Complex Analysis
Xue-Mei Li
May 16, 2016
Contents
1
2
3
4
5
Complex Differentiation
1.1 The Complex Plane . . . . . . . . .
1.2 Complex Functions of One Variables
1.3 Complex Linear Functions . . . . .
1.4 Complex Differentiation . . . . . .
1.5 The ∂ and ∂¯ operator . . . . . . . .
1.6 Harmonic Functions . . . . . . . . .
1.7 Holomorphic Functions . . . . . . .
1.8 Rules of Differentiation . . . . . . .
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6
6
6
10
10
13
13
14
14
The Riemann Sphere and Möbius Transforms
2.1 Conformal Mappings . . . . . . . . . . . . . . . . . . . . . .
2.2 Möbius Transforms (Lecture 5) . . . . . . . . . . . . . . . . .
2.2.1 The Extended Complex Plane . . . . . . . . . . . . .
2.2.2 Properties of Möbius Transforms . . . . . . . . . . .
2.3 The Riemann Sphere and Stereographic Projection (lecture 7) .
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16
16
17
18
18
22
Power Series
3.1 Power series is holomorphic in its disc of convergence . .
3.2 Analytic Continuation (Lecture 8) . . . . . . . . . . . . .
3.3 The Exponential and Trigonometric Functions . . . . . . .
3.4 The Logarithmic Function and Power Function (Lecture 9)
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25
25
27
29
30
Complex Integration
4.1 Complex Integration . . . . . . . . . . . . . . . . . . . . .
4.2 Integration Along a Curve (lecture 9) . . . . . . . . . . . . .
4.3 Existence of Primitives (Lecture 9) . . . . . . . . . . . . . .
4.4 Goursat’s Lemma . . . . . . . . . . . . . . . . . . . . . . .
4.5 Cauchy’s Theorem: simply connected domains (Lecture 12)
4.6 Supplementary . . . . . . . . . . . . . . . . . . . . . . . .
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31
31
32
34
37
39
41
Cauchy’s Integral Formula
5.1 Keyhole Operation and Other Techniques (Lecture 12) . . . . . .
5.2 Cauchy’s Integral Formula . . . . . . . . . . . . . . . . . . . . .
5.3 Taylor Expansion, Cauchy’s Derivative Formulas . . . . . . . . .
5.4 Estimates, Liouville’s Thm and Morera’s Thm . . . . . . . . . . .
5.4.1 Supplementary . . . . . . . . . . . . . . . . . . . . . . .
5.5 Locally Uniform Convergent Sequence of Holomorphic Functions
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42
42
44
45
46
48
48
1
5.6
5.7
5.8
5.9
5.10
6
7
Schwartz Reflection Principle (Lecture 15) . . . .
The Fundamental Theorem of Algebra . . . . . .
Zero’s of Analytic Functions . . . . . . . . . . .
Uniqueness of Analytic Continuation (Lecture 17)
Supplementary . . . . . . . . . . . . . . . . . .
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49
51
52
55
55
Laurent Series and Singularities
6.1 Laurent Series Development (Lecture 17-18) . . .
6.2 Classification of Isolated Singularities (Lecture 19)
6.2.1 Poles . . . . . . . . . . . . . . . . . . . .
6.2.2 Essential Singularities . . . . . . . . . . .
6.3 Meromorphic Function . . . . . . . . . . . . . . .
6.3.1 Supplementary . . . . . . . . . . . . . . .
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56
56
61
62
64
64
64
Winding Numbers and the Residue Theorem
7.1 The Index of a Closed Curve (Lecture 22)
7.1.1 Supplementary . . . . . . . . . .
7.2 The Residue Theorem (Lecture 23) . . . .
7.3 Compute Real Integrals . . . . . . . . . .
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66
67
69
71
72
Fundamental Theorems
8.1 The Argument Principle (Lecture 24) . . . . . . .
8.2 Rouché’s Theorem . . . . . . . . . . . . . . . .
8.2.1 Supplementary . . . . . . . . . . . . . .
8.3 The Open mapping Theorem (Lecture 25) . . . .
8.3.1 Supplementary . . . . . . . . . . . . . .
8.4 Bi-holomorphic Maps on the Disc (lecture 25-26)
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74
74
75
76
76
78
79
The Riemann Mapping Theorem
9.1 Hurwitz’s Theorem (lecture 26) . . . . . . . . . .
9.2 Family of Holomorphic Functions (Lecture 28) .
9.3 The Riemann Mapping Theorem (Lecture 28-29)
9.4 Supplementary . . . . . . . . . . . . . . . . . .
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81
81
82
84
86
10 Special Functions
10.1 Constructing Holomorphic Functions by Integration (Lecture 30) . . .
10.2 The Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 The zeta function(Lecture 30) . . . . . . . . . . . . . . . . . . . . .
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89
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9
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Prologue
This is the lecture notes for the third year undergraduate module: MA3B8. If you need
not be motivated, skip this section.
Complex Analysis is concerned with the study of complex number valued functions
with complex number as domain. Let f : C → C be such a function. What can we say
about it? Where do we use such
√ an analysis?
The complex number i = −1 appears in Fourier Transform, an important tool in
analysis and engineering, and in the Schrödinger equation,
~2 ∂ 2 ψ
∂ψ
=−
+ V (x)ψ(t, x),
∂t
2m ∂ 2 x
a fundamental equation of physics, that describes how a wave function of a physical
system evolves. Complex Differentiation is a very important concept, this is allured to
by the fact that a number of terminologies are associated with ‘complex differentiable’.
A function, complex differentiable on its domain, has two other names: a holomorphic map and an analytic function, reflecting the original approach. The first meant
the function is complex differentiable at every point, and the latter refers to functions
with a power series expansion at every point. The beauty is that the two concepts are
equivalent. A complex valued function defined on the whole complex domain is an
entire function. Quotients of entire functions are Meromorphic functions on the whole
plane.
A map is conformal at a point if it preserves the angle between two tangent vectors at that point. A complex differentiable function is conformal at any point where
its derivative does not vanish. Bi-holomorphic functions, a bi-jective holomorphic
function between two regions, are conformal in the sense they preserve angles. Often by conformal maps people mean bi-holomorphic maps. Conformal maps are the
building blocks in Conformal Field Theory. It is conjectured that 2D statistical models at criticality are conformal invariant. An exciting development is SLE, evolved
from the Loewner differential equation describing evolutions of conformal maps. The
Schramann-Loewner Evolution (also known as Stochastic Loewner Evolution, abbreviated as SLE ) has been identified to describe the limits of a number of lattice models
in statistical mechanics. Two mathematicians, W. Werner and S. Smirnov, have been
awarded the Fields medals for their works on and related to SLE.
Complex valued functions are built into the definition for Fourier transforms. For
f : R → R,
Z ∞
1
ˆ
f (k) = √
e−ikx dx, k ∈ R.
2π ∞
Fourier transform extends the concept of Fourier series for period functions, is an important tool in analysis and in image and sound processing, and is widely used in electrical engineering.
i~
3
A well known function in number theory is the Riemann zeta-function,
ζ(s) =
∞
X
1
.
s
n
n=1
The interests in the Riemann-zeta function began with Euler who discovered that the
Riemann zeta function can be related to the study of prime numbers.
ζ(s) = Π
1
.
1 − p−s
The product on the right hand side is over all prime numbers:
Π
1
1
1
1
1
1
1
=
·
·
·
·
...
....
1 − p−s
1 − 2−s 1 − 3−s 1 − 5−s 1 − 7−s 1 − 11−s
1 − p−s
The Riemann-zeta function is clearly well defined for s > 1 and extends to all complex
numbers except s = 1, a procedure known as the analytic /meromorphic continuation
of a real analytic function. Riemann was interested in the following question: how
many prime number are below a given number x? Denote this number π(x). Riemann
found an explicit formula for π(x) in his 1859 paper in terms of a sum over the zeros
of ζ. The Riemann hypothesis states that all non-trivial zeros of the Riemann zeta
function lie on the critical line s = 12 . The Clay institute in Canada has offered a prize
of 1 million dollars for solving this problem.
In symplectic geometry, symplectic manifolds are often studied together with a
complex structure. The space C is a role model for symplectic manifold. A 2-dimensional
symplectic manifold is a space that looks locally like a piece of R2 and has a symplectic
form, which we do not define here. We may impose in addition a complex structure Jx
at each point of x ∈ M . The complex structure Jx is essentially a matrix s.t. −Jx2 is the
identity and defines a complex structure and leads to the concept of Khäler manifolds.
Finally we should mention that complex analysis is an important tool in combinatorial enumeration problems: analysis of analytic or meromorphic generating functions
provides means for estimating the coefficients of its series expansions and estimates for
the size of discrete structures.
Topics
Holomorphic Functions, meromorphic functions, poles, zeros, winding numbers (rotation number/index) of a closed curve, closed curves homologous to zero, closed curves
homotopic to zero, classification of isolated singularities, analytical continuation, Conformal mappings, Riemann spheres, special functions and maps. Main Theorems:
Goursat’s theorem, Cauchy’s theorem, Cauchy’s derivative formulas, Cauchy’s integral formula for curves homologous to zero, Weirerstrass Theorem, The Argument
principle, Rouché’s theorem, Open Mapping Theorem, Maximum modulus principle,
Schwartz’s lemma, Mantel’s Theorem, Hürwitz’s theorem, and the Riemann Mapping
Theorem.
References
• L. V. Ahlfors, Complex Analysis, Third Edition, Mc Graw-Hill, Inc. (1979)
4
• J. B. Conway. Functions of one complex variables.
• T. Gamelin. Complex Analysis, Springer. (2001)
• E. Hairer, G. Wanner, Analyse Complexe et Séries de Fourier.
http://www.unige.ch/hairer/poly_complexe/complexe.pdf
• G. J. O. Jameson. A First course on complex functions. Chapman and Hall,
(1970).
• E. M. Stein and R. Shakarchi. Complex Analysis. Princeton University Press.
(2003)
Acknowledgement. I would like to thank E. Hairer and G. Wanner for the figures in
this note.
5
Chapter 1
Complex Differentiation
1.1
The Complex Plane
The complex plane C = {x + iy : x, y ∈ R} is a field with addition and multiplication,
on which is also defined the √
complexpconjugation x + iy = x − iy and modulus (also
called absolute value) |z| = z z̄ = x2 + y 2 . It is a vector space over R and over C
with the norm |z1 − z2 |.
We will frequently treat C as a metric space, with distance d(z1 , z2 ) = |z1 − z2 |,
and so we understand that a sequence of complex numbers zn converges to a complex
number z is meant by that the distance |zn − z| converges to zero. The space C with
the above mentioned distance is a complete metric space and so a sequence converges
if and only if it is a Cauchy sequence. Since
|zn − z|2 = |Re(zn ) − Re(z)|2 + |Im(zn ) − Im(z)|2 ,
zn converges to z if and only if the real parts of (zn ) converge to the real part of z and
the imaginary parts of (zn ) converge to the imaginary part of z.
In polar Coordinates z ∈ C can be written as z = reiθ where r = |z| and θ is a real
number, called the argument. We note specially Euler’s formula:
eiθ = cos(θ) + i sin(θ),
so arg z is a multi valued function. It is standard to take the principal value −π <
Argz ≤ π, a rather arbitrary choice.
Since e2πik = 1 for k an integer, the nth root function is multi-valued. If
ωk = e
2πk
n i
k = 0, 1, . . . , n − 1,
,
the nth roots of the unity, then
1
1
θ
(reiθ ) n = r n ei n ωk .
1.2
Complex Functions of One Variables
To discuss complex differentiation of a function, we request that it is defined on a subset
of the complex plane C which is open. By a set we would usually mean a subset of the
6
complex plane C. A set U is open if about every point in U there is a disc contained
entirely in U . We further assume that the set is connected, otherwise we could treat it
as a separate function on each connected subset.
A subset of C is connected if any two points from the subset can be connected by a
continuous curve which lies entirely within the subset.
An open set is connected if and only if it is not disconnected in the sense that it is
not the union of two disjoint open sets.
Definition 1.2.1 By a region we mean a connected open subset of C. By a proper
region we mean an open connected subset of C that is not the whole complex plane.
From now on, by a function we mean a function f : U → C where U is a region.
By an open disc we mean {z : |z − z0 | < r} where z0 ∈ C and r > 0. A closed
disc is {z : |z −z0 | ≤ r} where z0 ∈ C and r ≥ 0. The unit disc centred at 0 is denoted
by
D = {z : |z| < 1}.
Other frequently seen open sets are the deleted discs and the annulus
{z : 0 < |z − z0 | < r},
{z : r1 < |z − z0 | < r2 },
and polygons.
Example 1.2.1 Given z0 ∈ C, the function f : C → C given by the formula f (z) =
z + z0 is said to be a translation.
As a set we may wish to identify a complex number s + it with the pair of real
numbers (s, t), so C is identified with R2 . Since, for z = x + iy and c = s + it,
c(x + iy) = (sx − ty) + i(tx + sy),
the map z → cz is represented by a a linear map:
x
s −t
x
→
.
y
t s
y
Multiplication by i is the same as multiply by J on the left, where
0 −1
J=
.
1 0
Example 1.2.2 Given c ∈ C, the function f (z) = cz is of the form below. For z =
x + iy, c = |c| eiθ ,
x
cos(θ) − sin(θ)
x
7→ |c|
.
y
sin(θ) cos(θ)
y
This is the composition of a rotation by an angle θ and a scaling by |c|. This map
preserves the angle between two vectors, i.e. it is a conformal map.
7
w = cz
z
c
c
1
−1
1
1
−1
−1
1
−1
Figure 1.1: Graph by E. Hairer and G. Wanner
Example 1.2.3 Define f (z) = z 2 whose maximal domain of definition is C. Write
f = u + iv. Then
u(x, y) = x2 − y 2 , v(x, y) = 2xy.
It is easy to see that f takes the horizontal lines y = b where b 6= 0 to parabolas on
the w plane facing right. Solve the equations: x2 − b2 = u and v = 2xb to see
u=
1 2
v − b2 .
4b2
Also, f takes the vertical lines x = a where a 6= 0 to parabolas on the w plane facing
left.
1
u = 4a2 − 2 v 2 .
4a
These two sets of parabolas intersect at right angles, see Figure 1.2.
If b = 0, the line y = 0 is mapped to the right half of the real axis; the line x = 0 is
mapped to the left half of the real axis. We observe that at 0, these two curves, images
of the real and complex line from the domain space, fail to intersects with each other
at a right angle. (0 is the only point at which f 0 = 0, explaining the orthogonality and
failing of the orthogonality where the image curves meet, to which we return later.)
The function f (z) = z 2 is not injective. It takes the line y = b and y = −b to the
same image. To see this better let us use polar coordinates. Then f (reiθ ) = r2 e2iθ .
When restricted to the positive half complex plane, f is injective. In fact,
f : C+ → C \ [0, ∞)
√
is a bijection with inverse f −1 (w) = w. When restricted to the negative half plane
f : C− → C \ [0, ∞),
√
√
f is also injective with inverse f −1 (w) = ω2 w = − w.
8
z=
√
w = z2
w
1
−1
1
0
−1
−1
1
−1
Figure 1.2: Graph by E. Hairer and G. Wanner
√
Example 1.2.4 Define f on C \ (−∞,
0] by f (w) = w, the principal brach of the
√
square root function. So f (reiθ ) = reiθ/2 , −π < θ < π. It has another formula:
p
f (w) = |w|ei(Argw/2) , w ∈ C \ (−∞, 0].
It maps the slit w
√ plane into the right half of the z-plane. The other branch of the
square root is − w. It is possible to glue the two slit domains together to form a
complex manifold, known as a Riemann surface, so in one sheet (chart) the function
takes the value of one brach and in the other we use the other brach in a way f changes
continuously as w changes.
Example 1.2.5 The map f (z) = z1 , the inversion map, is defined on C \ {0}. It is easy
to see that f takes circles centred at the origin to circles centres at the origin. It take the
locus of the solutions of |z − z0 | = r to that of a circle in the w-plane, see Proposition
2.2.6 and Example sheets.
Example 1.2.6 (Möbius Transforms) Let a, b, c, d ∈ C where ad 6= bc. Define
f (z) =
az + b
.
cz + d
If c = 0 the domain of f is C otherwise it is C \ {− dc }.
Exercise 1.2.7
1. Prove that for any real number r, not 1, the equation
|z − z1 | = r|z − z2 |
determines a circle.
2. Prove that any Möbius transform is a composition of translations, scalings, and
inversions.
9
Later we will see that Möbius Transforms can be considered as maps on the extended
complex plane, the Riemann sphere.
1.3
Complex Linear Functions
We identify R2 with C. A function T : R2 → R2 is real linear if for all z1 , z2 , z ∈ R2 ,
T (z1 + z2 ) = T (z1 ) + T (z2 ),
T (rz) = rT (z),
∀r ∈ R
A map T : C → C is complex linear if for all z1 , z2 , z ∈ C,
T (z1 + z2 ) = T (z1 ) + T (z2 ),
T (kz) = kT (z),
∀k ∈ C
Proposition 1.3.1 A real linear function T : R2 → R2 is complex linear iff
T (i) = iT (1).
We now look at the matrix representations. Every real linear map is of the form
x
a b
x
→
.
y
c d
y
If k = s + it, the complex linear map T (z) = kz is given by
x
s −t
x
T
=
.
y
t s
y
For every real linear map T there exists a unique pair of complex numbers λ and µ
such that
T (z) = λz + µz̄,
which is complex linear if and only if µ = 0. Furthermore,
1
1
1
1
λ=
(a + ic) + (b + id) , µ =
(a + ic) − (b + id)
2
i
2
i
1.4
Complex Differentiation
Let f = u + iv, defined in a region U . When C is identified as R2 we may treat u and
v as real valued functions on R2 . In this way f is an R2 valued function of two real
variables x and y. Then f is (real) differentiable at (x0 , y0 ) if there exists a linear map
(df )(x0 ,y0 ) : R2 → R2 and a function φ such that
x − x0 x − x0
f (x, y) = f (x0 , y0 ) + (df )(x0 ,y0 )
+ φ(x, y) (1.4.1)
y − y0
y − y0 10
where φ satisfies φ(x0 , y0 ) = 0 and lim(x,y)→(x0 ,y0 ) φ(x, y) = 0. Note that
x − x0 y − y0 = |z − z0 |.
The linear function is represented by the Jacobian matrix.
∂x u ∂y u
Jf (x0 , y0 ) =
(x0 , y0 ).
∂x v ∂y v
The partial derivatives of f are denoted by
∂x u
∂y u
∂x f =
, ∂y f =
.
∂x v
∂y v
(1.4.2)
Treated as a complex function,
∂x f = ∂x u + i∂x v,
∂y f = ∂y u + i∂y v.
Definition 1.4.1 A function f : U → C is complex differentiable at z0 if there exists a
complex number f 0 (z0 ) and a function ψ with ψ(z0 ) = 0 and limz→z0 ψ(z) = 0, such
that
f (z) = f (z0 ) + (df )z0 (z − z0 ) + ψ(z) |z − z0 | .
(1.4.3)
The number f 0 (z0 ) is the derivative of f at z0 .
Equivalently, f is complex differentiable at z0 with derivative f 0 (z0 ) if and only if
f 0 (z0 ) = lim
w→0
f (w + z0 ) − f (z0 )
.
w
Example 1.4.1 f (z) = z is differentiable. So are any polynomials in z.
This follows from
f (z0 + z) − f (z0 )
=1
z
and chain rules.
Example 1.4.2 The function f (z) = z̄ is not complex differentiable.
Proof Note
f (z0 + z) − f (z0 )
z̄
= =
z
z
which means limz→0
f (z0 +z)−f (z0 )
z
1,
−1,
does not exist.
if Im(z) = 0
if Re(z) = 0,
Definition 1.4.2 A function is differentiable in U if it is differentiable everyewhere
in U .
Notation. A function f : Rn → Rm is C r if it is r times differentiable and its
partial derivatives of order less or equal to r are continuous.
11
Figure 1.3: Handwriting by Riemann
Theorem 1.4.3
1. If f : U → R is complex differentiable at z0 = x0 + iy0 then f
is real differentiable at (x0 , y0 ) and the Cauchy-Riemann Equations hold at z0 :
∂y u = −∂x v.
∂x u = ∂y v,
Also,
f 0 (z0 ) = ∂x u + i∂x v =
(1.4.4)
1
(∂y u + i∂y v).
i
2. If f : U → C is real differentiable and satisfies the Cauchy-Riemann equation
at a point (x0 , y0 ) ∈ U then f is complex differentiable at z0 = x0 + iy0 . In
particular, if u, v : R2 → R2 are C 1 functions satisfying the Cauchy-Riemann
equations in U then f = u + iv is complex differentiable in U .
Proof (1) Write f 0 (z0 ) = s + it. Then by the definition, (1.4.3),
s −t
f (z) = f (z0 ) +
(z − z0 ) + ψ(z) |z − z0 | .
t s
This is (1.4.1) with
(df )(x0 ,y0 ) =
So f is real differentiable with
∂x u
∂x v
s −t
.
t s
∂y u
s −t
=
.
∂y v
t s
Thus the Cauchy-Riemann equation follows and
f 0 (z0 ) = s + it = ∂x u + i∂x v = ∂y v − i∂y u.
(2) We have (1.4.1),
f (x, y) = f (x0 , y0 ) + (df )(x0 ,y0 )
x − x0
y − y0
x − x0 .
+ φ(x, y) y − y0 By the Cauchy-Riemann equation the Jacobian matrix is the following form
∂x u −∂x v
J=
,
∂x v ∂x u
12
and represent the complex linear map: multiplication by f 0 (z0 ) := ∂x u + i∂x v, Hence
f (z) = f (z0 ) + f 0 (z0 )(z − z0 ) + φ(z) |z − z0 | .
This implies f is complex differentiable at z0 . If u, v are C 1 , then f = (u, v) is
differentiable and the previous statement applies.
The Cauchy Riemann equation can also be written as ∂x f = 1i ∂y f .
1.5
The ∂ and ∂¯ operator
Given a function f , we have
f (x, y) = f
z + z̄ z − z̄
,
2
2i
.
This inspires the notation :
∂z =
1
1
(∂x + ∂y ),
2
i
∂z̄ =
1
1
(∂x − ∂y ).
2
i
(1.5.1)
¯ It is clear that ∂z̄ f = 0 is the CauchyIt is common to denote ∂z by ∂ and ∂z̄ by ∂.
Riemann equation
We can reformulate the earlier theorem using these notations. Suppose that f is
complex differentiable at z then f is real differentiable at z and,
¯ (z) = 0,
∂f
1.6
f 0 (z) = ∂z f (z).
Harmonic Functions
Definition 1.6.1 A real valued function u : R2 → R is a harmonic function if ∆u = 0
where ∆ = ∂xx + ∂yy is the Laplacian.
Proposition 1.6.1 If u, v are C 2 functions and satisfies the Cauchy-Riemann equations
∂y u = −∂x v,
∂x u = ∂y v,
then u, v are harmonic functions. Consequently u, v are C ∞ .
Proof We differentiate the Cauchy-Riemann equation to see
∂yy u = −∂yx v.
∂xx u = ∂xy v,
Consequently ∂xx u + ∂yy u = 0. Similarly, ∂xx v + ∂yy v = 0. From standard theory
in PDE, a solution of the elliptic equation ∆u = 0 is C ∞ .
