Math 116 Complex Analysis.
Yakov Eliashberg
October 3, 2022
2
Contents
I
Complex Analysis Basics
1
Linear algebra
2
3
4
9
11
1.1
Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.2
Complex linear function from the real perspective . . . . . . . . . . . . . . . . . . 14
Holomorphic functions
17
2.1
Differentiability and the differential . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2
Holomorphic functions, Cauchy-Riemann equations . . . . . . . . . . . . . . . . . 18
2.3
Complex derivative and directional derivatives . . . . . . . . . . . . . . . . . . . . 21
Differential 1-forms and their integration
23
3.1
Complex-valued differential 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.2
Holomorphic 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.3
Integration of differential 1-forms along curves . . . . . . . . . . . . . . . . . . . 27
3.4
Integrals of closed and exact differential 1-forms . . . . . . . . . . . . . . . . . . 29
Cauchy integral formula
33
4.1
Stokes/Green theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4.2
Area computation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.3
Cauchy theorem and Cauchy integral formula . . . . . . . . . . . . . . . . . . . . 35
4.4
Integral criterion for exactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3
4.4.1
5
6
7
8
II
9
Proof of Theorem 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Convergent power series and holomorphic functions
43
5.1
Recollection of basic facts about series . . . . . . . . . . . . . . . . . . . . . . . . 43
5.2
Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
5.3
Analytic vs holomorphic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Properties of holomorphic functions
51
6.1
Exponential function and its relatives . . . . . . . . . . . . . . . . . . . . . . . . . 51
6.2
Entire functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
6.3
Analytic continuation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
6.4
Complex logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
6.5
Schwarz reflection principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
Isolated singularities, residues and meromorphic functions
59
7.1
Holomorphic functions with isolated singularities . . . . . . . . . . . . . . . . . . 59
7.2
Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
7.3
Application of the residue theorem to computation of integrals . . . . . . . . . . . 63
7.4
Complex projective line or Riemann sphere . . . . . . . . . . . . . . . . . . . . . 67
7.5
Residue of meromorphic differential forms . . . . . . . . . . . . . . . . . . . . . . 70
7.6
Argument principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
7.7
Winding number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
Harmonic functions
77
8.1
Harmonic and holomorphic functions . . . . . . . . . . . . . . . . . . . . . . . . 77
8.2
Properties of harmonic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
Conformal mappings and Riemann mapping theorem
Conformal mappings and their properties
4
83
85
9.1
Biholomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
9.2
Conformal mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
9.3
Examples of conformal mappings . . . . . . . . . . . . . . . . . . . . . . . . . . 87
9.3.1
Unit disc and the upper-half plane . . . . . . . . . . . . . . . . . . . . . . 87
9.3.2
Strips and sectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
9.4
Schwarz lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
9.5
Automorphisms of the Riemann sphere, C, the unit disc and the upper-half plane . 89
9.5.1
GL(n, C), GL(n, R), PGL(n, C), PGL(n, R) and PGL+ (n, R) = PS L(n, R) . . 89
9.5.2
Automorphisms of CP1 and C . . . . . . . . . . . . . . . . . . . . . . . . 90
9.5.3
Automorphisms of H and D . . . . . . . . . . . . . . . . . . . . . . . . . 93
9.6
Summary of useful conformal maps . . . . . . . . . . . . . . . . . . . . . . . . . 95
9.7
Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
10 Riemann mapping theorem
99
10.1 Functional analytic background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
10.1.1 Arzelá-Ascoli theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
10.1.2 Montel’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
10.2 Proof of the Riemann mapping theorem . . . . . . . . . . . . . . . . . . . . . . . 104
10.2.1 Embedding into D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
10.2.2 Maximizing the derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 105
10.2.3 Preservation of injectivity . . . . . . . . . . . . . . . . . . . . . . . . . . 106
10.2.4 Surjectivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
10.2.5 Discussion: boundary regularity . . . . . . . . . . . . . . . . . . . . . . . 108
10.3 Annuli . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
10.3.1 Conformal classification of annuli . . . . . . . . . . . . . . . . . . . . . . 109
10.3.2 Laurent series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
10.3.3 Proof of Theorem 10.16 . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
10.4 Dirichlet problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
5
10.4.1 Poisson integral and Schwarz formula . . . . . . . . . . . . . . . . . . . . 114
10.4.2 Solution of the Dirichlet problem for the unit disc . . . . . . . . . . . . . . 116
10.4.3 Solving the Dirichlet problem for other domains . . . . . . . . . . . . . . 119
11 Riemann surfaces
123
11.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
11.2 Uniformization theorem (or strong Riemann mapping theorem) . . . . . . . . . . . 125
11.3 Quotient construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
11.4 Covering maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
11.5 Quotient construction and covering maps . . . . . . . . . . . . . . . . . . . . . . 128
11.6 Universal cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
11.7 Lattices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
11.8 Branched covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
11.9 Riemann surfaces as submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . 139
11.9.1 Affine case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
11.9.2 Projective case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
11.9.3 Projectivization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
11.10Linear projective transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
11.11Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
III
A few important meromorphic functions
12 Elliptic functions
149
153
12.1 Elliptic integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
12.1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
12.1.2 Elliptic curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
12.1.3 Projectivization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
12.1.4 From a cubic curve to a torus T (ω1 , ω2 ) . . . . . . . . . . . . . . . . . . . 158
6
12.1.5 Summary of the construction . . . . . . . . . . . . . . . . . . . . . . . . . 161
12.2 The Weierstrass ℘-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
12.2.1 The definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
12.2.2 Differential equation for ℘(u) . . . . . . . . . . . . . . . . . . . . . . . . . 163
12.2.3 Identifying Weierstrass ℘-function with the solution of the pendulum equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
12.2.4 More about geometry of the Weierstrass ℘-function . . . . . . . . . . . . . 167
13 The Gamma function
169
13.1 Product development . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
13.2 Meromorphic functions with prescribed poles and zeroes . . . . . . . . . . . . . . 172
13.3 Some product and series developments for trigonometric functions . . . . . . . . . 173
13.4 The Gamma function: definition and some properties . . . . . . . . . . . . . . . . 176
13.5 Integral representation of the Gamma function . . . . . . . . . . . . . . . . . . . . 179
14 The Riemann ζ-function
181
14.1 Product development fo ζ(s) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
14.2 Meromorphic extension of ζ(s) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
14.3 Zeroes of the ζ-function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
7
8
Part I
Complex Analysis Basics
9
Chapter 1
Linear algebra
1.1
Complex numbers
The space R2 can be endowed with an associative and commutative multiplication operation. This
operation is uniquely determined by three properties:
• it is a bilinear operation;
• the vector (1, 0) is the unit;
• the vector (0, 1) satisfies (0, 1)2 = −(1, 0).
The vector (0, 1) is usually denoted by i, and we will simply write 1 instead of the vector (1, 0).
Hence, any point (a, b) ∈ R2 can be written as a + bi, where a, b ∈ R, and the product of a + bi and
c + di is given by the formula
(a + bi)(c + di) = ac − bd + (ad + bc)i.
The plane R2 endowed with this multiplication is denoted by C and called the set of complex
numbers. The real line generated by 1 is called the real axis, the line generated by i is called the
imaginary axis. The set of real numbers R can be viewed as embedded into C as the real axis.
11
Given a complex number z = x + iy, the numbers x and y are called its real and imaginary parts,
respectively, and denoted by Re z and Im z, so that z = Re z + iIm z.
For any non-zero complex number z = a + bi there exists an inverse z−1 such that z−1 z = 1.
Indeed, we can set
z−1 =
a2
a
b
− 2
i.
2
+b
a + b2
The commutativity, associativity and existence of the inverse is easy to check, but it should not
be taken for granted: it is impossible to define a similar operation any Rn for n > 2.
Given z = a + bi ∈ C its conjugate is defined as z̄ = a − bi. The conjugation operation z 7→ z̄ is
the reflection of C with respect to the real axis R ⊂ C. Note that
1
1
Re z = (z + z̄), Im z = (z − z̄).
2
2i
Let us introduce the polar coordinates (r, φ) in R2 = C. Then a complex number z = x + yi
can be written as r cos φ + ir sin φ = r(cos φ + i sin φ). This form of writing a complex number
p
is called, sometimes, trigonometric. The number r = x2 + y2 is called the modulus of z and
denoted by |z| and φ is called the argument of φ and denoted by arg z. Note that the argument
is defined only mod 2π. The value of the argument in [0, 2π) is sometimes called the principal
value of the argument. When z is real than its modulus |z| is just the absolute value. We also not
√
that |z| = zz̄.
An important role plays the triangle inequality
|z1 | − |z2 | ≤ |z1 + z2 | ≤ |z1 | + |z2 |.
Exponential function of a complex variable
Recall that the exponential function e x has a Taylor expansion
∞
X
xn
x2 x3
e =
=1+x+
+
+ ... .
n!
2
6
0
x
We then define for a complex z the exponential function by the same formula
ez := 1 + z +
z2
zn
+ ··· + + ....
2!
n!
12
One can check that this power series absolutely converging 1 for all z and satisfies the formula
ez1 +z2 = ez1 ez2 .
In particular, we have
y3 y4
y2
− i + + ··· + ...
2!
3! 4!
∞
∞
2k
2k+1
X
X
ky
k y
(−1)
=
+ i (−1)
.
2k!
(2k + 1)!
k=0
k=0
eiy = 1 + iy −
But
∞
P
2k
y
= cos y and
(−1)k 2k!
k=0
∞
P
(1.1.1)
(1.1.2)
2k+1
y
(−1)k (2k+1)!
= sin y, and hence we get Euler’s formula
k=0
eiy = cos y + i sin y,
and furthermore,
e x+iy = e x eiy = e x (cos y + i sin y),
i.e. |e x+iy | = e x , arg(ez ) = y.
In particular, any complex number z = r(cos φ + i sin φ) can be rewritten in the form z = reiφ .
This is called the exponential form of the complex number z.
Given z1 = r1 eiφ1 , z2 = r2 eiφ2 we get
z1 z2 = r1 eiφ1 r2 eiφ2 = r1 r2 ei(φ1 +φ2 ) ,
i.e. when multiplying complex numbers their moduli are being multiplied and arguments added
(but be aware that arguments are defined mod 2π).
Note that
n
eiφ = einφ ,
and hence if z = reiφ then zn = rn einφ = rn (cos nφ + i sin nφ).
Note that the operation z 7→ iz is the rotation of C counterclockwise by the angle π2 . More
generally a multiplication operation z 7→ zw, where w = ρeiθ is the composition of a rotation by
the angle θ and a radial dilatation (homothety) in ρ times.
1
Convergence of power series will be discussed later in Chapter 5.
13
Exercise 1.1.
1. Compute
n
P
cos kθ and
0
2. Compute 1 +
1.2
n
4
+
n
8
+
n
12
n
P
sin kθ.
1
+ ....
Complex linear function from the real perspective
A linear map F : R2 → R2 is by definition is required to satisfy two conditions:
F(z1 + z2 ) = F(z1 ) + F(z2 );
F(λz) = λF(z),
where z1 , z2 , z are any vectors from R2 and λ ∈ R is a real number. Any such map is a multiplication
by a 2 × 2-matrix:



