < r. Set ε = r− d(x, a). By the triangle inequality, for any y ∈ Bε(x)

d(y, a) ≤ d(y, x) + d(x, a) < ε + d(x, a) < r

so that Bε(x) ⊂ Br(a). Br(a) thus is open by definition.

2. If U ⊂ Rn^ is open then U is a union of some open balls.

Proposition 1.1.8 The collection of all open sets (together with the empty set) in Rn^ is a topology in the following sense:

1. Empty set is open [you may considered it as part of the definition]. The whole space Rn itself is open.
2. The union of some (any many) open sets is open.
3. The intersection of finite many open sets is again open.

Proof. 2) Suppose U = ∪α∈ΛUα where all Uα ⊂ Rn^ are open. For a ∈ U then a ∈ Uα for some α ∈ Λ. Since Uα is open, there is ε > 0 such that Bε(a) ⊂ Uα ⊂ U , so that U is open.

3. Suppose U 1 , · · · , Un are open sets in Rn, and U = ∩ni=1Ui. If a ∈ U then a ∈ Ui for i = 1, · · · , n. Since each Ui is open, there is εi > 0 such that Bεi (a) ⊂ Ui, i = 1, · · · , n. Let ε = min{ε 1 , · · · , εn}. Then ε > 0 and Bε(a) ⊂ Bεi (a) ⊂ Ui for all i. Hence Bε(a) ⊂ ∩ni=1Ui and thus U is open. You should think about, why the proof of 3) does not work for infinite many open sets. For an abstract space in place of Rn, items 1) - 3) become the defining properties of a topology. It is enough to say that it is the topology which describes the continuity of mappings between spaces.

Definition 1.1.9 A subset F ⊂ Rn^ is closed (in Rn) if its complement F c^ = Rn^ \ F is open.

Due to the de Morgan law, we can translate the properties of topology for closed sets.

Proposition 1.1.10 The following properties:

1. Both empty set and the whole space Rn^ are closed.
2. The intersection of closed subsets is closed.
3. The union of finite many closed sets is again closed.

4 CHAPTER 1. TOPOLOGY OF EUCLIDEAN SPACES

There are many subsets which are neither open nor closed. For example, [0, 1] × (0, 1) is such an example in R^2. Let A ⊂ Rn, and a ∈ Rn. We say a is a limiting or accumulation point of A if for every ε > 0, Bε(a) ∩ (A{a}) 6 = ∅. Note that an accumulation point of A may not be in A.

Proposition 1.1.11 A subset F ⊂ Rn^ is closed, if and only if all accumulation points of F belong to F. [Exercise].

1.2 Topology on a subspace of Rn

The distance function d on Rn^ restricted on a subset X ⊂ Rn^ is a distance function on X:

1. d(x, y) ≥ 0 for any x, y ∈ X, and d(x, y) = 0 if and only if x = y.
2. d(x, y) = d(y, x) for x, y ∈ X.
3. d(x, y) ≤ d(x, z) + d(z, y) for any x, y, z ∈ X.

For r > 0 and a ∈ X, the ball (in X) centered at a with radius r is

BXr (a) = {x ∈ X : d(x, a) < r} = X ∩ Br(a). (1.2.1)

The topology on X (under the distance d) can be constructed in terms of balls in X. Namely, a subset U ⊂ X is open (in X) if for each a ∈ U there is ε > 0 such that BXε (a) ⊂ X.

Proposition 1.2.1 The collection of all open sets in X is called the sub-space topology on X ⊂ Rn, which satisfies the properties of topology:

1. The empty set and the whole space X are open (in X).
2. The union of some open sets is open.
3. The intersection of finite many open sets is open.

A subset F ⊂ X is closed (in X) if its complement (relative to X) F c^ = X \ F is open (in X).

Proposition 1.2.2 U ⊂ X is open in X if and only if U = X ∩ V for some open set V in Rn. Similarly F ⊂ X is closed in X if and only if F = X ∩ B for some closed set B in Rn.

