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MA3B80

THE UNIVERSITY OF WARWICK

FOURTH YEAR EXAMINATION: Summer 2020

Complex Analysis

Time Allowed: 3 hours

Read all instructions carefully. Please note also the guidance you have received in

advance on the departmental ‘Warwick Mathematics Exams 2020’ webpage.

Calculators, wikipedia and interactive internet resources are not needed and are not

permitted in this examination. You are not allowed to confer with other people. You

may use module materials and resources from the module webpage.

ANSWER COMPULSORY QUESTION 1 AND TWO FURTHER QUESTIONS out of

the four optional questions 2, 3, 4 and 5.

On completion of the assessment, you must upload your answer to Moodle as a single

PDF document if possible, although multiple files (2 or 3) are permitted. You have

an additional 45 minutes to make the upload, and instructions are available on the

departmental ‘Warwick Mathematics Exams 2020’ webpage.

You must not upload answers to more than 3 questions, including Question 1. If you

do, you will only be given credit for your Question 1 and the first two other answers.

The numbers in the margin indicate approximately how many marks are available for

each part of a question. The compulsory question is worth twice the number of marks

of each optional question. Note that the marks do not sum to 100.

COMPULSORY QUESTION

b :=

1. (a) (i) Define a Möbius transformation mapping the extended complex plane C

b

C ∪ {∞} to C.

[3]

(ii) Find a Möbius transformation that maps the left half-plane HL := {z ∈

C : R(z) < 0} onto the unit disk ∆ := {z ∈ C : |z| < 1}, and find a
Möbius transformation that maps ∆ onto the right half-plane HR := {z ∈
C : R(z) > 0}.

[4]

(b) (i) Give sufficient conditions on D ⊂ C and the curve γ that ensure that for

1

Question 1 continued overleaf

MA3B80

every holomorphic function f : D → C we have

Z

(♣)

f (z) dz = 0.

γ

[4]

(ii) Give an example of an open and connected set D, a closed piecewise C 1

curve γ : [a, b] → C, and a holomorphic function f : D → C such that (♣)

in part (i) does not hold (no proof required).

(c) (i) State the Removable Singularity Theorem (no proof required).

[3]

[4]

(ii) Show that for every holomorphic function f : D → C, D ⊂ C open, and

(z0 )

every z0 ∈ D the function F (z) = f (z)−f

defined on D {z0 } can be

z−z0

extended to a holomorphic function on all of D.

[3]

(d) (i) Let f : D → C, D ⊂ C open, be holomorphic. Show that the imaginary

part I(f ) is harmonic on D.

[4]

(ii) Give an example of an open and connected set D ⊂ C and a harmonic

function h : D → C such that there is no holomorphic function f : D → C

with I(f ) = h (no proof required).

[3]

(e) (i) State Cauchy’s Integral formulas for the derivatives of holomorphic functions.

[2]

(ii) Compute

Z

∂B2 (0)

z2

1

dz,

−1

∂B2 (0) = {z ∈ C : |z| = 2}.

[4]

(f) (i) State the Riemann Mapping Theorem.

[3]

(ii) Show that there is no holomorphic and bijective mapping from ∆ := {z ∈

C : |z| < 1} onto C.
[3]
OPTIONAL QUESTIONS
2. (a) (i) Define the four elementary Möbius transformations.
2
Question 2 continued overleaf
[4]
MA3B80
(ii) Show that every Möbius transformation f ,
f (z) :=
az + b
,
cz + d
a, b, c, d ∈ C,
can be written as a composition of the four elementary Möbius transformations in (i).
[2]
(b) Denote H+ := {z = x + iy ∈ C : y > 0} the upper half-plane and ∆ := {z ∈