 
a b
a b  x



  
F(z) = 
z = 
.


c d
c d y
A linear function of one complex variable is a linear map F : C → C satisfies in addition the
condition
F(λz) = λF(z), for any complex number λ.
(1.2.1)
Any such function has to satisfy F(z) = F(1)z = kz, where k = a + ib = F(1). Equivalently,
F(x + iy) = (a + ib)(x + iy) = ax − by + i(ay + bx).
 
 x
 
Thus, viewing a complex number z = x + iy as a vector   ∈ R2 we get
y

 
a −b  x

  
F(z) = 
b a  y
In other words, we proved the following


a b


Lemma 1.2. A real linear map F : R2 → R2 with a matrix A = 
is complex linear map
c d
C → C if and only if a = d and b = −c.
14


0 −1


In particular, the matric J = 
is the matrix of multiplication by i.
1 0 
Note that


a −b


det 
= a2 + b2 = |c|2 , where c = a + ib.
b a 
(1.2.2)
In other words, the (real) determinant of the matrix of the multiplication by a complex number c is
equal to |c|2 ,
We can also view a real linear map F : R2 → R2 as a map R2 → C, i.e. as a complex-valued


a b


linear (in a real sense) function F(x, y) = f1 (x, y) + i f2 (x, y). If F was given by a matrix 
then
c d
f1 (x, y) = ax + by, f2 (x, y) = cx + dy. We also have
F(x, y) = f1 (x, y) + i f2 (x, y) = ax + by + i(cx + dy) = (a + ic)x + (b + id)y = Ax + By.
(1.2.3)
Note that x = 12 (z + z̄), y = − 2i (z − z̄), and hence
F(x, y) = Ax + By =
A
Bi
1
1
(z + z̄) − (z − z̄) = (A − iB)z + (A + iB)z̄ = αz + βz̄,
2
2
2
2
(1.2.4)
where we denoted α := 12 (A − iB), β := 21 (A + iB).2 Note that the function l1 (z) = αz is complex
linear, while the function l2 (z) = βz̄ is complex anti-linear, which means that it is linear in the real
sense, but satisfied the condition l2 (λz) = λ̄l2 (z).
If F is a complex linear map, then F̄ is anti-linear and vice versa. In particular, every complex
anti-linear map F has the form F(z) = az̄ for a complex number a.
The following lemma summarizes the above discussion.
Lemma 1.3. Any linear in the real sense map F : C → C can be uniquely written as a sum
F = F1 + F2 , where F1 is complex