Proof. If U ⊂ X is open in X, then for each a ∈ U there is εa > 0 such that

BεXa (a) = X ∩ Br(a) ⊂ X.

Let V = ∪a∈U Br(a). Then V is open in Rn^ and

U = ∪a∈U BXεa (a) = ∪a∈U (X ∩ Br(a)) = X ∩ V.

Conversely, suppose U = X ∩ V for some open set V in Rn, then for any a ∈ U = X ∩ V , since V is open, there is ε > 0 such that Bε(a) ⊂ V , so that

BεX (a) = X ∩ Bε(a) ⊂ X ∩ V = U.

6 CHAPTER 1. TOPOLOGY OF EUCLIDEAN SPACES

Theorem 1.4.1 Any bounded sequence in Rn^ has a convergent subsequence.

Proof. We prove this for R^2. Let an = (a^1 n, a^2 n) be a bounded sequence in R^2. Then {a^1 n} is a bounded sequence of reals, so that there is convergent subsequence, denoted by {a^1 nk }. Then, {a^2 nk } is bounded in R, so that there is a convergent subsequence say {a^2 nk l }. Then {a^1 nk l } is a

subsequence of the convergent sequence {a^1 nk }, so that both {a^1 nk l } and {a^2 nk l } are convergent, so

is ankl = (a^1 nk l , a^2 nk l

Theorem 1.4.2 (Cantor’s Intersection Theorem) Let F 1 ⊃ F 2 ⊃ · · · ⊃ Fk ⊃ · · · be a decreasing sequence of non-empty, bounded and closed subsets in Rn. Then ∩∞ k=1Fk is non-empty.

Proof. Let F = ∩∞ k=1Fk. If there is N such that Fk = FN for all k ≥ N , then clearly F = ∩Nk=1Fk is non-empty. Otherwise, {Fk} has strictly decreasing subsequence {Fkl }, that is {Fkl } is decreasing and all Fkl are different. Thus chose akl ∈ Fkl but akl ∈/ Fkl+1. Since {akl } ⊂ F 1 is bounded, hence {akl } has a convergent subsequence. Without losing generality, we may assume that akl → a ∈ Rn. Since for any N , akl ∈ Fkl ⊂ FkN for all l ≥ N , and FkN is closed, so that a ∈ FkN , and therefore a ∈ ∩∞ N =1FkN = F. F is non-empty.

Theorem 1.4.3 Let A ⊂ Rn^ be bounded and closed. Suppose {Uα : α ∈ Λ} is a family of open subsets in Rn^ such that A ⊂ ∪α∈ΛUα [such a family {Uα : α ∈ Λ} is called an open covering of A]. Then there are finite many {Uαi : i = 1, · · · , k} such that A ⊂ ∪ki=1Uαi.

Proof. Since A is bounded, A ⊂ Br(0) for some r > 0. We first prove a weaker statement: there is a countable many {Uαi : i = 1, · · · } such that A ⊂ ∪∞ i=1Uαi [i.e. there is a countable sub-covering]. Let U = ∪α∈ΛUα. For any x ∈ U = ∪α∈ΛUα, so that x ∈ Uαx for some index αx. Since Uαx is open and Q is dense in R, we can find a cx ∈ Uαx with rational coordinates, and rational rx > 0 such that Brx (cx) ⊂ Uαx and x ∈ Brx (cx).

Hence U = ∪α∈ΛUα = ∪α∈ΛBrx (cx).

The collection of different balls Brx (cx) (as x runs through U ) is at most countable, therefore can be listed as {Bri (ci) : i = 1, 2 , · · · }

where for each i, Bri (ci) ⊂ Uαi for some αi ∈ Λ. By construction,

U = ∪α∈ΛBri (ci) ⊂ ∪∞ i=1Uαi

so that {Uαi : i = 1, 2. · · · } is a countable covering of A. Let Vk = ∪ki=1Uαi. Then V 1 ⊂ V 2 ⊂ · · · is a open covering of A. Suppose for any k, ∪ki=1Uαi ⊃ A is not true, then Fk = A \ Vk are closed, bounded and non-empty, and F 1 ⊃ F 2 ⊃ · · · , but ∩∞ k=1Fk is empty as {Vk : k = 1, 2 , · · · } is a open covering of A, a contradiction.

