Complex Analysis Lecturer: Ivan Smith · 2018. 12. 13. · Complex Analysis Lecturer: Ivan Smith...

Complex Analysis

Lecturer: Ivan Smith

Scribe: Paul Minter

Lent Term 2016

These notes are produced entirely from the course I took, and my subsequent thoughts.They are not necessarily an accurate representation of what was presented, and may have

in places been substantially edited. Please send any corrections to [email protected]

Complex Analysis Paul Minter

Contents

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1. Complex Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2. Conformal Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3. The Riemann Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4. Power Series and Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2. Integration along Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1. Cauchy’s Theorem and Cauchy’s Integral Formula . . . . . . . . . . . . . . . . . . . . 24

2.2. Liouville’s Theorem and Taylor’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 31

2.3. Identity Theorems and Analytic Continuation . . . . . . . . . . . . . . . . . . . . . . 35

3. Singularities and the Calculus of Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.1. Singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.2. Calculus of Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.3. Calculating Residues and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4. The Argument Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4.1. Local Degrees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

1


1. INTRODUCTION

1.1. Complex Differentiation.

Recall that a subset U ⊂ is open if: ∀x ∈ U , ∃ > 0 such that the open ball B(x) is a subset of U ,i.e. B(x) ⊂ U .

Recall also that a subset U ⊂ is said to be path-connected if for all x , y ∈ U , ∃ γ : [0, 1] → Ucontinuous such that γ(0) = x and γ(1) = y .

FIGURE 1. A path connected subset - any two points can be joined by a continuouspath in the subset.

Definition 1.1. A non-empty, open, path-connected subset of is called a domain in .

Consider U ⊂ open and a complex-valued function f : U → .

Definition 1.2. We say that f is (complex) differentiable at w ∈ U if:

f ′(w) := limz→w

f (z)− f (w)z − w exists.

Definition 1.3. We say that f is holomorphic at w ∈ U (or analytic at w) if f is differentiableon an open neighbourhood B(w) ⊂ U of w.

If f : → is defined on all of and is holomorphic on all of , then f is said to be entire.

The aim of the course will be to develop the theory of holomorphic functions, and integration suchfunctions along C1-smooth (i.e. continuously differentiable) paths.

Given f : U → , we can write f = u+ iv, where u, v : U → are real-valued.

Recall: A function u : U → , a real-valued function of 2 variables, is differentiable at (c, d) ∈ Uwith derivative Du|(c,d) = (λ,µ) if and only if:

u(x , y)− u(c, d)− (λ(x − c) +µ(y − d))(x , y)− (c, d) → 0 as (x , y)→ (c, d).

As above, function f : U → can be decomposed into two functions u, v : U → . How doescomplex differentiability of f relate to differentiability of u, v? This is explained by:

2


Proposition 1.1 (Cauchy-Riemann Conditions). Let f be defined on an open set U ⊂ and letw= c + id ∈ U, and write f = u+ iv. Then:

f is complex differentiable at w ⇔ u, v are differentiable at (c, d) and:ux = vy , uy = −vx at (c, d).

Moreover in this case we have f ′(w) = ux(c, d) + ivx(c, d).

Note: The conditions ux = vy and uy = −vx are called the Cauchy-Riemann conditions. Hereux =

∂ u∂ x , etc.

Proof. By definition, f is (complex) differentiable at w with f ′(w) = p+ iq if and only if:

(†) limz→w

f (z)− f (w)− f ′(w)(z − w)z − w = 0.

Now if z = x + i y , then

f ′(w)(z − w) = p(x − c)− q(y − d) + i(q(x − c) + p(y − d)).So hence we see, by considering real and imaginary parts of (†):

(†) ⇐⇒ both

lim(x ,y)→(c,d)u(x ,y)−u(c,d)−(p(x−c)−q(y−d))

(x−c)2+(y−d)2= 0

lim(x ,y)→(c,d)v(x ,y)−v(c,d)−(q(x−c)+p(y−d))

(x−c)2+(y−d)2= 0.

So hence comparing with what it meant for u, v to be differentiable, this tells us that

(†) ⇐⇒ u and v are differentiable at (c, d) with Du|(c,d) = (p,−q) and Dv|(c,d) = (q, p)which in particular shows ux = p = vy and uy = −q = vx , which are the Cauchy-Riemann conditions.So we are done.

□

Warning: (See question on Example Sheet 1) Given that f : U → can be written as f = u+ iv,where ux = vy and uy = −vx at (c, d) ∈ U , it does not follow that f is complex differentiable atw= c+ id. This is because we need u, v to be differentiable at (c, d) as well, and from Analysis II weknow that just existence of partial derivatives does not necessarily imply differentiability. Howeverif the partial derivatives exist and are continuous then it does (see Analysis II) and so we would getthat f is -differentiable.

Example 1.1. To get a feel for complex differentiability, note that:

(i) Any polynomial p : → is entire.(ii) A rational function p/q : U → with p, q polynomials and U ⊂ \{zeros of q} is holo-

morphic.

3


(iii) f (z) = |z| is not complex differentiable at any point in (note how this is different to in, where this was differentiable everywhere but zero).

To see this, note that for f (z) = |z| we have f = u+ iv where u(x , y) =

x2 + y2 andv(x , y) = 0. So if (x , y) ∕= (0, 0), then

ux =x

x2 + y2and uy =

yx2 + y2

and so at least one of these is non-zero and so the Cauchy-Riemann conditions do not hold.

Finally at (0, 0) ∈ we have the usual thing:f (h)− f (0)

h=|h|h= e−iθ

if h = |h|eiθ . So in particular if h ∈ + then θ = 0 and so f (h)− f (0)h = 1→ +1 as h→ 0.But if h ∈ − then θ = π and so f (h)− f (0)h = −1 → −1 as h → 0. So as these limitsdo not agree this shows that f is also not differentiable at (0, 0) and so f (z) = |z| is notdifferentiable anywhere.

□

Exercise: Show that the usual rules of differentiation (derivatives of sums, products, chain rule,derivative of the inverse function, etc) all hold for complex differentiation, with the same proofs asin the real-variable case.

1.2. Conformal Maps.

Definition 1.4. Let f : U → be holomorphic at w ∈ U. Then if f ′(w) ∕= 0 we say that f isconformal at w.

Remark: If f = u+ iv is holomorphic, then viewing f as a map 2 → 2 via x → (u(x), v(x)) wehave

D f =

ux uyvx vy

then we have by the Cauchy-Riemann conditions det(D f ) = ux vy − uy vx = u2x + v2x = | f ′|2. Soif f ′(w) ∕= 0, then det(D f |w) > 0 and so by the inverse function theorem f : 2 → 2 is locallyinvertible near w. So:

f conformal at w =⇒ f is locally invertible near w (with holomorphic inverse).So we see that locally near a point of conformality, f has an inverse which is also holomorphic andconformal at the point.

Remark: Conformal maps preserve angles in the following sense:

Let f : U → , w ∈ U and suppose f is conformal at w. Now let γi : [−1, 1]→ U with γi(0) = w becontinuously differentiable paths through w, and suppose γ′i(0) ∕= 0, for i = 1, 2.

4


FIGURE 2. Conformal maps preserve angles.

Then we can define the angle between the paths at w, angle(γ1,γ2), to be the difference:

angle(γ1,γ2) := arg(γ′1(0))− arg(γ′2(0))

i.e. this is just the difference in angle their direction vectors make at w.

Then if now we consider the image paths under f , f ◦ γi : [−1, 1]→ , we have:angle( f ◦ γ1, f ◦ γ2) = arg

( f ◦ γ1)′(0)− arg( f ◦ γ2)′(0)

= arg( f ◦ γ1)′(0)( f ◦ γ2)′(0)

as arg(z)− arg(w) = arg

zw

= arg

γ′1(0)

γ′2(0)

as ( f ◦ γi)′(0) = f ′(γi(0))γ′i(0) = f ′(w)γ′i(0) and f ′(w) ∕= 0

= angle(γ1,γ2)

i.e. the angle is preserved under the action of f . □

Fact: (to be proven later) If f : U → is holomorphic on an open set U , then f ′ : U → is alsoholomorphic.

Therefore if f = u + iv is holomorphic, applying the above inductively we see that f is infinitelydifferentiable, and thus so are u, v (i.e. they are smooth). In particular, differentiating the Cauchy-Riemann conditions we have:

ux = vyuy = −vx

=⇒ ux x = vy x = −uy y =⇒ ux x + uy y = 0 and vx x + vy y = 0

and so hence we see that the real and imaginary parts of a holomorphic function satisfy the Laplaceequation, and are called harmonic functions.

Definition 1.5. If U , V ⊂ are open, then we say f : U → V is a conformal equivalence if it isa bijection which is conformal everywhere.

Example 1.2. Any Möbius map A(z) = az+bcz+d with ad − bc ∕= 0 defines a conformal equivalence∪ {∞}→ ∪ {∞} in the obvious sense (A′(z) ∕= 0 follows from the chain rule and invertibilityof A(z)).

5


Similarly, if we consider the Möbius maps which preserve the open unit disc ⊂ , i.e.Möb() = { f ∈Möbius group : f () = }

=

a b−b a

∈Möbius Group : |a|2 + |b|2 = 1

these are conformal equivalences of the disc . [Recall that we can represent the Möbius mapz → az+bcz+d by the matrix

a bc d

.]

Example 1.3. The map f : z → zn, n ≥ 2, is holomorphic everywhere and conformal everywhereexcept at z = 0. This map gives a conformal equivalence between a slice and the half-plane, i.e.

z ∈ \{0} : 0< arg(z)< π2

and {z ∈ : Im(z)> 0}.

FIGURE 3. An illustration of the conformal equivalence between a wedge and theupper-half plane given in Example 1.3.

Notation: We write

• ∗ := \{0},• := {z ∈ : Im(z)> 0} for the upper-half-plane in ,• := {z ∈ : |z|< 1} for the open unit disc in .

Example 1.4. Note that:z ∈ ⇐⇒ z is closer to i than it is to − i

⇐⇒ |z − i|< |z + i|

⇐⇒z − iz + i

< 1.

So hence we see that f : z → z−iz+i defines a conformal equivalence between the unit disc and theupper-half plane .

6


Example 1.5. Consider the map f : ∗→ defined by f (z) = w := 12z + 1z

. Note that:

w+ 1w− 1 = 1+

2w− 1 = 1+

4

z + 1z − 2= 1+

4zz2 − 2z + 1 =

z2 + 2z + 1z2 − 2z + 1 =

z + 1z − 1

2.

This map f is holomoprhic on ∗ and f ′(z) = 1− z2+12z2 . Hence f is conformal everywhere exceptat z = ±1.