 linear and F2 is complex anti-linear.
a b


then F1 (z) = αz, F2 (z) = βz̄, where α := 21 (A−iB), β := 12 (A+iB),
If F is given by a matrix 
c d
A = a + ic, B = b + id.
2
Complex-valued linear (in the real sense) functions on R2 form a 2-dimensional complex vector space. Formulas
(1.2.3) and (1.2.4) say that the pairs of functions (x,y) as well as the pair (z, z̄) form a basis of this space.
15
16
Chapter 2
Holomorphic functions
2.1
Differentiability and the differential
For any point a = (a1 , a2 ) ∈ R2 we denote by R2a the space R2 with the origin shifted to the point a.
Though the parallel transport allows one to identify spaces R2 and R2a it will be important for us to
think about them as different spaces.
Let U be a domain in R2 . A vector-valued function f : U → R2 is called differentiable at a
point a ∈ U if near the point a it can be well approximated by a linear function. More precisely, if
there exists a linear map A : R2a → R2 such that
f (a + h) − f (a) = A(h) + o(||h||)
for any sufficiently small vector h = (h1 , h2 ) ∈ R2 , where the notation o(t) stands for any vectorvalued function such that
o(t)
→ 0.
t t→0
The linear function A is called the differential of the function
f at the point a and is denoted by da f . In other words, f is differentiable at a ∈ U if for any h ∈ R2a
there exists a limit
da f (h) = A(h) = lim
t→0
f (a + th) − f (a)
,
t
and the limit A(h) linearly depends on the vector h. By identifying R2a and with R2 via the parallel
transport we can associate with the linear map da f its matrix Ja ( f ), called the Jacobi matrix or
17
derivative of the map f . If we denote by u(x, y) and v(x, y) the coordinate functions of the map f ,
i.e.
f (x, y) = (u(x, y), v(x, y))
then

 ∂u (a)

Ja ( f ) =  ∂x
 ∂v (a)
∂x

∂u
(a)
∂y
 .