Definition 1.4.4 Let X ⊂ Rn. A family of open subsets {Uα : α ∈ Λ} is an open covering of X if ∪α∈ΛUα ⊃ X. {Uα : α ∈ Λ} is called a finite open covering if Λ is a finite set. X is called compact, if any open covering of X possesses a finite sub-covering.

1.4. BOUNDED SETS, COMPACT SUBSETS IN RN^7

Theorem 1.4.5 (Heine-Borel Theorem) A subset A ⊂ Rn^ is compact if and only if A is bounded and closed.

Proof. We have proven the “If Part”. “Only if” part. Since Rn^ = ∪∞ i=1Bi(0), so that A ⊂ ∪∞ i=1Bi(0). Thus there is finite many i 1 , · · · , ik ∈ N, such that A ⊂ ∪kl=1Bil (0). Let M = max{i 1 , · · · , ik}. Then A ⊂ BM (0), A is bounded. We next prove that A is closed. Suppose A is not closed, then there is an accumulation point a of A, such that a /∈ A. Therefore, for every ε > 0, Bε(a) ∩ A 6 = ∅. Let Uk = Rn^ \ Fk where Fk = {x : d(x, a) ≤ (^1) k } for k = 1, 2 , · · ·. Then

∪∞ k=1Uk = Rn^ \ ∩kFk = Rn^ \ {a} ⊃ A.

Since A is compact, so that there is N such that

∪Nk=1Uk = UN = Rn^ \ FN ⊃ A.

so that {x : d(x, a) ≤ (^) N^1 } ∩ A = ∅ which contradicts to the assumption that a is an accumulation point.

Theorem 1.4.6 If A is compact and f : A → Rm^ is continuous, then f (A) is compact, hence f (A) is closed and bounded.

This follows from the definition of compact sets and Theorem 1.3.3. A direct corollary is the following generalization of a result in Mod Analysis II: a continuous function f : A → Rm^ is bounded if A is compact. In fact, by the previous theorem, f (A) is compact in Rm, and therefore by the Heine-Borel theorem, f (A) must be bounded, that is, f is bounded on A. As an exercise, the reader may prove the following generalization of another result from Mod Analysis.

Exercise 1.4.7 If A is compact and f : A → Rm^ is continuous, then f is uniformly continuous on A. Here is a proof. Let ε > 0. For any x, by definition, there is δx > 0 such that

BδAx (x) ⊂ f −^1 (Bε/ 3 (f (x))).

Clearly {BAδx/ 3 (x) : x ∈ A} is an open covering of A, so that, as A is compact, there is x 1 , · · · , xk ∈ A such that ∪kj=1BδAxj / 3 (xj ) = A.

Let δ = 13 min{δx 1 , · · · , δxk }. Let x, y ∈ A such that d(x, y) < δ. Then there are i and j such that x ∈ BδAxi / 3 (xi) and y ∈ BδAxj / 3 (xj ), and

d(xi, xj ) ≤ d(xi, x) + d(x, y) + d(y, xj )

<

δxi 3

• δ +

δxj 3 < δxi

so that xj ∈ BδAxi (xi), and thus

d(f (xi), f (xj )) <

ε 3

1.5. CONNECTED AND PATH-CONNECTED SETS 9

Proposition 1.5.5 If X ⊂ R is connected, then X is an interval.