C : |z| < 1} the unit open disk.
(i) Show that the Möbius transformation h,
z − z
0
h(z) := eiθ0
, θ0 ∈ R ,
z − z̄0
maps H+ onto ∆ and the real line onto ∂∆ = {z ∈ C : |z| = 1} if z0 ∈ H+ .
[5]
(ii) Find the Möbius transformation g such that g(H+ ) = ∆ and g(i) = 0 and
g(∞) = −1.
[3]
(c) (i) Find the Möbius transformation f with
f
5
5
= 0, f (−4) = ∞, and f (0) = − .
2
4
[4]
(ii) Find all the fixed points of the Möbius transformation f in (i).
3. (a) Define the winding number (= index) of a curve about a point and the residue
of a function in a point. State the Residue Theorem.
[2]
[4]
(b) Let D ⊂ C be open and connected.
(i) Suppose that a holomorphic function f : D → C has an isolated singularity
of finite order n ∈ N in z0 . Show that
n
dn−1
1
z
−
z
f
(z)
.
Res (f, z0 ) =
0
(n − 1)! dz n−1
z=z0
[4]
g(z)
(ii) Suppose that f (z) = h(z)
with g, h : D → C holomorphic on D and h has a
zero at z0 ∈ D of order 1. Show that
Res (f, z0 ) =
g(z0 )
.
h0 (z0 )
[2]
3
CONTINUED
MA3B80
(c) Calculate
R∞
0
1
(1+x3 )
dx.
[8]
(d) Let γ : [0, 2π] → C be given by γ(t) =
denotes the Riemann zeta function.
1
2
+ 3ei3t . Calculate
R
γ
ζ(z) dz, where ζ
[2]
4. (a) Show that the sum
h(z) =
X
n∈Z
1
(z − n)2
describes a meromorphic function on C. Give the poles of h as well as the orders
of the poles.
[6]
(b) Show that the holomorphic function
π2
g(z) =
− h(z),
sin(πz)2
defined on C Z, can be extended to an entire function.
[5]
(c) Show that uniformly in x ∈ R we have
lim g(x + in) = 0
|n|→∞
Conclude that g is zero on all of C.
[5]
(d) Show that
∞
1
π2 X
=
.
2
8
(2n
−
1)
n=1
[4]
5. (a) State the Argument principle.
[4]
(b) Prove the following theorem (Rouché’s Theorem) using the Argument principle.
Theorem. Let D ⊂ C be open and connected and A ⊂ D be open, with piecewise C 1 boundary ∂A such that A ∪ ∂A ⊂ D. Let f, g : D → C be homomorphic
on D with |g(z)| < |f (z)| for all z ∈ ∂A. Then f and f + g have the same
number of zeros in A (counting multiplicity).
[6]
(c) Show that the polynomial f , defined by f (z) = z 4 − 5z + 1, has a zero z1 ∈ C
with |z1 | < 14 , and that the three remaining zeros zj , j = 2, 3, 4, satisfy
3
15
< |zj | <
.
2
8
[6]
(d) Show that the equation z sin(z) = 1, z ∈ C, has only real solutions.
4
[4]
END
MA3B8 Complex Analysis
University of Warwick
Spring 2021
Peter Topping
16 March 2021
https://homepages.warwick.ac.uk/˜maseq
Please alert me to any errors or typographical mistakes, however small, at:
p.m.topping@warwick.ac.uk
Contents
0
Preface
1
Review - Analysis 3 - portion 1
1.1 Complex differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Product and chain rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
5
7
10
2
Geometry of the complex plane
2.1 The Riemann sphere . . . . . . . . . . .
2.2 Stereographic projection . . . . . . . . .
2.3 Möbius transformations . . . . . . . . . .
2.4 P SL(2, C) . . . . . . . . . . . . . . . .
2.5 Decomposition of Möbius transformations
2.6 Three points determine a Möbius map . .
2.7 The cross ratio . . . . . . . . . . . . . . .
2.8 Examples of Möbius transformations . . .
2.9 Conformal maps . . . . . . . . . . . . . .
2.10 Exercises . . . . . . . . . . . . . . . . .
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11
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15
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19
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22
25
Review - Analysis 3 - portion 2
3.1 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Definitions of ez , sin(z), cos(z) etc. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
28
29
3
4
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CONTENTS
3.3
3.4
3.5
3.6
4
5
6
7
8
Euler’s formula; arg(z) and log(z) . . .
Complex integration . . . . . . . . . .
Anti-derivatives; baby Cauchy’s theorem
Exercises . . . . . . . . . . . . . . . .
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30
32
34
36
Winding numbers
4.1 Winding numbers of continuous closed paths . . .
4.2 Nearby closed paths have the same winding number
4.3 Winding number and simply connected domains . .
4.4 The winding number as an integral . . . . . . . . .
4.5 Exercises . . . . . . . . . . . . . . . . . . . . . .
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38
38
39
40
41
44
Cauchy’s Theorem
5.1 Preamble and connection with Analysis 3 . . . . .
5.2 Goursat’s theorem - Cauchy’s theorem on triangles
5.3 Goursat’s conclusion gives us an anti-derivative . .
5.4 Cauchy’s theorem on star-shaped domains . . . . .
5.5 Cauchy’s theorem on annuli . . . . . . . . . . . .
5.6 Cauchy’s integral formula . . . . . . . . . . . . . .
5.7 Exercises . . . . . . . . . . . . . . . . . . . . . .
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45
45
46
47
49
49
51
53
Review - Analysis 3 - portion 3
6.1 Taylor series - main result . . . . . . . . . .
6.2 Taylor’s theorem - proof . . . . . . . . . .
6.3 Basic consequences of Taylor’s theorem . .
6.4 Local invertibility of holomorphic functions
6.5 Exercises . . . . . . . . . . . . . . . . . .
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55
55
56
57
58
60
Further applications of Taylor Series
7.1 Morera’s theorem . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 The Schwarz reflection principle . . . . . . . . . . . . . . . . .
7.3 Zeros of holomorphic functions I . . . . . . . . . . . . . . . . .
7.4 The identity theorem . . . . . . . . . . . . . . . . . . . . . . .
7.5 Zeros of holomorphic functions II . . . . . . . . . . . . . . . .
7.6 Open mapping theorem . . . . . . . . . . . . . . . . . . . . . .
7.7 Injective holomorphic maps are biholomorphic onto their image
7.8 The Schwarz lemma . . . . . . . . . . . . . . . . . . . . . . .
7.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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61
61
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67
68
69
71
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73
73
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83
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Isolated singularities
8.1 Riemann’s removable singularity theorem . . . . . . . . . .
8.2 Isolated singularities: Classification and description of poles
8.3 Essential singularities; The Casorati-Weierstrass theorem . .
8.4 Laurent series I: Double-ended series . . . . . . . . . . . . .
8.5 Cauchy’s integral formula for annuli . . . . . . . . . . . . .
8.6 Laurent series II: Laurent’s theorem . . . . . . . . . . . . .
8.7 Classifying singularities in terms of Laurent series . . . . . .
8.8 Classification of injective entire functions . . . . . . . . . .
8.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
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CONTENTS
9
The general form of Cauchy’s theorem
9.1 Chains and cycles . . . . . . . . . . . . . . . . .
9.2 The homology version of Cauchy’s theorem . . .
9.3 The general version of Cauchy’s integral formula
9.4 The deformation theorem . . . . . . . . . . . . .
9.5 The Residue theorem . . . . . . . . . . . . . . .
9.6 Evaluation of residues . . . . . . . . . . . . . . .
9.7 Computation of real integrals . . . . . . . . . . .
9.8 The argument principle . . . . . . . . . . . . . .
9.9 Rouché’s theorem . . . . . . . . . . . . . . . . .
9.10 Exercises . . . . . . . . . . . . . . . . . . . . .
10 Sequences of holomorphic functions
10.1 Weierstrass convergence theorem
10.2 Hurwitz’s Theorem . . . . . . .
10.3 Compactness: Montel’s theorem
10.4 Exercises . . . . . . . . . . . .
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84
84
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87
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92
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96
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99
99
100
101
104
11 The Riemann mapping theorem
105
11.1 Riemann mapping theorem - statement and final ingredient . . . . . . . . . . . . . . . . . . . 105
11.2 Proof of the Riemann mapping theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
3
0
Preface
The starting point for this course is the basic grounding in complex analysis that you received in the
second half of the Analysis 3 course. These notes will assume specifically Section 4 of the Analysis
3 lecture notes of Jose Rodrigo from the year 2019/2020 that you can find online. You will find a link
on my moodle page. We will devote several sections to briefly reviewing that material, either without
proofs, or with different proofs, but those review sections will be interleaved with sections covering
new material.
I have used many sources for the material in this course. The classic book is that of Ahlfors, and this
was useful to give perspective. A few exercises were lifted from Priestley’s book. I learned my basic
complex analysis from Alan Beardon, and there may be elements of his lectures/exercises echoed in
these notes. According to the classic book of Ahlfors, the proof we give of the general homology form
of Cauchy’s theorem is due to Beardon. The basic chapter structure of these lecture notes is derived
from notes from previous years. The content is normally quite different, for example the treatment of
winding numbers and isolated singularities is very different, but in the first year 2020/2021 of these
notes there are a few passages that have been virtually lifted from previous years (e.g. Section 9.7
where real integrals are computed using the residue theorem) and others that arise as edits of previous
years. I am grateful to Stefan Adams for the latex file of his lecture notes from 2019/2020. The parts
we have adapted/incorporated seem to be taken from earlier years, dating at least as far back as the
notes of Hendrik Weber, and include quite a few displayed formulae. We had fewer hours than normal
to teach the material because of the blended learning approach we were asked to use in 2020/2021
because of covid, but most of this lost time has been made up by better integration with Analysis 3.
We even found some time to cover new topics, e.g. concerning Möbius transformations. The only
topics we had to abandon compared with previous years were the Gamma function and the Riemann
zeta function.
VIDEO: Introductory video
4
1
Review of material from MA244 Analysis 3 - first portion
Most of this material you will have seen, but we will interpret some of it differently by discussing
conformality.
1.1
Complex differentiability - definitions and intuition
VIDEO: Complex differentiability
By now you should have a good working understanding of complex numbers, how to add, multiply
and conjugate them. You will understand the topology of the complex plane, i.e. what it means
for a sequence {zn } to converge to some z ∈ C. You will understand what it means for a function
f : C → C to be continuous.
Reference: Section 4.1 from the Analysis 3 lecture notes, 2019/20
The notion of a function being complex differentiable is so fundamental that we repeat it.
Definition 1.1. Suppose Ω ⊂ C is open. Then a function f : Ω → C is complex differentiable at
z ∈ Ω if the limit
f (z + h) − f (z)
(1.1.1)
lim
h→0
h
exists. We denote the limit by f 0 (z) and call it the derivative of f at z.
In the discussion below, we sometimes view Ω as a subset of R2 rather than C in the obvious way
without change of notation, i.e. we consider
{(x, y) ∈ R2 : x + iy ∈ Ω},
and still write this as Ω. In this case we still use f to denote the resulting function. Sometimes (but
less and less as time goes on) we decompose f into real and imaginary parts:
f (x + iy) = f (x, y) = u(x, y) + iv(x, y),
for real valued functions u, v : Ω → R. That is, u = 1
or a contraction if λ < 1.
(iv) Complex inversion: This is the map f (z) = z1 . Its effect is most easily understood as a map from
S 2 to S 2 using the stereographic projection. In that viewpoint it is a rotation by 180 ◦ about the
x1 -axis in R3 . In other words, the map (x1 , x2 , x3 ) 7→ (x1 , −x2 , −x3 ).
Beware: the use of term ‘inversion’ is potentially misleading. For many purposes, inversion would
refer to the map z 7→ 1/z̄, which fixes the unit circle, sending reiθ to 1r eiθ .
The interpretations of translations, rotations and dilations that we have given are obvious. To verify
the interpretation of the complex inversion as the rotation by 180 ◦ , we can compute directly using the
formulae (2.2.1) and (2.2.1) for π and π −1 in Section 2.2. To see this, first, note that the map z 7→ 1/z,
in coordinates, is x + iy 7→ xx−iy
2 +y 2 . Also note that because (x1 , x2 , x3 ) lies in the unit sphere, we have
2
2
2
x1 + x2 = 1 − x3 = (1 − x3 )(1 + x3 ). Then as a map on S 2 we get
(x1 , x2 , x3 )
π
7−→
x1 + ix2
1 − x3
z7→1/z
7−−−−→
(1−x3 )
x1 − ix2
x1 − ix2
=
2
2
x1 + x2
1 + x3
16
π −1
7−−−→
(x1 , −x2 , −x3 ),
2.6
Three points determine a Möbius map
Lemma 2.9. Every Möbius transformation can be written as the composition of elementary Möbius
transformations.
Proof. First observe that given an arbitrary nonzero complex number, written as reiθ , with r > 0,

θ ∈ R, the transformation z 7→ reiθ z is a composition of a rotation by θ and a dilation by a factor r.

For c 6= 0 we write

az+b

cz+d

=

a

c

+

b− ad

c

.

cz+d

Then this map is obtained by

b − ad

1

a b − ad

c

c

z 7→ cz 7→ cz + d 7→

7

→

7→ +

.

cz + d

cz + d

c cz + d

The case where c = 0 is even easier.

(2.5.1)

We can often use this decomposition into elementary Möbius transformations in order to prove that

certain properties are preserved under general Möbius transformations: The proof is then reduced to

checking that the property is preserved for elementary Möbius transformations, as in the following.

Keeping in mind our definition of ‘circles’ in C∞ from Definition 2.3, we have

Theorem 2.10. The image of every circle in C∞ under any Möbius transformation is also a circle in

C∞ .

Proof. We only need check this property for each type of elementary Möbius transformation. The

property is obvious for translations, rotations and dilations. The property for the complex inversion

follows from its interpretation as a 180 ◦ rotation, together with the preservation of circles/lines by

stereographic projection given in Remark 2.2.

You proved essentially this theorem by direct calculation on the exercise sheets of Analysis 3.

2.6

Three points determine a Möbius transformation

VIDEO: Three points determine a Möbius transformation

In this section we will see the fundamental (and very useful) property that if we ask for three specific

distinct points in C∞ to be mapped to another three specific distinct points in C∞ , then we determine

a unique Möbius transformation. Precisely, we have:

Theorem 2.11. Given three distinct points z1 , z2 , z3 ∈ C∞ and three distinct points w1 , w2 , w3 ∈ C∞ ,

there exists a unique Möbius transformation f that satisfies f (zi ) = wi for i = 1, 2, 3.