∂v
(a)
∂y
The function f is called differentiable on the whole domain U if it is differentiable at each point
of U.
2.2
Holomorphic functions, Cauchy-Riemann equations
Let us now view the map f : U → R2 as a complex valued function f (z) = u(z) + iv(z), z = x + iy.
The function f is called differentiable at the point a in a complex sense, or holomorphic at a if
the differential da f is a complex linear map. The following theorem lists equivalent definitions of
holomorphicity.
Theorem 2.1. The function f = u + iv : U → C is holomorphic at a point a ∈ U if one of the
following equivalent conditions is satisfied:
(1) There exists a complex number c ∈ C such that for any h ∈ C with a + h ∈ U we have
f (a + h) − f (a) = ch + o(|h|);
(2) there exists a limit
f (a + h) − f (a)
,
h
denoted by f 0 (a) and called the complex derivative of f at the point a;
lim
h→0; h∈C
(3) f is differentiable in the real sense and the following Cauchy–Riemann equations are satisfied at the point a:
∂u ∂v
= ,
∂x ∂y
∂v
∂u
=− .
∂x
∂y
18
(2.2.1)
(4) f is differentiable in the real sense and
!
∂f
1 ∂f
∂f
(a) :=
(a) + i (a) = 0.
∂z̄
2 ∂x
∂y
Proof. Statement (1) is just a reformulation of the fact that the differential da f is a complex linear
map. Equivalence (1) and (2) is straightforward. According to Lemma 1.2 condition (3) just means
that the Jacobi matrix is a matrix of a complex linear map.
To deal with condition (4) let us recall that according to Lemma 1.3 for any differentiable at a
function f we can decompose the linear map da f into a complex linear and anti-linear:
da f = ∂a f + ∂¯ a f,
so that we have ∂a f (h) = αh, ∂¯ a f (h) = βh̄. Rephrasing Lemma 1.3 we have
!
!
1 ∂f
∂f
1 ∂f
∂f
(a) − i (a) h +
(a) + i (a) h̄.
α=
2 ∂x
∂y
2 ∂x
∂y
If we introduce the notation
∂f
(a) =
∂z
∂f
(a) =
∂z̄
1
2
1
2
then we can write
da f (h) =
!
∂f
∂f
(a) − i (a) ,
∂x
∂y
!
∂f
∂f
(a) + i (a) ,
∂x
∂y
∂f
∂f
(a)h +
(a)h̄.
∂z
∂z̄
Hence, f is holomorphic at a point a if and only if
∂f
(a) = 0,
∂z̄
(2.2.2)
and this condition is just another form of the Cauchy-Riemann equations (2.2.1).
It is important to note that if f is holomorphic at a, i.e.
∂f
(a)
∂z̄
= 0 then f 0 (a) =
∂f
(a).
∂z
Note that if f is holomorphic at the point a then the determinant of the real Jacobi matrix is
given by
det J( f )(a) =
∂u
(a)
∂x
∂u
(a)
∂y
∂v
(a)
∂x
∂v
(a)
∂y
19
= | f 0 (a)|2 ,
see formula (1.2.2)
The function f : U → C is called holomorphic in the domain U if it is holomorphic at every
point of U. This is equivalent to the condition
∂f
∂z̄
= 0 in U.
The following proposition summarizes property of complex differentiation which are analogous
to the corresponding facts in the real case.
Proposition 2.2.
(1) If f, g are holomorphic at a ∈ C then f ± g and f g are holomorphic at a
and
( f ± g)0 (a) = f 0 (a) ± g0 (a), ( f g)0 (a) = f 0 (a)g(a) + f (a)g0 (a);
if g(a) , 0 then
f
g
is holomorphic at a and
!0
f
f 0 (a)g(a) − f (a)g0 (a)
(a) =
;
g
g2 (a)
(2) If f is holomorphic at a and g is holomorphic at f (a) then the composition g ◦ f is holomorphic at the point a and
(g ◦ f )0 (a) = g0 ( f (a)) f 0 (a).
The proof of (1) repeats the corresponding proofs in the real case, while (2) is the chain rule
with an additional observation that a composition of complex linear maps is itself complex linear.
Proposition 2.2 implies, for instance that (zn )0 = nzn−1 for any integer n.
According to our definition of a holomorphic function it is not even clear whether this function
is C 1 -smooth, i.e. whether its derivative continuously depends on a point of the domain. It turns
out that this is automatically true, which is the subject of the following theorem.
Theorem 2.3 (H. Looman, D. Menchoff). Every holomorphic function in a domain U is of class
C 1 , i.e. its derivative continuously depends on the point of U.
The proof of this theorem is given in Section 4.4.1 below.
We will assume in what follows the conclusion of this theorem, i.e. that a holomorphic function
is of class C 1 and will show that this in turn implies that a holomorphic function is infinitely
differentiable, and moreover analytic, i.e it is equal to the sum of its converging Taylor series
expansion in a neighborhood of each point of U.
20
2.3
Complex derivative and directional derivatives
Let f : U → C, where U ⊂ C is an open domain, is a complex valued differentiable in a real sense,
but not necessarily holomorphic function. Recall that for any point a ∈ U and a vector v ∈ Ca the
directional derivative
∂f
(a)
∂v
by definition is the value of the differential da f on the vector v, i.e.
∂f
f (a + tv) − f (a)
(a) = da f (v) = lim
.
t→0
∂v
t
In particular,
∂f
∂f
(a) = da f (1),
(a) = da f (i).
∂x
∂y
The derivatives
∂f
∂z
and
∂f
∂z
cannot be interpreted as directional derivatives of the function f . How-
ever, it is possible, though maybe not necessarily very insightful, to interpret them as directional
derivatives of a function of two complex variables, as it is explained below.
Let us write z = x + iy, w = u + iv. Consider vector fields T = 21 (1, −i) and S = 12 (1, i).
Lemma 2.4. For a complex valued function f : U → C consider the function
F(z, w) = f (z) + i f (w), (z, w) ∈ U × U ⊂ C2
defined on the domain U × U ⊂ C2 . Then
∂f
∂f
(a) = d(a,a) F(T ),
(a) = d(a,a) F(S ).
∂z
∂z
Proof. We have
1
d(a,a) F(T ) = (da ( f )(1) − da ( f )(i)) =
2
1
d(a,a) F(S ) = (da ( f )(1) + da ( f )(i)) =
2
1
2
1
2
!
∂f
∂f
(a) − i (a) =
∂x
∂y
!
∂f
∂f
(a) + i (a) =
∂x
∂y
∂f
(a).
∂z
∂f
(a).
∂z
If f is holomorphic at a then for any vector v ∈ Ca we have
∂f
(a) = da f (v) = f 0 (a)v,
∂v
21
where we identify v with a complex number v ∈ C via a parallel transport. For instance,
∂f
(a) = da f (1) = f 0 (a) · 1 = f 0 (a),
∂x
∂f
(a) = da f (i) = i f 0 (a).
∂y
In this case
∂f
(a)
∂z
= f 0 (z) =
∂f
(a)
∂x
and
∂f
(a)
∂z
= 0.
Exercise 2.5. Justify the following way of computing the derivatives ∂∂zf and ∂∂zf . Express x, y
z−z
through z, z, and viewing g(z, z) := f z+z
,
as the function of two independent variables z and z
2
2i
compute formally its partial derivatives with respect to z, z.
Example 2.6. 1. Let f (z) = |z|2 . Compute
We have |z|2 = zz. Hence,
∂f
∂z
2. Let f (x, y) = xy. Compute
= z and