Proof. Let us consider the case that X has the greatest lower bound a and the least upper bound b, so that X ⊂ [a, b] [Unbounded case may be handled with a few modifications]. We want to show that (a, b) ⊂ X [so that X = [a, b], (a, b], · · · , or (a, b) depending a, b belongs to X or not]. Suppose there is c ∈ (a, b) but c /∈ X. Then X ⊂ [a, c) ∪ (c, b]. Let ε > 0 be any but fixed, and set V 1 = X ∩ [a, c) = X ∩ (a − ε, c)

and V 2 = X ∩ (c, b] = X ∩ (c, b + ε).

Both V 1 , V 2 are open in X [Proposition 1.2.2], non-empty (as c is neither the greatest lower bound nor least upper bound) and disjoint. Moreover X = V 1 ∪ V 2 so X is disconnected, a contradiction to the assumption.

Definition 1.5.6 X ⊂ Rn^ is path-connected, if for any a, b ∈ Rn^ there is a continuous mapping p : [0, 1] → X (such a mapping called a path in X) such that p(0) = a and p(1) = b.

By definition, it is obvious that any interval in R is path-connected, so the concepts of connectedness and path-connectedness for subspace of R are the same: X ⊂ R is path-connected if and only if X is an interval.

Proposition 1.5.7 If X ⊂ Rn^ is path-connected, then it is connected.

Proof. If X were not connected, then X = U 1 ∪ U 2 , Ui are open, disjoint and non-empty. Let a ∈ U 1 and b ∈ U 2. There is a path p : [0, 1] → X such that p(0) = a and p(1) = b. Let V 1 = p−^1 (U 1 ) and V 2 = p−^1 (U 2 ). Then V 1 , V 2 are open in [0, 1], disjoint and non-empty. Moreover [0, 1] = p−^1 (X) = V 1 ∪ V 2 , which contradicts to that [0, 1] is connected. However there are connected sets in Rn^ which are not path-connected. For example X = {(x, y) : x > 0, y = x sin (^1) x } ∪ {(0, 0)} is connected but not path-connected. On the other hand we have

Proposition 1.5.8 If X ⊂ Rn^ is open and connected, then X is path-connected.

Proof. Actually we prove that any two points in X can be joined by a polygon which lies in X. Let a ∈ X. Let A be the subset of X whose points can be joined to a by a polygon in X.

1. A is non empty. In fact, since X is open there is ε > 0 such that Bε(a) ⊂ X. Clearly any point in Bε(a) can be joined to a by a straight line in Bε(a). Thus Bε(a) ⊂ A.
2. A is open in Rn^ so is in X. Indeed, if x ∈ A, so that x is connected to a by a polygon in X. Since X is open, there is ε > 0 such that Bε(x) ∈ X. Any point in Bε(x) can be joined by a polygon to x hence to a. Therefore Bε(x) ⊂ A. Let B be the subset of X whose points can not be joined to a by a polygon. Similarly, we can show that B is open in X [Exercise]. By definition X = A ∪ B and A ∩ B = ∅, we thus must have B = ∅ as A is not empty, open and closed.

Definition 1.5.9 A connected open set in Rn^ is called a region.

Therefore any region is path-connected. A region in C is also called a domain by some authors.

10 CHAPTER 1. TOPOLOGY OF EUCLIDEAN SPACES

12 CHAPTER 2. HOLOMORPHIC FUNCTIONS

complex context as required in (2.1.2) is much more demanding than those in the real case. Indeed, f : G → C may be identified with a mapping f : G → R^2 and write as

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

for z = x + iy = (x, y) ∈ G. Then f : G → R^2 is differentiable in REAL SENSE, by definition, if

u(x + h 1 , y + h 2 ) − u(x, y) = A 11 h 1 + A 12 h 2 + o(

h^21 + h^22 )

v(x + h 1 , y + h 2 ) − v(x, y) = A 21 h 1 + A 22 h 2 + o(

h^21 + h^22 )

as

h^21 + h^22 → 0 (where h 1 , h 2 are reals). In this case one can identify A 11 = ∂u ∂x , A 12 = ∂u ∂y

etc. It is a theorem (Option: Multi-Val Calculus) that if u, v have continuous partial derivatives on G, then f (x, y) = (u(x, y), v(x, y)) is differentiable at any point in G. On the other hand, (2.1.1) implies much more: the partial derivatives of u and v must satisfy the Cauchy-Riemann equations (see below)!