Thus the group of Möbius transformations has three complex degrees of freedom. This illustrates

very clearly how it is a six real-parameter family.

The proof of Theorem 2.11 will be supported by a couple of sub-results. We start with an observation

about the fixed points of Möbius transformations.

17

2.6

Three points determine a Möbius map

Lemma 2.12. Every Möbius transformation other than the identity f (z) = z has at least one, but at

most two, fixed points. In particular, if f is a Möbius transformation and z1 , z2 , z3 ∈ C∞ are distinct

points such that f (zi ) = zi , then f is the identity.

Proof. As usual, we write

az + b

.

cz + d

Observe that f being the identity corresponds to the case a = d 6= 0 and b = c = 0, so if we assume

that this is not the case then we need to show that there can be at most two fixed points.

f (z) =

We split the proof into two cases: First, if c = 0, then f is just a linear transformation f (z) = ad z + db ,

where a, d 6= 0 by nondegeneracy of Möbius transformations. Such a linear transformation has a

b

if a 6= d.

fixed point at infinity, plus at most one further fixed point at z = d−a

In the remaining case that c 6= 0, we have

f (z) =

az + b

=z

cz + d

⇐⇒

(az + b) = (cz + d)z

⇐⇒

0 = cz 2 + (d − a)z − b .

The quadratic formula gives us two solutions, which might coincide.

Next, we give a special case of the desired Theorem 2.11 in which the points wi are explicit. In this

special case we also give formulae for the Möbius transformation, which will be useful later.

Proposition 2.13. Given three distinct points z1 , z2 , z3 ∈ C∞ , there exists a Möbius transformation

f that maps z1 , z2 , z3 to 1, 0, ∞ respectively. In the case that zi 6= ∞ for i = 1, 2, 3, then it is

(z − z2 )(z1 − z3 )

.

(z − z3 )(z1 − z2 )

(2.6.1)

f (z) =

z − z2

,

z − z3

(2.6.2)

f (z) =

z1 − z3

,

z − z3

(2.6.3)

f (z) =

z − z2

.

z1 − z2

(2.6.4)

f (z) :=

In the case that z1 = ∞ we set

in the case that z2 = ∞ we set

and in the case that z3 = ∞ we set

Proof. Clearly each of these functions f are Möbius transformations. By inspection, they send the

points zi to the required image points.

In the final three cases (2.6.2) to (2.6.4) in which one of the zi is ∞, the formulae for f arise from

(2.6.1) by dropping the factor on the numerator and the factor on the denominator containing that zi .

These two factors would cancel in the limit zi → ∞, so this makes sense.

We are finally in a position to prove the main result of this section.

18

2.7

The cross ratio

Proof of Theorem 2.11.

Existence: Let f1 be the function from Proposition 2.13 that sends z1 , z2 , z3 to 1, 0, ∞ respectively.

Let f2 be the function from Proposition 2.13 that sends w1 , w2 , w3 to 1, 0, ∞ respectively. The Möbius

transformation f we seek will simply be f2−1 ◦ f1 .

Uniqueness: Suppose that we have two Möbius transformations f and g, both of which send the

points zi to wi respectively. Then g −1 ◦ f is a Möbius transformation that has all three distinct points

zi as fixed points. Lemma 2.12 then tells us that g −1 ◦ f is the identity, i.e. f ≡ g.

2.7

The cross ratio

VIDEO: The cross ratio

Definition 2.14. The cross-ratio of pairwise distinct z0 , z1 , z2 , z3 ∈ C∞ , is the point in C that is the

image of z0 under the unique Möbius transformation that sends z1 , z2 , z3 to 1, 0, ∞ respectively.

Formulae for the Möbius transformation f , and hence for f (z0 ), are given by Proposition 2.13.

Theorem 2.15. The cross-ratio is invariant under Möbius transformations.

Proof. If f is any Möbius transformation, we have to show that the cross-ratio of (z0 , z1 , z2 , z3 ) is the

same as that of (f (z0 ), f (z1 ), f (z2 ), f (z3 )). If we denote by g the unique Möbius transformation that

sends z1 , z2 , z3 to 1, 0, ∞ respectively, then this cross ratio is g(z0 ) by definition. But then the the

unique Möbius transformation that sends f (z1 ), f (z2 ), f (z3 ) to 1, 0, ∞ respectively is g ◦ f −1 , and so

the cross-ratio of (f (z0 ), f (z1 ), f (z2 ), f (z3 )) is g ◦ f −1 (f (z0 )) = g(z0 ), which is the same.

Theorem 2.16. The cross ratio of (z0 , z1 , z2 , z3 ) is real-valued if and only if z0 , z1 , z2 , z3 all lie on a

common circle in C∞ .

Recall from Definition 2.3 that circles in C∞ restrict to circles and lines in C.

Proof. Since both the cross-ratio and the property of lying on a circle are invariant under Möbius

transformations, we may assume that z1 , z2 , z3 equal 1, 0, ∞ respectively. The cross ratio of (z0 , 1, 0, ∞)

is z0 by definition. But the circle passing through 1, 0 and ∞ is the real line.

Let’s digest a consequence of what we have proved. If z0 , z1 , z2 , z3 ∈ C, then they all lie on a common

line or circle in the plane if and only if

(z0 − z2 )(z1 − z3 )

∈ R.

(z0 − z3 )(z1 − z2 )

Amazing! Try it out on a few examples.

19

2.8

Examples of Möbius transformations

2.8

Examples and special classes of Möbius transformations

VIDEO: Examples of Möbius transformations

Around 23:10 when I said ‘reflection’, I meant ‘rotation’. I’ll explain more in the live lecture.

There are numerous distinguished examples and subclasses of Möbius transformations, of which we

give a few prominent examples.

Example 2.17 (Möbius transformation that gives a bijection from a disc to a half-space). We are

looking for a Möbius transformation that provides a bijection from the disc D := {z ∈ C : |z| <
1} to the upper half-plane H=>0 := {z ∈ C : =(z) > 0}. Since Möbius transformations are

homeomorphisms from C∞ to itself, this will map the boundary of D, i.e. the unit circle, to the

boundary of H=>0 , i.e. the real axis. There are many maps with this property. By Theorem 2.11 we

can pick any three distinct points on the unit circle and map them to 1, 0 and ∞ respectively. For

example, we could pick 1, −i and i on the unit circle, in which case the map is given by (2.6.1) as

f (z) =

z+i

(z − (−i))(1 − i)

=

.

(z − i)(1 − (−i))

iz + 1

By Theorem 2.10, this Möbius transformation will map the entire unit circle ∂D to the entire real

axis. Because f is a homeomorphism, this map must send the disc D either to the upper half-plane

H=>0 , or the lower half-plane. But since it sends 0 to i, it must be the former case, as required.

Example 2.18 (Möbius transformations that give bijections from the disc to itself). Consider the

Möbius transformations of the form

z−w

for w ∈ C, with |w| < 1.
(2.8.1)
f (z) =
w̄z − 1
We claim that Möbius transformations of this form map D onto itself, and map the boundary of D to
itself. To see this, we use the easily-checked identity
|z − w|2 = |w̄z − 1|2 − (1 − |z|2 )(1 − |w|2 )
to compute
|f (z)|2 =
|z − w|2
(1 − |z|2 )(1 − |w|2 )
=
1
−
.
|w̄z − 1|2
|w̄z − 1|2
Because we are assuming that 1 − |w|2 > 0, we see that |f (z)| < 1 if and only if |z| < 1, and
|f (z)| = 1 if and only if |z| = 1.
This argument also implies that f maps onto D, since f is a bijection from C∞ to itself. An alternative
argument to obtain the surjectivity would be to observe that the inverse of f is f itself!
Remark 2.19. We can slightly generalise the class of Möbius transformations from Example 2.18
that map the disc D to itself, by composing with a rotation about the origin, giving maps of the form
z−w
iθ
f (z) = e
,
(2.8.2)
w̄z − 1
20
2.8
Examples of Möbius transformations
still with w ∈ D, and now with θ ∈ (−π, π]. A stunning fact that will follow from the so-called
Schwarz lemma, Theorem 7.20, is that every holomorphic map D 7→ D that is a bijection is of the
form (2.8.2). We are not starting with the assumption that the map is a Möbius transformation here.
This is true in far greater generality! Think how different this is to what you have seen before. Just
imagine if there were only a finite-dimensional family of real differentiable bijective functions from
(−1, 1) to itself!
Example 2.20 (Möbius transformations that give bijections from H=>0 to itself). A Möbius transformation g(z) from D to itself can be converted into a Möbius transformation h := f ◦ g ◦ f −1 from

H=>0 to itself, where f : D → H=>0 is the Möbius transformation from Example 2.17. One can also

conjugate in the other direction to give g = f −1 ◦ h ◦ f . Because of this, it may seem pointless to consider Möbius transformations from H=>0 to itself after already considering Möbius transformations

from D to itself in Example 2.18.

However, working in the H=>0 viewpoint has some significant advantages in certain situations. To

see one, we return to the isomorphism between the group of Möbius transformations and the group

P SL(2, C) from Section 2.4. The key observation is that the subgroup

P SL(2, R) := SL(2, R)/{±I},

essentially restricting from complex matrices in SL(2, C) to real matrices in SL(2, R), but as before

identifying each pair A and −A, corresponds to an interesting subgroup of Möbius transformations.