Later we see that if f is differentiable in a region, it has derivatives of all orders.
So the conditions u, v ∈ C 2 can be reduced to C 1 .
13
1.7
Holomorphic Functions
Definition 1.7.1 A function f : U → C is said to be holomorphic on U if it is differentiable at every point of U ; it is holomorphic at z0 if it is holomorphic in a disc
containing z0 . A function f : C → C is said to be entire if it is complex differentiable
at every point of C.
For the next corollary we use the following, thia is also Corollary 4.3.2
Proposition 1.7.1 A holomorphic function in a region with vanishing derivative must
be a constant.
Proof To see this, we first note that f has vanishing Jacobian matrix, and so it derivatives along the coordinate directions vanishes. So f is constant along any line segment
parallel with x and y axis. But any two points in a region can be connected by piecewise line segments parallel to either x and y axis, and so the values of f at these two
points must be the same.
Example 1.7.2 Let U be a region in C. If f : U → R is a real valued function, then f
is not holomorphic in U unless f is a constant.
Proof Let f = u + iv where v vanishes identically. If f is holomorphic, by the
Cauchy-Riemann equation, ∂x u = ∂y u = 0 and f 0 (z) = ∂y v + i∂x v = 0, and f must
be a constant.
1.8
Rules of Differentiation
Theorem 1.8.1 If f, g are differentiable at z0 , the derivatives indicated below exist at
z0 and the relations stated below hold when evaluated at z0 .
1. (kf )0 = kf 0 , for any k ∈ C
2. (f + g)0 = f 0 + g 0
3. (f g)0 = f g 0 + f 0 g
4. (f /g)0 =
gf 0 −f g 0
g2
provided g(z0 ) 6= 0.
Theorem 1.8.2 Suppose that g is differentiable at z0 and f is differentiable at g(z0 )
then the composition f ◦ g is differentiable at z0 and
(f ◦ g)0 (z0 ) = f 0 (g(z0 )) g 0 (z0 ).
Observe that if the Jacobian matrix of a complex differentiable function f represents complex multiplication, i.e. it is of the form
∂x u ∂y u
J=
,
−∂y u ∂x u
14
then so is its inverse:
J −1 =
1
det J
∂x u −∂y u
.
∂y u ∂x u
Since f 0 (z0 ) = ∂x u(z0 ) − i∂y u(z0 ), and
1
1
∂x u(z0 ) + i∂y u(z0 )
=
=
(∂x u(z0 ) + i∂y u(z0 )).
f 0 (z0 )
(∂x u)2 (z0 ) + (∂y u(z0 ))2
det J(z0 )
In conclusion, J(z0 ) represents f 0 (z0 ), J −1 (z0 ) represents
This leads to the following theorem.
1
f 0 (z0 ) .
Theorem 1.8.3 Suppose that f : U → C is complex differentiable and u, v have
continuous partial derivatives. Suppose f 0 (z0 ) 6= 0 for some z0 ∈ U .
• Then there exists a disc U around z0 such that f : U → f (U ) is a bijection,
f (U ) is open and f −1 : f (U ) → U is continuous.
• Furthermore f −1 is complex differentiable on f (U ) and
(f −1 )0 (f (z)) =
1
,
f 0 (z)
z ∈ U.
Proof The first part of the statement follows from real analysis. To see f −1 is differentiable, write w0 = f (z0 ), w = f (z). Since f and f −1 are continuous, w → w0 is
equivalent to z → z0 . Since f 0 (z0 ) 6= 0,
f −1 (w) − f −1 (w0 )
=
w − w0
1
f (z)−f (z0 )
z−z0
.
Take w → w0 we see that the limit on the left hand side exists and
f −1 (w0 ) =
1
1
= 0 −1
.
f 0 (z0 )
f (f (w))
Remark 1.8.4 Later we see that if f is complex differentiable, it is infinitely differentiable. If f is one to one then f 0 does not vanish. See section 8.3.
15
Chapter 2
The Riemann Sphere and
Möbius Transforms
2.1
Conformal Mappings
Definition 2.1.1 A parameterized curve in the complex plane is a function z : [a, b] →
C where [a, b] is closed interval of R.
If z(t) = x(t) + iy(t) its derivative is z 0 (t) = x0 (t) + iy 0 (t).
Definition 2.1.2 The parameterized curve : [a, b] → C is smooth if z 0 (t) exists and is
continuous on [a, b]. We assume furthermore z 0 (t) 6= 0.
The derivatives at the ends are understood to be one sided derivatives. From now on by
a curve we mean a smooth curve.
If z 0 (t) does not vanish the curve has a tangent at this point, whose direction is
determined by arg(z 0 (t)).
Definition 2.1.3 Let z1 : [a1 , b1 ] → C and z2 : [a2 , b2 ] → C be two smooth curves
intersecting at z0 . The angle of the two curves is the angle of their derivatives at this
point.
They are given by the difference of the arguments of their derivatives. If z1 (t1 ) =
z2 (t2 ) = z0 , their angle at the point z0 is: arg(z20 (t2 )) − arg(z10 (t1 )).
Definition 2.1.4 A map f : U → C is conformal at z0 if it preserves angles, i.e. if z1
and z2 are two curves meeting at z0 , the angle from f ◦ z1 to f ◦ z2 at f (z0 ) are the
same as the angle from z1 to z2 at z0 .
Example 2.1.1 The linear map f (z) = kz where k 6= 0, is a conformal map as it is
composed of scaling by |k| and rotating by the angle arg(k). c.f.Example 1.2.2
Example 2.1.2 f (z) = z̄ is not a conformal map. This map reverses orientation.
Theorem 2.1.3 If f : U → C is holomorphic at z0 and f 0 (z0 ) 6= 0, then f is conformal
at z0 .
16
Proof Let z1 : [a1 , b1 ] → C and z2 : [a2 , b2 ] → C be two smooth curves intersecting
at z0 : for t1 ∈ [a1 , b1 ] and t2 ∈ [a2 , b2 ], z1 (t1 ) = z2 (t2 ) = z0 . Let γ1 (t) = f ◦ z1 (t)
and γ2 (t) = f ◦ z2 (t). Since f 0 (z0 ) does not vanish, γ1 and γ2 have well defined
tangents which are:
d
|t=t1 f ◦ z1 = f 0 (z1 (t1 ))z10 (t1 ) = f 0 (z0 )z10 (t1 )
dt
d
γ 0 (t2 ) = |t=t2 f ◦ z2 = f 0 (z2 (t2 ))z20 (t2 ) = f 0 (z0 )z20 (t2 ).
dt
γ 0 (t1 ) =
Multiply z10 (t1 ) and z10 (t1 ) by the non-zero complex number f 0 (z0 ) preserves angles
between the two vectors, as well as their orientation, c.f. Example 2.1.1, so the angle
from γ1 to γ2 at f (z0 ) is the same as the angle from z1 to z2 at z0 .
Note also,
arg(γ 0 (t1 )) = arg(f 0 (z0 )) + arg(z10 (t1 )).
2.2
Möbius Transforms (Lecture 5)
A polynomial P (z) = a0 + a1 z + . . . an z n , where ai ∈ C, is an entire function. The
roots zn are the zero’s of P . If there are exactly r roots coincide, this root is said to
have order r. In light of Theorem 2.1.3 it is interesting to know where lie the zero’s of
P 0 (z).
By the fundamental theorem of Algebra, which we prove later (Theorem 5.7.2),
P (z) = 0 has a complete factorisation:
P (z) = an (z − z1 ) . . . (z − zn ).
Suppose that P and Q are two polynomials without common factors and define the
rational function
P (z)
.
f (z) =
Q(z)
Then f is defined and is complex differentiable everywhere except at the zeros’ of Q.
The zero’s of Q are the poles of f . We now look at rational functions with one pole
and one zero.
Definition 2.2.1 The following collection of maps are Möbius transforms
az + b
:
ad − bc 6= 0, a, b, c, d, ∈ C .
cz + d
If ad − bc = 0, az+b
cz+d is a constant function, and are hence excluded. If f (z) =
is a Möbius transform, its maximal domain is: C \ {− dc }. Since
f 0 (z) =
ad − bc
6= 0,
(cz + d)2
f is a conformal map.
17
az+b
cz+d
2.2.1
The Extended Complex Plane
To make the statements neat we add a point at infinity to C and define the extended
complex plane to be
C∗ = C ∪ {∞}
with the convention:
1
= ∞,
0
1
= 0,
∞
a + ∞ = ∞,
a − ∞ = ∞,
and for a 6= 0, a · ∞ = ∞ · a = ∞.
2.2.2
Properties of Möbius Transforms
Let f (z) =
az+b
cz+d .
We extend the Möbius transform f from C to C∗ by defining:
d
f (− ) = ∞,
c
f (∞) =
a
c
if c 6= 0.
If c = 0, f (z) = f1 (az + b) is defined on the whole plane, then we define
f (∞) = ∞,
if c = 0.
The function f has an inverse
f −1 (w) =
dw − b
.
−cw + a
Note that multiply a, b, c, d, by a non-zero number λ does not change the function
f (z) =
azλ + bλ
az + b
=
.
cz + d
cλz + dλ
Hence we may eliminate one parameter and assume that ad − bc = 1. We define
az + b
M=
:
ad − bc = 1, a, b, c, d, ∈ C .
cz + d
Theorem 2.2.1 The set M of Möbius transforms is a group under composition. Each
Möbius transform is a composition of the following maps:
(1) translation: z 7→ z + a for some complex number a;
(2) composition of scaling and rotation:
z 7→ kz, some k ∈ C, k 6= 0.
(3) Inversion: z 7→ z1 .
Proof For the group we check the following:
• f (z) = z is the identity. (a = 1, b = c = 0, d = 1)
• If f (z) =
az+b
cz+d
∈ M, then
f −1 (w) =
dw − b
∈ M.
−cw + a
18
āz+b̄
• If f (z) = az+b
cz+d ∈ M and g(z) = c̄z+d¯ ∈ M. Then f ◦ g =
the complex numbers A, B, C, D are given by
A B
a b
ā b̄
=
.
C D
c d
c̄ d¯
For the second part of the statement, if c = 0,
f (z) =
az+b
d
Az+B
Cz+D
∈ M where
= ad z + db . If c 6= 0,
a bc − ad 1
az + b
= +
.
cz + d
c
c2 z + dc
z+1
is called the Cayley transform. It takes C \ {1}
Example 2.2.2 The map f (z) = z−1
to itself, f : C \ {1} → C \ {1} is a bijection and f −1 = f . Let us consider f as a map
on C ∗ by setting f (1) = ∞, f (∞) = 1. Note
f (x + iy) =
x2 + y 2 − 1
−2y
+
i.
(x − 1)2 + y 2
(x − 1)2 + y 2
Let γ = {x2 + y 2 = 1} with γ + and γ − denote respectively the upper and lower half
of the circle. Then,
• f sends {−1, 0, 1} to {0, −1, ∞} respectively.
• f sends the upper circle to the lower half of the imaginary axis.
• f sends x-axis to x-axis. It send the x-axis within the unit disc to the negative
x-axis.
• f sends the upper half of the unit disc to the third quadrant.
• f sends the lower circle to the upper half of the imaginary axis.
• f sends the lower half of the unit disc to the second quadrant.
• f sends the exterior of the unit circle to the right half of the plane.
If z2 , z3 , z4 are distinctive points in C∗ we associate to it the Möbius transform
f (z) =
z − z3 z2 − z3
(z − z3 )(z2 − z4 )
/
=
,
z − z4 z2 − z4
(z − z4 )(z2 − z3 )
if z2 , z3 , z4 ∈ C.
If one of these points is the point at infinity the map is interpreted as following:
 z − z3

,
if z2 = ∞


z − z4


 z −z
2
4
,
if z3 = ∞
f (z) =
z
−
z

4




 z − z3 ,
if z4 = ∞.
z2 − z3
Note that if z2 , z3 , z4 ∈ C,
f (z2 ) = 1,
f (z3 ) = 0,
19
f (z4 ) = ∞.
Also,
If z2 = ∞,
f (∞) = 1, f (z3 ) = 0, f (z4 ) = ∞,
If z3 = ∞,
f (z2 ) = 1, f (∞) = 0, f (z4 ) = ∞,
If z4 = ∞,
f (z2 ) = 1, f (z3 ) = 0, f (∞) = ∞.
We denote by Fz2 z3 z4 this map. Note it sends {z2 , z3 , z4 } to {1, 0, ∞}.
Lemma 2.2.3 A Möbius transform can have at most two fixed points unless f (z) is
the identity map.
2
Proof We solve for az+b
cz+d = z, equivalently cz + (d − a)z − b = 0. This has at most
two solutions(use polynomial long division/ the Euclidean algorithm).
Proposition 2.2.4 For any two sets of distinctive complex numbers {z2 , z3 , z4 } and
{w2 , w3 , w4 } in C∗ , there exists a unique Möbius transform taking zi to wi for i =
2, 3, 4.
Proof We know Fz2 z3 z4 takes {z2 , z3 , z4 } to {1, 0, ∞}, and the inverse map of Fw2 w3 w4
takes {1, 0, ∞} to {w2 , w3 , w4 }. The composition Fw−1
◦Fz2 z3 z4 takes {z2 , z3 , z4 }
2 w3 w4
to {w2 , w3 , w4 }.
To prove this map is unique, suppose f, g are two Möbiums transform sending
{z2 , z3 , z4 } to {w2 , w3 , w4 }. Then f ◦ g −1 (wi ) = f (zi ) = wi . The Möbius transform
f ◦ g −1 has three fixed points: w1 , w2 , w3 . By Lemma 2.2.3, f ◦ g −1 is the identity
map and f = g identically.
Corollary 2.2.5 For any distinctive complex numbers {z2 , z3 , z4 } in C ∗ , the Möbius
transform Fz2 ,z3 ,z4 is the only Möbius transform that takes {z2 , z3 , z4 } to {1, 0, ∞}.
Proposition 2.2.6 Let r, c be real numbers, k ∈ C. Then the equation
r|z|2 + k̄z + kz̄ + c = 0
• represents a line if r = 0 and k 6= 0.
• a circle if r 6= 0, and |k|2 ≥ rc.
• The circle equation is |z + kr | =
r 6= 0 and |k|2 < rc.
1
r
p
|k|2 − rc, whose locus is an emptyset if
This is clear by expanding z = x + iy in x and y.
Definition 2.2.2 The locus of the points of r|z|2 − k̄z − kz̄ + c = 0, if non-empty, is
called a circleline.
We see later this definition is not merely a simplification of terminologies. Both circles
and extended lines in the plane correspond to circles in the Riemann sphere.
Lemma 2.2.7 Let R be a real number, z1 , z2 complex numbers. The locus of the
equation
|z − z1 | = r|z − z2 |
20
represents a circle if r 6= 1. If z = x + iy, z1 = x1 + iy1 and z2 = x2 + iy2 then
(x − x1 )2 + (y − y1 )2 = r(x − x2 )2 + r(y − y2 )2 .
If r = 1, the equation is
2(x2 − x1 )x + 2(y2 − y1 )y = |z2 |2 − |z1 |2 ,
representing a line if z1 6= z2 . It is the set of points which are equi-distance from z1
and z2 , i.e. the line perpendicular to the line segment [a, b] and passing its mid-point.
Proposition 2.2.8 A Möbius transform maps a circleline to a circleline.
Proof Since a Möbius transform is the composition of translation, multiplication by
a non-zero complex number and inversion we only need to prove it for each of these
maps. The inverse of such transformations are of the same type.
A translation, z 7→ z + a, takes a circleline to a circleline: the image of
r|z|2 + k̄z + kz̄ + C = 0,
in the z-plane is precisely the locus of the equation below in the w-plane:
r|w − a|2 + k̄(w − a) + k(w̄ − ā) + C = 0,
i.e.
r|w|2 + (k − a)w + (k − a)w̄ + r|a|2 − (k̄a + kā) + C = 0.
Note that r|a|2 − (k̄a + kā) + C is a real number. Multiplication by a complex number
is a composition of scaling with rotation, it clearly takes a circleline to a circleline.
Finally we work with the inversion z 7→ z1 . It takes |z| = r to |z| = 1r trivially. Let
r
a 6= 0. If w is in the image of |z − a| = r then | w1 − a| = r, i.e. |w − a1 | = |a|
|w| which
is a circleline, by Lemma 2.2.7. Let us take a line k̄z + kz̄ + C = 0. The equation of its
image w = z1 satisfies k̄ w1 + k w̄1 + C = 0 which is equivalent to k̄ w̄ + kw + C|w|2 = 0
representing a circle if C 6= 0 and a line otherwise.
Exercise 2.2.9 Given r, c ∈ R and k ∈ C, and the equation
r|z|2 − k̄z − kz̄ + c = 0.
Identify its image under the transform w = k 0 z where k 0 is a non-zero complex number.
Definition 2.2.3 The cross ratio of z1 , z2 , z3 , z4 , denoted by [z1 , z2 , z3 , z4 ], is the complex number:
[z1 , z2 , z3 , z4 ] := Fz2 z3 z4 (z1 ).
In other words, it is
[z1 , z2 , z3 , z4 ] =
(z1 − z3 )(z2 − z4 )
,
(z1 − z4 )(z2 − z3 )
interpreted appropriately if one of them is ∞.
Proposition 2.2.10 Let z1 , z2 , z3 , z4 be distinct points in C∗ . Then [z1 , z2 , z3 , z4 ] is a
real number if the four points lie in a circle or on the extended line R ∪ {∞}.
21
Proof If [z1 , z2 , z3 , z4 ] is a real number, Fz2 ,z3 ,z4 maps the four points z1 , z2 , z3 , z4 to
respectively [z1 , z2 , z3 , z4 ], 1, 0, ∞, all on the extended x-axis. The map (Fz2 ,z3 ,z4 )−1
takes the 4 points [z1 , z2 , z3 , z4 ], 1, 0, ∞ back to z1 , z2 , z3 , z4 . Note a Möbius transform takes the extended line to a circleline, so the four points lie in a circleline.
If the four points lie in a circleline, then the map Fz2 ,z3 ,z4 takes the circleline
to a circleline. This will be the line determined by (1, 0, ∞), the x-axis. Hence
Fz2 ,z3 ,z4 (z1 ) must be a real number.
2.3
The Riemann Sphere and Stereographic Projection
(lecture 7)
The purpose of the section is to give a concrete geometric representation of the extended plane as the Riemann sphere. In particular we observe that the point at infinity
is just represented as a point in the sphere. Let us denote by S 2 the unit sphere in R3 :
S 2 = {(X, Y, Z) : X 2 + Y 2 + Z 2 = 1}.
We fix the north pole N = (0, 0, 1) and associate with each P on S 2 \{N } with a point
π(P ) on the plane which is the intersection of the line from N to P with the plane.
Proposition 2.3.1 The stereographic projection from S 2 → C∗ is :
X + iY
,
1−Z
π(N ) = ∞.
(2.3.1)
2Re(z) 2Im(z) |z|2 − 1
,
,
).
|z|2 + 1 |z|2 + 1 |z|2 + 1
(2.3.2)
π((X, Y, Z)) =
The inverse map is given by
π −1 (z) = (
Proof Suppose that P = (X, Y, Z), write (x, y, 0) = π(P ). The line equation connecting N, P and π(P ) is given by:
(x, y, z) = (0, 0, 1) + t(X − 0, Y − 0, Z − 1).
Setting z = 0 we see
t=
1
,
1−Z
x = tX =
X
,
1−Z
22
y = tY =
Y
,
1−Z
(2.3.3)
proving π((X, Y, Z)) =
π −1 (z). We use
X+iY
1−Z
. Let z = x + iy be a point in C, we find its inverse
t2 (X 2 + Y 2 + Z 2 ) = t2 .
Thus
x2 + y 2 = t2 (1 − Z 2 ) =
1 − Z2
1+Z
=
.
(1 − Z)2
1−Z
Finally
Z=
1
1−Z
By t =
x2 + y 2 − 1
|z|2 − 1
=
.
x2 + y 2 + 1
|z|2 + 1
and (2.3.3), we see X = x(1 − Z) =
2x
|z|2 +1 ,
Y = y(1 − Z) =
2y
|z|2 +1 .
The following is an easy exercise.
Proposition 2.3.2 The antipodal point to a point (X, Y, Z) in S 2 is (−X, −Y, −Z). If
z ∈ C∗ corresponds to a point in S 3 then − z̄1 corresponds to the antipodal point in S 2 .
Definition 2.3.1 If z1 , z2 ∈ C∗ we define the stereographic distance to be
d(z1 , z2 ) = |π −1 (z1 ) − π −1 (z2 )|.
If p = (X, Y, Z) and p0 = (X 0 , Y 0 , Z 0 ) are points in the sphere, their distance is:
|P − P 0 | = |X − X 0 |2 + |Y − Y 0 |2 + |Z − Z 0 |2 = 2 − 2(XX 0 + Y Y 0 + ZZ 0 ) (2.3.4)
If z 0 = ∞, then π −1 (z 0 ) = (0, 0, 1), X 0 = 0, Y 0 = 0 and Z 0 = 1. Consequently,
s
|z|2 − 1
2
=p
.
d(z, ∞) = 2 − 2 2
|z| + 1
|z|2 + 1
This agrees with the intuition, z → ∞ means |z| → ∞. If z, z 0 ∈ C, apply (2.3.4), and
use (2.3.2) we see
0
0
2
(d(z, z )) = 2 − 2
¯0
z−z̄
z −z
|z|2 − 1 |z 0 |2 − 1
(z + z̄) (z 0 + z 0 )
· 02
+ 2i
· 0 2i
+ 2
·
2
|z| + 1 |z | + 1 |z| + 1 |z | + 1 |z| + 1 |z 0 |2 + 1
Since
(|z|2 + 1)(|z 0 |2 + 1) − (|z|2 − 1)(|z 0 |2 − 1)
= 2|z| + 2|z | (z + z̄)(z 0 + z 0 ) − (z − z̄)(z 0 − z 0 ) = 2zz 0 + 2z̄z 0 .
2
0 2
Also, |z|2 + |z 0 |2 − zz 0 − z̄z 0 = (z − z 0 )(z̄ − z 0 ) = |z − z 0 |2 . Cleaning up the right
hand side we obtain:
2|z − z 0 |
p
d(z, z 0 ) = p
.
|z|2 + 1 |z 0 |2 + 1
Definition 2.3.2 The space S 2 is the Riemann sphere. A circle on S 2 is the intersection
of a plane with S 2 .
Proposition 2.3.3 A circle on S 2 corresponds to a circle or a line on C∗ .
23
!
Proof let us take a plane:
AX + BY + CZ + D = 0
where A, B, C, D are real numbers. Note that the north pole passes through the plane
if and only C + D = 0.
|z|2 −1
2y
2
Let z = x+iy ∈ C. Then a point π −1 (z) = ( |z|2x
2 +1 , |z|2 +1 , |z|2 +1 ) on S satisfies
the plane equation if and only if
2xA + 2BY + (|z|2 − 1)C + (|z|2 + 1)D = 0.
Rearrange the equation:
(C + D)(x2 + y 2 ) + 2xA + 2By + (D − C) = 0,
(2.3.5)
which represents a circle in the plane or an empty set when C + D 6= 0. If the plane
intersects with S 2 , it is not empty and so is a circle. If C + D = 0 this is a line on the
plane.