Now if we write z = reiθ , then f (z) = w= u+ iv, where

u=12

r +

1r

cos(θ ) and v =

12

r − 1

r

sin(θ ).

Thus if we fix one of r or θ and vary the other, we can paint a picture of the image of this map.

• If we fix r, say r = ρ (ρ ∕= 1), then we see:

{|z|= ρ} f−→

u2

14

ρ + 1ρ2 +

v2

14

ρ − 1ρ2 = 1

i.e. circles (about the origin) are mapped to ellipses under f ,

• If we fix θ = µ, then we see:

{arg(z) = µ} f−→

u2

cos2(µ)− v

2

sin2(µ)= 1

i.e. half-lines from the origin are mapped to hyperbola.

This map f is known as the Joukowski transform. If we instead consider a circle which passesthrough z = −1 and z = −i, then f maps this circle to a crude model of an aerofoil (show inFigure 4). This is used to model fluid flow over a wing in terms of the analytically simpler flowacross a circular cross-section (i.e. if we have a question/problem in terms of an aerofoil, then wecan transform it to a circle, where e.g. Laplace’s equation can easily be solved, and then transformthe solution back to get the actual solution on the aerofoil).

FIGURE 4. An illustration of the conformal equivalence given in Example 1.5 betweena circle and an aerofoil.

7


Remark: Recall that a helpful way to represent a region bound by a circular arc is as follows:

FIGURE 5. Illustration of finding an expression for the locus of a circular arc.

Along the arc shown in Figure 5, we have µ= θ −ϕ and so µ= arg(z −α)− arg(z −β) is constant.So the region bound by two such arcs can written:

z ∈ : arg

z −αz − β

∈ [µ−,µ+]

.

Example 1.6. If we consider the region U showed in Figure 6, then we have by the above that:

U =

z ∈ : arg

z − 1z + 1

∈π

2,π

i.e. U is the union of all arcs of circles which pass through +1 and −1 in .

FIGURE 6. U here is just the interior of the unit disc intersected with the upper-halfplane. However it is also the union of a continuum of arcs, which gives an alternativeexpression for this region.

Thus for instance, the map z → z−1

z+1

2 · (−1) is a conformal equivalence from U to . This can be seenby the composition in Figure 7. The composition of the three individual maps shown their is this map.So we are decomposing the map into separate steps to more easily see the effect that map has on a givenregion.

8


FIGURE 7. Mapping the upper-half circle U to the upper half plane using a series ofconformal equivalences. The arc-formulation of U makes the first map clear.

1.3. The Riemann Mapping Theorem.

The Riemann mapping theorem is an incredibly useful result to bear in mind, although we shall notprove it. It says that every simply connected proper subset of is conformally equivalent to the openunit disc.

Definition 1.6. A simple closed curve (in ) is the image of an injective continuous map S1→ ,

i.e. this is a path in which does not self-intersect (as injective) and starts and ends in the sameplace (by continuity, i.e. γ(2π) = γ(0)).

FIGURE 8. An example of a simple closed curve.

Definition 1.7. A domain U ⊂ is said to be simply connected if every contiuous map f : S1→ Ucan be extended to a continuous map F : → U with F |∂ = f .

Remark: Intuitively, being simply connected means that any loop in U can be shrunk to a pointwithout breaking the loop and without leaving U .

FIGURE 9. The region A is simply connected, because any loop in A can be shrunk to apoint without leaving A. However the region B is not simply connected, e.g. considerthe loop/map S1→ B shown by γ. We can shrink this to a point in B without leavingB.

9


Theorem 1.1 (Riemann Mapping Theorem). Let U ⊂ be the bounded region enclosed by asimple closed curve, or more generally, suppose U ⊊ is any simply connected subset. Then U isconformally equivalent to the open unit disc ⊂ .

Proof. None given. □

1.4. Power Series and Logarithms.

Example 1.7. The exponential function is defined by the power series:

ez ≡ exp(z) := 1+ z + z2

2!+

z3

3!+ · · ·

and defines a function → ∗ (one can check this is power series has infinite radius of convergenceand thus exp is defined on all of ).

Moreover, the exponential function is everywhere conformal, since ddz (ez) = ez ∕= 0.

The exponential map also provides conformal equivalences between vertical strips and annuli, asshown in Figure 10.

FIGURE 10. The exponential map sends a vertical strip to a annulus. Note that oneof these regions is simply connected, but the other is not. This is explained becauseexp is not a 1-1 map on this strip to this annulus (as it is not injective).

Recall: We shall make use of the following facts shown in Analysis II:

• A sequence of functions is said to converge uniformly on a subset U ⊂ C if ∀ > 0, ∃N suchthat ∀n≥ N we have | fn(z)− f (z)|< for all z ∈ U .• The uniform limit of a sequence of continuous functions is also continuous.

10


• (Weierstrass M -test). If (Mn)n=1 ⊂ ≥0 and functions ( fn)n are such that 0 ≤ | fn(z)| ≤ Mnfor all n and all z ∈ U , then:

∞

n=1

Mn converges =⇒

n≥1fn(x) converges uniformly on U

i.e. the partial sums gN =N

n=1 fn converge uniformly.

• Given (cn)n≥0 ⊂ a sequence of complex numbers, ∃! R ∈ [0,∞] such that:

z −→∞

n=0

cn(z − a)n converges absolutely if |z − a|< R and diverges if |z − a|> R.

Moreover if 0 < r < R then the convergence is uniform on {z ∈ : |z − a| ≤ r}, i.e. oncompact subsets of the domain the power series converges uniformly.

A key result about complex power series is that the are holomorphic within their radius of conver-gence, and are infinitely differentiable there as the next theorem shows.

Theorem 1.2. Let f (z) =∞

n=0 cn(z − a)n be a power series with radius of convergence R > 0.Then:

(i) f is holomorphic on BR(a), the open ball {z : |z − a|< R} ⊂ .(ii) f has derivative f ′(z) =

∞n=1 ncn(z − a)n−1, which also has radius of convergence R.

(iii) f is infinitely complex differentiable here, and therefore cn =f (n)(a)

n! .

Proof. Wlog take a = 0 (otherwise just let w= z − a and consider g(w) := f (w+ a)).

Certainly since |ncn| ≥ |cn| by the comparison test to the series for f , we see that the radius ofconvergence of

∞n=1 ncnz

n−1 is at most R.

But if |z|< ρ < R, then we can see that:

|ncnzn−1||cnρn−1|

= n

zρ

n−1→ 0 as n→∞.

So by comparison to

n≥1 cnρn−1 (which converges by the definition of radius of convergence for

f ) we see that the radius of convergence of∞

n=1 ncnzn−1 is at least ρ > 0. But ρ < R was arbitrary,

and so this shows that

n≥1 ncnzn−1 has radius of convergence at least R. Combining with the above

gives that it does have radius of convergence exactly R.

So now we want to show that f really is differentiable. Pick z, w with |z|, |w| ≤ ρ, where ρ < R.Then let:

ϕ(z, w) =∞

n=1

cn

n−1

j=0

z jwn− j−1

.

11


Noting that cnn−1

j=0

z jwn− j−1≤ n|cn|ρ

n−1

we see that the series defining ϕ converges, and converges uniformly on {(z, w) : |z| ≤ ρ, |w| ≤ ρ},and hence converges to a continuous limit. But by summing the geometric series we have:

• If z ∕= w, ϕ(z, w) =∞

n=1 cn zn−wn

z−w= f (z)− f (w)z−w .

• If z = w, ϕ(z, w) =∞

n=1 ncnzn−1.

So as ϕ(z, w) is continuous, we get that

f ′(z) = limz→w

f (z)− f (w)z − w = limz→wϕ(z, w) = ϕ(z, z) =

∞

n=1

ncnzn−1

as claimed. So hence we have proven claims (i) and (ii).

Part (iii) is then an immediate consequence of inductively applying (i) and (ii), since by (i) f ′(z) iscomplex differentiable, and so by induction it is infinitely complex differentiable.

The formula for cn then follows by differentiating n times and setting z = 0.

□

Corollary 1.1. Consider a power series f (z) =

n≥0 cn(z− a)n with radius of convergence R> 0.Then if ∃ ∈ (0, R) with f ≡ 0 on B(a), then f ≡ 0 on BR(a).

Proof. By Theorem 1.2 we have cn = 0 for all n. Hence f ≡ 0. □

Remark: Sometimes it is useful to have an explicit expression for the radius of convergence R. Someexpressions (which aren’t too hard to prove) are:

(i) R= sup{r > 0 : |cn|rn→ 0 as n→∞}.(ii) R= 1/λ, where λ = lim supn→∞

|cn|1/n.

Recall: The exponential function defined by:

ez ≡ exp(z) := 1+ z + z2

2!+

z3

3!+ · · ·

has radius of convergence∞, and so is an entire function. We know that ez+w = ez · ew for anyz, w ∈ [Exercise to show]. So in particular, if z = x + i y ,

ez = ex+i y = ex · ei y = ex [cos(y) + i sin(y)] .So given any w ∈ ∗ = \{0}, there are (infinitely many) solutions to ez = w. Different solutionsdiffer by integer multiples of 2πi. So we have to be careful when making inverse functions of exp(z)due to this multi-valuedness.

12


Definition 1.8. Let U ⊂ ∗ be open. Then a branch of the logarithm on U is a continuousfunction λ : U → such that:

eλ(z) = zfor all z ∈ U.

Example 1.8. Suppose U = \≤0. Then for z ∈ U we can write z = reiθ , with θ ∈ (−π,π)unique. Then define:

λ(z) := log(r) + iθ .This defines the principal branch of the logarithm (called so as it is the most ‘natural’ one).

Then on U there is a continuous function arg : U → (−π,π) measuring the argument. Contrastlywe would not expect there to exist such a continuous function on the unit circle about the origin,i.e. if U = S1 (since the argument would differ by two by upon returning to the starting positionafter travelling around the circle).

FIGURE 11. The “slit plane” or “cut plane” on which we define the principal branchof the logarithm.

Proposition 1.2. On U = {z ∈ : z ∕∈ ≤0}, the principal branch of the logarithm, denotedlog : U → , is holomorphic.

Moreover we have ddz (log(z)) =1z , and if |z|< 1, then:

log(1+ z) =

n≥1(−1)n−1 z

n

n= z − z

2

2+

z3− · · · .

Proof. The fact that log is holomorphic follows from the chain rule and the fact elog(z) = z. This alsoshows via the chain rule (since ddz (e

z) = ez) that ddz (log(z)) =1z .

Finally, the given power series has radius of convergence 1, and so by Theorem 1.2 it has derivative:

1− z + z2 − z3 + · · ·= 11+ z

13


via summing the geometric series. Therefore log(1 + z) and

n≥1(−1)n−1zn

n have equal derivatives(since we are working on |z|< 1, 1+z will remain in U for the logarithm) and so differ by a constant.Evaluating at z = 0 shows that this constant must be 0.