The definition of f ′(z) looks no difference from the real case, but, as we allow h → 0 in complex plane C, and therefore z + h approaches to z (though restricted in the open set G) in arbitrary ways (in contrast with the real case in which there are only two sided limits!). As a consequence, the very existence of the complex derivative (of first order) in G has far reaching consequences. For example, the existence of the first derivative of f at every point z ∈ G implies not only the existence of its all higher derivatives, but also the analyticity of f at any z ∈ G which says the Taylor theorem for f holds near z.

We will prove these claims by means of integration along paths. First we devise the Cauchy-Riemann equations. Let f : G → C, and f (z) = u(x, y) + iv(x, y) where z = x + iy and u and v are real and imaginary parts of f , and consider u and v as real functions on G ⊂ R^2. The partial derivatives ∂u ∂x etc., if exist, are denoted by^ ux^ etc.

Proposition 2.1.1 If f has [complex] derivative at a point z = x + iy ∈ G, then

1. All partial derivatives ux, uy, vx and vy at (x, y) exist.
2. The Cauchy-Riemann equations hold

ux = vy, uy = −vx. (2.1.3)

3. We have

f ′(z) =

∂f ∂x

≡ ux + ivx =

i

∂f ∂y

and

f ′(z) =

∂f ∂x

− i

∂f ∂y

∂x

− i

∂y

f. (2.1.4)

The differential operator (^12)

∂ ∂x −^ i^

∂ ∂y

is often denoted by (^) ∂z∂ , so that f ′(z) = ∂f ∂z.

2.1. THE CAUCHY-RIEMANN EQUATIONS 13

Proof. Since f ′(z) exists, so that

f ′(z) = lim h→ 0 h∈R

f (z + h) − f (z) h

= lim h→ 0 h∈R

u(x + h, y) − u(x, y) + i (v(x + h, y) − v(x, y)) h

exists, which implies both limits

ux = lim h→ 0 h∈R

u(x + h, y) − u(x, y) h

and vx = lim h→ 0 h∈R

v(x + h, y) − v(x, y) h

exist, and f ′(z) = ux + ivx. Similarly we also have

f ′(z) = lim h→ 0 h∈R

f (z + ih) − f (z) ih

i

lim h h→∈R 0

u(x, y + h) − u(x, y) + i (v(x, y + h) − v(x, y)) h

so that vy and uy exist, and

f ′(z) =

i

(uy + ivy) = vy − iuy.

We thus must have ux + ivx = vy − iuy

which implies the Cauchy-Riemann equations. Let us look at the equality (2.1.4) from another angle. Let z = x + iy and ¯z = x − iy so that x = 12 (z + ¯z) and y = (^21) i (z − z¯). Now let us forget the relationship between z and its conjugate ¯z, and consider z and ¯z as independent variables in C, apply the chain rule formally to f , to define ∂ ∂z and^

∂ ∂ z¯ as the following ∂ ∂z

f =

∂x ∂z

∂x

f +

∂y ∂z

∂y

f

∂x

− i

∂y

f

and

∂ ∂ z ¯

f =

∂x ∂ z¯

∂x

f +

∂y ∂ z¯

∂y

f

∂x

• i

∂y

f.