∂f
∂z
and
∂f
∂z
∂f
∂z
∂f
.
∂z
and
∂f
.
∂z
= z.
We have xy = − 4i (z + z)(z − z) = − 4i (z2 − z2 ). Hence,
22
∂f
∂z
= − iz2 ,
∂f
∂z
= iz2 .
Chapter 3
Differential 1-forms and their integration
3.1
Complex-valued differential 1-forms
Let us first recall some basics of the theory of real differential forms. For our purposes we will
need only 1-forms on domains in R2 . By definition a differential 1-form λ on a domain U ⊂ R2 is
a field of linear functions λz : R2z → R.
Thus, a differential 1-form is a function of arguments of 2 kind: of a point z ∈ U and a vector
h ∈ R2z . It depends linearly on h and arbitrarily (but usually continuously and even smoothly) on z,
i.e. we have
λz (h) = a1 (z)h1 + a2 (z)h2 ,
where h1 , h2 are Cartesian coordinates of h ∈ R2z . Any differential 1-form can be multiplied by a
function (”a field of scalars”): ( f λ)z (h) = f (z)λz (h).
Given a real-valued function f : U → R on U its differential d f is an example of a differential
form:
∂f
∂f
(z)h1 +
(z)h2 .
∂x
∂y
In particular, differentials dx and dy of the coordinate functions x, y are also differential 1-forms,
dz ( f )(h) =
and any other differential form can be written as a linear combination of dx and dy:
λ = Pdx + Qdy,
23
where P, Q : U → R are functions on the domain U. In particular, we have
df =
∂f
∂f
dx +
dy.
∂x
∂y
A differential 1-form λ is called exact if λ = d f . The function f is called the primitive of the
1-form λ. The primitive is defined uniquely up to adding a constant.
A differential form λ = Pdx + Qdy is called closed if
∂P
∂y
=
∂Q
.
∂x
An exact form is always closed because of the equality of mixed derivatives:
λ = df =
∂f
∂f
dx +
dy ⇒
∂x
∂y
!
!
∂ ∂f
∂ ∂f
∂2 f
.
=
=
∂y ∂x
∂x ∂y
∂x∂y
Not every closed differential 1-form λ = Pdx + Qdy is exact, as it is demonstrated by an
example of a 1-form dφ on R2 \ 0 (written in polar coordinates), or
dφ =
1
(xdy − ydx)
x2 + y2
in Cartesian coordinates.
We will discuss a bit later the precise argument for that, but it is intuitively clear that the
primitive of this form is not a univalent function on R2 \ 0. On the other hand, as we will see below,
any closed 1-form on R2 , or more generally on any simply connected domain in R2 is exact.
We will also consider complex-valued differential 1-forms. A C-valued differential 1-form is
a field of C-valued linear in the real sense functions, or simply it is an expression α + iβ, where
α, β are usual real-valued differential 1-forms. All usual operations on complex valued 1-forms are
defined in the same way as for real-valued forms, and in addition such forms can be multiplied not
only by real-valued but also by complex-valued functions.
Note that a complex-valued function (or 0-form) on a domain U ⊂ C is just a map f = u + iv :
U → C. Its differential d f is the same as the differential of this map, but it also can be viewed as a
C-valued differential 1-form d f = du + idv.
24
Lemma 3.1. Let f, g : U → C be two complex-valued functions which are differential in the real
sense. Then
d( f g) = f dg + gd f.
Proof. For any point a ∈ U, h ∈ Ca we have
f (a + h) = f (a) + da f (h) + o(h), and g(a + h) = g(a) + da g(h) + o(h).
Hence,
f (a + h)g(a + h) = ( f (a) + da f (h) + o(h))(g(a) + da g(h) + o(h))
= f (a)g(a) + f (a)da g(h) + g(a)da f (h) + (da f (h) + o(h))(da g(h) + o(h)) .
|
{z
}
o(h)
Example 3.2.
dz = dx + idy, dz̄ = dx − idy, zdz = (x + iy)(dx + idy) = xdx − ydy + i(xdy + ydx).
We have
1
i
dx = (dz + dz̄), dy = − (dz − dz̄).
2
2
Hence, any complex valued 1-form λ can be written as a linear combination of forms dz and dz̄:
λ = f dz + gdz̄,
which is a decomposition of λ into a sum of complex linear and complex anti-linear parts.
Lemma 3.3. The form λ = f dz + gdz̄ is closed if and only if
∂f
∂g
= .
∂z̄
∂z
Proof.
λ = f dz + gdz̄ = f (dx + idy) + g(dx − idy) = ( f + g)dx + i( f − g)dy.
25
The closedness of λ means by definition that
∂( f + g) ∂(i( f − g))
=
,
∂y
∂x
which is equivalent to
∂f
∂f
∂g
∂g
−i
=− −i .
∂y
∂x
∂y
∂x
Dividing both parts by (−i) we get
∂f
∂g
∂g
∂f
+i
=
−i ,
∂x
∂y ∂x
∂y
and hence
!
!
1 ∂g
∂g
1 ∂f
∂f
∂g
∂f
=
+i
−i
=
= .
∂z̄ 2 ∂x
∂y
2 ∂x
∂y
∂z
Let us express the differential d f of a complex valued function f as a combination of differential forms dz = dx + idy and dz̄ = dx − idy parts.
Lemma 3.4. For any complex valued function f = u + iv we have
df =
∂f
∂f
dz +
dz̄.
∂z
∂z̄
In particular, when f is holomorphic we have
df =
∂f
dz = f 0 (z)dz.
∂z
Proof. We have
∂f
∂f
1 ∂f
i ∂f
dx +
dy =
(dz + dz̄) −
(dz − dz̄)
∂x
∂y
2 ∂x
2 ∂y
!
!
1 ∂f
∂f
1 ∂f
∂f
∂f
∂f
−i
dz +
+i
dz̄ =
dz +
dz̄.
2 ∂x
∂y
2 ∂x
∂y
∂z
∂z̄
df =
26
3.2
Holomorphic 1-forms
A complex-valued 1-form λ is called holomorphic if λ = f dz for a holomorphic function f .
Corollary 3.5. The form f (z)dz is closed if and only if it is holomorphic.
Proof. According to Lemma 3.3 closedness of f dz is equivalent to
∂f
∂z̄
= 0, which, in turn, is
equivalent to the holomorphicity of f .
Example 3.6. The holomorphic form
dz
,
zn
n ≥ 1, on C \ 0 is always closed. It is exact if and only if
n > 1.
Indeed, If n > 1 then
!
dz
1
,
=d
zn
(1 − n)zn−1
i.e.
dz
zn
is exact. If n = 1 we have in polar coordinates
dz d(reiφ ) eiφ dr + ireiφ dφ dr
=
+ idφ = d(ln r) + idφ,
=
=
z
reiφ
reiφ
r
but we already discussed above that the form dφ is not exact.
3.3
Integration of differential 1-forms along curves
Curves as paths
A path, or parametrically given curve in a domain U ⊂ R2 , is a map γ : [a, b] → U. We will
assume in what follows that all considered paths are differentiable. Given a differential 1-form
α = Pdx + Qdy in U we define the integral of α over γ by the formula
Z
α=
γ
Zb
γ∗ α.
a
Denoting the coordinate functions of γ(t) by x(t) and y(t) (i.e. γ(t) = (x(t), y(t)) the pull-back
differential form γ∗ α is by definition equal to
γ∗ α = P(x(t), y(t))d(x(t)) + Q(x(t), y(t))d(y(t)) = (P(x(t), y(t))x0 (t) + Q(x(t), y(t))y0 (t))dt,
27
so that
Z
Zb
α=
γ
a
γ∗ α =
Zb
(P(x(t), y(t))x0 (t) + Q(x(t), y(t))y0 (t))dt.
a
An important property of the integral of a differential 1-form is that it does not depend on the
parameterization of the curve.
Proposition 3.7. Let a path e
γ be obtained from γ : [a, b] → U by a reparameterization, i.e.