Indeed, we claim that they send the upper half plane H=>0 to itself. To see this, we rewrite

f (z) =

(az + b)(cz̄ + d)

ac|z|2 + ad z + bc z̄ + bd

az + b

=

=

.

cz + d

(cz + d)(cz̄ + d)

|cz + d|2

Therefore, keeping in mind that =(z̄) = −=(z) and ad − bc = 1, we have

=(f (z)) =

(ad − bc)=(z)

=(z)

=(ad z + bc z̄)

=

=

2

2

|cz + d|

|cz + d|

|cz + d|2

In particular, if z ∈ H=>0 , equivalently =(z) > 0, if and only if =(f (z)) > 0, equivalently f (z) ∈

H=>0 . Thus f maps H=>0 to itself.

Similarly to before, we see that f also maps H=>0 onto itself. For example, one can observe that the

inverse of f is another Möbius transformation of the same form, and thus maps H=>0 to itself.

One can check that every Möbius transformation that maps H=>0 bijectively to itself arises in this

way. Indeed, any such map must send the real line to itself, or to ∞, and if we add the point ∞ ∈ C∞

to the real line to give a circle C, then the Möbius transformation will map this circle bijectively to

itself. If we now take the points x1 , x2 , x3 ∈ C that map to the points 1, 0, ∞ respectively, then we

can construct a Möbius transformation in P SL(2, R) using Proposition 2.13 that has the same effect

on these three points. By Theorem 2.11 itself, the two Möbius transformations must then coincide.

Example 2.21. Nonexaminable example: The rotations of the unit sphere S 2 ,→ R3 are Möbius

transformations when we identify S 2 and C∞ as above. By rotations, we mean elements of SO(3).

Since the group of Möbius transformations is isomorphic to P SL(2, C), by Lemma 2.7, it is natural

21

2.9

Conformal maps

to ask to which subgroup of P SL(2, C) these rotations correspond. It turns out that it is the subgroup

P SU (2). The corresponding Möbius transformations can be written

az − c

z 7→

,

cz + a

for a, c ∈ C with |a|2 + |c|2 = 1.

2.9

Conformal maps

VIDEO: Conformal maps

A general principle in mathematics is that one defines an object with some structure, for example a

vector space with its linear structure, and then one considers bijective maps between different objects

that preserves this structure, for example a bijection between vector spaces that preserves the linear

structure. Such bijections are typically called isomorphisms, and intuitively we view two objects that

are isomorphic as being the same. You will have seen many other examples of this viewpoint, for example isometries between metric spaces, isomorphisms between groups, homeomorphisms between

topological spaces or maybe diffeomorphisms between manifolds (if you have studied manifolds).

What is the right notion of equivalence for domains in C?

(A domain is an open and connected subset.)

Definition 2.22. Two domains Ω1 and Ω2 in C are said to be conformally equivalent if there exists a

bijective holomorphic function ϕ : Ω1 → Ω2 such that the inverse ϕ−1 is also holomorphic.

Such a map ϕ is called a conformal or biholomorphic map.

In due course, we will establish that the inverse ϕ−1 is automatically holomorphic for a bijective

holomorphic function ϕ between domains, but there is no point in fretting about that now.

This notion of equivalence gives us an equivalence relation. In particular, if Ω1 and Ω2 are conformally

equivalent via the conformal map ϕ : Ω1 → Ω2 , and Ω2 and Ω3 are conformally equivalent via the

conformal map ψ : Ω2 → Ω3 , then Ω1 and Ω3 are conformally equivalent via the conformal map

ψ ◦ ϕ : Ω1 → Ω3 . We are implicitly using the chain rule to be sure that the composition of these

conformal maps is conformal. Thus the notion of being conformally equivalent is an equivalence

relation.

The key property that is preserved by this notion of equivalence is the concept of a function on the

domain being holomorphic. More precisely, by the chain rule, a function f : Ω2 → C is holomorphic

if and only if the composition f ◦ ϕ : Ω1 → C is holomorphic.

For the rest of this section we try to get a feeling for which domains are conformally equivalent to

which other domains. Our knowledge of Möbius transformations will help. Throughout the discussion we write the unit disc as

D := {z ∈ C : |z| < 1}.
22
2.9
Conformal maps
Before we begin, let’s recall that we already showed in Example 2.17 that the upper half-space is
conformally equivalent to the disc D. Let’s find some more such domains.
Watch the video for pictures!
Example 2.23. We claim that the upper right quarter of the complex plane
Q := {z ∈ C | 0 and =(z) > 0}

is conformally equivalent to the disc D.

To see this, we first observe that Q is conformally equivalent to the upper half plane H=>0 by virtue

of the conformal map z 7→ z 2 .

The upper half plane is then conformally equivalent to D thanks to Example 2.17.

Example 2.24. We claim that the upper half disc

D=>0 := {z ∈ C : |z| < 1 and =(z) > 0}

is conformally equivalent to the whole disc D.

Danger: It is very tempting to try to use the conformal map z 7→ z 2 to do this job, but this actually

shows the quite different fact that D=>0 is conformally equivalent to the disc D with the real interval

[0, 1) removed!

Instead, we notice that D=>0 is conformally equivalent to Q via the unique Möbius transformation

that sends −1, 0 and 1 to 0, 1 and ∞ respectively. One could compute this explicitly, but we can argue

more geometrically as follows: First, we know from Theorem 2.11 that this Möbius transformation

exists. Second, by the preservation of circles in C∞ from Theorem 2.10 it sends the interval [−1, 1]

to the interval [0, ∞]. Third, by Theorem 2.10 again it must send the semicircle {z ∈ C : |z| =

1 and =(z) ≥ 0} to a half line starting at 0 and going in some direction off to infinity. Finally, the

right-angle in the boundary of D=>0 at −1 must be reflected in a right angle in the boundary of

the image of D=>0 under this Möbius transformation, because the Möbius transformation preserves

angles, and so the half line must be the positive imaginary axis.

Now we have shown that D=>0 is conformally equivalent to Q, the claim follows by Example 2.23.

By this point you may be getting the false impression that every domain is conformally equivalent to

D. But this is not true. One type of counterexample would be to take the domain that is the whole of

C. If we could find a conformal map from C to D then this would be a bounded holomorphic function

on C, and Liouville’s theorem that was mentioned in Analysis 3 would tell us that it would have to

be constant, and certainly not then surjective onto D. (We will revisit Liouville’s theorem later, in

Corollary 6.8, once we have rigorously proved Cauchy’s theorem.)

Another sort of domain that would certainly fail to be conformal to the disc D would be a domain