□

Note: If α ∈ and log : U → is a branch of the logarithm, we can define:zα := eα log(z) on U .

Remark: We can view log as a multivalued function on ∗.

Definition 1.9. We say that a point p ∈ is a branch point of a multivalued function if thefunction cannot be given a continuous, single-valued definition on any punctured disc about p, i.e.on any B(p)\{p} for any > 0.

Example 1.9. 0 is a branch point of log, since by moving around any circle about 0 we do not geta continuous argument (due to differing by integer multiples of 2πi).

-1 for example is not a branch point of log, since we can just vary the argument θ in [−π−δ,π+δ]instead of [−π,π].

Example 1.10. Consider f (z) =

z(z − 1). This has two branch points at z = 0 and z = 1 (sincethe square root

z can be uniquely defined in a punctured neighbourhood of any z ∈ ∗ but not at

z = 0).

Note that we can define a continuous branch of f on either of the subsets of shown in Figure 12.For the first this is just because no path can loop around either of the branch points. In the second,note that we can write

f (z) = e12 [log(z)+log(z−1)]

and as we move around the curve γ, the argument of each of log(z) and log(z − 1) jumps by 2πi,and so the expression for f (z) stays uniquely defined (simply because 12(2πi + 2πi) = 2πi ande2πi = 1), and so this second cut/subset does work as a suitable region to define a value of f .

[The intuition here is to imagine the Riemann sphere - the removed line in U1 corresponds to a line onthe Riemann sphere which passes through the north pole and joins back to the other point and so is anarc of a great circle. The removed line in U2 corresponds to the other arc of this great circle. This is“why” we use these lines/cuts.]

Remark: For a better way of thinking about branch points, see the Riemann Surfaces Part II course.

14


FIGURE 12. Possible branch cuts for f in Example 1.10. The regions U1, U2 allowa continuous definition of f , with the red lines indicating the “cuts”, i.e. lines weremove from . In U1 we have a finite slit, whilst in U2 we have a finite cut.

15


2. INTEGRATION ALONG PATHS

Our next task is to develop a theory of integration of -valued functions along paths and closedcurves.

Recall: We say that f : [a, b] → is Riemann integrable if Re( f ) and Im( f ) are both Riemannintegrable as -valued functions, and then:

b

af (t) dt :=

b

aRe( f (t)) dt + i

b

aIm( f (t)) dt.

Recall also that if f is continuous we know that f is integrable.

Lemma 2.1. Suppose f : [a, b]→ is continuous (and hence integrable). Then:

b

af (t) dt

≤ (b− a) supt∈[a,b]| f (t)|, with equality ⇐⇒ f is constant.

Proof. Let θ = arg b

a f (t) dt

and M = supt∈[a,b] | f (t)|. Then (using “z = reiθ ” for z = b

a f (t) dt):

b

af (t) dt =

b

af (t) dt

eiθ =⇒

b

af (t) dt

= b

ae−iθ f (t) dt

and so as the LHS of this equality is purely real, so must the RHS, so we get (from the definition ofthe integral of a complex valued function):

b

af (t) dt

= b

aRe(e−iθ f (t)) dt.

Then since we know for real valued functions g, b

a g ≤ b

a |g| ≤ (b − a) sup |g|, and we knowRe(e−iθ f (t))≤ |e−iθ f (t)|= | f (t)| for all t, we get:

b

af (t) dt

≤ (b− a) supt∈[a,b]| f (t)|

as required. Moreover, we have equality if and only if Re(e−iθ f (t)) = M for all t, i.e. if and only if| f (t)|= M for all t and arg( f (t)) = θ for all t, i.e. if and only if f is constant.

□

Definition 2.1. A path γ : [a, b]→ is simple if γ(t1) = γ(t2) ⇒ {t1, t2}= {a, b} (for t1 ∕= t2),i.e. if the path intersects itself, it must just be the endpoints.

If γ(a) = γ(b) then we say γ is a closed path.

16


FIGURE 13. For the three curves shown above, (i) is simple as it does not self-intersect, (ii) is simple because it only intersects at the endpoints, and (iii) is notsimple since it intersects not at the endpoints.

Definition 2.2. A contour is a simple closed path which is piecewise C1 (i.e.piecewise continuouslydifferentiable).

FIGURE 14. An example of a contour. γ is made up of four parts, each of which iscontinuously differentiable, but γ itself is not continuously differentiable.

Now if γ : [a, b]→ U ⊂ is a C1 curve and f : U → is continuous, then we can define the integralof f over γ by:

γ

f (z) dz :=

b

af (γ(t)) · γ′(t) dt

i.e. we parameterise γ by z = γ(t). By summing over the C1-pieces, this definition extends topiecewise C1 curves, and in particular to contours.

Properties of integrals over curves:

(i) The definition above is independent of the parameterisation of γ. Indeed, if ϕ : [a′, b′]→[a, b] is C1 with ϕ(a′) = a, ϕ(b′) = b, and if γ is a C1-path, then if we set δ = γ ◦ϕ (whichtraces out the same path as γ but is just a different parameterisation), then:

γ

f (z) dz =

δ

f (z) dz

since by changing variables u= ϕ(t) we have:

δ

f (z) dz =

b′

a′f (γ(ϕ(t)))γ′(ϕ(t))ϕ′(t) dt =

b

af (γ(u))ϕ′(u) du=

γ

f (z) dz.

(ii) If a < u< b, then:

γ

f (z) dz =

γ|[a,u]f (z) dz +

γ|[u,b]f (z) dz

17


from the additive properties of integrals. This tells us that we can sum over piecewise partsof a path if necessary, and so summing over subdomains (i.e. extending our definition topiecewise C1-paths) is fine.

(iii) If we denote the path γ transversed in the opposite direction to that of γ by −γ (i.e. movingbackwards along the path, not forwards), then:

−γf (z) dz = −

γ

f (z) dz

(this follows easily by the reparamerisation ϕ(t) = (b+ a)− t).(iv) If we define for γ : [a, b]→ the length of γ by

length(γ) :=

b

a|γ′(t)| dt

then we have:

γ

f (z) dz

≤ length(γ) · supt∈Image(γ)| f (t)|.

This follows exactly as in Lemma 2.1 as we get:

γ

f (z) dz

=

b

af (γ(t))γ′(t) dt

≤ supγ | f (t)| · b

a|γ′(t)| dt = sup

γ| f | · length(γ).

Example 2.1. Let U = ∗, and consider f (z) = zn. Let ϕ : [0, 2π]→ U be ϕ(θ ) = eiθ , i.e. thecontour of the unit circle. Then:

ϕ

f (z) dz =

2πi if n= −10 otherwise,

since

ϕ

f (z) dz =

2π

0

f (ϕ(θ ))ϕ′(θ ) = i

2π

0

eiθ (n+1) =

2πi if n= −10 if n ∕= −1

as if n ∕= −1 then we are just integrating trigonometric functions over an entire period.

Example 2.2. Take γ to be the half-circle contour as shown in Figure 15. Let f (z) = z2. Then

γ

f (z) dz =

integral on real axis R

−Rt2 dt +

1

0

R2e2πi t · iπReiπt dt

integral on upper half disc, i.e. γR : [0, 1]→ , γR(t) = Reiπt

=2R3

3+ iπR2 1

0

e3πi t dt

=2R3

3+ R3

13

eπi t1

0

= 0.

18


FIGURE 15. Illustration of semi-circle contour used in Example 2.2.

Definition 2.3. Suppose U ⊂ and f : U → is continuous. Then if ∃ a holomorphic functionF : U → with F ′(z) = f (z) on U, then we say f has anti-derivative F on U.

We then have the complex version of the fundamental theorem of calculus:

Theorem 2.1 (Fundamental Theorem of Calculus). Let f : U → be continuous. Suppose f hasanti-derivative F. Then if γ : [a, b]→ U be a piecewise C1 curve, we have:

γ

f (z) dz = F(γ(b))− F(γ(a)).

In particular, if γ is closed then

γ

f (z) dz = 0.

Proof. We simply have

γ

f (z) dz =

b

af (γ(t))γ′(t) dt =

b

aF ′(γ(t)) · γ′(t) dt

=

b

a(F ◦ γ)′(t) dt

= [(F ◦ γ)(t)]ba= F(γ(b))− F(γ(a))

as required. In the case γ is closed we have γ(b) = γ(a) and so this is zero.

□

19


Example 2.3 (Example 2.1 Revisited). Recall that in Example 2.1 we have f (z) = zn and ϕ :[0, 2π]→ U = ∗ defined by ϕ(t) := ei t . Then if n ∕= −1, we have

f =ddz

zn+1

n+ 1

and so f has an anti-derivative, and the fact that the integral vanishes as we found in Example 2.1is just an instance of the fundamental theorem of calculus.

If n= −1, then f (z) = ddz (log(z)), where log is defined on a slit plane. Hence sinceϕ

f (z) dz doesnot vanish here, this implies that ∕ ∃ a continuous branch of log on any set U such that {|z|= 1} ⊂ U(since otherwise f would have this branch of log as its anti-derivative and so the integral wouldvanish by the fundamental theorem of calculus).

Proposition 2.1. Let U ⊂ be a domain and let f : U → be continuous. Suppose that

γ

f (z) dz = 0 for any closed piecewise C1 path γ in U.

Then f has an antiderivative on U, i.e. ∃F : U → holomorphic with F ′(z) = f (z).

Proof. We expect as usual that F will be given by the integral of f in some form.

So pick a point a ∈ U (this will determine the “constant of integration” essentially). Then for anyw ∈ U , since U is a domain and so path connected, we can find a piecewise C1-path(i) γw : [0, 1]→ Uwith γw(0) = a and γw(1) = w. Then define:

F(w) :=

γw

f (z) dz.

First note that this definition of F is well-defined. To see this, we just need to show that it is inde-pendent of the choice of path γw. But this is true from our hypotheses: if we have another choiceof path γ̃w, then consider the path defined by γ = γw ∗ (−γ̃w) where ∗ denotes the concatenation ofthese paths, i.e. travel backwards along γ̃w from w to a, then back to w along γw. This γ is then aclosed piecewise C1-path in U , and so we have:

0=

γ

f (z) dz by hypothesis

=

γw

f (z) dz +

−γ̃wf (z) dz =

γw

f (z) dz −

γ̃w

f (z) dz

i.e.γw

f (z) dz =γ̃w

f (z) dz and so this is independent of the choice of γ. Hence F is well-defined.