R
R
e
γ = γ ◦ φ, where φ : [c, d] → [a, b] is an orientation preserving diffeomorphism. Then α = α.
e
γ
Thus the integral
R
γ
α depends only on the curve γ as an oriented submanifold and not on a
γ
particular parameterization which is compatible with the orientation.
Orientation and co-orientation. By an orientation of a a curve Γ ⊂ C we understand a choice of
direction of each its tangent line, continuously depending on a point. Similar co-orientation of Γ
is an orientation its normal line. Co-orientation defines an orientation by the following convention:
we orient tangent lines by rotating oriented normal lines 90 degrees counter-clockwise. If Γ bounds
a domain U then by default we will co-orient Γ = ∂U by an outward to U normal vector field.
For instance, the unit circle S 1 = {|z| = 1} viewed as a boundary of the unit disc {|z| < 1}
R
gets a counter-clockwise orientation. Computing the integral dφ with respect to this orientation
S1
we get 2π. Indeed, the circle can be parameterized by the angular coordinate φ ∈ [0, 2π], and this
parameterization is compatible with the counter-clockwise orientation.
Exercise 3.8. Compute
R
dz
z
S1
Solution. Let us parameterize the circle by polar coordinates: z = eiφ , φ ∈ [0, 2π]. Then
R
R2π
d(ln r) + idφ and dzz = i dφ = 2πi.
S1
0
28
dz
z
=
3.4
Integrals of closed and exact differential 1-forms
Theorem 3.9. Let α = d f be an exact 1-form in a domain U ⊂ C. Then for any path γ : [a, b] → U
which connects points A = γ(a) and B = γ(b) we have
Z
α = f (B) − f (A).
γ
In particular, if γ is a loop then
H
γ
α = 0.
Similarly for an oriented curve Γ ⊂ U with boundary ∂Γ = B − A we have
Z
α = f (B) − f (A).
Γ
Proof. We have
R
γ
df =
Rb
a
γ∗ d f =
Rb
d( f ◦ γ) = f (γ(b)) − f (γ(a)) = f (B) − f (A).
a
It turns out that closed forms are locally exact. A domain U ⊂ V is called star-shaped with
respect to a point a ∈ V if with any point x ∈ U it contains the whole interval Ia,x connecting a and
x, i.e. Ia,x = {a + t(x − a); t ∈ [0, 1]}. In particular, any convex domain is star-shaped.
Proposition 3.10. Let α = Pdx + Qdy be a closed 1-form in a star-shaped domain U ⊂ R2 . Then
it is exact.
Proof. Define a function F : U → R by the formula
Z
F(x, y) =
α, z = (x, y) ∈ U,
−→
Ia,z
where the intervals Ia,z are oriented from a to z.
We claim that dF = α. Let us choose a as the origin. Then I0,z can be parameterized by
t 7→ tz, t ∈ [0, 1].
Hence,
F(z) =
Z
−→
I0,z
α=
Z1
P(tx, ty)d(tx) + Q(tx, ty)d(ty) =
0
Z1
0
29
(xP(tx, ty) + yQ(tx, ty))dt
(3.4.1)
Taking partial derivatives of the integral with respect to x and y we get
∂F
(x, y) =
∂x
Z1
∂F
(x, y) =
∂y
Z1
!
Z1
∂P
∂Q
tx (tx, ty) + ty (tx, ty) dt +
P(tx, ty)dt;
∂x
∂x
0
0
!
Z1
∂P
∂Q
tx (tx, ty) + ty (tx, ty) dt +
Q(tx, ty)dt
∂y
∂y
0
0
By our assumption the form α is closed, and hence
∂F
(x, y) =
∂x
Z1
∂P
∂y
=
0
Z1
1
P(tx, ty)dt = tP(x, y) −
0
0
∂F
(x, y) =
∂y
=
Z1
P(tx, ty)dt +
0
Z1
P(tx, ty)dt = P(x, y);
0
!
Z1
∂Q
∂Q
tx (tx, ty) + ty (tx, ty) dt +
Q(tx, ty)dt
∂x
∂y
0
Z1
Using this we can further write
0
∂P
t (tx, ty)dt +
∂t
Z1
∂Q
.
∂x
!
Z1
∂P
∂P
P(tx, ty)dt;
tx (tx, ty) + ty (tx, ty) dt +
∂x
∂y
0
Z1
=
0
∂Q
t (tx, ty)dt +
∂t
0
Z1
1
Q(tx, ty)dt = tQ(x, y) −
0
0
Z1
Q(tx, ty)dt +
0
Z1
Q(tx, ty)dt = Q(x, y);
0
Thus
dF =Pdx + Qdy = α
Given a differential 1-form α = Pdx + Qdy we will define
Zb
Z
|α| :=
Γ
R
Γ
|α| as
|P(x(t), y(t))x0 (t) + Q(x(t), y(t))y0 (t)| dt,
a
where (x(t), y(t), t ∈ [a, b], is a parameterization of Γ. Unlike
R
Γ
and does not depend on the orientation of Γ. Clearly, we have
Z
α ≤
Γ
Z
|α|,
Γ
30
α the integral
R
Γ
|α| is non-negative
and
Z
.
Γ
|α + β| ≤
Z
Γ
31
|α| +
Z
|β|.
Γ
32
Chapter 4
Cauchy integral formula
4.1
Stokes/Green theorem
Given a bounded domain U ⊂ C with a smooth (or piece-wise smooth boundary) we will always
orient (unless it is explicitly stated otherwise) its boundary ∂U as follows: For each point p ∈ ∂U
take an outward normal vector ν. Then iν is tangent to ∂U and defines its orientation. For instance,
suppose A is the annulus 1 ≤ |z| ≤ 2. Its boundary is the union of two circles: S 1 = {|z| = 1} and
S 2 = {|z| = 2}. Then A induces on the outer circle S 2 the counter-clockwise orientation, and on the
inner circle S 1 the clockwise orientation.
The fundamentally important fact about integration of 1-forms is the following theorem which
belongs to George Green and it is a special case of a more general result, called Stokes’ theorem
(which was not actually proved by George Stokes!)
Theorem 4.1. Let U ⊂ C be a bounded domain with a piecewise smooth boundary ∂U, and α =
Pdx + Qdy a differential 1-form on U with coefficients which are C 1 -smooth in U and continuous
in the closure U. Let us orient the curve ∂U as the boundary of U. Then
!
Z
"
∂Q ∂P
Pdx + Qdy =
−
dxdy, .
∂x
∂y
∂U
U
33
Corollary 4.2. Suppose a 1-form α is closed in a domain U. Then
Z
α = 0.
∂U
Indeed, closedness of α = Pdx + Qdy just means that
4.2
∂Q
∂x
−
∂P
∂y
= 0.
Area computation
If for α = Pdx + Qdy we have
∂Q
∂x
− ∂P
= 1 in the closure U of the domain U, then Green’s formula
∂y
yields
Z
"
Pdx + Qdy =
∂U
dxdy = Area(U).
U
For instance this is the case for α = xdy, −ydx or 12 (xdy − ydx). Another example of such 1-form α
is the form
i
1
i
i
α = − zdz = − (x − iy)(dx + idy) = (xdy − ydx) − d(xy).
2
2
2
2
Proposition 4.3. Let Γ ∈ C be a piecewise smooth curve, Ω ⊃ Γ is a neighborhood of Γ and
f : Ω → C a holomorphic function such that f |Γ is injective, and f (Γ) ⊂ C bounds a domain
U ⊂ C. Suppose f (Γ) is oriented as the boundary of U and Γ is oriented in such a way that f
preserves the orientations. Then
i
Area(U) = −
2
Z
f (z) f 0 (z)dz.
Γ
Proof. Using Green’s formula together with the change of variable formula we get
i
Area( f (U)) = −
2
Z
f (Γ)
i
zdz = −
2
Z
Γ
i
f (z)d f (z) = −
2
Z
f (z) f 0 (z)dz.
Γ
34
4.3
Cauchy theorem and Cauchy integral formula
Corollary 4.4 (Cauchy theorem). Let f be a function, holomorphic in the bounded domain U and
continuous up to the boundary. Then
Z
f (z)dz = 0.
∂U
Proof. According to Corollary 3.5 the holomorphic differential 1-form f (z)dz is closed in U.
Example 4.5. Let U ∈ C be any domain such that 0 ∈ U. Then
Z
∂U
Indeed, for n > 1 the 1-form
dz
zn