‘with holes’ such as an annulus {z ∈ C : a < |z| < b}, where 0 < a < b < ∞. This domain is not
even homeomorphic to D so it is a bit much to ask for it to be homeomorphic via a homeomorphism
that is additionally a conformal map! To see that they are not homeomorphic, it suffices to notice
23
2.9
Conformal maps
that one is simply connected while the other is not. The notion of simply connected has been briefly
mentioned in Analysis 3. Informally it means that every loop in the space can be deformed to a point.
In order to make this precise we need to define the notion of homotopy.
You may have seen homotopies in greater generality in the course Introduction to Topology.
Definition 2.25 (Homotopic). Let Ω ⊂ C be open and let γ1 , γ2 : [a, b] → Ω be two continuous curves
(i.e. continuous maps from an interval) with the same endpoints γ1 (a) = γ2 (a) and γ1 (b) = γ2 (b).
Then γ1 and γ2 are called homotopic if there exists a continuous mapping h : [0, 1] × [a, b] → Ω such
that for all t ∈ [a, b] and all s ∈ [0, 1] we have
h(0, t) = γ1 (t),
h(1, t) = γ2 (t),
h(s, a) = γ1 (a),
and
h(s, b) = γ1 (b).
(2.9.1)
Such a mapping h is called a homotopy.
The notion of Ω being simply connected is intuitively that it does not contain any holes.
Definition 2.26 (Simply connected). An open set Ω ⊂ C is said to be simply connected if it is
connected and every closed1 continuous curve γ : [a, b] → Ω is homotopic to the constant curve
γ̃ : [a, b] → Ω defined by γ̃(t) = γ(a) = γ(b).
In more general situations the definition of simply connected would ask for path connectedness, but
path connectedness is equivalent to connectedness when considering an open set. In this very special
situation of open sets Ω in C, the notion of being simply connected is equivalent to both Ω being
connected, and its complement in the Riemann sphere C∞ being connected. We will not explicitly
use this formulation.
At this point we can look forward to the end of the course when we prove one of the greatest theorems
in the subject, namely the Riemann mapping theorem. That theorem will tell us that every simply
connected domain Ω ⊂ C other than Ω = C is conformally equivalent to the disc D. See Theorem
11.1.
1
γ : [a, b] → C is said to be closed if γ(a) = γ(b), i.e. the curve closes up.
24
1
1
b
a
r
Figure 2: Stereographic projection, cross-section
2.10
Exercises
2.1. By considering a triangle similar to the red triangle in Figure 2, prove that r =
formula (2.2.1) for stereographic projection.
a
.
1−b
Deduce the
Think of Figure 2 as a cross-section in the picture describing stereographic projection.
2.2. Invert formula (2.2.1) to give (2.2.2).
Hint: Solve first for x3 . Then get x1 and x2 .
2.3. We claimed in Remark 2.2 that straight lines in C are in one-to-one correspondence with circles
in S 2 that pass through the north pole N , with the correspondence being given by stereographic
projection π. By considering appropriate planes in R3 that pass through N , give a geometric
justification of this fact.
You could also check this correspondence by solving equations. See the next question.
2.4. We claimed in Remark 2.2 that circles in C are in one-to-one correspondence with circles in S 2
that don’t pass through the north pole N , with the correspondence being given by stereographic
projection π. Verify this by writing down the equations of the appropriate circles.
You might also try to give a geometric proof of this, but it’s trickier than the previous question!
Hint: Let’s remember some basic geometric facts from school mathematics. If n := (a, b, c) is
a unit vector, then the plane through the origin with normal vector n is given by (x1 , x2 , x3 ).n =
0. More generally if we shift that plane in the direction n by a distance d ∈ (−1, 1), then the
equation is (x1 , x2 , x3 ).n = d, i.e.
ax1 + bx2 + cx3 = d.
(2.10.1)
The case that N = (0, 0, 1) lies within this plane is then precisely that (0, 0, 1).n = d, i.e.
c = d. The intersection of any such plane with S 2 is a circle in S 2 . All such circles arise in this
way.
Now plug in the values (x1 , x2 , x3 ) given by the formula (2.2.2) into the formula (2.10.1), to
give a formula for the image of this circle under π.
2.10
Exercises
π −1 (z)
1
s
r
Figure 3: Antipodal points
In the case c = d you should end up with an equation of a line in the plane. (I obtained
ax + by = d.) This is the case of the last question.
In the case c 6= d, you should get the equation of a√ circle. By completing the square, you find
b
1−d2
a
, d−c
). The radius should be |d−c|
.
that the centre is ( d−c
2.5. The points 0 and ∞ in C∞ , when viewed as points in S 2 by applying π −1 , are the south and
north poles respectively. They are therefore antipodal points. Given any other point z ∈ C∞ ,
i.e. z ∈ C that is nonzero, show that −1/z̄ ∈ C corresponds to the antipodal point. More
precisely, show that π −1 (z) and π −1 (−1/z̄) are antipodal.
If you first convince yourself that z, 0 and −1/z̄ are co-linear in C, then you can work in a
cross-section as in Figure 3.
2.6. In Example 2.17 we mapped the unit disc to the upper half plane with a Möbius transformation.
Viewed as a transformation of the Riemann sphere, seen as a 2-sphere in R3 , what is this
transformation?
What is the square of this transformation? That is, which Möbius transformation is it, and what
does it look like as a transformation of the Riemann sphere seen as a 2-sphere in R3 ?
2.7. Where does the map
1+z
1−z
send the unit disc? What is the inverse of this map? What is it as a transformation of the
Riemann sphere S 2 ?
f (z) =
2.8. Show that the quarter disc
Q̂ := {z ∈ C : |z| < 1 and =(z) > 0 and 0}

is conformally equivalent to the unit disc D.

2.9. Show that the slit plane

S := C [0, ∞)

is conformally equivalent to the unit disc D.

26

2.10

Exercises

2.10. Construct explicitly the Möbius transformation that was constructed geometrically in Example

2.24.

2.11. Construct explicitly the unique Möbius transformation f that sends D to the half space H0 ,

while sending 0 to 12 and sending −1 to 0. By composing with the map z 7→ z 2 and then

z 7→ z − 41 , show that the map

z

K(z) :=

(1 − z)2

is a conformal map from D to the slit plane C (−∞, − 41 ].

This function is known as the Koebe function, and we will revisit it once we know a little more

theory. Amongst all the conformal maps from D to C(−∞, − 41 ], it is the unique one satisfying

the normalisation K(0) = 0 and K 0 (0) = 1.

27

3

Review of material from MA244 Analysis 3 – second portion

3.1

Power series

VIDEO: Power Series

In Section 4.2 of Analysis 3 you learned about power series, i.e. expressions of the form

∞

X

an z n ,

n=0

where an is a complex-valued sequence.

Theorem 3.1 (Theorem 4.13 from Analysis 3). Given a complex-valued sequence (an ), define the

so-called radius of convergence by

R :=

Then the power series

1

∈ [0, ∞].

lim sup |an |1/n

∞

X

an z n

n=0

converges for all |z| < R and diverges for all |z| > R.

Nothing is claimed in this theorem about z for which |z| = R. That question can be somewhat

delicate. The cases considered in the theorem follow easily from the root test.

Power series can be differentiated term-by-term within their radius of convergence:

Theorem 3.2 (Theorem 4.15 from Analysis 3). If the radius of convergence R of the power series

f (z) =

∞

X

an z n

n=0

is positive (including ∞), then within the disc BR := {z ∈ C : |z| < R}, the function f is
holomorphic, and
∞
X
0
f (z) =
nan z n−1 ,
n=1
with this new power series having the same radius of convergence R.
A function that can be written as a power series over some ball about each point in its domain is
called analytic. We see now that analytic implies holomorphic. Later we will see that holomorphic
implies analytic. Indeed, Taylor’s theorem will tell us that every holomorphic function can be written
as a power series locally. For this reason, many people use the terms analytic and holomorphic
interchangeably.
By repeatedly applying Theorem 3.2, we obtain
28
Definitions of ez , sin(z), cos(z) etc.
3.2
Corollary 3.3 (Corollary 4.16 from Analysis 3). If the radius of convergence R of the power series
f (z) =
∞
X
an z n
n=0
is positive, then within the disc BR := {z ∈ C : |z| < R}, the function f is infinitely differentiable,
and the n-th derivative of f at 0 ∈ C is given by
f (n) (0) = an n!.
(3.1.1)
Although the convergence of a power series is a bit delicate near the circle {z ∈ C : |z| = R}, if we
restrict to compact subsets of BR then we have uniform convergence:
Theorem 3.4 (Theorem 4.17 from Analysis 3). If the radius of convergence R of the power series
f (z) =
∞
X
an z n
n=0
is positive (including ∞), then for all r ∈ (0, R), the convergence
k
X
an z n →
n=0
∞
X
an z n
n=0
is uniform within the disc Br as k → ∞.
3.2
Definitions of ez , sin(z), cos(z), sinh(z) and cosh(z)
VIDEO: Exponentials, sine, cosine etc.
You’re already familiar with the idea of writing a complex number as reiθ . We’ve used that idea
already in the course. In the Foundations course (and probably at school before that) you defined
eiθ = cos θ + i sin θ and then defined eα+iθ = eα eiθ in order to make sense of a general complex
exponential ez , and gave ad hoc proofs that this gave something with the expected behaviour, with a
promise that you would see a better motivated definition later on.
The theory of power series described in the previous section allows us to give this better motivated
definition of exp : C → C, and also define other familiar functions such as sin and cos etc. as
functions from C rather than R. It does this by using the familiar Taylor series expansions of these
functions on R and using them instead on C.
Definition 3.5. We define exp : C → C, also written as z 7→ ez by
z
e :=
∞
X
zn
n=0
29
n!
.
Euler’s formula; arg(z) and log(z)
3.3
From the theory we have described, for this definition to make sense we have to verify that the radius
of convergence is R = ∞. From the definition of R, it suffices to establish that (n!)1/n → ∞, which
is a mini analysis exercise. (You can also use the ratio test to establish this without doing any exercise,
but we did not mention that.)
By differentiating term by term using Theorem 3.2, we find that the derivative of ez is ez itself:
(ez )0 = ez .
(3.2.1)
The familiar property of the exponential function that ex+y = ex ey extends to the complex case:
Lemma 3.6. For all z, w ∈ C we have
ez+w = ez ew .
(3.2.2)
Proof. In the Analysis 3 course this was proved by multiplying the power series together, which
requires some care. As an alternative, for arbitrary α ∈ C, consider the function f (z) := ez eα−z . By
the product and chain rules, f is entire and
f 0 (z) = −ez eα−z + ez eα−z = 0.
Therefore f is constant, with f (z) = f (0) = eα . That is, eα−z ez = eα . By setting α = z + w, we
obtain (3.2.2).
The definitions of sinh and cosh extend immediately to entire functions from C rather than just R:
sinh(z) :=
ez − e−z
,
2
cosh(z) :=
ez + e−z
,
2
and by (3.2.1) we have sinh0 (z) = cosh(z) and cosh0 (z) = sinh(z).
Moreover, if we define entire functions
eiz − e−iz
sin(z) :=
,
2i
eiz + e−iz
cos(z) :=
,
2
(3.2.3)
then these are the familiar sin and cos functions when restricted to R; one can see this by comparing
the power series, for example.
3.3
Euler’s formula; arg(z) and log(z)
VIDEO: Argument and logarithm
30
3.3
Euler’s formula; arg(z) and log(z)
Directly from the formulae for sin and cos, we obtain
eiz = cos z + i sin z.
(3.3.1)
In particular,
eiπ = −1,
and we recover the Euler formula that for θ ∈ R, we have
eiθ = cos θ + i sin θ,
thus telling us that eiθ represents a point on the unit circle, an angle θ anticlockwise from the positive
real axis (i.e. the x-axis). Thus we confirm that z ∈ C can be written in the form |z|eiθ , where θ is
determined up to the addition of 2π if z 6= 0 (and θ is arbitrary if z = 0).
This is essentially viewing z in polar coordinates (r, θ) with r = |z|.
For z 6= 0, this value θ, determined up to the addition of 2π, is known as the argument arg(z). Strictly
speaking it should be viewed as a function taking values in R/(2πZ).
If you understand the picture here, then you can read off properties of arg(z) at will. For example, if
z and w are nonzero complex numbers then
|zw|ei arg(zw) = zw = (|z|ei arg(z) )(|w|ei arg(w) ) = |zw|ei(arg(z)+arg(w)) ,
and so
arg(zw) = arg(z) + arg(w)
modulo 2π.
(3.3.2)
The multi-valued nature of arg can be annoying. There are several ways we can avoid it, and the best
strategy depends on the context. A simple idea would be to ask that it takes values in (−π, π], say,
yielding the principal value. This has a major disadvantage that it makes arg discontinuous along the
negative real axis. It does, however, give a nice continuous2 function on C − {x ∈ R : x < 0}.
This idea can be generalised. If we remove any ray {reiθ : r > 0} from C, for some θ ∈ R, then we