Therefore we define differential operators of first order

∂ ∂z

∂x

− i

∂y

∂ z¯

∂x

• i

∂y

2.2. POWER SERIES 15

Since ∂u ∂x and ∂u ∂y are continuous, so that

lim h→ 0

ε 1 (h) h

= lim h→ 0

∂u ∂x

(x+t∗ 1 h 1 ,y+t∗ 1 h 2 ) −^

∂u ∂x

(x,y)

h 1 h

• lim h→ 0

∂u ∂y

(x+t∗ 1 h 1 ,y+t∗ 1 h 2 )

− ∂u ∂y

(x,y)

h 2

h = 0

as

∣h hj

∣ ≤ 1. Similarly there is a t∗ 1 ∈ (0, 1) such that

v(x + h 1 , y + h 2 ) − v(x, y) = vx(x, y)h 1 + vy(x, y)h 2 + ε 2 (h)

and limh→ 0 ε^2 ( hh )= 0. Hence

f ′(z) = lim h→ 0

f (z + h) − f (z) h = lim h→ 0

uxh 1 + uyh 2 + ivxh 1 + ivyh 2 h = lim h→ 0

uxh 1 + iuxh 2 + ivxh 1 − vxh 2 h = lim h→ 0

uxh + ivxh h = ux + ivx

exists. Actually the continuity assumptions on the partial derivatives is redundant, and only exis- tence of partials at any point of G and the Cauchy-Riemann equations are necessary.

Theorem 2.1.5 (Looman-Menchoff ) If f = u + iv : G → C is continuous, if all partial deriva- tives ∂u ∂x , ∂v ∂x , ∂u ∂y , ∂v ∂y exist on G, and if the Cauchy-Riemanns are satisfied: ∂u ∂x = ∂v ∂y , ∂u ∂y = − ∂v ∂x , then f is holomorphic in G.

[This Theorem is not examinable] The proof is much more difficult, see pages 43, R. Narasimhan: Complex Analysis in One Variable. See also Theorem 2.7.15.

2.2 Power series

The basic examples of holomorphic functions are given by power series. The following elementary identity will be used in the proof of Theorem 2.2.2.

Lemma 2.2.1 For any n ≥ 2 and w 6 = z

wn^ − zn w − z

− nzn−^1 =

n∑− 1

k=

zn−^1 −k^

wk^ − zk

16 CHAPTER 2. HOLOMORPHIC FUNCTIONS

Proof. We have for z 6 = 1

1 − zn 1 − z

= 1 + z + z^2 + · · · + zn−^1 ∀n ≥ 1. (2.2.2)

Replace z by w/z or z/w in (2.2.2) to obtain

wn^ − zn w − z

− nzn−^1 = zn−^1 + zn−^2 w + · · · + zwn−^2 + wn−^1

−zn−^1 − zn−^1 − · · · − zn−^1 − zn−^1

=

∑^ n−^1

k=

zn−^1 −kwk^ − zn−^1

∑^ n−^1

k=

zn−^1 −k^

wk^ − zk

Let f (z) =

∑^ ∞

n=

anzn^ (2.2.3)

(where coefficients an are given) be a power series with convergence radius R > 0 (may be infinity). By formally differentiating (2.2.3) term by term one can form the derivative power series of f

g(z) =

∑^ ∞

n=

nanzn−^1 (2.2.4)

which has the same convergence radius R [Problem Sheet 1, Question 7].

Theorem 2.2.2 f defined by the power series 2.2.3 is holomorphic on BR(0). Moreover f ′^ = g on BR(0):

d dz

∑^ ∞

n=

anzn^ =

∑^ ∞

n=

d dz

anzn^ =

∑^ ∞

n=

nanzn−^1 ∀|z| < R. (2.2.5)

[We may differentiate a power series term by term]. Proof. We show that f ′(z) exists for z ∈ BR(0). Let |z| < r < R. For any h 6 = 0 such that |w| < r, where w = z + h, consider

f (z + h) − f (z) h

− g(z) =

∑^ ∞

n=

an

wn^ − zn w − z

− nzn−^1

∑^ ∞

n=

an

wn^ − zn w − z

− nzn−^1

where we have added the series f (w), f (z) and g(z) term by term, as all these series are absolutely convergent [Analysis 1: a power series converges absolutely inside the convergence disk]. Our aim is to show that f (z + h) − f (z) h

− g(z) → 0 as h → 0.