dz 
2πi,
=


zn 


0,
n = 1,
otherwise.
is exact in U \ 0,
!
dz
1
=d
,
zn
(1 − n)zn−1
and the integral of an exact form over any closed loop is equal to 0.
If n = 1 consider a disc D = {|z| < } ⊂ U. Then according to Corollary 4.2
Z
0=
∂(U\{D })
But we already computed that
R
dz
z
dz
=
z
Z
∂U
dz
−
z
Z
∂D
dz
.
z
= 2πi, and therefore
{|z|=})
Z
∂U
dz
= 2πi.
z
Theorem 4.6 (Cauchy integral formula). Suppose that U is a bounded domain, f : U → C is a
continuous function which is holomorphic in U. Then for any u ∈ U we have
1
2πi
Z
∂U
f (z)dz
= f (u).
z−u
35
Proof. Let Dδ (u) denote the disc {|z − u| < } centered at u, where 0 < δ < | − |u|. The function
f (z)
z−u
is holomorphic in U \ u, and therefore according to the Cauchy theorem 4.4 we have
Z
f (z)dz
= 0.
z−u
∂(U\Dδ (u))
Hence,
Z
∂U
f (z)dz
=
z−u
Z
∂Dδ (u)
f (z)dz
=
z−u
Z
f (u + w)dw
.
w
|w|=δ
The function f is continuous at the point u. Hence, for any there exists δ > 0 such that if |w| ≤ δ
then
| f (u + w) − f (u)| < .
Hence,
Z
f (u + w)dw
−
w
|w|=δ
Z
f (u)dw
≤
w
Z2π
| f (u + w) − f (u)|δdφ
≤ 2π.
δ
0
|w|=δ
Note that according to Example 4.5
Z
f (u)
= f (u)
w
|w|=δ
Z
dw
= 2πi f (u).
w
|w|=δ
Thus
Z
∂U
f (z)dz
− 2πi f (u) ≤ 2π
z−u
for any > 0. But the left-hand side is independent of , and therefore
1
2πi
Z
∂U
f (z)dz
= f (u).
z−u
As the first application of the Cauchy integral formula we prove the infinite differentiability of
a holomorphic function.
36
Corollary 4.7. Any holomorphic in a domain U function is infinitely differentiable at every point.
Its derivatives can be computed by the formula
Z
k!
f (u) =
2πi
f (z)dz
.
(z − u)k+1
(k)
Proof. The variable u enters the integral
R
∂D
f (z)dz
z−u
∂U
as a parameter. The integrand
f (z)dz
z−u
is differntiable
with respect to the parameter, and hence the integral itself is differentiable with respect to u and we
can compute the derivative f 0 (u) by the differentiating the integral with respect to the parameter,
i.e.


Z
Z


f
(z)dz
d
1
1


0
f (u) =

=
du  2πi
z − u  2πi
∂U
∂U
Applying the same argument to the integral
R
∂U
f (z)dz
(z−u)2
!
Z
f (z)dz
1
f (z)dz
=
.
z−u
2πi
(z − u)2
∂U
we compute f 00 (u), etc.
Corollary 4.8 (Cauchy inequality). Let f : U → C be a holomorphic function. Suppose that for a
point z0 the closed disc Dr (z0 ) = {|z − z0 | ≤ r} is contained in U. Then
| f (n) (z0 )| ≤
Mn!
,
rn
where M := max | f (z)|.
|z−z0 |=r
Proof. By the Cauchy integral formula we have
n!
f (z0 ) =
2πi
Z
(n)
f (z0 + ζ)dζ
.
ζ n+1
|ζ|=r
Therefore,
n!
| f (z0 )| ≤
2π
Z2π
(n)
Mrdθ Mn!
= n .
rn+1
r
0
37
4.4
Integral criterion for exactness
Theorem 4.9. Let α = Pdx + Qdy be a differential form with continuous coefficients in a domain
R
U ⊂ C. Suppose that for any piecewise smooth loop γ : [a, b] → U, γ(a) = γ(b), we have α = 0.
γ
Then the form α is exact.
Proof. We can assume that U is connected. Otherwise the same argument can be repeated for
each connected component. Choose any point a ∈ U. For any other point z ∈ U choose a path γz
connecting a to z and oriented from a to z. Define
Z
f (z) :=
α.
γz
Then f (z) is independent of the choice of the connecting path γz . Indeed, any other choice differs
by an integral over a loop, which is by assumption is equal to 0. We claim that f is differentiable
and d f = α. Indeed, for any point z = (x, y) ∈ U and a small t we have
f (x + t, y) − f (x, y) =
Z
α,
lt
where lt is a straight interval connecting the point (x, y) with the point (x + t, y). Hence
f (x + t, y) − f (x, y) =
Z
α=
lt
Zx+t
P(u)du.
x
Hence
 x+t

Z

∂f
f (x + t, y) − f (x, y)
d 

(x, y) = lim
=  P(u)du = P(x, y).