(i)Being path connected does not guarantee that the path we get is piecewise C1, just a continuous γ : [0, 1]→ U withγ(0) = a and γ(1) = w. But we can get a piecewise C1 path as follows. Since U is open, for all x ∈ Image(γ) we can find(x) > 0 such that B(x)(x) ⊂ U . But note that since γ is continuous and [0, 1] is compact, Image(γ) is a compact subsetof U . So as these balls cover Image(γ) we get by compactness a finite subcover. So the path is covered by finitely manyballs. Then we can connect a to w by piecewise linear line segments, which lie within these open balls (see Figure 18).This is then clearly a piecewise C1-path joining a to w so we get the path we were after.

20


FIGURE 16. Showing F is well-defined in Proposition 2.1.

Next we need to check that F is complex differentiable. Since U is open, for any w ∈ U we can find > 0 with B(w) ⊂ U . So let δh be a radial path in B(w) from w to w+ h, where |h| < . Then letγ be the closed path:

γ := γw ∗δh ∗ (−γw+h).

FIGURE 17. An illustration of δh.

Then we have by assumption thatγ

f (z) dz = 0, and so breaking this down we get:

F(w+ h) =

γw∗δhf (z) dz = F(w) +

δh

f (z) dz.

So hence we see

F(w+ h) = F(w) + hf (w) +

δh

[ f (z)− f (w)] dz

since f (w) is independent of z. So henceF(w+ h)− F(w)

h− f (w)=

1|h|

δh

f (z)− f (w) dz

≤ length(δh)h

· supz∈δh| f (z)− f (w)|

= supz∈δh| f (z)− f (w)|.

But then the RHS→ 0 as |h|→ 0 by continuity of f at w. Hence we see that F is holomorphic at wwith F ′(w) = f (w). But then as w ∈ U was arbitrary we are done.

□

So to summarise the last two results, we have shown that if f : U → is continuous, then:

21


FIGURE 18. An illustration of the footnote in Proposition 2.1 to show how we canfind a piecewise C1 path from γ. We take straight line segments which lie in thefinite cover of γ by balls.

• If f has an antiderivative F on U , thenγ

f (z) dz = 0 whenever γ ⊂ U is a closed piecewiseC1 path.

• If U is a domain and ifγ

f (z) dz = 0 for all closed piecewise C1 paths γ ⊂ U , then f has anantiderivative F on U .

So we see that integrating to 0 around closed loops is a useful property. We shall try to push this andsee what we can get.

Definition 2.4. A domain U is said to be star-shaped (or a star domain) if ∃p ∈ U such that forall w ∈ U, the straight line segment [p, w] joining p to w lies in U.

U is said to be star-shaped about such a p.

FIGURE 19. An example of a star domain.

Remark: We have:

U a disc =⇒ U is convex =⇒ U is star-shaped =⇒ U is path-connectedbut none of the reverse implications hold.

Recall now that in the proof of Proposition 2.1, all we needed to prove the antiderivative was complexdifferentiable was that for some choice of paths γw for all w ∈ U , we have

γ

f (z) dz = 0 for γ= γw ∗δh ∗ (−δw+h).

Note that this path γ looks like a triangle. This suggests that perhaps knowing just that the integralabout each “triangle” in U is zero is enough for existence of an antiderivative. We now try to makethis more precise.

22


FIGURE 20. In the proof of Proposition 2.1 the paths we considered looked as shown.These resemble a triangle which motivates us to look at the case of triangles in moredetail.

Definition 2.5. A triangle in a domain U is the Euclidean convex hull of 3 points in U, such thatthe triangle lies completely in U (i.e. vertices, boundary and interior all lie in U).

We write ∂ T for the boundary of a triangle, which we view as an oriented piecewise C1 path (inparticular, a contour).

FIGURE 21. Here we have three subsets U and three ‘triangles’. The leftmost diagramis a triangle, whilst the other two are not - this is because they do not lie wholly inU .

Our earlier result (Proposition 2.1) on constructing antiderivatives easily gives:

Proposition 2.2. Suppose U is a star-domain, and f : U → is continuous such that

∂ Tf (z) dz = 0

for all triangles T ⊂ U. Then f has an anti-derivative on U.

Proof. As before in the proof of Proposition 2.1, but now take a to be the/a point U is star-shapedabout, and take the paths γw to w ∈ U to be the straight line segments joining a to w. Then the prooffollows exactly as before (as γw ∗δh ∗ (−δw+h) is the boundary of a triangle).

□

23


2.1. Cauchy’s Theorem and Cauchy’s Integral Formula.

Cauchy’s theorem is a powerful result about the integral of holomorphic functions about contours ina domain U . To prove it we first work with triangles and then build up to the general result.

Theorem 2.2 (Cauchy’s Theorem for a triangle). Let U be a domain and let f : U → beholomorphic. Then for any triangle T ⊂ U we have

∂ Tf (z) dz = 0.

Proof. Momentarily. □

Remark: The conclusion is independent of the orientation of ∂ T (as this only changes a sign and sowe still get 0).

An important corollary is the generalisation to more general curves. We will see an even more generalversion latter on as well.

Corollary 2.1 (Convex Cauchy Theorem). Let U be a convex (or even star-shaped) domain, andlet f : U → be holomorphic. Then for any closed piecewise C1 path γ ⊂ U we have

γ

f (z) dz = 0.

Proof. If U is convex, then U is star-shaped. Cauchy’s theorem for triangles (Theorem 2.2) gives thatfor all triangles T ⊂ U ,

∂ Tf (z) dz = 0.

So by Proposition 2.2, we get that f has an anti-derivative on U . Then by the fundamental theoremof calculus (Theorem 2.1) we get

γ

f (z) dz = 0 for all closed pieceiwse C1 γ.

□

Proof of Cauchy’s Theorem for a triangle. Let T ⊂ U be a triangle. Then let η =∂ T f (z) dz and let

l = length(∂ T ) be its Euclidean length. We want to show η = 0.

Write T0 = T . Then subdivide T0 into triangles T0a , T0b , T

0c , T

0d by joining midpoints, and orient the

smaller triangles as shown in Figure 22. So then

∂ T0=

a,b,c,d

∂ T0•

f (z) dz

24


FIGURE 22. Subdividing T into four smaller triangles with the shown orientations.

if we orient the middle triangle (wlog T0d ) by the usual “anticlockwise” orientation, as then theinternal edge (i.e. edge joining midpoints) occurs twice in this sum, once with each orientation andso cancel. So as:

(†) η =

∂ Tf (z) dz

=

a,b,c,d

∂ T0•

f (z) dz

≤

a,b,c,d

∂ T0•

f (z) dz

∃ a subscript • ∈ {a, b, c, d} such that∂ T0•

f (z) dz≥ η4 , or else all of these terms would be < η/4,

which would give a contradiction in (†). So call this triangle (with this subscript) T1 := T0• . Thenthe length of ∂ T1 is:

length(∂ T1) =length(∂ T0)

2

just from properties of triangles in Euclidean space.

Iterating this process, we obtain triangles T0 ⊃ T1 ⊃ T2 ⊃ · · · such that

∂ T if (z) dz

≥η

4iand length(∂ T i) =

length(∂ T0)2i

.

This gives a decreasing nested sequence of closed sets T0 ⊃ T1 ⊃ T2 ⊃ · · · and we know by standardanalysis (see Metric and Topological Spaces or Analysis II) that we can find

z0 ∈

i=0T i ⊂ U

(this is essentially just the nested intervals property). By assumption f is holomorphic at z0. Sohence for any > 0, we can find δ > 0 such that if |w− z0|< δ then

| f (w)− f (z0)− (w− z0) f ′(z0)|≤ (w− z0).

We can now pick n such that T n ⊂ Bδ(z0) due to the length(∂ T i)→ 0 and z0 ∈ T i for all i.

Note that the functions 1 and z are definitely anti-derivatives on T n (as they are entire and T n ⊂ )and so∂ T n 1 dz = 0=∂ T n z dz. Therefore we can add constant multiples of

∂ T n 1 dz and∂ T n z dz

25


without changing anything, and so

∂ T nf (z) dz

=

∂ T n

f (z)− f (z0)− (z − z0) f ′(z0)

dz

≤

∂ T n|z − z0| dz

≤ · length(∂ T n) · supz∈∂ T n|z − z0|

≤ · [length(∂ T n)]2

as points z, z0 ∈ ∂ T n can never be more than length(∂ T n) apart, and so supz∈∂ T n |z−z0|≤ length(∂ T n).Thus we get

η

4n≤

∂ T nf (z) dz

≤ [length(∂ T n)]2 = ·length(∂ T0)2

4n

=⇒ η ≤ ·length(∂ T0)2

and this is true for any > 0. So taking → 0 we get η = 0, which is what we wanted.

□

Remark: We will see later that Cauchy’s theorem holds for simply connected domains.

Remark: (Addendum of Convex Cauchy Theorem) In fact we can give a stronger form of the convexCauchy theorem. Suppose in fact, in the situation above, that instead f : U → is continuous andf is holomorphic except at finitely many points, i.e. ∃ a finite set S ⊂ U such that f |U\S : U\S → is holomorphic. Then the same conclusion holds, i.e.

γ

f (z) dz = 0

for all closed piecewise C1 γ.

So why is this the case? Well in the proof of the convex Cauchy theorem, it was sufficient to focuson showing that

∂ Tf (z) dz = 0 for a triangle T ⊂ U .

We only need to consider the case where f is not holomorphic at a single point (since just worklocally about each point), and so let S = {a} ⊂ T lie in the interior of some triangle.

Suppose that we have a smaller triangle T ′ ⊂ T with a ∈ Int(T ′). Then for all other triangles in oursubdivision (shown in Figure 23) we know

∂ (triangle)f (z) dz = 0

as the triangle not equal to T ′ lies in a region where f is holomorphic (and then we can just do asbefore). Note then that this implies:

∂ Tf (z) dz =

∂ T ′f (z) dz if a ∈ T ′ ⊂ T

26


FIGURE 23. Dividing a triangle about a singular point. The integrals around eachblue triangle at all 0.

by choosing a suitable path around the triangle boundaries. This simply says that we can shrink thetriangle to a smaller one containing a without changing anything. But in general we have

∂ T ′f (z) dz

≤ length(∂ T ′) · supz∈∂ T ′| f (z)|

and as f is bounded (as it is continuous) we can divide T ′ and repeat, giving length(∂ T ′)→ 0 andso

∂ Tf (z) dz = 0

as required.

From here as mentioned above it is straightforward to conclude the general case (when S is not justa point) as we can just do the above for each a ∈ S.

□

Theorem 2.3 (Cauchy’s Integral Formula). Let U be a domain and let f : U → be holomorphic.Suppose Br(z0) ⊂ U for some z0 and r > 0. Then for all z ∈ Br(z0):

f (z) =1

2πi

∂ Br (z0)

f (w)w− z dw.