Let us consider a circle or an extended line in C ∗ . It is of the form:
Ã(x2 + y 2 ) + B̃x + C̃y + D̃ = 0
(2.3.6)
where Ã, B̃, C̃, D̃ are real numbers. Let us solve for A, B, C, D:
C + D = Ã, 2A = B̃, 2B = C̃, D − C = D̃.
Then (2.3.6) is equivalent to (2.3.5) which means the corresponding points of the circleline on the plane satisfies
AX + BY + CZ + D = 0,
and their image by π −1 line on a circle in S 2 .
If the plane passes through the origin we have a great circle. This is so if and only
if D = 0 and we have
2B
2A
x+
y = 1.
(x2 + y 2 ) +
C
C
The plane passes through the north pole if and only if C + D = 0 in which case
the circle projects to a line.
24
Chapter 3
Power Series
P∞
Definition 3.0.1 A series of complex numbers n=0 an is said to converge if the parPN
P∞
tial sum n=0 an converge. It is said to converge absolutely if n=0 |an | converges.
P∞
P∞
P∞
Evidently n=0 an is convergent is equivalent to both n=0 Re(an ) and n=0 Im(an )
converge.
Proposition 3.0.1 If
P∞
n=0
an converges absolutely, then it is convergent.
Just note that |Re(an )| ≤ |an | and |Im(an )| ≤ |an |. Follow this with the standard
comparison test.
3.1
Power series is holomorphic in its disc of convergence
P∞
Let us consider a power series n=0 an (z − z0 )n where an , z0 and z are complex
numbers. For simplicity let us take z0 = 0.
P∞
Theorem 3.1.1 Let n=0 an z n be a power series where an ∈ C. There exists R ∈
[0, ∞], such that the following holds:
(1) If |z| < R, the series converges absolutely.
(2) If |z| > R, the series diverges.
Moreover, there is Hadamard’s formula:
1
1
= lim sup(|an |) n
R
n→∞
1
with the convention ∞
= 0 and 10 = ∞. The region {|z| < R} is called the disc of
convergence and R its radius of convergence.
1
Proof Suppose A = lim supn→∞ (|an |) n is such that 0 < A < ∞. Then there exists
1
N such that for n ≥ N , |an | n ≤ A. If |z| < A1 , then there exists δ > 0 such that
P∞
1
1
A
|z| < A+δ
and for n ≥ N , |an | n |z| ≤ A+δ
< 1, and n=0 |an ||z|n is convergent. If
25
1
1
A
and |an | n |z| ≥ A−δ
> 1 for
|z| > A1 , there exists 0 < δ < A such that |z| > A−δ
P∞
n
n ≥ N , it follows that n=0 an z is divergent.
1
If lim supn→∞ (|an |) n = ∞, then for any non-zero z, |an ||z|n does not converge
to 0 as n → ∞ and the power series is divergent for all z 6= 0. ( There is a sequence
1
2
ank with |ank | nk > |z|
).
1
If lim supn→∞ (|an |) n = 0, then for any |z| and for any 0 < <
N such that |an |
1
n
1
2|z|
there exists
≤ , for all n ≥ N , and
|an ||z|n ≤ n |z|n ≤
1
.
2n
The power series converges absolutely for any z.
By composing with translation
statement of the
P∞z → z − z0 , we may translatePthe
∞
theorem from the power series n=0 an |z|n to the power series n=0 an |z − z0 |n .
P∞
Theorem 3.1.2 The power series f (z) = n=0 an (z − z0 )n defines a holomorphic
function in its disc of convergence. Furthermore,
f 0 (z) =
∞
X
n an (z − z0 )n−1
n=1
and f 0 has the same radius of convergence as f .
Proof If R is the radius of convergence
for f , then using Hadamard’s formula we see
P∞
the radius of convergence for n=1 n an (z − z0 )n−1 is R. Take z from its disc of
convergence, {z : |z − z0 | < R}. Define
fN (z) =
N
X
an (z − z0 )n ,
0
fN
(z) =
n=0
N
−1
X
n an (z − z0 )n−1 .
n=1
Then for h ∈ C,
∞
f (z + h) − f (z) X
n−1 −
n an (z − z0 )
h
n=1
∞
X an (z + h − z0 )n − an (z − z0 )n fN (z + h) − fN (z)
0
− fN
(z) +
≤ h
h
n=N +1
+
∞
X
n|an (z − z0 )n−1 |.
n=N
The
right hand side is the remainder term of the convergent series
P∞ last term on then−1
n
a
(z
−
z
)
. Given > 0, there exists N0 such that if n ≥ N0 , this last
n
0
n=1
term is less than /3. Furthermore there exists a number δ0 > 0 and 0 < A < R such
that |z + h − z0 | < A for |h| ≤ δ0 . In the following we use the identity:
an − bn = (a − b)(an−1 + an−2 b + . . . + bn−1 ),
∞
∞
X
X
(z + h − z0 )n − (z − z0 )n an (z + h − z0 )n − an (z − z0 )n ≤
|an | h
h
n=N +1
n=N +1
26
≤
≤
∞
X
|an | (z + h − z0 )n−1 + (z + h − z0 )n−2 (z − z0 ) + . . . + (z − z0 )n−1 n=N +1
∞
X
|an |nAn .
n=N +1
The right P
hand side is again the tail of a convergent series, hence there exists N1 >
∞
N0 such that n=N1 |an |nAn−1 ≤ /3. Finally, fN1 is a differentiable function,
fN1 (z + h) − fN1 (z)
lim
− SN (z) = 0.
h→0 h
f (z+h)−fN1 (z)
By choosing h sufficiently small, N1
−
S
(z)
< 13 . The proof is
N
h
complete.
Corollary 3.1.3 A power series function f (z) =
P∞
n=0
an (z − z0 )n is infinitely dif-
ferentiable in its disc of convergence. Furthermore an =
3.2
f (n) (z0 )
.
n!
Analytic Continuation (Lecture 8)
Definition 3.2.1 A function f : U → C is said to be analytic, or has a power series
expansion, at z0 ∈ U , if there exists a power series with positive radius of convergence
such that
f (z) =
∞
X
an (z − z0 )n ,
for z in a neighbourhood of z0 .
n=0
We say f is analytic on U if it has a power series expansion at every point of U .
Example 3.2.1 The function f (z) = z1 is analytic on C \ {0}. Note
for all |z| < 1. It is clear f has the power series expansion at z = 1:
∞
X
(1 − z)n ,
1
1−z
=
P∞
n=0
zn
|z − 1| < 1.
n=0
If z0 is any non-zero number, take z with |z0 − z| < |z0 |, then
∞
X
1
1
1
1
=
=
·
=
(−1)n (z0 )−n−1 (z − z0 )n .
z
z0 − (z0 − z)
z0 1 − z0z−z
n=0
0
The power series converges for any z with |z0 − z| < |z0 | (in particular z 6= 0). Hence
f is analytic.
P∞
Theorem 3.2.2 The power series function f (z) = n=0 an (z − z0 )n is analytic in its
disc of convergence D = {z : |z − z0 | < R}. In fact for w ∈ D,
f (z) =
∞
X
f (n) (w)
(z − w)n ,
n!
n=0
27
∀z ∈ D(w, R − |w − z0 |).
Proof Take w ∈ D. Note that |w − z0 | < R and let z satisfy |z − w| < R − |w − z0 |.
We expand the power series
!
∞
∞
n X
X
X
n
n
k
n−k
f (z) =
an (z − w + w − z0 ) =
an
(z − w) (w − z0 )
k
n=0
n=0
k=0
!
∞
∞ X
X
n
=
a (w − z0 )n−k (z − w)k .
k n
k=0
n=k
To justify the exchange of the order in the above computation, we bound the the partial
sum and use the rearrangement of double series lemma below:
N X
n
X
|an |
n=0 k=0
N
X
n
|z − w|k |w − z0 |n−k =
|an |(|z − w| + |w − z0 |)n
k
n=0
≤
∞
X
n
|an |(|z − w| + |w − z0 |) <
n=0
∞
X
|an |rN < ∞,
n=0
where r is a number smaller than R. Hence the partial sum up to N is bounded by
P
∞
N
n=0 |an |r . Then
f (z) =
∞
X
k
bk (z − w) ,
∞
X
where bk =
k=0
n=k
n
an
(w − z0 )n−k .
k
It is clear f (k) (w) = bk k!.
Lemma 3.2.3 (Double Series Lemma) Suppose there exists a number M such that
N X
N
X
|aij | ≤ M
i=0 j=0
Then all linear arrangements of the double series converge absolutely to the same number.
a00
+
a10
+
a20
+
a30
+
:
=
v0
+ a01
+
+ a11
+
+ a21
+
+ a31
+
:
=
+ v1
+ a02
+
+ a12
+
+ a22
+
+ a32
+
:
=
+ v2
+ a03
+
+ a13
+
+ a23
+
+ a33
+
:
=
+ v3
+...
=
+...
=
+...
=
+...
=
+...
=
s0
+
s1
+
s2
+
s3
+
:
=
???
Figure 3.1: Figure From E. Hairer and G. Wanner
28
(3.2.1)
Definition 3.2.2 If f : V → C is a function where V is a subset of C and g : U → C
is an analytic function in a region U with V ⊂ U . If f, g agree on V we say g is an
analytic continuation of f into the region U .
The set V is not required to be a region. We may wonder which functions has an
analytic continuation.
P
Example 3.2.4 If f (x) =
an (x − P
x0 )n is a real power series function with radius
of convergence R. We define g(z) =
an (z − x0 )n . Then g has the same radius of
convergence R ( use Hadamard’s formula for R). So g is an analytic continuation (also
known as analytic extension) of f from (−R, R) to the disc {z : |z| < R}.
Example 3.2.5 If P (x) is a polynomial with one real variable then P (z) is its analytic
continuation into C.
P∞
Example 3.2.6 Let h(z) = n=0 (1 − z)n , |z − 1| < 1. Then f (z) =
continuation of h into the punctured complex plane C \ {0}.
1
z
is an analytic
Later we will study zeros of analytic function and conclude a function defined
on any connected set containing an accumulation point can have only one analytic
continuation.
3.3
The Exponential and Trigonometric Functions
The exponential functions and trigonometric functions are analytic continuations of
their corresponding functions on the real line.
Definition 3.3.1 We define the following function by power series:
∞
X
zn
e =
,
n!
n=0
∞
X
z 2n+1
sin(z) =
(−1)
,
(2n + 1)!
n=0
z
n
cos(z) =
∞
X
(−1)n
n=0
z 2n
.
(2n)!
They are entire functions. By adding two series together we see that
sin(z) =
eiz − e−iz
,
2i
cos(z) =
eiz + e−iz
,
2
and Euler’s formula:
eiz = cos(z) + i sin(z).
We also define the following functions: sinh(z) =
z 2n
n=0 (2n)! .
P∞
z
−z
z 2n+1
n=0 (2n+1)! and cosh(z) =
z
−z
= e +e
. Most properties
2
P∞
Note that sinh(z) = e −e
and cosh(z)
2
for the corresponding real trigonometric functions are inherited by the complex valued
trigonometric functions. For example the zeros of sin(z) are at nπ. But sin(z) is not a
bounded function, nor is cos(z).
29
Theorem 3.3.1 A power series f (z) =
zn
n=0 n!
P∞
satisfies
f (z + w) = f (z)f (w)
for all z, w, z + w. In particular,
e2kπi = 1,
ez+2kπi = ez ,
k = 0, ±1, ±2, . . . .
Note that ex+iy = ex eiy and eiy for y ∈ (−π, π) traces out a circle without the
point on the left real axis. Let
U = {z : −π < Imz < π}
Then
ez : U → C \ {re±iπ : r ≥ 0}
is a bijection.
Figure 3.2: Graph by E. Hairer and G. Wanner
3.4
The Logarithmic Function and Power Function (Lecture 9)
Definition 3.4.1 The principal branch of the logarithm is the inverse of ez on the slit
plane C \ {re±iπ : r ≥ 0}. log(z) = log |z| + i arg(z),
arg(z) ∈ (−π, π).
Theorem 3.4.1 The logarithmic function defined above is holomorphic on its domain
of definition and (log z)0 = z1 .
Proof Apply Theorem 1.8.2 to the exponential function from U to the slit domain. Other branches of log z include: log z = log |z|+i arg(z)+2kπ, arg(z) ∈ (−π, π).
Definition 3.4.2 For λ ∈ C we define z λ = eλ log z .
30
Chapter 4
Complex Integration
If a continuous function has a primitive in a region, then its integral along any closed
piecewise smooth curve vanishes. The converse holds in a star region (more generally,
in a simply connected region): if a continuous function integrate to zero along any
triangle inside the region, then it has a primitive. Goursat’s theorem states that the integral of a holomorphic function in a region indeed integrate to zero along any triangle
who and whose interior is contained in the region. From this we see Cauchy’s theorem
for a star region: if f is holomorphic in a star region, then it integrates to zero along
any closed smooth curve γ whose interior is contained entirely in the region. Cauchy’s
theorem is in fact valid for any simply connected region.
Every point in a region U has a disc around it, contained entirely in U . So every point in U has a star region neighbourhood (i.e. a region is ‘locally simply connected’). In another word, a holomorphic function integrates to zero along any closed
smooth curve with sufficiently small enclosure. The distinction between the local and
the global null integral property relates to homotopy theory as well as to de Rham’s
cohomology theory built on closed and exact differential forms, both are related to the
concept of ‘simply connectedness of a region’. On the complex plane, the simply connectedness can be explained visually which is also responsible for the beauty of the
theory on the plane. It is perhaps confusing in the beginning when confronted with
different version’s and various forms of Cauchy’s formulas, the rule of thumb is the
following. That a function is differentiable at a point is a local property, we could pick
our disc as small as we like. A value for an integral along a closed interval is a global
property: it depends on the region enclosed by the curve.
4.1
Complex Integration
By a C 1 function we mean a differentiable function with continuous derivative. The
derivatives at the ends of an interval are one sided derivatives.
Definition 4.1.1 A parameterized smooth curve is a C 1 function, z : [a, b] → C. It is
piecewise smooth if z is continuous on [a, b] and there exist ti s.t. a = t0 < t1 < · · · <
tn = b s.t. z is smooth on each sub-interval [ti , ti+1 ].
A piecewise smooth curve consisting of a finite number of smooth pieces, joined at the
ends. We sometimes abbreviate ‘a piecewise smooth curve’ to ‘a smooth curve’. A
31
parameterized curve has an orientation: it is the direction a point on the curve travels
as the parameter t increases.
Definition 4.1.2 Two parameterizations γ : [a, b] → C and γ 0 : [a0 , b0 ] → C are
equivalent if there exists a C 1 bijection α : [a, b] → [a0 , b0 ] such that α0 (t) > 0 and
γ = γ 0 ◦ α.
The condition α0 (t) > 0 means orientation is preserved. The family of all equivalent
parameterizations determine a smooth oriented curve.
Example 4.1.1 The circle {|z − z0 | = r} has the obvious parameterizations:
z = z0 + eiθ ,
0 ≤ θ ≤ 2π.
The orientation of the curve is anticlockwise (it is a positively oriented curve). The
curve
z = z0 + e−iθ ,
0 ≤ θ ≤ 2π
is negatively oriented.
If f : [a, b] → C is a continuous function and f (t) = u(t) + iv(t) then
Z
b
Z
f (t)dt =
a
Lemma 4.1.2
b
Z
u(t)dt + i
a
b
v(t)dt.
a
Z
Z
b
b
f (t)dt ≤
|f (t)|dt.
a
a
Rb
Proof Let θ be the principle argument of the complex number a f (t)dt. Then
Z
!
Z b
Z b
Z b
b
−iθ
−iθ
−iθ
f (t)dt = e
f (t)dt =
e f (t)dt = Re
e f (t)dt
a
a
a
a
Z b
Z b
=
Re e−iθ f (t) dt ≤
|f (t)|dt.
a
a
4.2
Integration Along a Curve (lecture 9)
Definition 4.2.1 Let f : U → C be a continuous function. Let γ be a smooth curve
contained in U with parameterization z : [a, b] → C. We define the integral of f along
γ to be:
Z
Z b
f (z)dz =
f (z(t))ż(t)dt.
γ
a
32
Let us write f = u + iv and z(t) = x(t) + iy(t). Then
f (z(t))ż(t) = u(z(t))ẋ(t) − v(z(t))ẏ(t) + i (u(z(t))ẏ(t) + v(z(t))ẋ(t)) .
Hence
Z
b
Z
f (z)dz =
γ
u(x(t), y(t))ẋ(t) − v(x(t), y(t))ẏ(t) dt +
a
b
Z
i
u(x(t), y(t))ẏ(t) + v(x(t), y(t))ẋ(t) dt.
a
The integral
R
γ
f (z)dz can also be defined directly by the Riemann sums:
n−1
X
f (z(si ))(z(si+1 ) − z(si ))
i=0
where a = s0 < s1 < · · · < sn = b is a partition on [a, b]. If the Riemann sum
Rb
converges as the size of the partition converges to zero, the limit is a f (z)dz.
Proposition 4.2.1 The integral
R
γ
f (z)dz is independent of the parameterization.
Proof Let z : [a, b] → C and z̃ : [a0 , b0 ] → C be two parameterizations of γ. Let
α : [a0 , b0 ] → [a, b] be a C 1 bijection such that α0 (t) > 0 and z̃ = z ◦ α. Then
b0
Z
f (z̃(t))
a0
Z
d
z̃(t)dt =
dt
Z
b0
f (z ◦ α(t))
a0
b0
Z
f (z ◦ α(t))ż(α(t))α̇(t)dt =
=
a0
d
z(α(t))dt
dt
b
f (z(s))ż(s)ds.
a
Definition 4.2.2 The length of the curve γ is:
Z
b
|z 0 (t)|dt.
length(γ) =
a
The length of the curve is also independent of the parameterization, be an argument
similar to that in the proposition above.
Definition 4.2.3 On a piecewise smooth curve γ, consisting of a finite number of
smooth curves γi , we define
Z
XZ
f (z)dz =
f (z)dz.
γ
γi
i
Theorem 4.2.2 The following properties hold.
(1)
Z
Z
(k1 f + k2 g)(z)dz = k1
γ
f (z)dz + k2
γ
33
Z
g(z)dz.
γ
(2)
Z
Z
f (z)dz = −
f (z)dz,
γ−
γ
where γ − is the curve γ with reversed orientation, e.g. z − (t) = z(a + b − t).
(3)
Z
f (z)dz ≤ sup |f (z)| · length of (γ).
z∈γ
γ
A curve is simple if it does not intersect with itself except at the end points. We
would be interested in the contour of a region, e.g. the contour of a disc is the circle, traveled anticlockwise once. By a circle, we usually mean traveling along it, anti
clockwise, once.
Example 4.2.3 Let γ be the circle |z − z0 | = r with positive orientation. Let n be an
integer. Then
Z
0,
n 6= −1
n
(z − z0 ) dz =
2πi,
n = −1.
γ
Proof Let us take the parameterization z = z0 + reit , 0 ≤ t ≤ 2π.
Z
(z − z0 )n dz =
γ
Z
2π
(reit )n
0
n+1
=r
(
=
Z
i
d
(reit )dt
dt
2π
ei(n+1)t dt
0
R 2π
R 2π
rn+1 i 0 cos((n + 1)t)dt − rn+1 0 sin((n + 1)t)dt = 0, n 6= −1
R 2π
i 0 dt = 2πi,
n = −1.
4.3
Existence of Primitives (Lecture 9)
Definition 4.3.1 A curve γ : [a, b] → C is closed if γ(a) = γ(b).
Definition 4.3.2 A function f : U → C where U is a region is said to have a primitive
F if F : U → C is holomorphic on U and F 0 = f .
The Fundamental theorem of Calculus statesRthat if g : [a, b] → R is a continuous
x
functions, then G0 (x) = g(x) where G(x) = a g(t)dt. Unlike for real differentiability, the existence of a primitive is a much stronger property. To begin with, given two
points on a plane, there are many paths leading from one to the other. This is related
to a path independent property. Later we see that if f has a primitive in a region, it is
itself complex differentiable in this region.
Rule of thumb. If you have a real valued function g of one variable x with explicit
formulation which does not involve the number i, denote f the function obtained by
replacing x with z. Then f is a ‘candidate analytic continuation’ of g. Compute the
formal anti-derivative of g and denote it by G. Now in G replace x by z and denote
34
the function by F . Then F is a ‘candidate primitive’ for f . Once you have a primitive,
check it out: is it actually complex differentiable in the desired region? If so, compute
its derivative by the rule of calculus.
Theorem 4.3.1 If f : U → C is continuous and has a primitive F , if γ is a piecewise
C 1 curve in U begins at w1 and ends at w2 then
Z
f (z)dz = F (w2 ) − F (w1 ).
γ
In particular, if γ is a closed curve then
Z
f (z)dz = 0.
γ
Proof Let z : [a, b] → C be a parameterization of γ. We first assume that it is C 1 and
discuss the piecewise C 1 curve later. Then
Z
Z b
Z b
0
f (z(t))z (t)dt =
F 0 (z(t))z 0 (t)dt
f (z)dz =
a
γ
Z
=
a
a
b
d
F (z(t))dt = F (z(b)) − F (z(a)) = F (w2 ) − F (w1 ).
dt
If γ is a piecewise smooth curve, joined by smooth curves on each subinterval of t0 =
a < t1 < · · · < tn = b, then we have a telescopic sum:
Z
f (z)dz
γ
=[F (z(tn )) − F (z(tn−1 ))] + · · · + [F (z(t2 )) − F (z(t1 ))] + [F (z(t1 )) − F (z(t0 ))]
=F (z(b)) − F (z(a)) = F (w2 ) − F (w1 ).
Corollary 4.3.2 If f : U → C is holomorphic, where U is a connected open set, and
f 0 = 0 identically on U then f is a constant.
Proof Let us fix a point z0 ∈ U . Let z ∈ U be any other point. Recall U is path
connected: we can connect z0 to z by a piecewise smooth curve γ. Since f 0 = 0 is a
continuous function,
Z
f 0 (z)dz = 0.
f (z) − f (z0 ) =
γ
The converse to the vanishing of integral theorem is at the heart of complex integration, for which we must restrict the region. We work with a sub class of regions of
the simply connected region, called the star regions.
Definition 4.3.3 A region U is said to be a star region if it has a centre C, by which
we mean for all z ∈ U , the line segment
{(1 − t)C + tz : 0 ≤ t ≤ 1}
belongs to U .
35
Note that z(t) = (1 − t)C + tz is a prameterization of the line segment from C to z.
Star regions include discs, triangles, rectangles and more generally convex sets. A
star region is simply connected, by which we mean any simple closed curve in that
region can be continuously deformed to a point. Many regions such as polygons can
be divided into star regions, which allow us to conclude statements for star regions for
more general regions.
Figure 4.1: Graph by E. Hairer and G. Wanner
Let us denote by [C, z] the line segment from C the z.
Theorem 4.3.3 (Integrability Criterion) Let U be a star region with a centre C and
f : U → C a continuous function. Suppose that
Z
f (ζ)dζ = 0
T
for any triangle T contained entirely in U , and with C one of its vertex. Then f has a
primitive in U . In particular
Z
F (z) =
f (ζ)dζ, z ∈ U.