can define a single valued choice of arg(z), although different people may make a choice of arg(z)

that differs by a fixed constant multiple of 2π. The ray here is known as a branch cut.

A particularly useful resolution of this 2π ambiguity will be given in Lemma 4.1.

The (complex) logarithm is defined for z 6= 0 by

log(z) = log |z| + i arg(z).

Here the logarithm on the right-hand side is taking real values, so it is uniquely defined. But the

argument is only defined up to the addition of an integer multiple of 2π, so log(z) is only defined

up to the addition of an integer multiple of 2πi. As for the argument arg(z), we are sometimes

content with this state of affairs, and sometimes we ask arg(z) to take its principal value in (−π, π],

in which case log(z) is a well defined function that extends the usual logarithm. Unfortunately it is

2

even smooth, i.e. infinity real differentiable

31

3.4

Complex integration

discontinuous across the negative real axis {x < 0} ⊂ C, and we sometimes make a branch cut by
removing the half-line.
The complex logarithm inherits many of the useful properties of the real logarithm. For example, it is
the inverse of the exponential function in the sense that
elog z = elog |z|+i arg(z) = |z|ei arg(z) = z,
and
log(ez ) = log |ez | + i arg(ez )
= log ex + i arg(eiy )
= x + iy
= z,
(3.3.3)
where the latter computation is carried out modulo 2πi.
By (3.3.2) we have the familiar identity
log(zw) = log |zw| + i arg(zw) = log |z| + log |w| + i(arg(z) + arg(w))
= log z + log w,
(3.3.4)
again modulo 2πi. Beware that identities that are claimed modulo 2πi should not be expected to work
if we take the principal values of the functions log or arg.
As you saw in Analysis 3, the function log(z) allows us to define what it means to raise a complex
number to a complex power, albeit with complications arising from log(z) only being defined up to a
multiple of 2πi.
3.4
Complex integration
VIDEO: Complex integration
Recall that in Analysis 3 you defined what it means to integrate a suitable (e.g. continuous) function
f : [a, b] → C. You defined
Z b
Z b
Z b
f (t)dt :=
0 we could write
Z
Z
f (z)dz := f (z)dz,
∂Br (a)
γ
where γ : [0, 2π] → C is defined by γ(θ) = a + reiθ . Alternatively, if
A := {z ∈ C : R1 < |z| < R2 }
is an annulus, then we have two boundary components that are parametrised by curves γ1 , γ2 :
[0, 2π] → C defined by γ2 (θ) = R2 eiθ and γ1 (θ) = R1 e−iθ , and we analogously define
Z
Z
Z
f (z)dz :=
f (z)dz +
f (z)dz.
∂A
γ1
γ2
In both Definitions 3.7 and 3.9, we say that the curve is closed if γ(a) = γ(b). Informally this means
that it closes up into a loop. The curve is a simple closed curve if whenever a ≤ x < y ≤ b and
γ(x) = γ(y), we have x = a and y = b.
3.5
Anti-derivatives, and a baby version of Cauchy’s theorem
VIDEO: Anti-derivatives; baby Cauchy’s theorem
Later, in Section 5, we will turn our attention to a theorem that is at the heart of complex analysis,
namely Cauchy’s theorem. Loosely speaking it will tell us that in certain situations when we integrate
holomorphic functions around closed curves we obtain zero. In this section we see a baby version of
this theory in which our holomorphic function f is the derivative of some other holomorphic function
F.
34
3.5
Anti-derivatives; baby Cauchy’s theorem
Lemma 3.10. Suppose Ω ⊂ C is open. Suppose further that f : Ω → C and F : Ω → C are
holomorphic with F 0 (z) = f (z). If γ is a piecewise C 1 closed curve in Ω. Then
Z
f (z)dz = 0.
γ
This lemma follows immediately from the following type of fundamental theorem of calculus.
Lemma 3.11. Suppose that F : Ω → C is holomorphic, with F 0 continuous. Suppose further that
γ : [a, b] → Ω is a piecewise C 1 curve. Then
Z
F 0 (z) dz = F (γ(b)) − F (γ(a)).
(3.5.1)
γ
In particular, if γ is a closed curve (i.e. γ(b) = γ(a)) then we have
R
γ
F 0 (z) dz = 0.
Proof. By the definition of contour integration and the chain rule of Lemma 1.5, we have
Z
Z b
Z b
d
0
0
0
F (z) dz =
F (γ(t)) γ (t) dt =
F (γ(t)) dt = F (γ(b)) − F (γ(a)).
γ
a
a dt
Note that to be able to apply the usual fundamental theorem of calculus in the last equality, we
need some regularity on the integrand such as continuity, which is why we are assuming that F 0 is
continuous.
The result assumes that the derivative F 0 of the holomorphic function F is continuous. Later we will
see that this is always true, but we can’t assume that now or our arguments will be circular.
Corollary 3.12. Suppose n ∈ Z does not equal −1. Then for γ : [a, b] → C {0} any piecewise C 1
closed curve, we have
Z
z n dz = 0.
γ
Proof. If we define F (z) := z n+1 /(n + 1), then F 0 (z) = z n , so the result follows from Lemma
3.10.
This corollary fails in a very important way if n = −1! In Q. 3.8 you will compute:
Example 3.13. For r > 0 and k ∈ Z let γ : [0, 2π] → C be the closed C 1 curve γ(θ) = reikθ that

travels anticlockwise k times around the circle of radius r. Then

Z

dz

= 2πi k.

γ z

35

3.6

Exercises

P

n

3.1. Suppose Alice has P

a power series ∞

n=0 an z with radius of convergence R1 > 0, and Bob

∞

has a power series n=0 bn z n with radius of convergence R2 ≥ R1 . Suppose we are told that

these power series give the same function on the ball BR1 (0) where they are both converging.

Prove that an = bn for every n ∈ N0 := {0, 1, 2, . . .}, i.e., the power series and their radii of

convergence agree.

P

1

k

3.2. Write the function z 7→ 1−z

as a power series ∞

k=0 ak z and give its radius of convergence.

3.3. By using the previous question and a theorem from Section 3.1, write down the power series of

1

the function z 7→ (1−z)

2 and its radius of convergence.

We will use this exercise when we take a closer look at the Koebe function.

P

1

k

3.4. Suppose w ∈ C {0}. Write the function z 7→ w−z

as a power series ∞

k=0 ak z and give its

radius of convergence.

We will use this fact when we prove Taylor’s theorem.

P

k

3.5. Suppose that n ∈ N, and that the power series ∞

k=n ak z , which omits the first n terms of a

general power series, has radius of convergence R > 0 and thus defines a holomorphic function

f : BR (0) → C. Prove that there exists a holomorphic function g : BR (0) → C such that

f (z) = z n g(z) for all z ∈ BR (0),

and that g can be written as a power series with radius of convergence R.

We’ll use this fact when we study the zeros of holomorphic functions.

3.6. Consider the function f : C {0} → C defined by

f (z) =

sin z

.

z

Prove that we can extend f to a function on the whole of C (by defining f (0) to be a suitable

value in C) that is entire, i.e. holomorphic on the whole of C.

The ‘singularity’ of f at 0 in this example will be known as a ‘removable singularity’.

3.7.

(a) For R > 0, Compute

1

2i

Z

z̄ dz,

∂BR (0)

and show that it agrees with the area of the ball BR (0).

(b) For R := {z ∈ C : 0,

compute

Z

1

z̄ dz,

2i ∂R

and show that it agrees with the area of R.

This is not a fluke. We are seeing a couple of instances of a general fact that could be derived

from an appropriate form of Stokes’ theorem.

3.6

Exercises

3.8. For r > 0 and k ∈ Z let γ : [0, 2π] → C be the closed C 1 curve γ(θ) = reikθ that travels

anticlockwise k times around the circle of radius r. Prove that

Z

dz

= 2πi k.