Proof. Firstly recall that we know (from Example 2.1) that∂ B1(0)

1z dz = 2πi.

Since U is open, ∃δ > 0 such that Br+δ(z0) ⊂ U . Fix z ∈ Br(z0). Then define g : Br+δ(z0)→ via:

g(w) =

f (w)− f (z)

w−z if w ∕= zf ′(z) if w= z.

Note that g is holomorphic as a function of w ∈ Br+δ(z0), except perhaps at w = z. But as f isholomorphic we know g is continuous everywhere on Br+δ(z0) (as it is therefore also continuous atz). So the addendum of the convex Cauchy theorem says:

∂ Br (z0)g(w) dw= 0 =⇒

∂ Br (z0)

f (w)w− z dw=

∂ Br (z0)

f (z)w− z dw.

27


FIGURE 24. Finding δ > 0 with Br+δ(z0) ⊂ U .

Now rewrite:

1w− z =

1w− z0

· 11− z−z0w−z0

=1

w− z0

∞

n=0

(z − z0)n(w− z0)n

=∞

n=0

(z − z0)n(w− z0)n+1

via Taylor expanding 11− z−z0w−z0

, noting that this converges uniformly on ∂ Br(z0) since for w on this

circle we have z−z0w−z0< 1 (as z ∈ Br(z0)).

So, by uniform convergence we can interchange the summation and integration, and so:

∂ Br (z0)

f (z)w− z dw=

∞

n=0

∂ Br (z0)f (z) · (z − z0)

n

(w− z0)n+1dw.

But we know that∂ Br (z0)

(w−z0)k dw= 0 unless k = −1, and then it is 2πi. So hence only the n= 0term survives, and so

∞

n=0

∂ Br (z0)f (z) · (z − z0)

n

(w− z0)n+1dw= 2πi f (z)

which gives the result.

□

Corollary 2.2 (Local Maximum Principle). Let f : Br(z) → be holomorphic, and suppose| f (w)|≤ | f (z)| for all w ∈ Br(z). Then f is constant.

i.e. a non-constant holomorphic function cannot achieve an interior local maximum.

Proof. Let 0< ρ < r. Applying the Cauchy Integral Formula (Theorem 2.3) we have

| f (z)|=

12πi

∂ Bρ(z)

f (w)w− z dw

=

1

0

fz +ρe2πiθ

dθ

via setting w= z +ρe2πiθ

≤ sup|z−w|=ρ

| f (w)| · length([0, 1])

≤ | f (z)|

28


by assumption. So hence we need all the inequalities in the above to be inequalities. But we haveinequality in our integral inequality if and only if the integrand is constant (see Lemma 2.1), i.e.| f (z+ρe2πiθ )| is constant on [0, 1]. This is just saynig that | f (w)| is constant on the circle |z−w|= ρ.But then this is true for all 0< ρ < r, and so | f | is constant on each circle. By continuity of f theseconstants must all be the same, and so | f | is constant on Br(z).

Now writing f (w) f (w) = | f (w)|2 = constant, the Cauchy-Riemann equations imply that f itself isconstant here [see Example Sheet 1].

□

Remark: The local maximum principle is an extremely useful result, and occurs in many differentareas of mathematics (particularly in the study of elliptic PDEs).

We now give another proof of the Cauchy Integral formula (Theorem 2.3) before moving onto someconsequences of these results.

2nd Proof of Cauchy’s Integral Formula. The second strategy of the proof is to change the contour ofintegration instead of changing the integrand (which is what we did in the first proof).

Given > 0, pick δ > 0 such that Bδ(z) ⊂ Br(z0), and such that if |w−z|= δ, then | f (w)− f (z)|< (this is possible since f is uniformly continuous on a (compact) neighbourhood of z).

We consider the contours shown in Figure 25. f (w)w−z is holomorphic on discs which are sufficientlysmall open neighbourhoods of the half contours indicated (i.e. the two curves shown).

FIGURE 25. An illustration of how to change the integral over the larger ball to oneover the small ball about z. By the convex Cauchy theorem, the integrals over theorange and blue contours are zero.

By dividing once more if necessary, the integral of f (w)w−z around the half-contours vanish by Cauchy’stheorem (Corollary 2.1). Adding these integrals together, we get:

∂ Br (z0)

f (w)w− z dw=

∂ Bδ(z)

f (w)w− z dw

where the boundaries of the balls are oriented anticlockwise (the other lines cancel as they are insame but with opposite orientations).

29


Now, f (z)−1

2πi

∂ Br (z0)

f (w)w− z dw= f (z)−

12πi

∂ Bδ(z)

f (w)w− z dw

=

1

2πi

∂ Bδ(z)

f (z)− f (w)w− z dw since

∂ Bδ(z)

1w− z dw= 2πi

≤ 12π· length(∂ Bδ(z)) · sup

w:|w−z|=δ

f (z)− f (w)

w− z

≤ 12π· 2πδ ·

δ= .

Then taking → 0 we see that the Cauchy Integral Formula holds.

□

Note: Given piecewise C1 closed paths ϕ,ψ : [0, 1] → U , we say that ψ is an elementary defor-mation of ϕ if ∃ convex open sets C1, . . . , Cn ⊂ U and 0 = x0 < x1 < · · · < xn = 1 such that on[x i−1, x i] ⊂ [0, 1], ϕ and ψ have images in Ci .

If this is true, then if f is holomorphic on U , by considering the shaded square in Figure 26 we have

ϕ

f (z) dz =

ψ

f (z) dz

when ϕ and ψ are in elementary deformation (this is essentially saying if we can piecewise contin-uously deform ϕ into ψ in a convex set without leaving U , then the integral along ϕ is the same asthat along ψ for holomorphic f ).

FIGURE 26. An illustration of how the integral over curves which are elementarydeformations of one another is the same. By applying the convex Cauchy theoremto the blue region, we get the integral over the blue region is zero. This relates theintegral along each part of the curve, but we need to remove the integral over thevertical lines. This happens when we sum over i, as then the vertical parts cancelsas we include them twice, once with each orientation.

30


2.2. Liouville’s Theorem and Taylor’s Theorem.

Theorem 2.4 (Liouville’s Theorem). Suppose f : → is entire (i.e. holomorphic everywhere)and bounded. Then f is constant.

Proof. Suppose we have | f (z)| ≤ M for all z ∈ . We will estimate | f (z1) − f (z2)| for a fixed(arbitrary) pair z1, z2 ∈ .

Let R> 2 max{|z1|, |z2|}. Then by the Cauchy integral formula,

| f (z1)− f (z2)|=

12πi

∂ BR(0)

f (w)

w− z1− f (w)

w− z2

dw

=

1

2πi

∂ BR(0)

f (w)(z1 − z2)(w− z1)(w− z2)

dw

≤ 12π· length(∂ BR(0)) · sup

w∈∂ BR(0)

f (w)|z1 − z2||w− z1| · |w− z2|

≤ 12π· 2πR ·M · |z1 − z2|

(R/2) · (R/2)

=4M |z1 − z2|

Rsince for w ∈ ∂ BR(0) we have |w− zi |≥ R/2 for each i = 1, 2 by choice of R.

But R was arbitrary, and so letting R→∞ we get f (z1)− f (z2) = 0. So as z1, z2 were arbitrary in, we have that f is constant on .

□

Corollary 2.3 (The Fundamental Theorem of Algebra). A non-constant polynomial has a root in.

[Iterating this, every polynomial over can be written as a product of linear factors.]

Proof. Let p(z) = anzn + an−1zn−1 + · · ·+ a0 have an ∕= 0, n> 0 (so p is non-constant).

Being non-constant therefore means |p(z)|→∞ as |z|→∞. So in particular ∃R > 0 such that if|z|> R then |p(z)|> 1.

Suppose for contradiction that p(z) has no root in . Then we can consider:

f (z) :=1

p(z)

which is then entire on , as p(z) ∕= 0 for all z ∈ . But then on BR(0) f is certainly continuous andhence bounded (as this is compact). But on \BR(0) by construction |p(z)|≥ 1 and so | f (z)|≤ 1. So

31


f is in fact bounded on all of , and so by Liouville’s theorem f is constant. But p was not constant,and so we have a contradiction. So p must have a root in .

□

Remark: All proofs of the fundamental theorem of algebra involve analysis or topology. This is notsurprising since the construction of (and hence ) is intrinsically analytic and not algebraic(ii).

Recall that our first source of interesting holomorphic functions came from power series within theirradius of convergence. But is the converse true, i.e. do holomorphic functions give power series?This is exactly Taylor’s theorem.

Theorem 2.5 (Taylor’s Theorem). Let f : Br(a)→ be holomorphic. Then f has a convergentpower series representation:

f (z) =∞

n=0

cn(z − a)n for all z ∈ Br(a).

Moreover we have:

cn =f (n)(a)

n!=

12πi

∂ Bρ(a)

f (z)(z − a)n+1 dz

for any ρ ∈ (0, r).

Proof. To prove this we will again use the Cauchy integral formula. If |w−a|< ρ < r, then we knowthat

f (w) =1

2πi

∂ Bρ(a)

f (z)z − w dz.

Now (recalling the first proof of Cauchy’s Integral Formula), we have

1z − w =

1z − a ·

1

1−w−a

z−a =

∞

n=0

(w− a)n(z − a)n+1

sincew−a

z−a< 1, and moreover this series is uniformly convergent on |z−a|≤ ρ (as this is a compact

subset - see the recalled facts at the start of §1.4).

So due to the uniform convergence we can interchange summation and integration, and so substi-tuting this into Cauchy’s integral formula above we get:

f (w) =1

2πi

∂ Bρ(a)

f (z)(w− a)n(z − a)n+1 dz =

∞

n=0

cn(w− a)n

where

cn =1

2πi

∂ Bρ(a)

f (z)(z − a)n+1 dz.

(ii)i.e. we form from by forcing additive inverses - this is an algebraic structure. We then form from by forcingmultiplicative inverses - again an algebraic structure. Then we form from by using Cauchy sequences, which is ananalytic structure! To get to (and hence the fundamental theorem of algebra) we always need some sort of analyticstructure, and so we cannot hope to do it on purely algebraic structures.

32


So we have a convergent power series representation, and this works on any disc Bρ(a) ⊂ Br(a).

For the alternative expression for cn, just note that the power series is infinitely differentiable here(and hence so is f ) and so differentiating n times and setting z = a gives that also cn = f (n)(a)/n!.

□

Note: This shows that we have the following generalisation of Cauchy’s integral formula:

f (n)(a) =n!

2πi

∂ Br (a)

f (w)(w− a)n+1 dw.

Remark: Contrast Theorem 2.5 with functions like e−1

x2 on which have no power series (about0).

From Taylor’s theorem we immediately get holomorphic functions are infinitely differentiable.