[c,z]
R
Proof The line segment [c, z] is contained in U , so F (z) = [c,z] f (ζ)dζ makes
sense. Let z0 ∈ U . Take z close to z0 so that the triangle region with vertex z0 ,
z and C are contained entirely in U . Denote T this triangle. By the assumption,
36
Z
0=
Z
Z
f (ζ)dζ =
T
Z
f (ζ)dζ +
f (ζ)dζ +
[z0 ,C]
[C,z]
Thus
f (ζ)dζ.
[z,z0 ]
Z
F (z) = F (z0 ) −
f (ζ)dζ.
[z,z0 ]
Since −
R
[z,z0 ]
f (z0 )dζ =
R
[z0 ,z]
f (z0 )dζ = f (z0 )(z − z0 ),
Z
F (z) = F (z0 ) + f (z0 )(z − z0 ) +
(f (ζ) − f (z0 ))dζ.
[z0 ,z]
Define ψ(z0 ) = 0 and for z 6= z0 ,
R
ψ(z) =
[z0 ,z]
(f (ζ) − f (z0 ))dζ
|z − z0 |
.
Then
R
[z0 ,z] (f (ζ) − f (z0 ))dζ |ψ(z)| = ≤ max |(f (ζ) − f (z0 ))|.
ζ∈[z0 ,z]
|z − z0 |
Since f is continuous, limz→z0 maxζ∈[z0 ,z] |(f (ζ) − f (z0 ))| = 0. This proves that F
is differentiable at z0 with F 0 (z0 ) = f (z0 ). Since z0 is an arbitrary point in U , F is
holomorphic in U and a primitive of f .
4.4
Goursat’s Lemma
We begin with proving a special version of Cauchy’s theorem for triangles whose proof
is historical and illuminating. We assume the triangle and its interior is contained in an
open set on which f is holomorphic, hence we do not need to assume U itself is simply
connected.
Theorem 4.4.1 (Goursat’ Lemma/Goursat’s Theorem) If U is an open set of C and
f : U → C is holomorphic, then for any triangle T , whose interior and the triangle
itself is contained in U ,
Z
f (z)dz = 0.
T
37
Figure 4.2: Graph by E. Hairer and G. Wanner
Proof We may assume that U is connected. Denote T (0) = T . Let us take the middle
point on each sides of T (0) and obtain four triangles covering T whose vertices are the
(1)
vertices of T (0) and the middle points, which we denote by Ti , i = 1, 2, 3, 4 and
Z
f (z)dz =
4 Z
X
f (z)dz.
(1)
T (0)
Ti
i=1
The extra sides of the smaller triangles are traversed twice, in opposite directions and
(1)
hence the integral
denote
R of f along them canceled out. Let T
R one of the four
triangles s.t. | T (1) f (z)dz| is the largest among the four values | T (1) f (z)dz|. We
i
have
Z
Z
f (z)dz ≤ 4 (1) f (z)dz .
T
T
(1)
We continue with this procedure, divide T
to four triangles from which we select
a triangle T (2) . In this way we obtain a sequence of triangles T (i) whose enclosed
regions are nested and
Z
Z
Z
n
f (z)dz ≤ 4 f (z)dz ≤ · · · ≤ 4 f (z)dz .
T
T (1)
T (n)
Note that
T ⊃ T (1) ⊃ T (2) ⊃ . . . .
The sequence of nested triangles contains a common point which we denote by z0 . The
length between z0 and any other points in the triangle T (n) is less than 2−n L where L
is perimeter of the original triangle. Also the side of T (n) is also less than 2−n L.
Since f is holomorphic there exists a function ψ with limz→z0 ψ(z) = 0 and
f (z) = f (z0 ) + f 0 (z0 )(z − z0 ) + ψ(z)|z − z0 |.
38
Since f (z0 )+f 0 (z0 )(z−z0 ) has a primitive its integral along any closed curve vanishes.
Hence
Z
Z
f (z)dz =
ψ(z)|z − z0 |dz.
T (n)
T (n)
−n
Since |z − z0 | ≤ 2 L and
Z
n
4 f (z)dz ≤ 4n length(T (n) )2−n L max |ψ(z)| ≤ 4n (2−n L)2 max |ψ(z)| → 0.
z∈T (n)
T (n)
Consequently,
R
T
z∈T (n)
f (z)dz = 0.
Goursat (1884) actually proved the above for rectangles by sub-dividing rectangles.
Corollary 4.4.2 Suppose that f is holomorphic in a region U , then for any rectangle
R who and whose interior contained entirely in U ,
Z
f (z)dz = 0
R
Proof Just divide R into two triangles T1 and T2 whose share a side which are transversed twice in opposite directions. Hence
Z
Z
Z
f (z)dz.
f (z)dz +
f (z)dz =
R
T2
T1
A similar proof shows that in Goursat’s theorem we may replace the triangle by
any polygons. Argue by induction on the number of sides in a polygon, once see that
Goursat’s theorem holds for polygons.
Theorem 4.4.3 (Cauchy’s Theorem, existence of primitive) A holomorphic function
in a star region U has
Z
F (z) =
f (ζ)dζ,
z∈U
[c,z]
as its primitive, where C is a centre of U . If γ is a closed curve in U then
Z
f (z)dz = 0.
γ
R
Proof By Goursat’s Lemma, T f (z)dz = 0 for any triangles in U . By Theorem 4.3.3
on the existence of primitives in a star region, f has a primitive F given by the stated
formula. Apply Theorem 4.3.1 to conclude the vanishing of the curve integral.
4.5
Cauchy’s Theorem: simply connected domains (Lecture 12)
Definition 4.5.1 A connected set is simply connected if any continuous curve can be
continuously deformed into a point.
39
The precise meaning of the continuous deformation is the following: there exists a
continuous map F (s, t) : [0, 1] × [a, b] → C such that F (0, ·) : [a, b] → C is a
parameterization for γ, F (1, ·) : [a, b] → C is a constant curve, i.e. F (1, t) = F (1, 0)
for all t.
Remark 4.5.1 A star region is simply connected. Let γ1 , γ2 be two continuous curves
with the same end points. Let z : [a, b] → C be a parameterization of γ1 . Then a line
connecting C to z(t) pass through a point on γ2 which we denote by z̃(t). Set
F (s) = (1 − s)z(t) + sz̃(t).
Remark 4.5.2 A region in C is simply connected if and only if its complement is
connected in the extended plane C∗ . A bounded region in C is simply connected if and
only if its complement is connected in C.
The strip {x + iy : 1 < x < 2} is simply connected. Whenever a closed simple curve
lies in a simply connected region, its interior lies also in the region. If U is a region, it
is easy to spot a subset of C \ U that is a positive distance from the rest. Such subsets
are ‘holes’ in U .
Theorem 4.5.3 (Cauchy’s Theorem, c.f. Theorem 4.4.3) If f is a holomorphic function in a simply connected domain U then
Z
f (z)dz = 0
γ
for any closed curve in U .
The proof for this theorem is in Section 4.6. For now we do not prove this theorem, and
hence will limit the use of Cauchy’s Theorem for simply connected domain to enhance
understanding.
The following are non-simply connected regions:
{z : r1 < |z − z0 | < r2 },
{z : 0 < |z − z0 | < r2 }
Also, not simply connected:
{|z| < 20} \ {{|z − i| < 2} ∪ {|z − 1| < 1}}.
We are forced to consider non simply connected domains if a function has singularities.
Definition 4.5.2 A point z0 is an isolated singularity of f if f is holomorphic in some
punctured disk {0 < |z − z0 | < r}, not differentiable or not defined at z0 .
The function f (z) =
1
z
has an isolated singularity at 0.
Example 4.5.4 The function sin1 z has singularities at nπ where n ∈ Z.
For this we need to check the function sin(z) = 0 at and only at z = nπ. Firstly
iz
−iz
we see that
using the formula sin(z) = e −e
2i
sin(x + iy) = sin x cosh y + i cos x sinh y
40
where cosh z =
ez +e−z
2
and sinh z =
ez −e−z
.
2
sin x cosh y = 0,
So sin(z) vanishes if and only if
cos x sinh y = 0.
Since cosh y > 0 for y ∈ R, we have sin x = 0 and x = nπ. For x = nπ,
cos x sinh y = ± sinh y which vanishes only if y = 0. Thus the zero’s of sin z are
nπ.
√
Example 4.5.5 f (z) = z does not have an isolated singularity at 0. It cannot be
defined on any punctured disc around 0.
In the next chapter we discuss the integral of a function f along a contour curve
which enclose an isolated singularity of f . We learn how to divide a circle or more generally a star region with punctured holes into star regions / simply connected regions.
4.6
Supplementary
We give a proof for Cauchy’s theorem for simply connected regions.
Theorem 4.6.1 Let U be a simply connected region. Let f : U → C be holomorphic.
If γ1 and γ2 are homotopic then
Z
Z
f (z)dz =
f (z)dz.
γ1
γ2
Proof Let F : [0, 1] × [a, b] → U be a continuous map such that F (0, ·) is a parameterization for γ1 and F (1, ·) a parameterization for γ2 .
Suppose that s1 − s2 is sufficiently small, the two curves γ̃1 = F (s1 , ·) and γ̃2 =
F2 (s2 , ·) are close to each other:
max |F (s1 , t) − F (s2 , t)| ≤ .
a≤t≤b
Then there exists a finite division of [a, b], a = t0 < t1 < · · · < tn+1 = b such
(k)
that γ̃1 is the union of a finite number of curves γ̃1 (t), t ∈ [tk , tk+1 ]. Similarly γ̃2 is
(k)
the union of γ̃2 (t), t ∈ [tk , tk+1 ]. Then it is clear
Z
n Z
X
F (z)dz =
F (z)dz, i = 1, 2.
(k)
γ̃i
(k)
k=0
γ̃i
(k)
We connect γ̃1 with γ̃2 with two line segments to obtain a closed curve Γ̃k . We
choose small so each region enclosed by Γ̃k lie entirely in a disc contained in U .
Then by Cauchy’s theorem in a disc,
Z
f (z)dz = 0.
Γk
Summing up over k, we see
n Z
X
F (z)dz =
n Z
X
(k)
k=0
F (z)dz.
(k)
γ̃1
k=0
γ̃2
This completes the proof.
41
Chapter 5
Cauchy’s Integral Formula
If f is holomorphic in a region U then f has derivatives of all orders. Furthermore if z0
is such that a disc D centred at z0 is contained entirely in U , then it has a power series
expansion around z0 on D. Furthermore,
Z
f (ζ)
n!
f (n) (z0 ) =
dζ.
2πi |z−z0 |=r (ζ − z0 )n+1
This is intimately related to the properties of the function
5.1
1
z−z0 .
Keyhole Operation and Other Techniques (Lecture
12)
In this section we learn how to operate with domains with holes and in doing so we
prove a simpler form of the Cauchy integral formula whose general form will be proved
using a different technique later.
It is much easier to describe a curve as the boundary of a region (the contour). The
boundary curve of a region may not be very nice. We always assume that the curves
is piecewise smooth and oriented positively, so the the enclosed region stays on its left
as we travel in the positive direction along the boundary. If U is a region the boundary
curve of U is denoted by ∂U .
One of the techniques is called the key hole operation.
D(z0 , r) = {z : |z − z0 | < r},
C(z0 , r) = {z : |z − z0 | = r}.
Lemma 5.1.1 (Keyhole Lemma) Let γ be the piecewise smooth and positively oriented boundary of a star region U . Suppose that z0 ∈ U and f is holomorphic at every
point of an open set containing U with the exception z0 . Let r be sufficiently small so
the closed disc centred at z0 of radius r is contained in U . Then
Z
Z
f (z)dz =
f (z)dz.
γ
C(z0 ,r)
We give two proofs, representing two techniques.
Proof This technique is called key hole operation. Denote C the circle |z − z0 | = r.
Let us consider the line from the centre of U to z0 whose continuation intersects with
42
γ at a point on the boundary which we denote by a. Take any line through z0 if the
centre coincide with z0 . Then Ũ = U \ {[z0 , a]} is a star region with the same centre,
on which f is holomorphic.
We take a narrow corridor of width along the segment [z0 , a], connecting to the
circle C. This givesRa closed path Γ contained in Ũ . We now apply Cauchy’s theorem
to f on Ũ and get Γ f (z)dz = 0. Since f (z) is bounded, the sum of the integrals
of f (z) along the two sides of the corridors vanishes,
R as → 0. Its
R integral along the
missing arc of γ vanishes as converges to 0. Thus γ f (z)dz = C f (z)dz as stated.
Proof This proof is based on Cauchy’s theorem for simply connected domains. The
line segment passing through the centre of U and z0 , (take any line through z0 if the
centre coincide with z0 ), splits the region U \ {|z − z0 | = r} into two parts. Let γ1 and
γ2 denote the boundaries of the two parts (each curve contains one part of C and one
part of γ). Then γ1 (also γ2 ) is contained in a simply connected domain in which f (z)
is holomorphic. By Cauchy’s theorem for simply connected domains,
Z
f (z)dz = 0, i = 1, 2.
γi
Also
Z
Z
γ
Z
Z
f (z)dz −
C
f (z)dz = 0.
f (z)dz +
f (z)dz =
γ1
γ2
Corollary 5.1.2 Let U be a star region containing z0 with piecewise smooth boundary
γ, f a holomorphic function in an open set containing U , then
Z
1
dz = 2πi.
z
−
z0
γ
Also for n 6= −1,
Z
γ
1
dz = 0.
(z − z0 )n
Proof For any r,
Z
|z−z0 |=r
1
dz =
z − z0
Z
0
2π
1 d
(z0 + reit )dt = i
reit dt
Z
2π
dt = 2πi.
0
Also for an integer n 6= 1,
Z
|z−z0 |=r
Z 2π
1 d
1
it
1−n
(z0 + re )dt = ir
dt
n
itn
it(n−1)
r e dt
e
0
0
Z 2π
= ir1−n
(cos((1 − n)t) + i sin((1 − n)t)) dt = 0.
1
dz =
(z − z0 )n
Z
2π
0
43
5.2
Cauchy’s Integral Formula
Theorem 5.2.1 (Cauchy’s Integral formula) Suppose U is a star region with piecewise smooth positively oriented smooth boundary γ. Suppose f is holomorphic in an
open set containing U . Then for every z0 in U ,
Z
1
f (z)
f (z0 ) =
dz.
2πi γ z − z0
Proof Let C = {|z − z0 | = } where is sufficiently small. Then by Lemma 5.1.1,
Z
Z
f (z)
f (z)
dz =
dz.
C z − z0
γ z − z0
Now
Z
C
Z
f (z)
dz =
z − z0
C
Z
=
C
Since f is holomorphic,
Z
C
Z
f (z) − f (z0 )
1
dz + f (z0 )
dz
z − z0
z
−
z0
C
f (z) − f (z0 )
dz + f (z0 )2πi.
z − z0
f (z)−f (z0 )
z−z0
is bounded by a number M on a disc,
f (z) − f (z0 ) dz ≤ M 2π → 0, as → 0.
z − z0
Finally
Z
γ
f (z)
dz = lim
→0
z − z0
Z
C
f (z) − f (z0 )
dz + f (z0 )2πi = f (z0 )2πi,
z − z0
completing the proof.
Corollary 5.2.2 (Cauchy’s Integral formula for a disc) If f is holomorphic in an open
set containing the disc D̄ = {z : |z − z0 | ≤ r} then for every z in {z : |z − z0 | < r},
Z
f (ζ)
1
dζ,
f (z) =
2πi C ζ − z
where C = {z : |z − z0 | = r} is the boundary circle, positively oriented.
The corollary says that the value of a holomorphic function in a disc is determined by
its values on the boundary circle. In particular,
1
f (z0 ) =
2πi
Z
C
f (ζ)
1
dζ =
ζ − z0
2π
44
Z
0
2π
f (z0 + reit )dt.
5.3
Taylor Expansion, Cauchy’s Derivative Formulas
We prove a holomorphic function has a power series expansion around every point, and
is hence analytic. There are integral formulas for its derivatives.
Theorem 5.3.1 (Cauchy’s Theorem on Taylor expansion) If f is holomorphic on an
open disc D(z0 , R) = {z : |z − z0 | < R}, then there exists a sequence of complex
numbers an such that
f (z) =
∞
X
an (z − z0 )n ,
z ∈ D(z0 , R).
n=0
The an ’s are given by
1
an =
2πi
Z
C
f (z)
dz,
(z − z0 )n+1
where C is a circle centred at z0 with radius r < R. In particular f is infinite times
differentiable and is analytic on D(z0 , R).
Proof Let z ∈ D(z0 , R). Choose a number r between |z − z0 | and R. By Cauchy’s
integral formula, Corollary 5.2.2,
Z
1
f (ζ)
f (z) =
dζ.
2πi |ζ−z0 |=r ζ − z
Now
0
Since z−z
ζ−z0 =
1
1
1
1
=
=
.
0
ζ −z
(ζ − z0 ) − (z − z0 )
ζ − z0 1 − z−z
ζ−z0
|z−z0 |
r
< 1,
∞
f (ζ)
f (ζ) X (z − z0 )n
=
.
ζ −z
(ζ − z0 ) n=0 (ζ − z0 )n
On |ζ − z0 | = r,
f (ζ) (z − z0 )n max|ζ−z0 |≤r |f (ζ)| |z − z0 |n
.
(ζ − z0 ) (ζ − z0 )n ≤
r
rn
n
P∞
0|
Since n=0 |z−z
< ∞, by the dominated convergence theorem, we may integrate
rn
term by term,
Z
Z
∞
X
1
f (ζ)
f (ζ)
n 1
f (z) =
dζ =
(z − z0 )
dζ.
2πi |ζ−z0 |=r ζ − z
2πi
(ζ
−
z0 )n+1
|ζ−z0 |=r
n=0
P∞
Then f (z) = n=0 an (z − z0 )n . By Lemma 5.1.1, the value
Z
f (ζ)
dζ
n+1
|ζ−z0 |=r̃ (ζ − z0 )
makes sense for all 0 < r̃ < R and is independent of r̃. By Theorem 3.2.2, a power
series function is analytic in its disc of convergence .
45
Remark 5.3.2 Note if f (z) =
P∞
n=0
an (z − z0 )n then an =
f (n) (z0 )
.
n!
By the theorem above if f is holomorphic on a disc around z0 it has a power series
expansion whose disc of convergence, if not the whole complex plane, must meet a
point where f is not differentiable (singularity). If f is holomorphic on D(z0 , R), its
power series expansion has radius of convergence R.
Corollary 5.3.3 Let f be a holomorphic function on D(z0 , R). Let z ∈ D and r be
such that |z − z0 | + r < R, then
Z
f (ζ)
n!
dζ.
f (n) (z) =
2πi |ζ−z|=r (ζ − z)n+1
Proof By the standard theory on power series, and the last paragraph in the proof we
see that f is infinitely times differentiable. And, (using for example Theorem 3.1.2),
Z
n!
f (ζ)
(n)
f (z) =
dζ,
2πi |ζ−z|=r (ζ − z)n+1
where r is a number such that {|ζ − z| ≤ r} is contained in the disc D(z0 , R), concluding the formula.
Under the assumption of the Corollary, if |z − z0 | < δ < R, the following also
holds:
Z
n!
f (ζ)
f (n) (z) =
dζ.
2πi |ζ−z0 |=δ (ζ − z)n+1
Indeed, as |z − z0 | < δ, D(z0 , r) contains z, we may use Lemma 5.1.1 to change the
contour curve, to obtain
Z
Z
f (ζ)
f (ζ)
dζ =
dζ.
n+1
(ζ
−
z)
(ζ
−
z)n+1
|ζ−z0 |=δ
|ζ−z|=r
Theorem 5.3.4 If f is holomorphic in an open set U , then f is infinitely times differentiable and is analytic. If C ⊂ U is a circle whose interior is also contained in U
then
Z
n!
f (ζ)
f (n) (z) =
dζ
2πi C (ζ − z)n+1
for all z in the interior of C.
Proof For each z0 ∈ U , simply apply the earlier theorem to a small disc D(z0 , R) that
is contained in U .
5.4
Estimates, Liouville’s Thm and Morera’s Thm
Denote by C(z0 , R) the circle centred at z0 of radius R.
Theorem 5.4.1 (Cauchy’s Inequality) If f is holomorphic in an open set U and if the
closed disc {z : |z − z0 | ≤ R} is contained in U , then
|f (n) (z0 )| ≤
sup
z∈C(Z0 ,R)
46
|f (z)| ·
n!
.
Rn
Also if f (z) =
then
P∞
n=0
an (z − z0 )n on an open set containing the closed disc D(z0 , R),
|an | ≤
1
· sup |f (z)|.
Rn z∈C(Z0 ,R)
Proof By Cauchy’s derivative formula,
Z
n!
f (ζ)
(n)
f (z0 ) =
dζ.
2πi C(z0 ,R) (ζ − z0 )n+1
Taking the parameterization, Reit ,
Z
n! 2π f (z0 + Reit )
it |f (n) (z0 ) ≤
iRe
dt
it
n+1
2π 0
(Re )
Z 2π
n!
1
n!
≤
dt =
sup |f (z)| · n .
sup |f (z0 + Reit )|
n
2π t∈[0,2π]
R
R
z∈C(Z0 ,R)
0
The second part can be proved similarly,
Z
1
1
f (z)
|an | = sup |f (z)| · n .
dz ≤
n+1
2πi C (z − z0 )
R
z∈C(Z0 ,R)
Corollary 5.4.2 (Liouville’s Theorem) A bounded entire function is a constant.
Proof Let |f |∞ = supz∈C |f (z)|. By Cauchy’s inequality, for any z0 , and any R,
|f 0 (z0 )| ≤ |f |∞
1
.
R
Taking R to infinity to see that f 0 (z0 ) = 0. Since z0 is an arbitrary point in C, f 0
vanishes identically and f is a constant (by Corollary 4.3.2).
Example 5.4.3 Since sin(z) is an entire function, by Liouville’s theorem it cannot be
bounded on C, a fact can also easily be deduced from the formula sin(x + iy) =
y
−y
y
−y
sin x e +e
+ i cos x e −e
.
2
2
Corollary 5.4.4 (Morera’s Theorem) If f is continuous in an open disc D s.t.
Z
f (ζ)dζ = 0
T
for any triangle T contained in this disc, then f is holomorphic on D.
Proof By the integrability criterion, Theorem 4.3.3, f has a primitive F in D. By
Cauchy’s theorem for Taylor expansions, F is infinite many times differentiable, and
so in particular f = F 0 is holomorphic.
47
5.4.1
Supplementary
An extension to Liouville’s theorem is:
Example 5.4.5 If f is an entire function with limz→∞
f (z)
z
= 0, then f is a constant.
Proof Let z0 ∈ C, R > 0. By Cauchy’s derivative formula,
f 0 (z0 ) =
1
2πi
Z
|ζ−z0 |=R
1
f (ζ)
dζ =
(ζ − z0 )2
2π
Z
2π
0
f (z0 + Reit )
dt.
Reit
Hence
Z
f (z0 + Reit ) |z0 | + R
1 2π f (z0 + Reit ) z0 + Reit |f (z0 )| ≤
dt ≤ sup .
2π 0
z0 + Reit
Reit
z0 + Reit R
0≤t≤2π
0
The right hand side converges to 0 as R → ∞. Hence f 0 = 0 identically and f is a
constant.
Example 5.4.6 Let f be an entire function such that there exist real numbers R and M
such that |f (z)| ≤ M |z|n outside of the disc {z : |z| ≤ R}. Then f is a polynomial of
degree at most n.