γ z

37

4

Winding numbers

4.1

Winding numbers of continuous closed paths

VIDEO: Winding numbers of continuous closed paths

Watch the video for an instant explanation-by-pictures of what the winding number is!

In Section 3.3 we have discussed the function arg(z), and the issue that it is only defined modulo an

integer multiple of 2π. The following lemma will tell us that if we decide on a choice of arg(z) at one

point z, and then move along a continuous path/curve that stays away from 0, then this determines a

unique continuously varying choice of the argument along this path.

Lemma 4.1 (Lifting lemma). Suppose γ : [a, b] → C {0} is continuous, and fix θ0 ∈ R such that

γ(a) = |γ(a)|eiθ0 . Then there exists a unique continuous function θ : [a, b] → R such that θ(a) = θ0

and γ(t) = |γ(t)|eiθ(t) for all t ∈ [a, b].

For example, for a curve γ : [0, 2π] → C given by γ(t) = eit , if we choose θ0 = 0 rather than any

other value in 2πZ then θ(t) = t. In particular, even though the start and end points are the same, the

argument differs by 2π.

Those of you who have already studied Introduction to Topology will know how to prove this lemma.

There is an obvious ‘covering map’ R 7→ R/(2πZ) given by θ 7→ θ + 2πZ, and we are taking a lift of

the function arg ◦ γ : [a, b] → R/(2πZ).

For the Introduction to Topology course in 2020/21, Section 8.3 talks about the homotopy lifting

property, and its special case, Definition 8.6, the path lifting property. In Section 10, Proposition

10.1, you saw that covering maps satisfy the homotopy lifting property.

For those who have not taken the course Introduction to Topology, here is a self-contained proof.

The video could be useful in order to understand this proof!

Proof. First observe that if γ avoids a slit {−reiθ0 : r > 0} on the opposite side of the starting point,

then the existence of θ(t) is clear: We just ask θ(t) to take values in the interval (θ0 − π, θ0 + π).

In the general case, we are free to replace γ(t) by the curve γ̃(t) := γ(t)/|γ(t)| without changing the

argument. Since γ̃ is a continuous function from a closed interval, it is uniformly continuous, and by

dividing up [a, b] into a large enough number of equal intervals, we can be sure that γ̃ does not move

too far when restricted to each of these sub-intervals. More precisely, by taking n ∈ N large enough,

we can be sure that for each k ∈ {0, 1, . . . , n − 1} and each t ∈ [ck , ck+1 ], where ck := a + nk (b − a),

we have |γ̃(t) − γ̃(ck )| < 1.
The idea then is to do the lifting on each of these sub-intervals in turn. Indeed, the restriction of γ̃
to [c0 , c1 ] must avoid a slit {−reiθ0 : r > 0} on the opposite side of the starting point, so by the

38

4.2

Nearby closed paths have the same winding number

comment at the start of the proof we can find our function θ(t) at least for t ∈ [c0 , c1 ], with θ(c0 ) = θ0 .

At this point we can use θ(c1 ) as a new starting argument analogous to θ0 , and do our lifting on the

next interval [c1 , c2 ], where γ̃(t) avoids the new opposite slit {−reiθ(c1 ) : r > 0}. This extends

θ(t) to the interval [c0 , c2 ]. By repeating this process a total of n times, we obtain a lift to the whole

interval [c0 , cn−1 ] = [a, b].

To establish uniqueness of θ(t), suppose we have a second continuous lift θ̂(t) also with θ̂(a) = θ0 .

Then t 7→ θ(t) − θ̂(t) is a continuous function that vanishes at t = a, and takes values in 2πZ because

both θ(t) and θ̂(t) represent the argument, modulo 2π. Thus θ(t) − θ̂(t) = 0 for all t ∈ [a, b], as

required.

Lemma 4.1 allows us to unambigously define the change in argument as we move along a continuous

path.

Definition 4.2. Suppose γ : [a, b] → C {0} is continuous, and let θ : [a, b] → R be a function

arising in Lemma 4.1. We define

](γ) := θ(b) − θ(a).

The function θ was only defined up to a constant multiple of 2π that was determined by θ0 . However,

when we subtract θ(a) from θ(b) this unknown multiple of 2π will disappear.

Definition 4.3. Suppose γ : [a, b] → C {0} is a closed continuous path. Then we define the index

or winding number of γ around 0 to be

I(γ, 0) :=

1

](γ) ∈ Z.

2π

More generally, if w ∈ C and γ : [a, b] → C {w} is a closed continuous path then we define the

index of γ around w to be

1

I(γ, w) :=

](γw ),

2π

where γw : [a, b] → C {0} is the path γ translated to send w to the origin, i.e. γw (t) := γ(t) − w.

Example 4.4. For n ∈ Z, consider the curve γ : [0, 2π] → C defined by γ(θ) = reinθ , for some

r > 0, which winds around the origin n times in an anticlockwise direction. Then

I(γ, 0) = n.

Remark 4.5. In the case that γ maps into a region of C on which we can make a choice of arg(z),

for example if γ : [a, b] → C − {reiθ : r > 0} is a continuous closed path (see Section 3.3 for a

discussion of branch cuts) then we can choose θ(t) = arg(γ(t)), and we see that I(γ, 0) = 0. The

branch cut prevents γ from winding around the origin.

4.2

Nearby closed paths have the same winding number

39

4.3

Winding number and simply connected domains

VIDEO: Nearby closed paths have the same winding number

In this section we prove that if we have a closed path γ : [a, b] → C {0}, then a small-enough

perturbation of γ will wind round 0 the same number of times as γ itself.

In the video this will be completely self evident!

However, a little care is required. If γ goes very close to 0 then we must ensure that we perturb very

little so the path does not jump to the other side.

Lemma 4.6. Suppose γ : [a, b] → C Bε (0) is a continuous closed path, and γ̃ : [a, b] → C {0} is

a continuous closed path with |γ(t) − γ̃(t)| < ε for every t ∈ [a, b]. Then
I(γ, 0) = I(γ̃, 0).
Note that if γ : [a, b] → C {0} is a continuous closed path then we can always find some ε > 0 such

that the hypotheses of the lemma are satisfied. The smaller that is, the less we can perturb.

Proof. Let θ(t) and θ̃(t) be lifts of the arguments of γ(t) and γ̃(t) respectively, as given by Lemma

4.1. Define a continuous function α : [a, b] → R by α(t) := θ̃(t) − θ(t). By definition of winding

number, we have

1

1

[θ̃(b) − θ̃(a)] −

[θ(b) − θ(a)]

I(γ̃, 0) − I(γ, 0) =

2π

2π

(4.2.1)

1

=

(α(b) − α(a)) .

2π

Because the winding number is always an integer, the difference α(b) − α(a) must be an integer

multiple of 2π, and our task is to show that it is zero.

If this were not the case, then α(t) would either increase or decrease by at least 2π as t varied from

a to b. By continuity of α, we could be sure that there would exist some t0 ∈ [a, b] with α(t0 ) = π2 ,

modulo 2π, and we will use this to derive a contradiction.

By dividing the hypothesis |γ(t) − γ̃(t)| < ε by |γ(t)|, which is at least as large as ε, we find that
1−
which forces
γ̃(t)
γ(t)
γ̃(t)
< 1,
γ(t)
to have positive real part. Draw a picture or watch the video! But
iπ
γ̃(t)
γ(t)
= | γ̃(t)
|eiα(t) ,
γ(t)
and at t = t0 this will be purely imaginary because eiα(t0 ) = e 2 = i, giving a contradiction.
4.3
Winding number and simply connected domains
40
4.4
The winding number as an integral
VIDEO: Winding number and simply connected domains
In Definition 2.26 we recalled what it meant for an open subset Ω ⊂ C to be simply connected. The
most useful consequence of being simply connected for us will be:
Theorem 4.7. If an open set Ω ⊂ C is simply connected then for every w ∈ C Ω and every
continuous closed path γ : [a, b] → Ω, we have I(γ, w) = 0.
In the video we will draw a picture that makes this seem completely obvious.
Proof. In order to simply notation, we translate Ω and γ by −w in order to reduce to the case that
w = 0. Note that then 0 ∈
/ Ω.
Because Ω is simply connected, there exists a homotopy of γ to a constant path. More precisely, there
exists a continuous map h : [0, 1] × [a, b] → Ω such that for all t ∈ [a, b] and all s ∈ [0, 1] we have
h(0, t) = γ(t),
h(1, t) = γ(a),
h(s, a) = γ(a),
and
h(s, b) = γ(a).
(4.3.1)
In particular, for each s ∈ [0, 1], we have a simple closed curve γs : [a, b] → Ω defined by γs (t) :=
h(s, t). The theorem will be proved if we can show that the index I(γs , 0) is the same for each
s ∈ [0, 1], because I(γ0 , 0) = I(γ, 0) and I(γ1 , 0) = 0 since γ1 is a constant path.
Because h is continuous, and [0, 1] × [a, b] is closed, the image of h will be closed in Ω ⊂ C {0},
and in particular we can find ε > 0 so that this image does not intersect Bε (0).