Corollary 2.4. If f : Br(a)→ is holomorphic on a disc, then f is infinitely differentiable on thedisc.

Proof. Follows from Theorem 2.5 and Theorem 1.2. □

This corollary therefore justifies our claim from the start of the course about Re( f ) and Im( f ) beingharmonic functions.

Corollary 2.5. Suppose f : U → is complex-valued. Then:Re( f ) and Im( f ) satisfy the Cauchy-Riemann equations

f is holomorphic at p ∈ U ⇐⇒ and if we write f = u+ iv then ux , uy , vx , vy are continuousin an open neighbourhood of p.

Proof. (⇐): If the partial derivatives exist and are continuous in a neighbourhood of p, then (fromAnalysis II) we know u and v are differentiable as functions of 2. Then we proved in Proposition1.1 that this and the Cauchy-Riemann equations implied holomorphicity.

(⇒): If f is holomorphic, then we know by Corollary 2.4 that f ′(z) is also holomorphic, and so indeedux , uy , vx , vy are continuous, and they satisfy the Cauchy-Riemann equations from Proposition 1.1.

□

The following theorem provides a partial converse to the (convex) Cauchy theorem.

33


Corollary 2.6 (Morera’s Theorem). Let U ⊂ be a domain. Let f : U → be continuous andsuppose that:

γ

f (z) dz = 0 for all piecewise C1 closed curves γ ∈ U .

Then f is holomorphic in U.

Proof. From Proposition 2.1 we know that f has an antiderivative F : U → , i.e. a holomorphicfunction F such that F ′ = f . But then since F is holomorphic, from Corollary 2.4 we know that F isinfinitely differentiable, and hence F ′ = f is differentiable and so is holomorphic.

□

Note that Morera’s theorem does not require any geometric property of the domain U . This can bethought of because holomorphicity is a local property, and so we may just consider discs.

Corollary 2.7 (Corollary of Morera’s Theorem). Suppose U ⊂ is a domain and we have a se-quence ( fn)n of holomorphic functions in U. Suppose fn→ f uniformly in U. Then f is holomorphicin U, and f ′(z) = limn→∞ f ′n(z) for all z ∈ U

i.e. the uniform limit of holomorphic functions is holomorphic.

Proof. We essentially just use the fact that the values of integrals is preserved by uniform limits.

So given a piecewise C1 closed curve γ, uniform convergence tells us that:

γ

fn→

γ

f

as n→∞. Note that being holomorphic is a local condition, and so we can fix p ∈ U and work ona small (convex) disc B(p) ⊂ U (exists as U is open). Then for any such closed γ ⊂ B(p) we have

γ

fn(z) dz = 0

by the (convex) Cauchy theorem. So hence by the above we have

γ

f (z) dz = 0.

Hence as γ was an arbitrary such curve, Morera’s theorem gives that f is holomorphic on B(p).Hence as p was arbitrary, f is holomorphic on U .

□

34


2.3. Identity Theorems and Analytic Continuation.

Let f : Br(a)→ be holomorphic. Then we know that we can write (from Taylor’s theorem)

f (z) =∞

n=0

cn(z − a)n

as a complex power series. We have two cases:

• Either cn = 0 for all n, and f ≡ 0 on Br(a), or• There is a least N such that CN ∕= 0. Then N =min{n | f (n)(a) ∕= 0}. If N > 0 we say that f

has a zero of order N at a.

In this case we can write f (z) = (z − a)N g(z) on Br(a), where g(a) = cN ∕= 0.

Lemma 2.2 (Principle of Isolated Zeros). Let f : Br(a)→ be holomorphic and not identicallyzero. Then, ∃ρ ∈ (0, r) such that f (z) ∕= 0 in Bρ(a)\{a},

i.e. if a non-zero, holomorphic function has a zero at a, it non-zero on a punctured neighbourhoodof a.

Proof. If f (a) ∕= 0, then the result is obvious by continuity of a.

Otherwise, if f has a zero of order N at a, we can write f (z) = (z − a)N g(z) with g(a) ∕= 0 and gholomorphic (as above). Then by continuity of g, g(z) ∕= 0 on some Bρ(a) and so f ∕= 0 here as well.[If N =∞ then f is identically zero.]

□

Definition 2.6. We say that a point w ∈ S is a non-isolated point of S if ∀ > 0, S∩B(w)\{w} ∕=,

i.e. if S contains points arbitrarily close to w.

Corollary 2.8 (The Identity Theorem). Let U ⊂ be a domain and let f , g : U → be holomor-phic. Let S = {z ∈ U : f (z) = g(z)}.

Then if S contains a non-isolated point, then f = g on U.

Proof. Let h := f − g. If S has a non-isolated point w, then h has a non-isolated zero at w, i.e. ∀ > 0,S ∩ B(w)\{w} ∕= .

35


Hence applying the contrapositive of Lemma 2.2 to h, we see that ∃ρ > 0 such that h ≡ 0 onBρ(w) ⊂ U (i.e. because we can never find a disc where h is non-zero on the disc, it must beidentically zero on some disc).

Now set:U0 = {a ∈ U : h≡ 0 on some neighbourhood Bρ(a) of a in U}U1 = {a ∈ U : ∃n≥ 0 such that h(n)(a) ∕= 0}.

Then clearly U0 ∩ U1 = and U0 ∪ U1 = U . U0 is open by definition, and U1 is open since h(n)(z)is continuous near any given a ∈ U1. But since U is connected (as it is path connected and this isequivalent to connectedness on ) we must have that one of U0 or U1 is empty. But w ∈ U0 by theabove, and so we must have U1 = , U0 = U , i.e. h(z)≡ 0 on U , and so f ≡ g on U .

□

Definition 2.7. Let U0 ⊂ U ⊂ be domains and let f : U0→ be holomorphic. Then a holomor-phic function h : U → such that h|U0 = f is called an analytic continuation of f .

Remark: The identity theorem implies that such an analytic continuation h onto U is unique if itexists. Put differently, given f : Br(a)→ holomorphic, we can ask for the largest domain U ⊃ Br(a)on which an analytic continuation of f exists (we could take U = Br(a), but this isn’t very exciting).

Example 2.4. Let f (z) =∞

n=0 zn = 1 + z + z2 + · · · , which is holomorphic on B1(0). Then by

summing the series, we also know that f (z) = 11−z on B1(0). So we see that f has an analyticcontinuation to \{1}, that being h(z) = 11−z .

Example 2.5. If f (z) =

n≥0 z2n , then this converges on B1(0), but has no analytic continuation

to any larger domain (see Example Sheet 2).

Example 2.6. Define the zeta-function via ζ(z) :=∞

n=1 n−z . This defines a holomorphic function

on {z : Re(z) > 1} ⊂ . Indeed to see that, just note that |nz | = |nRe(z)|, and

n≥1 n−t converges

for t > 1.

Now since

n≥1 n−t converges uniformly on compact subsets of >1, we see by Morera’s theorem

that ζ(z) is holomorphic on Re(z)> 1.

Then it is a fact that ζ(z) has an analytic continuation to \{1}. At least formally, we can show

ζ(z) =

primes p

1+ p−z + p−2z + · · ·

=

primes p

1

1− p−z

by the fundamental theorem of arithmetic. The behaviour of ζ(z) as z → 1 can then be used toshow that there are infinitely many primes.

36


3. SINGULARITIES AND THE CALCULUS OF RESIDUES

3.1. Singularities.

Proposition 3.1 (Removal of Singularities). Let U be a domain and z0 ∈ U. Then if f : U\{z0}→ is holomorphic, and if f is bounded near z0, then ∃a such that f (z)→ a as z→ z0.

Furthermore, if we define

g(z) =

f (z) if z ∈ U\{z0}a if z = z0

then g is holomorphic on all of U.

Remark: This proposition essentially says that if a function is holomorphic on an open punctureddisc, the only thing can go wrong is if f is discontinuous at the puncture (or unbounded), i.e. if wedefined f in a silly way, such as

f (z) =

0 at 01 elsewhere.

So this g we define in Proposition 2.3 fixes this problem, and is the “correct definition” for holomor-phicity.

Proof. Let h : U → be defined by:

h(z) =

(z − z0)2 f (z) if z ∕= z00 if z = z0.

We define this h because as f is bounded, the behaviour near z0 is cancelled out by the (z − z0)2term.

Clearly h is holomorphic away from z0, and at z = z0,h(z)− h(z0)

z − z0

≤ M |z − z0|

where M is such that supBρ(z0) | f |≤ M for some ρ > 0. So in fact we see that h is also differentiableat z0, and h(z0) = 0= h′(z0).

So near z0, h(z) has a Taylor series, say h(z) =

n≥0 an(z − z0)n. Since h(z0) = 0 = h′(z0) we havea0 = a1 = 0, and so we can write

h(z) = (z − z0)2 g(z)where g(z) =

n≥0 an+2(z − z0)n is defined on some ball Bρ̃(z0) where the Taylor series for h isdefined.

So by construction we have:

• g(z) = f (z) on Bρ(z0)\{z0}• g(z)→ a2 as z→ z0, so g has a well-defined limit.

37


So since g agrees with f on Bρ̃(z0)\{z0}, we have f (z) → a2 as z → z0 as well. We also have gdefined as we required, and g is holomorphic since it is a power series and h was holomorphic. Sothe result follows.

□

Let U ⊂ be a domain and z0 ∈ U . Suppose f : U\{z0}→ is holomorphic. Proposition 3.1 tellsus that if f is bounded near z0, then in fact f admits a holomorphic extension over z0. Next weconsider the case when f is not bounded near z0.

Proposition 3.2. Let U be a domain and z0 ∈ U. Suppose f : U\{z0} → is holomorphic.Suppose | f (z)|→∞ as z → z0. Then ∃! k ∈ ≥1 and a unique holomorphic function g : U → such that g(z0) ∕= 0 and

f (z) =g(z)

(z − z0)kfor all z ∈ U\{z0}.

Proof. Pick δ > 0 such that | f (z)| ≥ 1 for all z ∈ Bδ(z0)\{z0}, i.e. a punctured neighbourhood of z0.Then define h : Bδ(z0)→ by:

h(z) =

1

f (z) if z ∈ Bδ(z0)\{z0}0 if z = z0.

Clearly h is holomorphic on Bδ(z0)\{z0} as f is, and h is bounded as z→ z0, since | f |≥ 1 here.

So by Proposition 3.1, as h(z0) = 0, ∃! integer k ≥ 1 such that h has a zero of order k at z0. Thereforewe have

h(z) = (z − z0)k l(z)for some l : Bδ(z)→ holomorphic with l(z0) ∕= 0. So by continuity of l, ∃ some 0 < < δ suchthat l(z) ∕= 0 for all z ∈ B(z0), and hence we can consider 1l here. So define g : B(z0)→ by

g(z) :=1

l(z).