Proof An entire function f has a power series expansion at 0:
f (z) =
∞
X
an z n ,
z ∈ C.
n=0
For m > n, and for any r > R,
Z
Z
1 1
1 f (z) f (z)
=
|am | =
dz
dz
2πi |z|=r z m+1 2πi |z|=r z n z m+1−n ≤ sup
|z|=r
|f (z)|
1
(2πr) → 0,
n
m+1−n
|z | r
as r → ∞. Thus am = 0 for all m ≥ n and f is a polynomial of at most degree n. 5.5
Locally Uniform Convergent Sequence of Holomorphic Functions
The limit of a sequence of continuous functions, converging uniformly, is continuous.
Continuity is a concept which is the same weather we treat f as a complex valued
function of one complex variable or as a R2 valued function of two real variables.
Theorem 5.5.1 (The Uniform Convergence Theorem/Weierstrass Theorem) Let fn
be a sequence of holomorphic function on an open set U . Suppose that fn converges
uniformly to a function f on any compact subsets of U , then f is holomorphic. Furthermore fn0 converges to f 0 uniformly on compact subsets of U .
48
Proof Part (1). Let z0 ∈ U and D = D(z0 , R) a disc with its closure contained
entirely in U . By Goursat’s lemma for holomorphic functions on a disc, for each n,
Z
fn (z)dz = 0
T
for any triangle in D. Since fn → f uniformly on D̄, f is continuous on D and
Z
Z
fn (z)dz →
f (z)dz
T
T
and the latter vanishes. By Morera’s theorem, f is holomorphic on D(z0 , R). Since z0
is an arbitrary point in U , we may conclude that f is holomorphic.
Part (2). Let K be a compact subset of U . Then there exists R > 0 such that for
each z0 ∈ K, the disc D(z0 , R) is contained in U , and such that for any r < R, K can
be covered by a finite number of ball of radius r. To prove fn0 → f 0 uniformly on K,
we only need to prove fn0 converges uniformly for every disc D(z0 , r) where r < R/2.
Let z0 ∈ U , we prove that fn0 → f 0 uniformly on D(z0 , r) where 2r < R and
D(z0 , R) ⊂ U . Let
C = {z : |z − z0 | = 2r}.
By Cauchy’s derivative formula, for any z in the interior of C,
Z
fn (ζ) − f (ζ)
1
dζ.
fn0 (z) − f 0 (z) =
2πi C (ζ − z0 )2
For any > 0 there exists N s.t. for n > N ,
sup
|fn (z) − f (z)| ≤ r.
z∈D̄(z0 ,2r)
Hence for n > N and z ∈ {|z − z0 | ≤ r},
|fn0 (z) − f 0 (z)| ≤
1 r
(4πr) ≤ ,
2π (2r)2
where (4πr) is the length of the circle C ( c.f. Theorem 4.2.2). We have proved that
fn0 → f 0 uniformly on compact sets.
5.6
Schwartz Reflection Principle (Lecture 15)
This section can be considered as an application of Morera’s Theorem. Schwartz Reflection Principle is a method of extending a holomorphic function over and to the other
side of the real axis.
Let U be an open subset of C, symmetric with respect to the real axis, i.e.
z∈U
iff
z̄ ∈ U.
Let I = U ∩ R,
U − = {z ∈ U, Im(z) < 0}.
U + = {z ∈ U, Im(z) > 0},
Then U = U + ∪ I ∪ U − .
49
Theorem 5.6.1 (Symmetry Principle) If f + and f − are respectively holomorphic
functions on U + and U − , that extends continuously to I, and f + (x) = f − (x) for
any x ∈ I, define
 +
z ∈ U+
 f (z),
+
−
f (z) = f (z), z ∈ I
f (z) =
 −
f (z),
z ∈ U −.
Then f is holomorphic on U = U + ∪ I ∪ U − .
Proof It is clear that f is continuous on U , holomorphic on U + and on RU − . Let z0 ∈ I
and D a disc in U centred at z0 . Let T be a triangle in D we prove that T f (z)dz = 0.
There are three possibilities:
R
(a) T lies entirely within U + or entirely within U − , then T f (z)dz = 0 by Goursat’s lemma.
(b) One side of T is on I. For example assume T is in U + ∪ I, in this case we take a
triangle T by moving its side on I upward, horizontally, with its distance from
I to be . Then T → T and
Z
Z
f (z)dz →
f (z)dz = 0.
T
By part (a),
R
T
T
f (z)dz vanishes for any , hence
R
T
f (z)dz = 0.
(c) T lies cross I in this case we may divide the triangles, T1 lies entirely on side of
the I axis,
R each of the other triangles T2 and T3 have one side on I. By part (a)
and (b), Ti f (z)dz = 0 for i = 1, 2, 3 and
Z
f (z)dz =
T
3 Z
X
i=1
f (z)dz = 0.
T
We apply Morera’s theorem to conclude that f is holomorphic on D. Since z0 is an
aribitary point in I, f is holomorphic at each point of I.
The following Lemma is given in Example Sheet 1.
Theorem 5.6.2 (Schwartz Reflection Principle) Suppose f is a holomorphic function on U + extending continuously to I, and f takes real value on I. Then there exists
a holomorphic function F on U such that F and f agree on U + ∪ I.
Proof We define
F (z) =

 f (z),

f (z̄)
z ∈ U+ ∪ I
z ∈ U −.
For z ∈ I, F (z) = F (z̄) and so the function is well defined. We have seen that if
g : U − → C given by g(z) = f (z̄) is holomorphic. By the reflection principle, F is
holomorphic on U .
There are two proof for the complex differentiability of g : U − → C given by
g(z) = f (z̄) where f is holomorphic on U + . Proof 1. (On Example sheet 1) Let
50
z0 ∈ U − and z ∈ U − . Since f is differentiable in z̄0 , there exists a function ψ with
limζ→z̄0 ψ(ζ) = 0 and
f (z̄) = f (z̄0 ) + f 0 (z̄0 )(z̄ − z̄0 ) + ψ(z̄)|z̄ − z̄0 |.
Taking conjugate of both side we see that
g(z) = g(z0 ) + f 0 (z̄0 )(z − z0 ) + ψ(z̄)|z − z0 |.
Since limz→z0 ψ(z) = 0 this proves that g is complex differentiable at z0 and g 0 (z0 ) =
f 0 (z̄0 ).
Proof 2. Let z0 ∈ U − . We make a Taylor expansion of f at z̄0 around a disc D
contained in U + :
∞
X
f (z̄) =
an (z̄ − z̄0 )n .
n=0
Then taking conjugate of both sides we see
g(z) =
∞
X
an (z − z0 )n ,
n=0
hence g is analytic on U − .
5.7
The Fundamental Theorem of Algebra
The zero of a function f is a solution to f (z) = 0.
Lemma 5.7.1 Every non-constant polynomial p(z) = a0 + a1 z + · · · + an z n has at
least one zero.
1
Proof Suppose that p(z) has no root then p(z)
is an entire function. We only need to
prove it is bounded. A bonded entire function is a constant.
1
The entire function p(z)
is certainly bounded on any compact set. At far away it
converges to zero and so is also bounded on C \ K where K is a compact set.
1
Detail for boundedness of p(z)
. Suppose an 6= 0 for n ≥ 1. If z 6= 0,
p(z)
a0
a1
an−1
= ( n + n−1 + · · · +
) + an .
zn
z
z
z
|an |
The first term converges to 0 as z → ∞ and p(z)
z n ≥ 2 for |z| sufficiently large and
1 |z n | 1
2 1
p(z) = |p(z)| |z|n ≤ |an | |z|n ,
which converges to zero as z → ∞. Hence
1
p(z)
is a bounded.
Theorem 5.7.2 (The Fundamental Theorem of Algebra) Every polynomial p(z) =
a0 + a1 z + · · · + an z n of degree n ≥ 1 has precisely n zeros in C. Furthermore if
z1 , . . . , zn are the roots then
p(z) = an (z − z1 )(z − z2 ) . . . (z − zn ).
51
Proof By the previous lemma p(z) = 0 has at least one root, z1 . Write z = (z−z1 )+z1
and expand p(z) in z − z1 . There exist b0 , b1 , . . . , bn such that
p(z) = b0 + b1 (z − z1 ) + · · · + bn (z − z1 )n .
Evaluate p(z) at z1 giving b0 = 0. Hence
p(z) = (z − z1 )(b1 + b2 (z − z1 ) + . . . bn (z − z1 )n−1 = (z − z1 )Q(z).
By induction on the degree of the polynomial, Q(z) = 0 has n − 1 roots and
p(z) = A(z − z1 ) . . . (z − zn ).
The number A is the coefficients in front of z n and is hence an .
5.8
Zero’s of Analytic Functions
Zero’s of analytic functions are useful for a number of reasons: they are closely related
to poles of meromorphic functions; the uniqueness of analytic continuations, and the
distributions of eigenvalues of a dynamical system. Or simply one wants to count the
zero’s, e.g. that of the zeta function!
The humblest application is to know where are the zero’s of the derivative of a
holomorphic function, which is also a holomorphic function. c.f. The Inverse Function
Theorem and Conformal Mappings.
Let pn (z) be a polynomial of degree n and z0 a zero of pn (z). Then there exists a
natural number r and w1 , . . . , wn−r , not equal to z0 such that
pn (z) = an (z − z0 )r (z − w1 ) . . . (z − wn−r ).
The holomorphic function h(z) = an (z − w1 ) . . . (z − wn−r ) does not vanish in a disc
containing z0 . Furthermore,
r
1
1
0
pn (z) = pn (z)
+
+ ··· +
.
z − z0
z − w1
z − wn−r
If r > 1 it is clear p0n (z0 ) = 0. If r = 1,
p0n (z) = an Πn−1
(z
−
w
)
+
p
(z)
i
n
i=1
1
1
+ ··· +
z − w1
z − wn−1
,
and p0n (z0 ) = an Πn−1
i=1 (z0 − wi ) 6= 0.
By induction on r we conclude that if z0 is a zero, or order r, of pn (z) then the
following holds:
(k)
• For k = 0, 1, . . . , r − 1, pn (z0 ) = 0 (the k th derivatives).
(r)
• pn (z0 ) 6= 0.
• There exists a holomorphic function h such that h(z) 6= 0 on a disc D containing
z0 and
pn (z) = (z − z0 )r h(z), z ∈ D.
52
Definition 5.8.1 Let f be a holomorphic function on a region U . The order of a zero
z0 of f is defined to be:
ord(f, z0 ) = inf{k : f (k) (z0 ) 6= 0}.
A trivial function is one which vanishes identically. A non-vanishing function is a
function that does not vanish anywhere.
We prove a multiplicative statement for a general holomorphic function.
Lemma 5.8.1 (Multiplicative Statement) Suppose that f is a holomorphic function
in a region U with a zero at z0 . There exists a neighbourhood D of z0 such that either
f is identically zero on D or there exists a unique non-vanishing holomorphic function
h on D, and a unique natural number k such that
f (z) = (z − z0 )k h(z),
∀z ∈ D.
This number k is the order of z0 .
Proof (1) Since f is holomorphic, it has a power series expansion at z0 , on a disc
D = D(z0 , r):
∞
X
f (z) =
an (z − z0 )n , ∀z ∈ D.
n=0
Firstly a0 = f (z0 ) = 0 by the assumption. Suppose that f does not vanish identically
on D. Let k ≥ 1 be the first non-zero term in the power series expansion. Then
!
∞
X
a
n
(z − z0 )n−k−1 .
f (z) = (z − z0 )k ak 1 + (z − z0 )
ak
n=k+1
Define
h(z) = ak
∞
X
an
1 + (z − z0 )
(z − z0 )n−k−1
ak
!
.
n=k+1
There exists M such that
∞
X
n−k−1 sup an (z − z0 )
≤ M,
|z|≤r/2 n=k+1
as the power series inside the modulus sign converges on D(z0 , r/2) and is a con1
tinuous function on the disc D. Let r̃ = min(r/2, 2M
). Then h is holomorphic
and does not vanish on the open disc {|z − z0 | ≤ r̃}. (3) Uniqueness. Suppose
f (z) = (z − z0 )k h(z) = (z − z0 )k̃ h̃(z). If k̃ > k then h(z) = (z − z0 )k̃−k h̃(z) and
h(z0 ) = 0, which contradicts with the definition of h. Consequently, k = k̃, and also
h = h̃.
(4) Finally, f 0 (z) = k(z − z0 )k−1 h(z) + (z − z0 )k h0 (z). If k = 1, f 0 (z0 ) =
h(z0 ) 6= 0. If k > 1, f 0 (z0 ) = 0. Induction on the degree k,
f (j) (z) = k(k − 1) . . . (k − j + 1)(z − z0 )k−j h(z) + (z − z0 )k−j+1 G(z)
for a holomorphic function G. Thus f (j) (z0 ) = 0 for any j < k and f (k) (z0 ) 6= 0. 53
Theorem 5.8.2 If f is a holomorphic function with f (z0 ) = 0. Then there exists a
disc D(z0 , r) on which either f vanishes identically or f has no other zero’s.
Proof Suppose that f (z) = (z − z0 )k h(z) on {|z − z0 | < r} where h is holomorphic
and does not vanish. Then f (z) does not vanish on {z : 0 < |z − z0 | < r}.
Definition 5.8.2 A zero of f , z0 , is said to be isolated if there exists an open disc D
around z0 such that f (z) 6= 0 for any z ∈ D \ z0 .
Every zero of a holomorphic function is isolated unless f is trivial on a disc near z0 .
Theorem 5.8.3 Let U be a region in C.
• Suppose g1 and g2 are holomorphic in U and g1 , g2 agree on a sequence of zk in
U who has an accumulation point z0 in U and zk 6= z0 for any k. Then g1 = g2
identically on U .
• In particular, if f is a holomorphic function in U with a sequence of zeros’s
zk ∈ U converging to z0 in U , and z0 6= zk for any k. Then f (z) is identically 0
on U .
Proof Setting f = g1 − g2 we only need to prove the special case. By the continuity,
f (z0 ) = limk→∞ f (zk ) = 0. Within every disc D(z0 , n1 ) f has a zero other than z0 .
By Theorem 5.8.2, f ≡ 0 on a disc around z0 . We proceed to prove that f is identically
zero on U .
Let V be the interior of the set of zeros of f . Then V is non empty, it contains z0 ,
and is an open set by the construction. Let zn ∈ V → z0 . Then f (z0 ) = 0. Since
z0 is the accumulation point of distinct zero’s, f vanishes in a disc around z0 and z0 is
contained in V . Hence z0 ∈ V . This proves V is closed.
Thus U is the union of two open sets:
U = (U \ V ) ∪ V.
If U \ V is not empty, U is the disjoint union of nonempty open sets, contradicting with
the fact that U is connected. If U \ V is empty, then f is identically zero.
The multiplicative statement can now be strengthened as below.
Theorem 5.8.4 (Multiplicative Statement) Suppose that f is a non-trivial holomorphic function in a region U and has a zero z0 . There exists a neighbourhood D of z0 ,
a unique non-vanishing holomorphic function h on D, and a unique natural number k
such that
f (z) = (z − z0 )k h(z), ∀z ∈ D.
This number k is the order of z0 .
Proof By the Lemma 5.8.1, there exists a disc D around z0 on which either f vanishes or f (z) = (z − z0 )k h(z). The former implies that f vanishes identically on U
contradicting the assumption f is not trivial (i.e. not everywhere 0).
54
5.9
Uniqueness of Analytic Continuation (Lecture 17)
Let γ be a simple closed curve and U the interior region determined (enclosed in ) by
γ. Let f : U → C be an analytic function in U . It is an interesting question whether f
ca be analytically continued from a boundary point.
Definition 5.9.1 Suppose f is holomorphic on U . Let z0 be a point on the boundary
of U . We say f is analytically continuable at z0 if there exists a function g : U 0 → R
where U 0 is an open set containing z0 such that g and f agree in U ∩ U 0 .
Corollary 5.9.1 Suppose that f1 , f2 are defined and holomorphic on U1 and U2 respectively. If U1 ∩ U2 is connected and f1 = f2 on a sequence of point zk ∈ U1 ∩ U2
with a limit z0 s.t. zk 6= z0 for any k then f1 = f2 on U1 ∩ U2 .
Corollary 5.9.2 Let f be a holomorphic function in a region U . Let V ⊂ U be a
closed bounded subset of U . Then for any w ∈ C, {z ∈ V : f (z) = w} has a finite
number of elements or f is a constant on V .
Proof Let g(z) = f (z) − w. For any z ∈ V there exists a number r(z) > 0 such
that w is not in the image of D0 (z, r(z)) or f (z) ≡ w on this disc, in which case f is
identically w on U . Since V is compact there exists a finite number of points z, . . . , zm
in V such that
0
V ⊂ ∪m
i=1 D (zi , zi (r)).
It is clear that {z ∈ V : f (z) = w} ⊂ {z1 , . . . , zm }.
5.10
Supplementary
The Lacunary function is:
f (z) =
∞
X
n
z2 = z + z2 + z4 + z8 + . . . .
n=1
The radius of convergence of the power series is R = 1. The largest open set on which
it can be extended analytically is the unit disc. In fact it has singularity at all 2n th
roots of the unity. Since the singularity points on the unit circle is dense in the circle,
it cannot be analytically continued beyond the disc.
55
Chapter 6
Laurent Series and Singularities
A series of the form
b1 z −1 + b2 z −2 + . . .
can be considered as a power series in z1 . Let R denote its radius of convergence R. If
R is finite number, the power series converge absolutely for z with |z| > R1 . As usual,
1
∞ is interpreted as 0.
If this series is combined with an ordinary power series we have a general series of
the form
∞
X
an z n .
n=−∞
Definition 6.0.1 The series is said to converge, if both its positive power part
P−∞
and negative power part n=−1 an z n are separately convergent.
P∞
n=0
an z n
Suppose that the positive part of the power series has a radius of convergence R2 > 0
and the negative part of the power series converges when |z| > R1 . If R1 < R2 ,
they have a common region of convergence: R1 < |z| < R2 . Conversely if f is
holomorphic function in a region contains the annulus R1 < |z − z0 | < R2 , it has a
general series expansion at z0 :
f (z) =
∞
X
an (z − z0 )n .
n=−∞
This is called a Laurent series (development) for f .
6.1
Laurent Series Development (Lecture 17-18)
Lemma 6.1.1 Let f be a holomorphic function on a disc D(z0 , r). Define
g(z) =
f (z)−f (z0 )
,
z−z0
f 0 (z0 ),
Then g is holomorphic on D(z0 , r).
56
z=
6 z0
z = z0 .
Proof As the quotient of two holomorphic functions, g is differentiable on D \ {z0 }.
We only need to prove g is differentiable at z0 . Firstly, there is a Taylor series expansion
for the holomorphic function f in a neighbourhood of z0
0
f (z) = f (z0 ) + f (z0 )(z − z0 ) +
∞
X
an (z − z0 )n .
n=2
Hence for z 6= z0 ,
g(z) − g(z0 ) =
And
∞
X
f (z) − f (z0 )
− f 0 (z0 ) =
an (z − z0 )n−1 .
z − z0
n=2
∞
g(z) − g(z0 ) X
=
an (z − z0 )n−2 ,
z − z0
n=2
The right hand side, a holomorphic function, has a limit as z → z0 and g is differentiable at z0 .
Lemma 6.1.2 (Cauchy’s Integral formula for the annulus) Let f be a holomorphic
function in an open set containing the annulus A = {z : r1 < |z − z0 | < r2 } where
0 < r1 < r2 < ∞. Then for all z ∈ A,
Z
Z
f (ζ)
1
f (ζ)
1
dζ −
dζ.
f (z) =
2πi C(z0 ,r2 ) ζ − z
2πi C(z0 ,r1 ) ζ − z
Proof We define on A,
(
g(ζ) =
f (ζ)−f (z)
,
ζ−z
0
f (z),
ζ=
6 z
ζ = z.
Then g is holomorphic on the annulus. ( Take a disc D around z, which is completely
contained in A. Then f is holomorphic on D, apply the previous lemma.)
We apply the keyhole technique Lemma 5.1.1 to the function g:
Z
Z
g(ζ)dζ =
g(ζ)dζ.
C(z0 ,r1 )
C(z0 ,r2 )
(c.f. Example sheet 6, Problem 4 where we take z1 = z2 and Lemma 5.1.1 in case
r1 = 0) We observe further,
Z
Z
1
1
dζ = 0,
dζ = 2πi,
ζ
−
z
ζ
−
z
C(z0 ,r1 )
C(z0 ,r2 )
as z is outside of the circle of integration C(z0 , r1 ) in the first case, and inside the circle
of integration D(z0 , r2 ) in the second case.
Z
Z
Z
Z
f (ζ)
f (z)
f (ζ) − f (z)
dζ =
dζ +
dζ
g(ζ)dζ =
ζ −z
C(z0 ,r1 ) ζ − z
C(z0 ,r1 ) ζ − z
C(z0 ,r1 )
C(z0 ,r1 )
Z
f (ζ)
=
dζ.
ζ
−z
C(z0 ,r1 )
57
Consequently,
Z
C(z0 ,r2 )
f (ζ)
dζ =
ζ −z
Z
C(z0 ,r2 )
Z
=
C(z0 ,r1 )
Z
=
C(z0 ,r1 )
f (ζ) − f (z)
dζ +
ζ −z
Z
C(z0 ,r2 )
f (z)
dζ
ζ −z
f (ζ) − f (z)
dζ + 2πif (z)
ζ −z
f (ζ)
dζ + 2πif (z).
ζ −z
The conclusion follows.
Let r1 < r < r2 . If f is holomorphic on the entire disc D(z0 , r2 ), then for z in
(ζ)
is holomorphic in a disc containing D(z0 , r1 ), its integral along the
the annulus, fζ−z
circle |z − z0 | = r1 is zero by Cauchy’s theorem:
Z
1
f (ζ)
dζ = 0.
2πi C(z0 ,r1 ) ζ − z
Theorem 6.1.3 Let r1 , r2 be two real numbers with 0 ≤ r1 < r2 ≤ ∞. Suppose that
f is holomorphic in a neighbourhood of the annulus {z : r1 < |z − z0 | < r2 }. Then
there exists a unique sequence of complex numbers an such that
∞
X
f (z) =
an (z − z0 )n ,
r1 < |z − z0 | < r2 .
n=−∞
For any r1 < r < r2 , for each n,
1
an =
2πi
Z
C(z0 ,r)
f (ζ)
dζ.
(ζ − z0 )n+1
(6.1.1)
Proof Let z ∈ {z : r1 < |z − z0 | < r2 }. Then by Lemma 6.1.2,
Z
Z
f (ζ)
1
f (ζ)
1
f (z) =
dζ −
dζ.
2πi C(z0 ,r2 ) ζ − z
2πi C(z0 ,r1 ) ζ − z
By the same proof as in the proof for Cauchy’s theorem on Taylor formula, c.f. Theo1
rem 5.3.1, we expand ζ−z
to obtain a power series expansion for the first term
1
2πi
Z
C(z0 ,r2 )
∞
X
f (ζ)
dζ =
an (z − z0 )n
ζ −z
n=0
where an is as in formula (6.1.1). We work on
Z
1
f (ζ)
−
dζ.
2πi C(z0 ,r1 ) ζ − z
We observe that
1
1
1
1
=
=−
.