Because h is continuous on its closed domain, it is also uniformly continuous. In particular, we can

pick δ > 0 so that whenever t ∈ [a, b] and s1 , s2 ∈ [0, 1] with |s1 − s2 | < δ, we must have
|γs1 (t) − γs2 (t)| = |h(s1 , t) − h(s2 , t)| < ε.
By Lemma 4.6, we then have I(γs1 , 0) = I(γs2 , 0). This implies that s 7→ I(γs , 0) is locally constant,
and then constant for all s ∈ [0, 1].
4.4
The winding number as an integral
VIDEO: The winding number as an integral
In the last minute or two of the video I refer to something that is ‘coming in the next section’. But
I subsequently decided to move that material to Corollary 3.12 back in Section 3.5. This makes the
ordering more logical.
If we have a simple closed path that is not just continuous but also piecewise C 1 , then we can characterise the winding number as an integral:
41
4.4
The winding number as an integral
Lemma 4.8. If w ∈ C and γ : [a, b] → C {w} is a closed piecewise C 1 curve, then
Z
1
dz
I(γ, w) =
.
2πi γ z − w
This expression for the winding number is often used as the definition. It is often easier to use in
rigorous proofs, but the definition we gave is more immediately visual, more general, and is useful
when considering homotopies.
In Example 3.13 and Q. 3.8 we showed that for r > 0, n ∈ Z, and γ : [0, 2π] → C defined by

γ(θ) = reinθ , which winds around the origin n times in an anticlockwise direction, we have

Z

1

1

dz = n.

2πi γ z

Thus by Example 4.4 the claimed formula at least works for I(γ, 0).

Proof of Lemma 4.8. By translation, we may assume that w = 0. We give the proof assuming that γ

is C 1 . The modifications to handle piecewise C 1 curves are straightforward. We must control

Z

Z b 0

dz

γ (t)

=

dt.

γ z

a γ(t)

Taking any function θ(t) from the Lifting lemma 4.1, so γ(t) = |γ(t)|eiθ(t) , we notice that because γ

is C 1 and keeps away from 0, the function θ(t) is also C 1 and we can compute

γ 0 (t) = eiθ(t)

and so

d

|γ(t)| + |γ(t)|iθ0 (t)eiθ(t) ,

dt

γ 0 (t)

d

=

log |γ(t)| + iθ0 (t).

γ(t)

dt

Integrating gives

Z

Z b

dz

d

0

=

log |γ(t)| + iθ (t) dt = 0 + i[θ(b) − θ(a)] = i](γ) = 2πiI(γ, 0)

dt

γ z

a

because γ is closed.

Given our definition of the winding number I(γ, w) as the number of times a curve γ winds around a

point w, it is intuitively obvious that it will be constant as we vary w continuously without touching

the image of γ. We formulate and prove a precise version of this assertion in the case that γ is

piecewise C 1 .

Lemma 4.9. Suppose γ : [a, b] → C is a piecewise C 1 closed curve. Then on each connected

component of C γ([a, b]), the function w 7→ I(γ, w) is constant.

42

4.4

The winding number as an integral

Note that as the continuous image of a closed set, we know that γ([a, b]) is closed, and so C γ([a, b])

is open.

Proof. It suffices to prove that the map w 7→ I(γ, w) is holomorphic, with derivative zero, for w not

in the image of γ (so that I(γ, w) is defined). That then implies that on each component of Cγ([a, b])

the function is constant, as required. To see this last implication, one can recall that a holomorphic

function with zero derivative has all its partial derivatives equal to zero, so it is locally constant and

hence constant on each connected component.

From the definition of

∂

∂ w̄

we have

∂

1

I(γ, w) =

∂ w̄

2πi

Z

γ

∂

∂ w̄

1

z−w

dz = 0.

Here we are using the fact that we can differentiate under the integral sign because the partial derivatives of the integrand with respect to the real and imaginary components of w are uniformly continuous

for z = γ(t), t ∈ [a, b], and w in a small neighbourhood of its starting point.

Differentiation under the integral sign was covered in Analysis 3, and I am not planning to examine

you on it in this context.

On the other hand

1

∂

I(γ, w) =

∂w

2πi

Z

γ

∂

∂w

1

z−w

by Corollary 3.12.

43

1

dz =

2πi

Z

γ

1

dz = 0

(z − w)2

4.5

Exercises

4.1. Given a continuous path γ : [a, b] → C {0}, we write −γ for the continuous path [a, b] 7→

C {0} defined by t 7→ γ(a + b − t), which reverses the direction.

(a) Verify that ](−γ) = −](γ).

(b) Now suppose additionally that γ is closed. What is I(−γ, w) in terms of I(γ, w)?

4.2. Suppose that Ω ⊂ C contains the closure of the ball Br (a) of radius r > 0 centred at a ∈ Ω.

Prove that for all z0 ∈ Br (a) we have

I(∂Br (a), z0 ) = 1

by showing that we may as well take z0 = a and computing.

Recall that ∂Br (a) refers to a curve γ passing once around ∂Br (a) in an anticlockwise direction. Please use the integral formulation of index, i.e. Lemma 4.8, to make it easier to formulate

a rigorous argument.

4.3. Prove that if γ : [a, b] → C is a closed piecewise C 1 curve then the set of points w ∈ Cγ([a, b])

for which I(γ, w) 6= 0 is bounded.

1

4.4. In Q. 3.4, hopefully you wrote (for w ∈ C {0}) the function z 7→ w−z

as a power series

P∞

−k−1 k

z valid for |z| < |w|. By replacing z by 1/z and setting w = 1/z0 , where
k=0 w
1
z0 ∈ Br (0) for some r > 0, obtain an expansion for z−z

that is valid for |z| > |z0 | and

0

converges uniformly for z in ∂Br . By integrating around ∂Br (0), prove that

I(∂Br (0), z0 ) = 1

thus reproving Q. 4.2.

5

Cauchy’s Theorem

Baron Augustin-Louis Cauchy (1789 – 1857).

5.1

Preamble and connection with Analysis 3

VIDEO: Cauchy’s theorem: Analysis 3 reminder

In Analysis 3 (Theorem 4.29 in 2019/20) you heard about the iconic theorem of Cauchy from which a

spectacular amount of wonderful theory gushes forth. To restate the particular form of the result that

you saw, we need the notion of simply connected from Definition 2.26.

Theorem 5.1 (Cauchy’s theorem on simply connected domains). Suppose Ω ⊂ C is open and simply

connected. Suppose further that f : Ω → C is holomorphic and γ is a piecewise C 1 closed curve in

Ω. Then

Z

f (z)dz = 0.

γ

Cauchy’s theorem has a huge number of applications. An example application that you saw in Analysis 3 (and that we’ll review shortly) is that a function f : Ω → C that is holomorphic is necessarily

infinitely differentiable. Amazing. This is nothing like what happens for real differentiable functions.

Just think of the function f : R → R that is zero for x < 0 and equal to x2 for x ≥ 0.
You also saw, in Analysis 3, a heuristic proof of Theorem 5.1 that, strictly speaking, requires a few
extra hypotheses. Indeed, you saw a proof somewhat along the lines that Cauchy originally gave. That
proof requires extra regularity for the curve γ (it needs to be simple and have nonvanishing derivative)
and extra regularity for the holomorphic function f (it needs to have continuous derivative in order to
apply Stokes’ theorem in its standard form). The required extra regularity for f will always hold, but
in order to prove this we will need Cauchy’s theorem itself!
In this section we would like to sketch the layout of the theory required to give a rigorous proof of a
slightly restricted form of this theorem. Indeed, in Theorem 5.7 we will prove it in the special case of
so-called star-shaped domains. In due course we will also see a much more general form of Cauchy’s
theorem that includes Theorem 5.1 as a special case.
Before doing that, we recall the following special case of Example 3.13 and Q. 3.8, which shows that
Cauchy’s theorem fails on C {0}, so the requirement that Ω is simply connected cannot simply be
dropped.
Example 5.2. Consider the holomorphic function f (z) = z1 on Ω := C{0}, and for r > 0 let

γ : [0, 2π] → C be the simple closed C 1 curve γ(θ) = reiθ that travels anticlockwise around the circle

of radius r. Then

Z

f (z)dz = 2πi.

γ

45

5.2

Goursat’s theorem – Cauchy’s theorem on triangles

Although you have already done a more complicated computation in Q. 3.8, let’s redo it in this special

case. Note that f (γ(θ)) = f (reiθ ) = r−1 e−iθ , and γ 0 (θ) = ireiθ , and so

Z

Z 2π

Z 2π

−1 −iθ

iθ

r e ire dθ = i

dθ = 2πi.

f (z)dz =

0

γ

5.2

0

Goursat’s theorem – Cauchy’s theorem on triangles

VIDEO: Goursat’s theorem: Cauchy’s theorem on triangles

Édouard Jean-Baptiste Goursat (1858-1936). Known for his contribution to the current rigorous

theory for Cauchy’s theorem and his spectacular moustache.

A first situation in which one rigorously proves Cauchy’s theorem for general holomorphic f is when

one heavily restricts the curves γ one allows and integrates around the boundary of a triangle. This

will then later be used as a tool in order to prove more general results such as Theorem 5.1. This

special case is named after Goursat, who came long after Cauchy.

Theorem 5.3 (Gour…

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