Then g is holomorphic on B(z0) since l is. Note also that, at least away from z0 we have

g(z) =1

l(z)=(z − z0)k

h(z)= f (z)(z − z0)k.

This expression shows that g admits an analytic continuation from B(z0) to the whole of U , asf (z)(z − z0)k is well-defined on U\{z0}. Therefore this g does what we desire.

□

Using Proposition 3.1 and 3.2 above, we can classify the different types of behaviour f can have ata singularity (note that a singularity is just a point in the domain where f is not defined).

Definition 3.1. Given a domain U, z0 ∈ U, and f : U\{z0}→ holomorphic, we say that z0 isan isolated singularity of f . We also say:

38


(i) z0 is a removable singularity if f is bounded near z0,

(ii) z0 is a pole if | f (z0)|→∞ as z→ z0,(iii) z0 is an isolated essential singularity if neither (i) or (ii) holds.

Note: Due to Proposition’s 3.1 and 3.2 we know that:

(i) If f has a removable singularity at z0, we may as well just define f at z0 via its limit asz→ z0. By Proposition 3.1 we know this will be holomorphic at z0.

(ii) If f has a pole at z0 then by Proposition 3.2 we know that ∃! k ∈ ≥1 and g holomorphicwith g(z0) ∕= 0 with f (z) = (z − z0)−k g(z) near z0. In this case we say that k is the order ofthe pole at z0. If k = 1, we say that the pole is a simple pole.

Note: These definitions depend on f and not necessarily the point. If f : \{0}→ is defined byf (z) = z, 0 is still an isolated singularity of f . But can we “fix” this? If the singularity is a removableone, this is easy to fix as we can just define our function at the singularity as what it “should” be.

Example 3.1. Consider f (z) := e1/z . This has an isolated essential singularity at z = 0. Indeed,clearly f is not bounded near z = 0 (as f (1/n) = en→∞), but it is not true that |e1/z |→∞ asz→ 0, as if we take z purely imaginary then |e1/z |= 1 always.

Remark: If f : B(z0)\{z0} → has a pole of order k at z0, then f naturally defines a map f̃ :B(z0)→ ∞ ≡ ∪ {∞} (i.e. the Riemann sphere) via z0 →∞ and if z ∕= z0, f̂ (z) = f (z). In thiscase, in a local coordinate w= 1/z on ∞, the function has a zero of order k at w= 0.

So what we are saying is that, just like how removable singularities aren’t really singularities on ,poles are also not really singularities from the viewpoint of the Riemann sphere (as∞ is no differentto any other point).

Definition 3.2. If U is a domain and S ⊂ is a finite or discrete set, a function f : U\S → which is holomorphic and has (at worst) poles on S is said to be meromorphic on U.

So meromorphic functions are those which would be well-defined holomorphic maps onto the Rie-mann sphere ∞.

Example 3.2. A rational function P(z)/Q(z) with P,Q polynomials is holomorphic on\{zeroes of Q}, but is meromorphic on .

So far we have looked at the behaviour of holomorphic functions near removable singularities andpoles. So what happens if there is an isolated essential singularity? It turns out that the image onany neighbourhood of such a point is dense in .

39


Theorem 3.1 (Casorati-Weierstrass Theorem). Let U be a domain and z0 ∈ U. Suppose f :U\{z0}→ has an isolated essential singularity at z0. Then for all w ∈ , ∃ a sequence zn → zsuch that f (zn)→ w,

i.e. on any punctured neighbourhood B(z0)\{z0}, the image of f is dense in .

Proof. An exercise on Example Sheet 2. □

Note that this does not guarantee that the function takes every value in . For example, e1/z nevertakes the value 0, even though it has an isolated essential singularity. However there is a strongerresult than Casorati-Weierstrass:

Theorem 3.2 (Picard’s Theorem). If f has an isolated singularity at z0, then ∃b ∈ such thaton each punctured neighbourhood B(z0)\{z0}, the image of f contains \{b}.

Proof. None given - beyond the scope of the course. □

So what Picard’s Theorem says is that at worst, the function can only miss at most one point in about an isolated essential singularity.

Recall that we know from Taylor’s theorem that if f is holomorphic about z0, then we have a localpower series expansion f (z) =

∞n=0 cn(z− z0)n about z0. Our next result gives us an analogue near

an isolated singularity.

Theorem 3.3 (Laurent Series). Let 0 ≤ r < R


(iii) If cn ∕= 0 for arbitrary large negative values of n, this is the case of an (isolated) essentialsingularity at a.

Remark: The Laurent theorem says that if f is holomorphic on an annulus A, then we can write

f (z) = fin(z) + fout(z)

where:

• fin is holomorphic on |z − a| < R (this corresponds to the n ≥ 0 terms in the Laurent seriesexpansion)

• fout is holomorphic on |z − a| > r (this corresponds to the n < 0 terms in the Laurent seriesexpansion).

Geometrically, each is holomorphic on one of the two hemispheres on the Riemann sphere, as shownin Figure 27.

FIGURE 27. Riemann sphere viewpoint of Laurent series.

Proof of Laurent Series. THe proof is essentially a blend of our two proofs for the Cauchy integralformula.

Let w ∈ A, and let r < ρ′ < |w− a| < ρ′′ < R. Then consider contours γ̃ and ˜̃γ in A which look asshown in Figure 28. If we apply the Cauchy integral formula to each of these contours we get

f (w) =1

2πi

γ̃

f (z)z − w dz and 0=

12πi

˜̃γ

f (z)z − w dz

(where the second integral is 0 since w is not in ˜̃γ and so the integrand is holomorphic). So addingwe see:

f (w) =1

2πi

γ̃

f (z)z − w dz +

12πi

˜̃γ

f (z)z − w dz

=1

2πi

∂ Bρ′′ (a)

f (z)z − w dw−

12πi

∂ Bρ′ (a)

f (z)z − w dw

since moving around γ̃ and ˜̃γ is the same as moving around both of these contours (other parts canceldue to opposite orientation). We have a negative sign on the second integral as we travel around theinner circle in the opposite orientation (as we travel around circles anticlockwise usually).

41


FIGURE 28. An illustration of the contours we take in the proof of the Laurent seriesformula. Note the due to opposite orientations, the straight parts of each contourcancel out. Note how w does not lie in the region with boundary ˜̃γ.

Now as in the first proof of the Cauchy integral formula, we use the expansions:

• First integral: 1z−w = 1z−a · 11−( w−az−a ) =∞

n=0(w−a)n(z−a)n+1 , which is uniformly convergent for z ∈

∂ Bρ′′(a) as then |w− a|< ρ′′ and |z − a|= ρ′′ and sow−a

z−a< 1.

• Second integral: − 1z−w = 1w−a · 11−( z−aw−a ) =∞

m=1(z−a)m−1(w−a)m , which is uniformly convergent for

z ∈ ∂ Bρ′(a) since then |w− a|> ρ′ and |z − a|= ρ′ and so z−a

w−a< 1.

Inputting these and using the uniform convergence to swap the order of integration and summation,we get:

f (w) =∞

n=0

1

2πi

∂ Bρ′′ (a)

f (z)(z − a)n+1 dz

(w− a)n +

∞

m=1

1

2πi

∂ Bρ′ (a)

f (z)(z − a)1−m dz

(w− a)−m.

Now substitute n= −m in the second sum to get:

f (w) =∞

n=−∞c̃n(w− a)n

for integrals c̃n as above. So we are now almost done: the only thing left is that current our integralsc̃n are around circles of radii ρ

′ and ρ′′, whereas we want just one circle for all the integrals.

But for any r < ρ < R, these two ρ′ and ρ′′ radius circles are just convex deformations of the|z − a| = ρ circle inside the annulus A (recall the remark about elementary deformations at the endof §2.1). Thus,

∂ Bρ(a)

f (z)(z−a)n+1 dz is independent of ρ for ρ ∈ (r, R), and so we can take each c̃n to

be this and so we are done.

□

42


Notation: If f : Br(a)\{a} is holomorphic, and if f has a Laurent series f (z) =∞

n=−∞ cn(z − a)n,then the principal part of f at a is:

fprincipal(z) :=−1

n=−∞cn(z − a)n.

So f − fprincipal is holomorphic near a, and fprincipal carries the information about what kind of sin-gularity f has at a.

Remark: If f : Br(a) → is holomorphic with Taylor series f (z) =∞

n=0 cn(z − a)n then we sawthat:

cn =1

2πi

∂ Bρ(a)

f (z)(z − a)n+1 dz =

f (n)(a)n!

and the second expression in particular shows that the coefficients cn are uniquely determined by f .But what about those of a Laurent series, or even just those in the principal part?

Lemma 3.1. Suppose f : A → is holomorphic and A = {z : r < |z − a| < R} is an annulus.Suppose f has Laurent series

f (z) =∞

n=−∞cn(z − a)n

on A. Then the cn are uniquely determined by f .

Proof. Suppose we also had f (z) =∞

n=−∞ bn(z − a)n for some bn’s. Then we know:

2πick =

∂ Bρ(a)

f (z)(z − a)k+1 dz from the Laurent series proof

=

∂ Bρ(a)

n∈bn(z − a)n−k−1

dz

=

n∈bn

∂ Bρ(a)(z − a)n−k−1 dz by uniform convergence on compact sets

= 2πi bkwhere in the last line we have used the fact that the only term which survives is when n= k.

So hence we see ck = bk for all k, and hence the Laurent series coefficients are uniquely determinedby f (and hence the Laurent series is also unique).

□

Example 3.3. Let sin(z) = z − z33! + z5

5! − · · · , which defines an entire function (as this series hasinfinite radius of convergence).

Now consider cosec(z) := 1sin(z) , which is holomorphic except at the zeros of sin(z), i.e. except atz = kπ for k ∈ . So this is holomorphic on a punctured neighbourhood of 0, and hence has a

43


Laurent series near z = 0. Using the power series for sin(z) we get:

cosec(z) =1

z1− z23! + z

4

5! − · · · = 1

z

1+

z2

3!+O(z4)

where we have used the expansion of (1− z)−1 (and this must be the Laurent series by uniquenessof the Laurent series). So hence the Laurent series has cn = 0 for all n ≤ −2, c−1 = 1, c1 = 1/3!,etc. So cosec(z) has a simple pole at z = 0.

By periodicity we can see that cosec has simple poles at each z = kπ, k ∈ (as cosec looks the sameabout all of these points).

Example 3.4. From the series expansion of sin we have:

sin(1/z) =1z− 1

3!·

1z

+

15!·

1z5

− · · ·

and we know this is holomorphic on ∗. But we can see that cn ∕= 0 for infinitely many n < 0, sosin(1/z) has an isolated essential singularity at z = 0.