ζ−z0
ζ −z
(ζ − z0 ) − (z − z0 )
z − z0 1 − z−z
0
58
Since r1 = |ζ − z0 | < |z − z0 |, we expand
1
ζ−z
1− z−z0
as a power series and obtain:
0
∞
∞
X
1 X (ζ − z0 )n
(ζ − z0 )n−1
1
=
−
.
=−
ζ −z
z − z0 n=0 (z − z0 )n
(z − z0 )n
n=1
By the uniform convergence of the series on {z : r1 < |z − z0 |} (see remark below),
we may exchange the order of taking limit and summing up and obtain
1
−
2πi
Z
C(z0 ,r1 )
Z
∞
X
f (ζ)
−n 1
dζ =
f (ζ)(ζ − z0 )n−1 dζ.
(z − z0 )
ζ −z
2πi
C(z
,r
)
0
1
n=1
Let m = −n and
am =
1
2πi
Z
|ζ−z0 |=r
f (ζ)
dζ.
(ζ − z0 )m+1
Note the integration is independent of the radius of the circle, where r1 < r < r2 , to
conclude the theorem.
To prove the uniqueness suppose that we have another sequence of numbers such
that
∞
X
f (z) =
bn (z − z0 )n , r1 < |z − z0 | < r2 .
n=−∞
Then
∞
X
f (z)
=
bn (z − z0 )n−k .
(z − z0 )k
n=−∞
Integrate both sides on the circle C(z0 , r), with r ∈ (r1 , r2 ), the only non-zero contribution comes from the term bk−1 /(z − z0 ) which is:
Z
bk−1 (z − z0 )−1 dz = 2πibk−1 .
C(z0 ,r)
Consequently,
bk =
1
2πi
Z
C(z0 ,r)
f (z)
dz.
(z − z0 )k+1
Remark 6.1.4 Since the annulus and its closure are contained in contained in the region on which f is holomorphic, by slightly enlarging the annulus we may also take
r = r1 ot r = r2 in (6.1.1).
P∞
n
Remark 6.1.5 Suppose that the power series function
n=0 bn (z − z0 ) has
P∞ f (z) = −n
radius of convergence R > 0. Then g(z) = n=0 bn (z − z0 )
is convergent for
|z − z0 | > R1 , and the convergence is uniform on any set |z − z0 | ≥ 1r for any r < R.
1
Just note that g(z) = f ( z−z
).
0
Note that the radius of convergence for the positive part of the Laurent series, in Theorem 6.1.3, is greater or equal to r2 , the negative part converges on |z − z0 | > r1 .
59
Example 6.1.6 Let
1
1 1
1
= (
−
).
(z − 1)(z − 3)
2 z−3 z−1
f (z) =
Develop the Laurent series for f on the following regions:
{|z| < 1},
{1 < |z| < 3},
{|z| > 3}.
We observe that z0 = 0 and
 P∞ n
 − n=0 z ,
1
=
z − 1  P∞ −n
,
n=1 z

1

 = −3 ·
1
=
z−3 

1
z
·
1
1− z3
1
1− z3
=
=−
P∞
n=1
P∞
|z| < 1
|z| > 1.
n=0
3−n−1 z n ,
3n−1 z −n ,
|z| < 3
|z| > 3.
Hence on {|z| < 1}, where f is holomorphic,
1
f (z) =
2
−
∞
X
3
−n−1 n
z +
n=0
∞
X
!
z
n
=
n=0
∞
1X
1 − 3−n−1 z n .
2 n=0
On {1 < |z| < 3},
1
f (z) =
2
−
∞
X
−n−1 n
z −
3
n=0
∞
X
!
z
−n
=−
n=1
∞
∞
1 X −n 1 X −n−1 n
z −
3
z .
2 n=1
2 n=0
On |z| > 3,
1
f (z) =
2
∞
X
n−1 −n
3
z
−
n=1
∞
X
!
z
−n
=
n=1
−1
1 X −n−1
(3
− 1)z n .
2 n=−∞
Corollary 6.1.7
P∞(Cauchy’s Inequality) The coefficients in the Lauren series expansion f (z) = n=−∞ an (z − z0 )n satisfies:
|ak | ≤
sup
|f (z)|
|z−z0 |=r
1
rk
for any r ∈ (r1 , r2 ).
Proof Just use the formula
1 Z
f (ζ)
1
|ak | = dζ
≤
2πi C(z0 ,r) (ζ − z0 )k+1 2π
k+1
1
sup |f (z)|
L
r
|z−z0 |=r
where L = 2πr is the length of the circle. The conclusion follows.
60
6.2
Classification of Isolated Singularities (Lecture 19)
Denote D0 (z0 , r) = {z : 0 < |z − z0 | < r} the deleted disc. Such a disc is a deleted
neighbourhood of z0 . A full disc D(z0 , r) = {z : |z − z0 | < r} is a neighbourhood of
z0 . Both are assumed to be fully contained in the open set on which a function under
discussion is defined.
In this section we discuss isolated singularities. A point z0 is an isolated singularity
for f , if f is defined and holomorphic on a deleted disc D0 (z0 , R), in which case f has
a Laurent series at z0 :
f (z) =
∞
X
an (z − z0 )n ,
z ∈ D0 (z0 , R).
n=−∞
Definition 6.2.1
1. If ak 6= 0 for an infinite number of negative integers, we say f
has an essential singularity at z0 .
2. If −n is the least integer for which a−n 6= 0, and n is positive, we say f has a
pole at z0 of order n.
3. If ak = 0 for all negative k, we say z0 is a removable singularity.
We have not assumed that f is defined at z0 . If an = 0 for all negative integers,
we extend f to f˜ by defining f˜(z0 ) = a0 . Then the extended function is holomorphic,
hence the terminology f has a removable singularity at z0 .
The equivalence of (1) with (4) in the theorem below is called Riemann’s theorem
on removable singularity.
Theorem 6.2.1 Let f be a holomorphic function on the deleted disc D0 (z0 , R). The
following statements are equivalent.
1. ** f has a removable singularity at z0 ;
2. f can be analytically continued to the whole disc D(z0 , R);
3. limz→z0 f (z) exists in C (i.e. the limit is a complex number);
4. ** f is bounded in a deleted neighbourhood of z0 . (i.e. there exists δ such that
sup0<|z−z0 |<δ |f (z)| < ∞).
P∞
Proof (1)=⇒ (2). Suppose f (z) = n=0 an (z − z0 )n . We define f (z0 ) = a0 and f
extends analytically to the whole disc.
(2) =⇒ (3). If f˜ is the analytic continuation, then limz→z0 f (z) = f˜(z0 ), a finite
number;
(3)=⇒ (4) If limz→z0 f (z) = c, then it is clear f (z) is bounded in a neighbourhood
of z0 . (Take = 1, there exists R > δ > 0 such that if |z − z0 | < δ, |f (z) − c| < 1
and so |f (z)| ≤ |c| + 1.)
(4)=⇒ (1) We assume that sup|z−z0 |≤δ0 |f (z)| ≤ M for some positive numbers
δ0 and M . By Cauchy’s inequality, on D(z0 , δ) where δ < δ0 , the coefficients is the
Laurent series expansion of f satisfy:
|a−k | ≤ M
1
= M δk .
δ −k
We take δ → 0. If k ≥ 1, a−k = 0. Thus f has removable singularity.
61
6.2.1
Poles
Recall that f has a pole at z0 if it has Laurent expansion at z0 of the form below, where
n is a natural number and a−n 6= 0:
f (z) =
a−n
a−n+1
a−1
+
+ ...
+ G(z)
n
n−1
(z − z0 )
(z − z0 )
(z − z0 )
where G is a holomorphic function. The order of f at z0 , denoted by ord(f ; z0 ), is n.
Definition 6.2.2 If ord(f ; z0 ) = −1, the pole z0 is said to be simple.
Remark 6.2.2 If f has a pole of order n ∈ N at z0 , then for any k > n
(z − z0 )k
(z − z0 )k
(z − z0 )k
+
a
+
.
.
.
a
= 0.
lim (z−z0 )k f (z) = lim a−n
−n+1
−1
z→z0
z→z0
(z − z0 )n
(z − z0 )n−1
(z − z0 )1
If k = n, limz→z0 (z − z0 )k f (z) = a−n , which does not vanish by the assumption. If
k < n, limz→z0 (z − z0 )k f (z) = ∞. Thus,
ord(f ; z0 ) = inf{n : lim (z − z0 )n f (z) exists and is finite}.
z→z0
If n > ord(f ; z0 ) then limz→z0 (z − z0 )n f (z) = 0.
Remark. If we wish to treat zeros and poles together, without discrimination, it is
advantage to define the order of a pole to be a negative integer. I am happy with either
definitions, as far as in the same line of computation, we are consistent.
Definition 6.2.3 If {zk } is a sequence of complex numbers, we say limk→∞ zk = ∞
if limk→∞ z1k = 0.
Theorem 6.2.3 Let f be a holomorphic function on the deleted disc D0 (z0 , R). The
following statements are equivalent.
1. f has a pole at z0 .
2. limz→z0 f (z) = ∞ (i.e. limz→z0 |f (z)| = ∞).
Proof (1) =⇒ (2). If f has a pole of order n at z0 , by the definition in a disc D0 (z0 , R),
f (z) = (z − z0 )−n
∞
X
ak (z − z0 )k+n = (z − z0 )−n h(z).
k=−n
P∞
where h(z) = k=−n ak (z−z0 )k+n = j=0 aj−n (z−z0 )j has an analytic extension
to D(z0 , R). Since a−n 6= 0, it is routine to prove h(z0 ) 6= 0 and hence does not
n
1
0)
= limz→z0 (z−z
= 0, and
vanish in a small disc around z0 , thus limz→z0 f (z)
h(z)
limz→z0 f (z) = ∞.
(2) =⇒(1). Since limz→z0 f (z) = ∞, f (z) does not vanish in a sufficiently small
1
disc D(z0 , δ), where we define g(z) = f (z)
. Then limz→z0 g(z) = 0 and by Riemann’s theorem on removable singularity (Theorem 6.2.1), g extends to a holomorphic
function on D(z0 , δ) with g(z0 ) = 0 and g is not identically zero. Then there exists a
positive integer n and a holomorphic function H (see Lemma 5.8.1) with H(z0 ) 6= 0
P∞
62
1
. Then h has a Taylor expansion
such thatP
g(z) = (z − z0 )n H(z). Let h = H
∞
n
h(z) =
b
(z
−
z
)
and
h(z
)
=
6
0.
Consequently
f has a Laurent series
0
0
z=0 n
expansion
f (z) =
h(z)
= (z − z0 )−n (b0 + b1 (z − z0 ) + . . . ),
(z − z0 )n
and f has a pole at z0 .
Remark 6.2.4 A holomorphic function f on D0 (z0 , R) has a pole of order n at z0 if
and only if there exists a holomorphic function on a disc D(z0 , δ) with h(z0 ) 6= 0 s.t.
f (z) = (z − z0 )−n h(z),
z ∈ D0 (z0 , δ),
i.e. f (z)(z − z0 )n is holomorphic and does not vanish at z0 .
Definition 6.2.4 If f has a pole at z0 , the number a−1 is the residue of f at the pole z0
and is denoted by Res(f ; z0 ).
Corollary 6.2.5 Let f be a holomorphic function on D0 (z0 , R) having a pole at z0 ,
then
Z
1
f (ζ)dζ,
Res(f ; z0 ) =
2πi C
for any circle centred at z0 with radius r < R.
If the order of the pole z0 is one then
Res(f ; z0 ) = lim (z − z0 )f (z).
z→z0
Theorem 6.2.6 If f has a pole of order n at z0 , then
Res(f ; z0 ) = lim
z→z0
Proof Since
f (z) =
∞
X
1
d
( )n−1 (z − z0 )n f (z).
(n − 1)! dz
ak (z − z0 )k ,
z ∈ D0 (z0 , R),
k=−n
n
(z − z0 ) f (z) = a−n + a−n+1 (z − z0 ) + · · · + a−1 (z − z0 )n−1 + G(z)(z − z0 )n
where G is a holomorphic function, the positive part of the Laurent expansion.
(
d n−1
d
)
(z − z0 )n f (z) = (n − 1)!a−1 + ( )n−1 (G(z)(z − z0 )n ) .
dz
dz
The last term vanishes as z → z0 .
63
6.2.2
Essential Singularities
Theorem 6.2.7 (Casorati-Weierstrass) Let f be a holomorphic function on the deleted
disc D0 (z0 , R). The following statements are equivalent.
1. **f has an essential singularity at z0 ;
2. for all 0 ≤ r ≤ R, the image f (D0 (z0 , r)) is dense in C.
3. ** limz→z0 f (z) does not exist on the extended plane C∗ .
Proof (2) =⇒(3): (3) follows trivially from (2).
(3) =⇒(1). If (3) holds, z0 is not a removable singularity nor a pole for f . Hence
(1) holds.
(1) =⇒(2). Let z0 be an essential singularity of f . Suppose (2) fails. There exists
r > 0 such that the image of D0 (z0 , r) by f is not dense in C. Then there exists a point
w ∈ C and δ > 0 such that |f (z) − w| > δ for all z ∈ D0 (z0 , r). Define
g(z) =
1
.
f (z) − w
Then g is holomorphic and bounded by 1δ on D0 (z0 , r). By Theorem 6.2.1, z0 is a
removable singularity for g. If g(z0 ) 6= 0 then
lim f (z) = w + lim
z→z0
z→z0
1
g(z)
exists, then z0 would be a removable singularity for f ! If, otherwise g(z0 ) = 0, then
limz→z0 f (z) = ∞ and z0 is a pole for f ! Both scenario lead to contradictions and so
(2) must holds.
6.3
Meromorphic Function
Definition 6.3.1 A meromorphic function on an open set U is a function which are
complex differentiable everywhere with the exception of isolated singularities that are
poles.
Remark 6.3.1 A meromorphic function has only a finite number of poles on a bounded
closed subset of U .
Proof Indeed, if there is an infinite number of singularities in a closed subset A, it
has an accumulation point z0 in A which cannot be a singularity by the assumption.
However f cannot be differentiable at z0 either, for otherwise it would be bounded in a
neighbourhood D of z0 on one hand, and approaches infinity near the poles within D
on the other hand.
6.3.1
Supplementary
Remark 6.3.2 The set of meromorphic functions, with function addition, multiplication, division, is a group.
64
Proof It is clear that the addition, subtraction, and multiplication of two meromorphic
functions are meromorphic. Let us consider the quotient of two holomorphic functions
f and g. Suppose that f or g has an isolated zero or an isolated pole at z0 . Then near
z0 ,
f (z) = (z − z0 )m h(z), g(z) = (z − z0 )n k(z)
where h, k are holomorphic functions. Then
h(z)
f (z)
= (z − z0 )m−n
g(z)
k(z)
is holomorphic in a deleted disc about z0 .
65
Chapter 7
Winding Numbers and the
Residue Theorem
The function f (z) = z1 does not have a primitive in an open set of C containing a
disc about the origin, for its integral along the boundary of the disc is 2πi 6= 0. If we
integrate the z1 along a square centred at the origin, the integral is also 2πi (on each
side of the line we may choose a brach of the logarithm). If we integrate it along the
circle concatenated with itself n − 1 times we see 2πni.
Lemma 7.0.1 If a piecewise smooth closed curve γ does not pass through a point z0
then
Z
dz
γ z − z0
is a multiple of 2πi.
Proof Let z : [a, b] → C be a parameterization of γ, then
Z
Z b
dz
z 0 (s)
=
ds.
γ z − z0
a z(s) − z0
Define
Z
t
g(t) =
a
−g(b)
z 0 (s)
ds,
z(s) − z0
t ∈ [a, b].
It is sufficient to prove that e
= 1, from which g(t) = 2kπi where k ∈ Z.
z(t)−z0
An obvious candidate for g(t) is ln z(a)−z
. Define
0
F (t) = e−g(t)
z(t) − z0
.
z(a) − z0
Then for t ∈ [a, b] where z is C 1 , we have
F 0 (t) = e−g(t) (−
z 0 (t)
z(t) − z0
z 0 (t)
)
+ e−g(t)
= 0.
z(t) − z0 z(a) − z0
z(a) − z0
Since F is continuous in t, and piecewise a constant , then F must be a constant on
[a, b]. Since g(a) = 0, F (a) = 1. Note z(b) = z(a),
F (b) = e−g(b)
z(b) − z0
= e−g(b) = 1.
z(a) − z0
66
7.1
The Index of a Closed Curve (Lecture 22)
Definition 7.1.1 If a piecewise smooth closed curve γ does not pass through a point
z0 then the index of γ with respect to z0 is:
Z
1
dz
ind(γ, z0 ) =
.
2πi γ z − z0
It is also called the winding number or rotation number of the curve γ about z0 .
We recall the following facts. Any curve z : [a, b] → C can be reparameterized to
[0, 1]. z̃(t) = t(b − a) + a, t ∈ [0, 1]. If γ has a parameterization z : [0, 1] → C then
−γ is the curve defined by z̃(t) = z(1 − t). Let i = 1, 2 and γi be two curves with
respectively the parameterizations zi : [0, 1] → C with z1 (1) = z2 (0). Then we define
γ be the concatenated curve given by
z1 (2t),
0 ≤ t ≤ 21
z(t) =
z2 (1 − 2t), 12 ≤ t ≤ 1.
This curve is denoted by γ1 ∪ γ2 , otherwise denoted by γ1 + γ2 .
Remark 7.1.1
1. For any z0 not passing through the curve,
ind(−γ, z0 ) = − ind(γ, z0 ).
2. If γ1 and γ2 are two smooth closed curves with the same initial points, then for
z0 not passing through γ1 nor γ2 ,
ind(γ1 ∪ γ2 , z0 ) = ind(γ1 , z0 ) + ind(γ2 , z0 ).
Definition 7.1.2 A curve γ : [a, b] → C is simple if it does not have self intersection.
That is: γ(t) 6= γ(s) unless s = t or a = b.
By the boundary of a star region we mean the curve which travels along the boundary
once, with positive orientation. This is a simple curve.
Example
7.1.2 Let R denote the square centred at the origin. By standard theorem,
R 1
dz
=
2πi. However we compute the integral side by side to see the ‘winding’.
R z
We work with the top horizontal side, which is contained in a slit domain where it
has a primitive. Evaluate the difference of the end points, we find the change in the
argument: π2 . By symmetry this is π/2 for the other sides. All together we obtain:
Z
1
1
ind(R, 0) =
dz = 1.
2πi R z
The curve in the example below is not a simple curve.
Example 7.1.3 The curve z(t) = z0 + eint , 0 ≤ t ≤ 2π is the circle concatenated
with itself n − 1 times, it is not simple.
Z
1
dz
= n.
ind(γ, z0 ) =
2πi γ z − z0
67
Let γ be a smooth curve with parameterization z : [a, b] → C. Then its complement
is an open set which we denote by U . Any (connected) component of an open set is an
open set and there are only a countable number of components. Indeed, if z belongs to
an open connected component V of U , let D(z, r) be a disc in U , then V ∪ D(z, r) is
connected. Hence V is open. Also, the set {q1 + q2 i ∈ U : q1 , q2 ∈ Q} is countable,
each connected component contains a disc and hence one of these points.
Let M = supt∈[a,b] |f (z(t))|, which is finite. Then for any R > M , the set
{z : z| > R} belongs to the complement of γ in C. Hence γ has only one unbounded
component.
Denote by {γ} the set of points on the curve γ.
Theorem 7.1.4 Let γ be a closed piecewise smooth curve and U = C \ {γ}. Then
ind(γ, z) is a constant of z when restricted to a component of U . If z belongs to the
unbounded component of U then ind(γ, z) = 0.
Proof We prove that z 7→ ind(γ, z) is a continuous from U to C. Since ind(γ, ·) is
integer valued, it has to be a constant on each connected component of U .
Let V be a connected component of U and z0 ∈ V and R the distance from z0 to
γ. Let δ = R3 , then V contains the D(z0 , δ). For any z with |z − z0 | ≤ δ,
Z 1
1
1
ind(γ, z) − n(γ, z0 ) =
−
dζ.
2πi γ ζ − z
ζ − z0
Then
Z
l(γ) 1
z − z0
1 dζ ≤
|z − z0 |,
| ind(γ, z) − n(γ, z0 )| =
2π γ (ζ − z)(ζ − z0 )
2π δ 2
as |ζ −z0 | ≥ δ and |ζ −z0 | ≥ δ for ζ ∈ γ and z ∈ D(z0 , r). This proves the continuity.
Let z0 be a point in the unbounded component of U , which contains any z with
|z| > supw∈γ |f (w)|. For such z,
Z
1
1
ind(γ, z0 ) = ind(γ, z) =
dζ.
2πi γ ζ − z
But by taking z far away, ind(γ, z0 ) can be take to be arbitrarily close to 0 and is
therefore zero.
Example 7.1.5 If U is a simply connected domain with γ its smooth boundary with
positive orientation, then
1, if z0 ∈ U
ind(γ, z0 ) =
0, If z0 ∈ C \ Ū .
Definition 7.1.3 A closed piecewise smooth curve γ in an open set U is said to be
homologous to zero in U , if ind(γ, z0 ) = 0 for all z0 ∈ C \ U . This is denoted by
γ ≈ 0.
Example 7.1.6 If γ is a closed piecewise smooth curve in U , homotopic to 0, then
γ ≈ 0.
68
The curve γ can be deformed continuously to a curve γ̃ in a disc D, the disc
1
is a holomorphic function for any z0 6∈ D,
is contained entirely in U . But z−z
0
and ind(γ̃, z0 ) = 0 by Cauchy’s theorem. Since the index of a curve is an integer,
ind(γ, w) = ind(γ̃, w), and ind(γ, z) = 0 for z ∈ C \ U . This is not a proof, as the
intermediate deformation from γ to the constant curve may not be piecewise C 1 .
Theorem 7.1.7 (Cauchy’s integral formula) Let f be a holomorphic function in an
open set U . If γ1 , . . . , γm are closed piecewise C 1 curves s.t.
m
X
ind(γk , w) = 0,
∀w ∈ C \ U,
k=1
then for any z0 ∈ U \ ∪∞
k=1 {γk },
f (z0 )
m
X
ind(γk , z0 ) =
k=1
Z
m
X
f (z)
1
dz.
2πi γk z − z0
k=1
The proof is given in the supplementary section.
Theorem 7.1.8 (Cauchy’s Theorem) Let f be a holomorphic function in an open set
U . If γ1 , . . . , γm are closed piecewise C 1 curves s.t.
m
X
ind(γk , w) = 0,
∀ w ∈ C \ U,
k=1
then
m Z
X
k=1
f (z)dz = 0.
γk
Proof In Cauchy’s Integral formula substitute f (z)(z − z0 ) for f (z) at a point z0 ∈
U \ ∪∞
k=1 {γk }.
7.1.1
1. If γ is homologous to zero,
R
f (z)dz = 0.
R
2. If γ is a closed curve in a simply connected region then γ f (z)dz = 0.
Example 7.1.9
γ
Supplementary
Lemma 7.1.10 For f defined and continuous on a piecewise C 1 curve γ, the function
R f (w)−f (z)
dw is holomorphic on C \ {γ}. Furthermore for each integer m ≥ 1,
w−z
γ
define for z 6∈ {γ},
Z
f (w)
Fm (z) =
dw
(w
− z)m
γ
0
then each Fm is holomorphic on C \ {γ} and Fm
(z) = mFm+1 (z).