Example 3.5. cosec(1/z) has singularities at z = 1kπ , k ∈ . So this is not holomorphic in anypunctured neighbourhood of 0 (i.e. any Br(0)\{0}) as there will always be a singularity within theneighbourhood Br(0)\{0}. So cosec(1/z) has a non-isolated singularity at 0, and hence has noLaurent series in a neighbourhood of 0.

Definition 3.3. A non-isolated singularity z ∈ is one where ∃ no punctured neighbourhoodabout z such that f is holomorphic in that region, i.e. f has singularities arbitrarily close to z.

A major application of Laurent series is in the exact summation of complicated series:

Example 3.6 (Series Summation). In this example, we shall show that

(a) f (z) =∞

n=−∞1

(z−n)2 is holomorphic on \, and

(b) f (z) = π2

sin2(πz).

(a): For w ∈ \ we can use the comparison test with∞

n=11n2 to show convergence. Indeed, pick

r > 0 such that |w− n|> 2r for all n ∈ (i.e. w is at least 2r from every integer).

Then for all z ∈ Br(w) we have |z − n|≥max{r, n− |w|+ r}, and so hence:1

|z − n|2 ≤min{1r2

,1

(n− |w|− r)2 }=: Mn.

44


Then

n Mn converges by comparison with

n≥11n2 (as eventually always Mn =

1(n−|w|−r)2 ) and

so by the Weierstrass M-test, our series f (z) converges uniformly on Br(w). Then by our resultsaround Morera’s theorem, we see that f is a uniform limit of holomorphic functions (as each finitesum is holomorphic away from , so is holomorphic on Br(z)), and so f is holomorphic at w.

So as w ∈ \ was arbitrary, this shows f is holomorphic on \.

(b): To establish the alternate expression for f , first note that f has period 1, i.e. f (z + 1) = f (z)for all z ∈ \.

So therefore, as at z = 0 f has a pole of order two (as f (z) = 1z2 +(holomorphic thing) near z = 0),by periodicity f has a double pole at each k ∈ .

Note that from Example 3.3 we know that

1sin(πz)2

has a double pole at each k ∈ .

Consider now the principal parts of both of these functions at k ∈ . We know that f has principalpart 1(z−k)2 , and one sees that if g(z) =

π

sin(πz)

2, then limz→0 z

2 g(z) = 1, so g has principal part1z2 at z = 0, and hence by periodicity has principal part

1(z−k)2 at each k ∈ (can check as in

Example 3.3 that there is no (z − k)−1 term).

So g has the same principal part as f at each k ∈ . So f − g is holomorphic on \ and has noprincipal part at each k ∈ , and hence has at worst a removable singularity at each k ∈ . So ifwe define h := f − g we know h is entire.

By Liouville’s theorem, to show that h ≡ 0 (and hence that f ≡ g) it suffices to show that h isbounded and that h vanishes at a point (so that it is constant and then the constant must be zero).

To see that h is bounded, note that since f , g have period 1, it suffices to show that h is boundedon a vertical strip of width 1. So we focus our attention on the strip S = {z ∈ : −1/2 ≤ Re(z) ≤1/2} ⊂ , as shown in Figure 29.

On any box [−1/2, 1/2] × [−Y, Y ] we get a bound on h by continuity of h as this is a compactsubset of . Now if z = x + i y with |x |≤ 1/2 and (wlog) y > 0, then:

|g(z)|≤ 4π2

|eπy − e−πy |2 → 0 as y →∞

and

| f (z)|≤

n∈

1|x + i y − n|2 ≤

1y2+ 2

∞

n=1

1(n− 1/2)2 + y2 → 0 as y →∞.

So hence combining we see that we can choose Y sufficiently large so that |h| ≤ 1 on {z = x + i y ∈ : |x |≤ 1/2, |y|> Y }, and so as h is bounded on |x |≤ 1/2, |y|≤ Y , this shows that h is bounded onthe strip S. Hence by periodicity h is bounded on all of , and so is constant by Liouville’s theorem.

45


But since |h(z)|→ 0 as z →∞ (seen by the above bounds), the constant value of h is 0. So henceh≡ 0, and so f ≡ g, as we wanted. □

FIGURE 29. An illustration of the strip S we consider in Example 3.6.

The above Example 3.6 gives us lots of explicit values for series, such as∞

n=−∞

1(n− 1/2)2 = π

2

which upon rearranging gives

n odd

1n2=π2

8.

So if we set C =

n≥11n2 , we have

C =

n odd

1n2+∞

n even

1n2=π2

8+

14

C =⇒

n≥1

1n2= C =

π2

6

a result we know and love.

3.2. Calculus of Residues.

Definition 3.4. Let f : Br(a)\{a}→ be holomorphic with Laurent series

f (z) =∞

n=−∞cn(z − a)n.

Then the coefficient c−1 ∈ is called the residue of f at a, written Res( f , a) or Res f (a).

The idea here is that when we consider integrals of functions with a Laurent series around closedloops containing a, if we could exchange the sum with the integral the only term which survives is then = −1 term (recall Example 2.1). Hence knowledge of this coefficient is all we need to determineintegrals - this is exactly the idea behind the calculus of residues.

46


But more formally, if ρ < r, then we know that

∂ Bρ(a)f (z) dz = 2πic−1 since

∂ Bρ(a)(z − a)n dz ∕= 0 ⇐⇒ n= −1.

So we see:

Res f (a) =1

2πi

∂ Bρ(a)f (z) dz

(which we could take as the definition of the residue if we wanted, but it tends to be easier tocalculate the Laurent series to find the residue).

Now for a C1 curve γ as shown in Figure 30, we would see by the additivity of the integral (as wecan write γ= γ1 ∗ γ2 where the γi are the two individual circles) that

12πi

γ

f (z) dz = 2×Res f (a).

So clearly we need to be careful about how many times the curve we integrate over “circles around”the point a. We therefore want a general notion of how a given closed curve (not necessarily a simplecurve) “winds” around a point a not lying on the curve.

FIGURE 30. The curve wraps or “winds” around the point a twice.

So let γ : [a, b]→ be continuous and suppose w ∕∈ Image(γ), and that γ is closed, i.e. γ(a) = γ(b).Then we have the following:

Lemma 3.2 (Existence of winding number). There are continuous functions r : [a, b] → >0and θ : [a, b]→ such that:

γ(t) = w+ r(t)eiθ (t).

Proof. Clearly r(t) := |γ(t)− w| exists, is unique, and is continuous, since γ is.

To define θ (t), let us for simplicity translate so that w = 0. Moreover, replacing γ by γ(t)/r(t) asnecessary, we have wlog that |γ(t)|= 1 for all t (note that since w ∕∈ Im(γ) so r ∕= 0 for all t).

Recall now the principal branch of log, and hence of the argument Im(log), which takes values in(−π,π) and is defined on \≤0. Then if say Im(γ) ⊂ {z : Re(z)> 0}, then just let θ (t) = arg(γ(t))for arg the principal branch.

Similarly, if Im(γ) ⊂ {z : Re(z/eiα) > 0}, i.e. γ lies in a half-plane at angle α (as shown in Figure31), we can just define θ (t) = α+ arg

γ(t)eiα

.

So we have dealt with the cases when γ lies in a half-plane. In general, the idea is just to subdivideγ into regions where it lies in a half-plane and patch them together.

47


FIGURE 31. Half-planes on which we can easily define a continuous θ .

So since γ : [a, b]→ is continuous, it is uniformly continuous and hence we can find a subdivisionof [a, b], a = a0 < a1 < · · · < an = b such that, if s, t ∈ [ai−1, ai] then |γ(s)− γ(t)| <

2. Hence as

γ lies on the unit circle, this guarantees that γ(s) and γ(t) belong to one such half-plane as above.

So we can define θ j : [a j−1, a j]→ such that: γ(t) = eiθ j(t) for t ∈ [a j−1, a j], for each 1≤ j ≤ n−1.

Then θ j(a j) and θ j+1(a j) are both values of the argument for a fixed complex number (namely γ(a j)),and hence differ by an element of 2π. We can choose θ1 arbitrarily (for an arbitrary multiple of2πi), and then for j > 1, we can recursively re-define θ j such that the resulting map θ is continuous(i.e. use the value θ j−1(a j) to determine the starting value of θ j(a j) and go from there. Then defineθ (t) = θ j(t) for t ∈ [a j−1, a j]).

□

Equipped with this, we can define:

Definition 3.5. Given a continuous closed path γ : [a, b] → and w ∕∈ Image(γ), the windingnumber of γ about w is:

I(γ, w)≡ nγ(w) :=θ (b)− θ (a)

2πfor θ (t) as in Lemma 3.2.

So the winding number I(γ, w) simply tells us how many times γ wound around w.

Note: We always have I(γ, w) ∈ , since we have γ(a) = γ(b) describing the same point in , andso θ (b) and θ (a) differ by an integer multiple of 2π.

However we do need to check that γ is well-defined, since we made an arbitrary choice in the proofof Lemma 3.2 to define θ1. So suppose we have

γ(t) = r(t)eiθ1(t) = r(t)eiθ2(t)

for some continuous θ1,θ2 : [a, b]→ (not that we already know r is unique).

Then θ1−θ2 : [a, b]→ is continuous, but only takes values in the discrete set 2π (as the argumentat any point on γ can differ only by an integer multiple of 2π). So hence θ2 − θ1 is constant as a

48


function of t, and so θ2(b)− θ2(a) = θ1(b)− θ1(a). Hence the winding number of these agrees andso is well-defined.

We can also calculate the winding number from the following integral expression:

Lemma 3.3. Suppose γ : [a, b] → is a piecewise C1 smooth closed path with w ∕∈ Image(γ).Then:

I(γ, w) =1

2πi

γ

1z − w dz.

Proof. Let γ(t) = w+ r(t)eiθ (t), with now r,θ piecewise C1. Then we have:

γ

1z − w dz = b

a

γ′(t)γ(t)− w dt

=

b

a

r ′(t)r(t)

+ iθ ′(t)

dt

= [log(r(t)) + iθ (t)]ba= i(θ (b)− θ (a)) since γ is closed and so r(a) = r(b)= 2πi I(γ, w).

□

Remark: Sometimes the integral expression is given as the definition of the winding number, butthen one has to argue that I(γ, w) ∈ . However the integral definition does have its advantages.

For example, the expression 12πiγ

1z−w dz is continuous when considered as a function of w ∈

\Image(γ), since it is even holomorphic as a

Date post:	01-Feb-2021
Category:	Documents
Upload:	others
View:	0 times
Download:	0 times

Complex Analysis Lecturer: Ivan Smith · 2018. 12. 13. · Complex Analysis Lecturer: Ivan Smith...

Documents