Proof This can be proved by the definition of complex differentiation, and by induction
on m.
We supply one proof for Cauchy’s integral formula.
69
Theorem 7.1.11 (Cauchy’s integral formula) Let f be a holomorphic function in an
open set U . If γ1 , . . . , γm are closed piecewise C 1 curves s.t.
m
X
ind(γk , w) = 0,
∀w ∈ C \ U,
k=1
then for any z0 ∈ U \ ∪∞
k=1 {γk },
f (z0 )
m
X
k=1
Z
m
X
f (z)
1
dz.
ind(γk , z0 ) =
2πi γk z − z0
k=1
Proof We prove the case where there is only one curve γ, the proof for m ≥ 2 is
analogous. We wish to prove that for z0 ∈ U \ {γ},
Z
f (w) − f (z0 )
dw = 0.
w − z0
γ
If so,
Z
γ
f (w)
dw =
w − z0
Z
γ
f (z0 )
dw = f (z0 ) ind(γ, z0 )2πi.
w − z0
Define g : C \ {γ} → C by
Z
g(z) =
γ
f (w) − f (z)
dw.
w−z
By Lemma 7.1.10, g is holomorphic on C \ {γ}. We construct an analytic continuation
of g to the whole plane and prove the resulting entire function is a constant.
Let
H = {z ∈ C \ {γ} : ind(γ, z) = 0}.
If z ∈ H,
Z
γ
f (z)
dw = f (z) ind(γ, z)2πi = 0,
w−z
Z
g(z) =
γ
f (w)
dw.
w−z
(7.1.1)
Since the index is integer valued and ind(γ, z) is continuous in z, then H is an open set
(as H = {z ∈ C : ind(γ, z) < 21 }). Also C = U ∪ H following from the assumption
that γ is homologous to zero (and so C \ U is a subset of H).
It remains to define g on {γ}. We define φ : U × U → C by
f (w)−f (z)
, z 6= w
w−z
φ(z, w) =
.
f 0 (z),
z=w
Then φ is continuous. For each w ∈ C, φ(·, w) is holomorphic (Lemma 6.1.1). Define
( R
φ(z, w)dw, z ∈ U
Rγ f (w)
g(z) =
.
dw,
z∈H
γ w−z
If z ∈ U ∩ H, z ∈
6 {γ}, the two formulas agree:
Z
Z
Z
f (w)
f (w) − f (z)
φ(z, w)dw =
dw =
dw,
w
−
z
w
−z
γ
γ
γ
70
by (7.1.1), and g is well defined. Furthermore g is an entire function.
We observe H contains the unbounded component of C\{γ} and thus a neighbour1
=
hood of ∞ in the extended complex plane. Since f is bounded on γ and limz→∞ z−w
0 uniformly on γ,
Z
f (w)
dw = 0.
(7.1.2)
lim g(z) = lim
z→∞
z→∞ γ w − z
By the extension to Liouville’s theorem, Example 5.4.5, g is a constant on C. Since
limz→∞ g(z) = 0 then g is identically zero. The proof is complete.
7.2
The Residue Theorem (Lecture 23)
Recall that a meromorphic function has only a finite number of poles on a bounded
subset of U . Also if w belongs the unbounded component of C\{γ} where γ is a closed
piecewise C 1 smooth curve, ind(γ, w) = 0. In particular the set {w : ind(γ; w) 6= 0}
is a bounded set.
Theorem 7.2.1 (The Residue Theorem) Let f be a meromorphic function in a region
U with a finite number of poles {z1 , . . . , zm }. If γ is a closed piecewise C 1 smooth
curve in U , not passing through any of the poles {z1 , . . . , zm }, and γ ≈ 0 in U . Then
1
2πi
Z
f (z)dz =
γ
m
X
ind(γ, zk ) Res(f ; zk ).
k=1
Proof Let V = U \ {z1 , . . . , zm } then f is holomorphic on V . We choose positive
numbers δk such that the discs D(zk , δ) are contained in U and no two discs intersect.
Define
γk (t) = zk + δk e−it·ind(γ,zk ) ,
0 ≤ t ≤ 2π.
Then they are all homologous to 0 in U and
ind(γk , zj ) = − ind(γ, zk )δkj .
We thus have
ind(γ, w) +
m
X
ind(γk , w) = 0,
∀ w ∈ C \ V.
k=1
By Cauchy’s theorem (Theorem 7.1.8),
Z
f (z)dz +
γ
m Z
X
k=1
f (z)dz = 0.
γk
Now by the Laurent’s expansion for f in the deleted disc D0 (zk , δk ),
Z
Z
Res(f ; zk )
f (z)dz =
dz = 2πi Res(f ; zk ) ind(γk , zk )
z − zk
γk
γk
= −2π Res(f ; zk ) ind(γ, zk ).
Pm R
Since γ f (z)dz = − k=1 γk f (z)dz, this completes the proof.
R
71
7.3
Compute Real Integrals
Rule of Thumb. Let P and Q be polynomial of degree n and m respectively with
m ≥ n + 2. We wish to compute real integrals of the form
Z ∞
P (x)
dx.
−∞ Q(x)
+
+
Let us take the contour curve γR = [−R, R] ∪ CR
, where CR
is the upper half of the
circle of radius R centred at 0. Then on one hand, we may apply the Residue theorem,
to obtain for sufficiently large R,
Z
X
P
P (z)
dz = 2πi
Res( ; ζ).
Q
γR Q(z)
ζ
The summation is over poles inside the contour. On the other hand, as R → ∞.
Z
P (z) C
dz ≤
→ 0.
CR+ Q(z) R
This means
Z
∞
−∞
X
P (x)
P
dx = 2πi
Res( ; ζ).
Q(x)
Q
ζ
Example 7.3.1 Suppose that p, q are holomorphic functions in a neighbourhood of z0
and f = pq has a simple pole at z0 . If p(z0 ) 6= 0, then
Res(f ; z0 ) =
p(z0 )
.
q 0 (z0 )
(7.3.1)
In fact,
Res(f ; z0 ) = lim (z − z0 )
z→z0
z − z0
p(z)
p(z0 )
= p(z0 ) lim
= 0
,
z→z0 q(z)
q(z)
q (z0 )
This follows from the Taylor expansion:
q(z) = q(z0 ) + q 0 (z0 )(z − z0 ) + (z − z0 )2
∞
X
ak (z − z0 )k−2 .
k=2
Example 7.3.2 Prove that
Z
∞
−∞
dx
2π
=
.
+1
3
x6
Let us consider the
f (z) =
z6
1
.
+1
On the region bounded by the line segment [−R, R] and the semi-circle C + , f has
the following singularities:
iπ
c1 = e 6 ,
c2 = e
3πi
6
72
= i,
c3 = e
5πi
6
.
The residues at these points are, using formula (7.3.1):
1 − 5πi
e 6 ,
6
Res(f ; c1 ) =
1
Res(f ; c2 ) = − i,
6
Res(f ; c3 ) =
1 −iπ
e 6 .
6
By the residue theorem,
Z
Z
R
f (z)dz =
−R
γ
But (2πi)
P3
i=1
dx
+
6
x +1
Z
0
3
X
iReit
dt
=
(2πi)
Res(f ; ci ).
R6 ei6t + 1
i=1
Res(f ; ci ) = (2πi)(− 2i
6)=
Z
lim
R→∞
Consequently,
2π
R∞
dx
−∞ x6 +1
=
0
2π
2π
3
and
iReit
dt = 0.
R6 ei6t + 1
2π
3 .
73
Chapter 8
Fundamental Theorems
Any non-zero number z has a logarithm, so elog z = z. As z varies, log z would define
an inverse function to ez , which we hope to be holomorphic. For z ∈ C \ {(−∞, 0)},
we may write z = reiθ where θ ∈ (−π, π) and define the principal branch of the
logarithm, on the slit plane C \ {(−∞, 0)}, by
log z = log r + iθ.
If f (z) is a never vanishing function in a region U , for which domains could we
choose the logarithm of f (z) in a way so that the resulting function in holomorphic?
Theorem 8.0.1 Let U be a star region (or more generally a simply connected region).
Suppose that f (z) is holomorphic and never vanishing in U . Then there exists a holomorphic function h : U → C such that eh(z) = f (z).
0
Proof Since f (z) 6= 0 for any z ∈ U , ff is holomorphic in U and hence has a
primitive, by theorem 4.4.3). (More generally, a holomorphic function in a simply
connected region has a primitive).
0
0
Let h be a primitive of ff , i.e. a holomorphic function such that h0 = ff . Set
g(z) = e−h(z) f (z), then g 0 (z) = −h0 (z)e−h(z) f (z) + e−h(z) f 0 (z) = 0. So g is a
constant, say k, and k = e−h(z0 ) f (z0 ) for a given point z0 ∈ U . Choose a primitive h
0
of ff such that eh(z0 ) = f (z0 ) to conclude.
8.1
The Argument Principle (Lecture 24)
The logarithm of a holomorphic never vanishing function f is mutli-valued, log f (z) =
0
(z)
log |f (z)| + i arg(z), its derivative is single-valued (log f (z))0 = ff (z)
.
If a holomorphic function f has a zero of order m at z0 , then f (z) = (z −z0 )m g(z)
where g is holomorphic and g(z0 ) 6= 0,
f 0 (z)
m
g 0 (z)
=
+
.
f (z)
z − z0
g(z)
The function g is holomorphic near z0 . Suppose f is meromorphic and has a pole of
order m at z0 . Then f (z) = (z − z0 )−m g(z) where g is holomorphic and g(z0 ) 6= 0.
74
Also,
f 0 (z)
m
g 0 (z)
=−
+
.
f (z)
z − z0
g(z)
Theorem 8.1.1 Let f be a meromorphic function in a region U with poles p1 , . . . , pn
and zeros z1 , . . . , zm , counted according to multiplicity. Let γ be a closed piecewise
smooth curve γ in U with γ ≈ 0, and not passing through p1 , . . . , pn , z1 , . . . , zm . Then
1
2πi
Z
γ
m
n
X
X
f 0 (z)
dz =
ind(γ, zj ) −
ind(γ, pi ).
f (z)
j=1
i=1
Proof By repeated application of the discussion above,


Z
Z 0
n
m
0
X
X
g
(z)
1
1
f (z)

 dz
dz =
−
+
z − zk j=1 z − pj
g(z)
γ
γ f (z)
k=1
where g is holomorphic, g 0 /g is holomorphic. By Cauchy’s theorem, since γ ≈ 0,
Z 0
g (z)
dz = 0.
γ g(z)
The rest follows from the definition of the index.
Apply the above theorem to a holomorphic function we obtain the following:
Theorem 8.1.2 Let U be a simply connected region with contour γ. Let f be a meromorphic function in an open set containing the closure Ū with poles p1 , . . . , pn and zeros z1 , . . . , zm , counted according to their multiplicity, none of which passing through
the curve γ. Then
Z 0
f (z)
dz = Z(f ) − P (f )
γ f (z)
where Z(f ) and P (f ) are respectively the number of zeros and poles of f in U .
Interpretation of the theorem. Suppose f is holomorphic. If γ a closed piecewise
C 1 curve, so is the pushed forward curve f ◦ γ, and it is easy to see (c.f. exercise 3,
sheet 4),
Z 0
Z
1
f (z)
1
1
dz =
dz = ind(f ◦ γ, 0).
2πi γ f (z)
2πi f ◦γ z
8.2
Rouché’s Theorem
Rouché’s theorem states that the number of zeros minus the numbers of poles, Z(f ) −
P (f ), of a function f is unchanged if we modify f slightly. If on a closed curve, the
distance of the change g(z) to the origin is smaller than the distance of f (z) to the
origin, then f and f + g have the same number of zeros minus poles inside the curve.
We state a simple version of the theorem.
75
Theorem 8.2.1 Suppose that f, g are holomorphic in an open set containing a closed
disc D(z0 , R). Let Z(f ) and Z(g) denote respectively the number of zeros of f and g
inside the circle γ = C(z0 , R), counted according to their multiplicity. Suppose that
|g(z)| < |f (z)|,
∀z ∈ {γ}.
(8.2.1)
Then Z(f + g) = Z(f ).
Proof Let h(z) = f (z)+g(z)
= 1 + fg(z)
f (z)
(z) . By (8.2.1), f has no zero on {γ}, f + g has
no zero on {γ} and h is a holomorphic around γ. Indeed, On {γ}, |f (z)| > |g(z)| > 0
and |f + g| ≥ |f | − |g| > 0. Furthermore,
h0
(f + g)0
f0
=
+
h
f +g
f
Integrate the above identity and apply Theorem 8.1.2 to f + g and f to obtain:
Z 0
h
= Z(f + g) − Z(f ).
γ h
But
1
2πi
Z
γ
h0 (z)
1
dz =
h(z)
2πi
Z
h◦γ
1
dz.
z
But there exists δ < 1 such that |h(z) − 1| < δ for any z ∈ {γ}. This means {h(γ)}
is contained in the disc D(1, δ) which does not contain 0 and ind(h ◦ γ, 0) = 0. This
concludes Z(f + g) − Z(f ) = 0.
Example 8.2.2 This gives another proof of the fundamental theorem of algebra. If
pn (z) is a polynomial of degree n. Write pn (z) = an z n + Qn−1 (z). Then for R
sufficiently large, |Q(z)| ≤ |an ||z n |, so pn (z) has the same number of zero’s as z n
inside |z| ≤ R, while z n = 0 has zero or order n at 0.
8.2.1
Supplementary
Example 8.2.3 If f is holomorphic in a neighbourhood of the closed disc D̄ = {z : |z| ≤ 1}
and |f (z)| < 1 for |z| = 1. There there exists exactly one solution of f (z) = z in
D = {z : |z| < 1}.
This is an easy consequence of Rouché’s theorem: treat f (z) as a perturbation to
z, |f (z)| < |z| on the circle {z : |z| = 1}, hence z − f (z) has one solutions in D, just
as g(z) = z.
8.3
The Open mapping Theorem (Lecture 25)
Recall that f has a zero of order m ≥ 1 at z0 if in a neighbourhood of z0 , f (z) =
P
∞
k
(m−1)
(z0 ) = 0, and f (m) (z0 ) 6= 0.
k=m ak (z −z0 ) . Equivalently, f (z0 ) = 0, . . . , f
In particular f is not a constant.
Theorem 8.3.1 Let f be a non-constant holomorphic function in D(z0 , R). Suppose
that f has a zero of order m ≥ 1 at z0 . If > 0 is sufficiently small, then there
exists a corresponding positive number δ such that for w ∈ D0 (f (z0 ), δ) the equation
76
f (z) = w has exactly m solutions in z ∈ D0 (z0 , ). In particular D(f (z0 ), ) is
contained in the image of D(z0 , ) by f , i.e. D(f (z0 ), ) ⊂ f (D(z0 , )). Each zero of
f (z) − w, for w ∈ D0 (f (z0 ), δ), is a simple zero.
Proof For w ∈ C define Fw (z) = f (z) − w. Set w0 = f (z0 ). Since z0 is an isolated
zero of Fw0 (z) = f (z) − w0 , there exists a number > 0 such that 3 < R and on
D0 (z0 , 3) := {z : 0 < |z − z0 | < 3} the following statements hold:
1. Fw0 (z) ≡ f (z) − w0 6= 0,
2. Fw0 0 (z) ≡ f 0 (z) 6= 0.
(If m = 1, then Fw0 0 (z0 ) = f 0 (z0 ) 6= 0. Since Fw0 0 is continuous, it does not vanish in
a neighbourhood of z0 .
If m ≥ 1, then Fw0 0 (z0 ) = 0 in which case z0 is an isolated zero for the holomorphic
function Fw0 0 .)
We observe that Fw (z) = f (z) − w0 − (w − w0 ) = Fw0 (z) − (w − w0 ). As
|f (z) − w0 | is continuous and non-zero on the circle, δ = inf |z−z0 |= |f (z) − w0 | > 0,
by the compactness of the circle. If |w − w0 | < δ, so in particular |w − w0 | < |Fw0 (z)|
on the circle, we apply Rouché’s theorem to see Fw and Fw0 have the same number of
zeros’s inside the disc {z : |z − z0 | < }.
In particular each w ∈ {w : |w − w0 | < δ} has a pre-image in {z : |z − z0 | < }.
For w with 0 < |w − w0 | < δ, let z be a-pre-image of w, then Fw0 (z) 6= 0 and z must
be a simple root of Fw .
A map f is open if it takes an open set to an open set. Theorem 8.3.1 leads, trivially,
to the following corollary.
Theorem 8.3.2 (The Open Mapping Theorem) A non-constant holomorphic map in
a region is an open map.
Remark 8.3.3 Suppose that f : U → C is holomorphic. Then f is injective in a
neighbourhood of z0 ∈ U if and only if f 0 (z0 ) 6= 0. (This is a local statement, only.
The holomorphic function f : C \ {0} → C given by f (z) = z 2 is not globally
injective, f 0 (z) 6= 0.) In particular if f is one to one then f 0 never vanishes and
f −1 : f (U ) → U is holomorphic (by Theorem 1.8.3, The Inverse Function Theorem);
consequently f preserves angle at each point.
Lemma 8.3.4 Suppose f is a holomorphic and non-constant in a region U . Then for
any z0 ∈ U and any R > 0 such that D(z0 , R) ⊂ U , there exists z ∈ D(z0 , R) such
that |f (z)| > |f (z0 )|.
Proof By the open mapping theorem, f (D(z0 , R)) contains a disc D(f (z0 ), δ), in the
latter disc there exists w such that |w| ≥ |f (z0 )|. (e.g. if w0 = r0 eiθ0 take w = r0 eiθ0 +
δ iθ0
, |w| = |w0 | + 2δ ). The pre-image of w in D(z0 , R) satisfies |f (z)| > |f (z0 )|. 2e
Theorem 8.3.5 (Maximal Modulus Principle) If f is a non-constant holomorphic function in a region U , then |f | cannot attain a local maximum in U .
Proof If f attains its maximum at z0 ∈ U , it is a maximal in a disc D(z0 , r), where r
is small so that D(z0 , r) ⊂ U . This is impossible from the previous lemma.
77
Corollary 8.3.6 Let U be a region with compact closure Ū . Let f : Ū → C be
continuous and holomorphic on U . Then supz∈U |f (z)| is attained on the boundary of
U.
Proof Since f is continuous on the compact set Ū , it attains its maximum in Ū , which
cannot be attained in the interior.
8.3.1
Supplementary
Why the argument for proving Maximal Modulus Principle might fail for the infimum,
inf |f (z)|? Think of the case w0 = 0! In fact if the function f never vanishes, we
would have no problem, just apply the Maximal Modulus Principle to f1 .
Corollary 8.3.7 If f is a holomorphic function on a disc D(z0 , R) and f (z0 ) 6= 0.
Then for all 0 < r < R there exists z such that |f (z)| < |f (z0 )|.
1
Proof Since f (a) 6= 0, f does not vanish in a disc D(z0 , r0 ), r0 < R then f (z)
is
on D(z0 , r0 ). There exists a point z in the disc D(z0 , r0 ) such that
holomorphic
1 1 f (z) > f (z0 ) .
Example 8.3.8 Let f be a not constant holomorphic function in a neighbourhood of
the closed unit disk D := {z : |z| ≤ 1}. Assume that |f (z)| is constant on ∂D :=
{z : |z| = 1}. Then f must attain at least one zero in D.
Suppose |f (z)| = 0 on the unit circle, then |f (z)| = 0 inside the disc, by the
maximal principle, and the assertion holds trivially. Suppose |f (z)| = a > 0 on
the unit circle then |f (z)| is non-zero in a neighbourhood of the unit circle. Assume
that f does not have a zero in D. Then f1 is holomorphic in D := {z : |z| < 1}
and continuous on D. Then the supremum and infimum of |f (z)| are attained on the
boundary: supz∈D |f (z)| = inf z∈D |f (z)| = a. Hence |f (z)| is a constant on the
whole disc. By the Maximal modulus theorem this implies that f is constant.
The following is an alternative proof for the maximal modulus principle.
Theorem 8.3.9 Suppose that f is a non-constant holomorphic function in an open set
containing z0 and a disc D(z0 , R). Then for each 0 < r < R, there exists a point in
D(z0 , r) s.t. |f (z)| > |f (z0 )|.
Proof Firstly, f is not a constant on D(z0 , r), otherwise f would be a constant everywhere, c.f. Theorem 5.8.3. If f (z0 ) = 0 the conclusion holds ( there is a point z with
f (z) 6= 0.) By Cauchy’s formula, for every z ∈ D(z0 , r),
Z
Z 2π
1
f (z)
1
f (z0 ) =
dz =
f (z0 + eit )dt.
2πi C(z0 ,r) z − z0
2π 0
Suppose f (z0 ) 6= 0 and |f (z)| ≤ |f (z0 )| on D(z0 , r). Then |f (z)| ≤ f (z0 )| on
C(z0 , r). We divide the equation above by f (z0 ) and split the real and pure imaginary
(z)
= u(z) + iv(z). Then u2 (z) + v 2 (z) ≤ 1, and
parts of the integrand: ff(z
0)
Z
2π
u(z0 + eit )dt = 2π.
0
78
This can happen if and only if u is identically 1. (If u(z0 ) < 1 at one point there
exists
a real number
M < 1 s.t. f (z) ≤ M on an interval of non-zero length and
R
2π
it
0 u(z0 + e )dt ≤ 2π.)
Thus u ≡ 1 and v = 0 and f is a constant on the disc D(z0 , r), a contradiction. We
conclude that there exists z ∈ D(z0 , r) such that |f (z)| > |f (z0 )|.
8.4
Bi-holomorphic Maps on the Disc (lecture 25-26)
We would like to understand how does a holomorphic function look like.
Lemma 8.4.1 (Schwartz’s Lemma) Let D = {z : |z| < 1}. Suppose that f is a
holomorphic function on D, mapping D into D, and f (0) = 0. Then f satisfies one of
the following descriptions:
• For all z, f (z) = cz for a complex number c with |c| = 1.
• |f (z)| < |z| for z with 0 < |z| < 1.
Proof Since f is holomorphic on D and f (0) = 0, we apply Lemma 5.8.1: there exists
a holomorphic function h such that
f (z) = zh(z),
z ∈ D.
Since |f (z)| ≤ 1, for any number 0 < r < 1,
1 ≥ sup |f (z)| = sup |z||h(z)| = r sup |h(z)|.
|z|=r
Thus sup|z|=r |h(z)| ≤
8.3.9,
1
r
|z|=r
|z|=r
for any r < 1. By the maximum modulus principle, Theorem
sup |h(z)| = sup |h(z)| ≤
|z|≤r
|z|=r
1
.
r
Taking r → 1 we see that,
sup |h(z)| ≤ 1.
|z|<1
If there exists z0 such that |h(z0 )| = 1, then h has a local maximum at z0 . Thus
h(z) = h(z0 ) a constant and f (z) = zh(z0 ).
Otherwise |g(z)| < 1 for any z ∈ D. Then for z 6= 0 and z ∈ D, |f (z)| =
|z||g(z)| < |z|.
Definition 8.4.1 A bijective holomorphic function f : U → V is said to be biholomorphic (In this case its inverse is also holomorphic.) We say U and V are
conformally equivalent, if a bi-holomorphic map f : U → V exists.
Proposition 8.4.2 If f : D → D is a bi-holomorphic function with f (0) = 0. Then
f (z) = kz for some complex number k with |k| = 1.
79
Proof We apply Schwartz’s lemma to both f and f −1 : |f (z)| ≤ |z| and |f −1 (z)| ≤ |z|
on D. Thus |f (z)| = |z| on D. By the same lemma, f (z) = kz for some complex
number k.
z−b
Denote Gb (z) = 1−
the Möbius transform, where |b| < 1. Then Gb takes D
b̄z
−1
onto D, and (Gb ) = G−b . (See Example sheet 2)
Theorem 8.4.3 If f : D → D is a bi-holomorphic function, then there exist complex
numbers k, b with |k| = 1 and |b| < 1 such that f = kGb .
Proof There exists a unique point b such that f (b) = 0. Then f ◦ (Gb )−1 : D → D
is bi-holomorphic and takes 0 to 0. By Proposition 8.4.2, there exists k such that
f ◦ (Gb )−1 (z) = kz and f (z) = kGb (z).
Remark 8.4.4 Let H := {z : Im(z) > 0}. Then f : H → H given by the formula
f (z) = az+b
cz+d , where a, b, c, d ∈ R with ad−bc = 1, are bi-holomorphic. (c.f. Example
sheet 2) All bi-holomorphic maps from H to H are of the above form. This can be
proved by transferring our problem to D by the bi-holomorphic map F : H → D,
i+z
F (z) = i−z
. Its inverse is G(w) = i 1−w
1+w .
80
Chapter 9
The Riemann Mapping
Theorem
Given any two open sets, does there exist a bi-holomorphic function between them? If
U = C we know this is not possible if V is a bounded region, not a singleton. Such
maps are bounded entire and so a constant by Liouville’s theorem. If f : U → V is biholomorphic (or a homeomorphism) and U is simply connected then V is also simply
connected. If f : U → V is continuous, its image will be contained in a connected
sub-set of V . It is perhaps surprising that these properties determine all open sets that
are conformally equivalent to D. Recall that a region is an open connected subset of
C.
Theorem 9.0.1 (The Riemann Mapping Theorem) A simply connected region U , which
is not the whole plane C, is conformally equivalent to the disc D = {z : |z| < 1}.
Main Idea. Let z0 ∈ U . Let
F = {f : U → D holomorphic, one to one, f (z0 ) = 0, f 0 (z0 ) > 0}.
We prove that (1) F is not empty. (2) Prove B = sup{|f 0 (z0 )| : f ∈ F } < ∞. There
exist a sequence of functions fn such that limn→∞ fn0 (z0 ) = B. We prove that fn has
a subsequence, converging to a function f uniformly on compact subsets of U . (3) If
f ∈ F satisfies f 0 (z0 ) = B, it is onto.
9.1
Hurwitz’s Theorem (lecture 26)
Pn
k
Example 9.1.1 Let fn (z) = k=1 zk! . The entire functions fn are mapped onto C by
the Fundamental Theorem of Algebra. They converge uniformly on compact subsets
of C to ez . But ez does not take the value 0.
Example 9.1.2 let fn (z) = z0 + nz . These are entire bijective functions, converge
uniformly on compact subsets of C to the constant function z0 , failing injectivity.
Theorem 9.1.3 (Hurwitz’s Theorem) Suppose fn is a sequence of holomorphic functions in a region U , converging to a function f uniformly on compact subsets of U .
Suppose f is not a constant. Then the following statements hold.
81
1. Let w0 = f (z0 ) where z0 ∈ U . Then for sufficiently small , there exists a
natural number N () such that for each n ≥ N (), there exists zn ∈ D(z0 , )
such that fn (zn ) = w0 .
2. If each fn is one to one, so is f .
Proof By the uniform convergence theorem (Thm 5.5.1), f is holomorphic. Now,
fn (z) − w0 = (f (z) − w0 ) + (fn (z) − f (z)).
Since z0 is an isolated zero of f −w0 , there exists a number > 0 such that D(z0 , 3) ⊂
U and f (z) 6= w0 on D0 (z0 , 3). Let
δ=
inf
z:|z−z0 |=
|f (z) − w0 | > 0.
Since fn converges uniformly to f on D(z0 , 2), there exists N such that for n ≥ N ,
|fn (z) − f (z)| < δ. For such n, we apply Rouché’s theorem: fn − w0 and f − w0
have the same number of zero’s inside D(z0 , ), in particular fn − w0 has at least one
zero. This completes (1).
Suppose f (z1 ) = f (z2 ) for z1 6= z2 . Denote w0 = f (z1 ). There exists >
0 such that the discs D(z1 , ) and D(z2 , ) are disjoint and contained in U , and for
n sufficiently large, w0 has a pre-image by fn in each of the discs. Such fn is not
injective, giving a contradiction, proving (2).
9.2
Family of Holomorphic Functions (Lecture 28)
If U is a region in C, we denote by H(U ) the family of holomorphic functions on U .
Definition 9.2.1 Let U be a region in C and F a family of holomorphic functions on
U . We say
1. F is normal or relatively compact, if every sequence from F has a sub-sequence
that converges uniformly on every compact subset of U .
2. F is locally uniformly bounded, if for any compact subset K of U , there exists
M > 0 such that |f (z)| ≤ M for any f ∈ F and z ∈ K.
3. F is equi-continuous on a compact set K of U if for every > 0 there exists
δ > 0 such that if |z − w| < δ, z, w, ∈ K,
|f (z) − f (w)| < ,
∀, f ∈ F.
If there exists M such that |f 0 | ≤ M for all f ∈ F then F is equi-continuous.
The functions {fn (x) = xn : x ∈ [0, 1]} is not equi-continuous, nor has uniform
convergent subsequences (the limit is a discontinuous function).
Theorem 9.2.1 (Mantel’s Theorem) Suppose F is a locally uniformly bounded family of holomorphic functions in a region U , then F is normal.
The proof for Mantel’s theorem is split into two lemmas. The first one is specific to
holomorphic functions, while the second is part of an Arzela-Ascoli theorem.
82
Lemma 9.2.2 Suppose F is a family of holomorphic functions uniformly bounded on
compact subsets of U , then F is equi-continuous on every compact subset of U .
Proof Let K be a compact subset of U . Then there exists a positive number r and an
enlargement of K which is contained in U:
K̃ = {z : the distance from z to K is less or equal to 4r} ⊂ U.
Let M = supf ∈F ,z∈K̃ |f (z)|. Let z, w ∈ K with |z − w| < r. Then the circle
γ = C(z, 2r) and its interior is contained in U .
Z Z
1
f (ζ)
f (ζ)
1
f (ζ)(z − w)
f (z) − f (w) =
−
dζ =
dζ.
2πi γ ζ − z
ζ −w
2πi γ (ζ − z)(ζ − w)
r
For a given > 0, let δ = min( 2M
, r) and |z − w| < δ, z, w ∈ K,
|f (z) − f (w)| ≤
2M
4πr
1
{ sup
|f (ζ)| 2 |z − w| ≤
|z − w| < .
2π ζ:|ζ−z|=2r}
r
r
This proves f is equi-continuous on K.
A sequence {Kl }∞
l=1 of compact subsets of an open set U is an exhaustion if (1)
Each Kl is contained in the interior of Kl+1 , (2) Any compact subset K of U is contained in Kl for some l. Such an exhaustion always exists: take Kl = {z ∈ U : |z| ≤
l, distance(z, ∂U ) ≥ 1l }.
Lemma 9.2.3 Suppose F is a locally uniformly bounded family of holomorphic functions in a region U which are equi-continuous on compact subsets of U , then F is
normal.
Proof Let {fn } be a sequence from F. Given a compact subset K of U let {wj } be
a dense set of K. The sequence of numbers {fn (w1 )} is bounded and hence has a
convergent subsequence given by a sub-sequence of function {fn,1 }. From {fn,1 } we
extract another sequence {fn,2 } such that fn,2 (w2 ) converges as n → ∞. We repeat
this procedure and obtain nested sequence {{fn,k }, k = 1, 2, . . . , } with the property
that for each k ≥ 1, fn,k (wk ) converges. Let gn = fn,n be the diagonal sequence of
functions.
Since F is equi-continuous on K, for > 0 there exists δ > 0 such that |f (z) −
f (w)| < 3 whenever |z − w| < δ and z, w ∈ K. Since K is compact we may
select a finite number of points from {wj } which we denote by z1 , . . . , zÑ such that
K ⊂ ∪Ñ
k=1 D(zj , δ). There is N () > 0 such that for n, m ≥ N (),
|gn (zj ) − gm (zj )| <
,
3
∀j = 1, . . . , Ñ .
If w ∈ K, there is a number j0 such that w ∈ D(zj0 , δ). Then
|gn (w) − gm (w)| ≤ |gn (w) − gn (zj0 )| + |gn (zj0 ) − gm (zj0 )| + |gm (zj0 ) − gm (w)|.
If n, m ≥ N (), observing |zj0 − w| < δ,
|gn (w) − gm (w)| ≤
Thus gn converges, uniformly on K.
83
+ + .
3 3 3
We proceed to proceed that there is a sequence of gn which converges uniformly
on every compact set of U , by another diagonal argument. Let {Kn } be an exhaustion
of U . Let {gn,1 } be a subsequence of {gn } which converges uniformly on K1 , {gn,2 }
a sub-sequence of {gn,1 } converging uniformly on K2 . Then the diagonal sequence
{gn,n } converges uniformly on each Kn . Since every compact set is included in one
of the set from {Kl }, the proof is complete.
9.3
The Riemann Mapping Theorem (Lecture 28-29)
There are homeomorphisms (bijection, continuous with inverse continuous) from D to
z
; there are no bi-holomorphic functions from D to C.
C, e.g. take f (z) = 1−|z|
Theorem 9.3.1 (The Riemann Mapping Theorem) If U is a simply connected region U , which is not the whole plane C, there exists a bi-holomorphic map from U
to the disc D = {z : |z| < 1}.
The bi-holomporphic maps from U to D is unique after we fix some parameters.
Proposition 9.3.2 Suppose U is a simply connected region which is not the whole C,
and z0 ∈ U then there is at most one bi-holomorphic map f : U → D s.t. f (z0 ) = 0
and f 0 (z0 ) > 0.
Proof If f, g are two such maps, f ◦g −1 : D → D is holomorphic with f ◦g −1 (0) = 0.
Hence f = cg for a complex number c with |c| = 1. Now f 0 (z0 ) = cg 0 (z0 ), both
derivatives are real numbers, then c is a real number and c = 1.
Let us consider the following family of holomorphic functions:
F = {f : U → D holomorphic, one to one, f (z0 ) = 0}.
A function f on U is said to be non-vanishing if f (z) 6= 0 for any z ∈ U .
Lemma 9.3.3 If U is a simply connected and connected set which is not the whole
plane, there exists a bi-holomorphic function h : U → h(U ) such that h(U ) does not
intersect a disc D(w0 , δ) for some δ > 0 and some w0 ∈ C.
Proof Since U is a proper subset of C, there exists a ∈ C \ U . Then z − a is a
non-vanishing holomorphic function on the simply connected region U . There exists a
holomorphic function h̃ : U → C such that z − a = eh̃(z) (see Theorem 8.0.1). Define
1
h(z) = e 2 h̃(z) , a holomorphic function with
h2 (z) = z − a.
We observe that h does not take the same value twice nor opposite values. (If h(z1 ) =
±h(z2 ) then z1 − a = z2 − a.)
By the open mapping theorem, h(U ) contains a disc D(h(z0 ), δ), it therefore cannot intersect D(−h(z0 ), δ), which contains opposite values. Let w0 = −h(z0 ).
Lemma 9.3.4 F is not empty.
84
Proof By Lemma 9.3.3, we may assume that D(w0 , δ) ∩ U = ∅ for some δ > 0,
δ
for z ∈ U . Since |z − w0 | > δ for all z ∈ U ,
and w0 ∈ C and define f (z) = z−w
0
f is holomorphic. It is clearly one to one, and |f (z)| < 1. If Gb : D → D is the
z−b
bi-holomorphic map taking b to zero, Gb (z) = 1−
, then Gf (z0 ) ◦ f ∈ F.
b̄z
Since G0b (b) =
1
1−|b|2 ,
(Gf (z0 ) ◦ f )0 (z0 ) =
1
0
1−|f (z0 )|2 f (z0 )
6= 0.
Lemma 9.3.5 There exist f ∈ F s.t. |f 0 (z0 )| = supf ∈F |f 0 (z0 )|.
Proof Take a disc D(z0 , r) ⊂ U . By Cauchy’s inequality,
|f 0 (z0 )| ≤
1
1
sup |f (z)| ≤ ,
r |z−z0 |=r
r
∀f ∈ F.
The set {|f 0 (z0 )|, f ∈ F} is bounded and has a supremum which we denote by B,
B = supf ∈F |f 0 (z0 )|. There exists a sequence fn ∈ F s.t. limn→∞ |fn0 (z0 )| = B. By
Montel’s theorem, since |f | ≤ 1 for all f ∈ F, there exists a subsequence {fnk } of
{fn } converging to a function f uniformly on compact subsets of U . By Weierstrass
Theorem ( i.e. the Uniform Convergence Theorem, Theorem 5.5.1), f is holomorphic.
Since |f 0 (z0 )| = limk→∞ |fn0 k (z0 )| =
6 0, f is not a constant function and is therefore
one to one by Hürwitz’s theorem. Also f (z0 ) = limk→∞ fnk (z0 ) = 0. So f ∈ F. Let us review properties of the basic bi-holomorphic maps on D. For b ∈ D
z−b
, a bi-holomorphic function from D to D with Gb (z) = 0 and
define Gb (z) = 1−
b̄z
Gb (0) = −b, G0b (z) =
1−|b|2
,
(1−b̄z)2
G0b (0) = 1 − |b|2 and G0b (b) =
1
1−|b|2 .
Lemma 9.3.6 If f ∈ F, f is not onto, there exists g ∈ F s.t. |g 0 (z0 )| > |f 0 (z0 )|.
Proof Let w0 ∈ D, w0 6∈ f (U ). The Möbius transform Gw0 has only one zero at
w0 . Hence Gw0 ◦ f : U → D is one to one, holomorphic, and never vanishing. Since
Gw0 ◦ f is defined on a simply connected domain there exists a holomorphic function
h : U → C, a square root function, such that
(h(z))2 = Gw0 ◦ f (z) =
f (z) − w0
,
1 − w¯0 f (z)
(9.3.1)
and (h(z0 ))2 = −w0 . See the proof of Lemma 9.3.
We now take h(z0 ) back to 0 by another möbius transform: g = Gh(z0 ) ◦ h. Then
g ∈ F and |g 0 (z0 )| > |f 0 (z0 )|. To check the inequality holds, we first differentiate
(9.3.1):
2h(z0 )h0 (z0 ) = Gw0 (f (z0 ))f 0 (z0 ) = (1 − |w0 |2 )f 0 (z0 ),
On the other hand,
|g 0 (z0 )| = |G0h(z0 ) (h(z0 )) · h0 (z0 )| =
1
(1 − |w0 |2 )f 0 (z0 )
2
1 − |h(z0 )|
2|h(z0 )|
1 + |w0 |
= |f 0 (z0 )| p
< |f 0 (z0 )|.
2 |w0 |
This completes the proof.
We have competed the proof for the Riemann mapping Theorem.
Remark. Let U be an open connected set. The property that every a holomorphic
never vanishing function on U has a holomorphic square root characterises the property
that U a simply connected.
85
9.4
Supplementary
The following concepts are equivalent for an open connected subset U of C.
1. U is simply connected;
2. C ∗ \ U is connected.
3. Every holomorphic function on U has a primitive;
4. Any non-vanishing holomorphic function on U has a holomorphic square root.
5. Any non-vanishing holomorphic function on U has a holomorphic logarithm, i.e.
a holomorphic function h such that eh = f .
6. U is homeomorphic to a disc.
86
Chapter 10
Special Functions
10.1
Constructing Holomorphic Functions by Integration (Lecture 30)
The following theorem is a simple corollary of Mantel’s Theorem. We will give instead
an elementary proof for it.
Theorem 10.1.1 Let U be a region in C and G : U × [0, 1] → C a continuous function
R1
s.t. G(·, s) is holomorphic for each s ∈ [0, 1]. Then the function f (z) = 0 G(z, s)ds
is holomorphic.
Proof We define a family of holomorphic functions:
fn (z) =
n Z
X
k=1
n
k
n
G(z,
k−1
n
1X
k
k
)ds =
G(z, ).
n
n
n
k=1
Let z0 ∈ U and D(z0 , 2r) a disc in U . Then G is uniformly continuous on the compact
set D(z0 , r). Indeed, for any > 0, there exists δ > 0 such that if |s − t| < δ,
|G(z, s) − G(z, t)| < for all z ∈ D(z0 , r). If n > 1δ , and z ∈ D(z0 , r)
n Z k
X
n
k
|fn (z) − f (z)| = (G(z, ) − G(z, s))ds < .
k−1
n
k=1
n
Hence fn → f converges uniformly on D(z0 , r). By Theorem 5.5.1, f is holomorphic
on D(z0 , r). Since z0 is an arbitrary point of U , f is holomorphic in U .
10.2
The Gamma function
A complex differentiable (holomorphic functions) has a power series expansion around
every point (analytic). In this section we switch terminology and use analytic for holomorphic.
For any positive real number s, the Gamma function
Z ∞
Γ(s) =
e−t ts−1 dt
(10.2.1)
0
87
R1
is well defined. At 0, ts−1 is integrable for s > 1. Let Γ (s) = e−t ts−1 dt. For
t large e−t dominates and is integrable. Hence Γ() converges if s is real and greater
than 1.
Lemma 10.2.1 Formula (10.2.1) defines an analytic function on {s : Re(s) > 0}.
Proof Firstly |ts−1 | = |e(Re(s)+iIm(s)−1) ln t = tRe(s)−1 . Let
Z
1
Γ =
e−t ts−1 dt,
each of which is holomorphic by Theorem 10.1.1. They converge uniformly on the
strips An = {s : n1 < Re(s) < n}. (Check that |Γ (s) − Γ(s)| → 0 uniformly as
→ 0.) Thus Γ is an analytic function.
Lemma 10.2.2 If Re(s) > 0, Γ(s + 1) = sΓ(s). Consequently Γ(n + 1) = n! for
n = 0, 1, 2, . . . .
Proof For any > 0,
1
Z
But lim→0
R
1
d −t s
(e t )dt = −
dt
d
−t s
dt (e t )dt
Z
1
e−t ts dt + s
Z
1
e−t ts−1 dt.
1
= lim→0 e− ( 1 )s − e− s = 0. The right hand side
converges to −Γs+1 + sΓ(s).
Theorem 10.2.3 The function Γ extends to a meromorphic function into C whose only
singularity are simple poles at s = 0, −1, −2, . . . .
Proof On {s : Re(s) > −1}, we define
Γ1 (s) =
Γ(s + 1)
.
s
Then Γ1 agree with Γ for s > 1, and has a simple pole at s = 0. Indeed, since
Γ(1) = 1, lims→0 Γ1 (s) = lims→0 Γ(s+1)
= ∞ proving 0 is a pole.
s
On {s : Re(s) > −2} we define
Γ2 (s) =
Γ1 (s + 1)
Γ(s + 2)
=
.
s
s(s + 1)
Inductively we define on {s : Re(s) > −n}
Γn (s) =
Γn−1 (s + 1)
Γ(s + n)
=
,
s
s(s + 1) . . . (s + n − 1)
where Γ(s + n) is holomorphic on {s : Re(s) > −n}. It is clear that Γn (s) has simple
poles at 0, −1, . . . , −(n − 1), also Γn agree with Γn+1 where they overlap. Hence Γ
has a continuation given by the above formula on each strip {s : Re(s) > −n}.
88
n
1
It is easy to compute the residues: Res(Γ, −n) = (−1)
n! . Also, Γ(s) is an entire function whose only zero’s are simple zeros at 0, −1, −2, . . . , following from the
identity
π
, ∀s ∈ C.
Γ(s)Γ(1 − s) =
sin(πs)
π
1
= Γ(1−s) sin(πs)
. The simple poles of Γ(1−s), at s = 1, 2, 3, . . .
In other words, Γ(s)
and the simple zero’s of sin(πs) cancel.
10.3
The zeta function(Lecture 30)
Let Re(s) > 1, the following infinite sum converges
ζ(s) =
∞
X
1
s
n
n=1
and we have an analytic function on the strip {s : Re(s) > 1}.
Theorem 10.3.1 The zeta function has a meromorphic continuation to C whose only
singularity is a simple pole at s = 1.
Idea of the Proof.
P∞
2
1. Define the theta function θ(t) = n=−∞ e−πn t for t > 0. It has the following
1
properties: |θ(t) − 1| ≤ ce−πt for t ≥ 1 and |θ(t)| ≤ Ct− 2 as t → 0.
Define the xi function:
Z
1 ∞ s −1
ξ(s) =
r 2 (θ(r) − 1)dr.
2 0
Then ξ is holomorphic for Re(s) > 1 and has an meromorphic extension to C
with simple poles at 0 and −1, Moreover ξ(s) = ξ(1 − s) for all s ∈ C.
2. The following holds for Re(s) > 1,
s
ζ(s) =
ξ(s)π 2
.
Γ(s/2)
Since ξ has an meromorphic extension to C whose only singularity are simple
poles at 0 and 1, ζ has a meromorphic extension to C with a simple pole at 1.
The simple pole of ξ(s) at s = 0 cancels with the simple zero of Γ(s/2) at s = 0.
The Euler’s identity states:
ζ(s) = Πp
1
1 − p−s
where the product is taken over all prime number. The heuristic argument for a proof
that there
an infinite number of prime numbers is as following. If we take s = 0,
Pare
∞
we see n=1 n1 = Πp 1−p1 −1 . Since the harmonic series diverges, there must be an
infinite number of terms on the right hand side.
Euler’s trick is to consider both sides as functions on the complex plane.
89
Theorem 10.3.2 There are an infinite number of prime numbers.
Proof If there are only a finite number of prime numbers, then x := Πp 1−p1 −1 > 0 is
a finite number. But by monotonicity, for any s > 1,
ζ(s) = Πp
1
< Πp 1 − p−1 = x.
1 − p−s
But lims→1 ζ(s) = ∞ since it has a pole at 1, giving a contradiction.
Note that ζ(s) has simple zeros at s = −2, −4, . . . , inherited from
it is fairly easy to see ζ(s) does not vanish if Re(s) > 1.
ζ(s) = π s−1/2
1
Γ(s/2) .
In fact
Γ( 1−s
2 )
ζ(1 − s).
Γ( 2s )
Let us consider the region Re(s) < 0 where Re(1 − s) > 1 and ζ(1 − s) has no zero,
1
Γ( 1−s
2 ) has no zero, Γ( s ) has zero at 0, −2, −4, . . . . From this we conclude that
2
Theorem 10.3.3 The only zero’s of the zeta function outside of the critical strip {s : 0 ≤ Re(s) ≤ 1}
are at the negative even integers −2, −4, −6, . . . . Furthermore, there are no zeros on
the vertical line Re(s) = 1.
The negative even integers are called the trivial zero’s of the zeta function.
Riemann’s Hypothesis. The zeros of the zeta function in the critical strip {s : 0 ≤ Re(s) ≤ 1}
lie on the vertical line Re(s) = 21 .
90