t→0
∂x
t
dt 
x
Similarly, we get
∂f
(x, y) = Q(x, y).
∂y
Thus the function f has both partial derivatives, which are by our assumption are continuous. This
38
implies differentiability of f . Indeed, for any point a = (x, y) and a vector h = (h1 , h2 ) we have
f (a + h) − f (a) = ( f (x + h1 , y + h2 ) − f (x + h1 , y)) + ( f (x + h1 , y) − f (x, y))
= Q(x + h1 , y)h2 + o(h2 ) + P(x, y)h1 + o(h1 ) = (Q(x, y) + o(h1 )) h2 + P(x, y)h1 + o(h1 )
= P(x, y)h1 + Q(x, y)h2 + (o(h1 )h2 + o(h1 ) + o(h2 )) = P(x, y)h1 + Q(x, y)h2 + +o(|h|).
And this is by definition means that f is differentiable and that d f =
∂f
dx+ ∂∂yf dy
∂x
= Pdx+ Qdy = α.
Remark 4.10. The analysis of the above proof shows that it is enough to assume that
R
γ
α = 0 only
for piece-wise linear loops formed with only horizontal and vertical intervals. Moreover, to prove
local exactness, or equivalently closedness of the form α it is sufficient to prove the statement only
for boundaries of rectangulars. We will use this remark in the proof of the Looman-Menchoff’s
theorem in Section 4.4.1 below.
Corollary 4.11. Let f (z) be a continuous function in U. Suppose that
R
γ
f (z)dz = 0 for any piece-
wise smooth (or piecewise linear) loop γ in U. Then the function f is holomorphic.
Proof. According to Theorem 4.9 the differential 1-form f (z)dz is exact in U. Hence, f (z)dz = dg,
or g0 (z) = f (z). Thus g is holomorphic and so is g0 = f .
4.4.1
Proof of Theorem 2.3
Green’s theorem required the form to be C 1 -smooth. Hence, in order to deduce from it Cauchy’s
theorem we had to assume C 1 -smoothness of a holomorphic function. We prove below LoomanMenchoff’s theorem 2.3, which shows that the C 1 -condition follows from holomorphicity, i.e. complex differentiability at every point of the domain. The key in the proof is the following Lemma
4.12 which is due to R. Narasimhan, see Section 1.6 in the book “Complex Analysis on One Variable” by R. Narasimhan and Y. Nievergelt, Birkhäuser, 2001.
Lemma 4.12. Suppose the function f is holomorphic in the domain U, i.e. it is differentiable in the
R
complex sense at every point of the domain U. Let Q ⊂ U be any rectangular. Then f (z)dz = 0.
∂Q
39
Proof. Denote the perimeter of the rectangular Q, i.e. the total length of its sides by L. Note that
the diameter d of the rectangular, i.e. the maximal distance between its points satisfies the the
ingequality 2d ≤ L.
f (z)dz = a > 0. Let us partition the rectangular Q into 4
R
Suppose the converse, i.e. that
∂Q
rectangulars Q1 , Q2 , Q3 and Q4 of perimeter L2 . Then
Z
4 Z
X
f (z)dz =
f (z)dz.
1
∂Q
(4.4.1)
∂Q j
Indeed, each side S of Q j which is not contained in ∂Q is also a side of another rectangular
Q j0 , j0 , j. As the part of boundaries of Q j and Q j0 the side S has opposite orientations, and hence
the integrals over all parts of boundaries of Q j which are not in ∂Q cancel, and hence we get
equality (4.4.2). Then
a=
Z
f (z)dz ≤
4 Z
X
1
∂Q
f (z)dz .
(4.4.2)
∂Q j
Hence, there exists j0 ∈ {1, 2, 3, 4} we have
Z
a
f (z)dz ≥ .
4
∂Q j0
Now we partition the rectangular Q j0 into 4 rectangulars Q j0 1 , Q j0 2 , Q j0 3 , Q j0 4 of perimeter
L
4
and
again conclude that there exists j1 ∈ {1, 2, 3, 4} such that
Z
f (z)dz ≥
∂Q j0 j1
a
.
16
Continuing this process we find a sequence of rectangulars
Q j0 ⊃ Q j0 j1 ⊃ · · · ⊃ Q j0 j1 ... jk ⊃ . . . ,
such the perimeter of ∂Q j0 j1 ... jk is equal to
L
2k
and
Z
f (z)dz ≥
∂Q j0 j1 ... jk
40
a
.
4k
(4.4.3)
The intersection
Q j0 ∩ Q j0 j1 ∩ · · · ∩ Q j0 j1 ... jk ∩ . . .
consists of a unique point z0 ∈ Q. The function f is holomorphic at z0 and hence for any > 0
there exists K such that for k > K and any point z ∈ ∂Q j0 j1 ... jk we have
| f (z) − f (z0 ) − f 0 (z0 )(z − z0 )| < |z − z0 | <
because z − z0 | ≤
L
2
< L. Note that
Z
L
,
2k
(4.4.4)
( f (z0 ) + f 0 (z0 )(z − z0 ))dz = 0,
∂Q j0 j1 ... jk
because the inhomogeneous linear function f (z0 ) + f 0 (z0 )(z − z0 ) is holomorphic and C 1 -smooth.
Therefore, (4.4.5) implies that
Z
∂Q j0 j1 ... jk
Choosing <
a
L2
L2
f (z)dz ≤ k .
4
we get a contradiction with (4.4.3).
(4.4.5)
Proof of Theorem 2.3. In view of Theorem 4.9 and Remark 4.10 Lemma 4.12 implies that the
form f (z)dz is locally exact, i.e. in a neighborhood of each point there exists a function g(z) such
that dg(z) = f (z)dz, and hence g is holomorphic. But then we can apply Corollary 4.7 to conclude
that the function f = g0 is holomorphic as well.
41
42
Chapter 5
Convergent power series and holomorphic
functions
5.1
Recollection of basic facts about series
Let us recall that a series
∞
P
bk , where b j are complex numbers, is called converging if there exists
0
a finite limit of partial sums
S := lim S N = lim
N→∞
In this case we write
∞
P
0
N→∞
N
X
bk .
k=0
bk = S . A necessary and sufficient condition for convergence is given by
Cauchy criterion:
For any > 0 there exists N such that for any n ≥ N and m > 0 we have
A series
∞
P
bk is called absolutely converging if the series
0
∞
P
n+m
P
bk < .
k=n
|bk | is converging. Absolute con-
0
vergence implies convergence, as it it immediately follows from the Cauchy criterion and the inequality
n+m
X
bk ≤
k=n
n+m
X
|bk |.
k=n
An important tool for establishing an absolute convergence (or divergence) is the following comparison criterion:
43
bn be two series such that an , bn ≥ 0. Suppose that there exists N
P
P
P
such that for n ≥ N we have an ≤ bn . Then if bn is converging then so does an , and if an is
P
diverging then so does bn .
Lemma 5.1. Let
5.2
P
an and
P
Power series
A power series is a series of the form
∞
P
0
an zn , an , z ∈ C.
A remarkable fact about power series is existence of a radius of convergence.
Proposition 5.2. For any power series
∞
P
0
an zn there exists R (which could be = 0 or ∞) such that
for |z| < R the power series is absolutely converging and for |z| > R it is diverging. The radius of
covergence R can be computed by the following formula (due to Jacques Hadamard):
1
1
= lim sup |an | n .
R
Proof. The proof follows from the comparison with a geometric series
P
rn which converges when
r < 1 and diverges when r ≥ 1. Indeed, for any r < R we have |an | < r1n for a sufficiently large n.
n
∞
P
Hence, if |z| < r then |an ||z|n < |z|r , and therefore the power series an zn is absolutely converging
0
∞ n
P
|z|
. But r is any number < R, and hence
due to the comparison with the geometric series
r
∞
P
0
an z is absolutely converging for all |z| < R. If |z| > R then there exists infinitely many n such
n
0
that |an |n >
1
,
|z|
and hence for these values of n we have |an ||z|n > 1. This implies that
∞
P
an zn is
0
diverging because the common term of a converging series must converge to 0.
The disc {|z| < R} is called the disc of convergence.
Exercise 5.3. Verify the following statements.
1. The radius of convergence of the geometric series
∞
P
0
zn is equal to 1. On the boundary of
{|z| = 1} of the disc of convergence the series diverges at every point.
44
2. The radius of convergence of the geometric series
∞ n
P
z
is also equal to 1. However, the
n
1
behavior on the boundary of the disc of convergence is different: the series is convergent at
every point except z = 1.
3. The radius of convergence of the exponential series
∞ n
P
z
0
n!
is equal to ∞, i.e. the series is
absolutely converging on the whole C.
4. The radius of convergence of the series
∞
P
n!zn is equal to 0, i.e. the series is divergent for
0
any z , 0.
Exercise 5.4 (Operations on converging power series). Suppose power series
∞
P
∞
P
are converging for |z| < R. Denote A(z) := an z , B(z) :=
0
!
∞ P
n
P
and
ak bn−k zn are also converging for |z| < R and
n
∞
P
an zn and
0
n
bn z . Then the series
0
∞
P
∞
P
bn zn
0
(an + bn )zn
n=0
n=0 k=0
A(z) + B(z) =
∞
X
(an + bn )zn ,
n=0
 n

∞ X
X


 ak bn−k  zn .
A(z)B(z) =
n=0
Proposition 5.5. Suppose the series f (z) =
a) The series
∞
P
∞
P
k=0
an zn is converging in the disc DR := {|z| < R}. Then
0
nan zn−1 is also converging in DR and
1
∞
P
1
nan zn−1 = f 0 (z). Thus, the sum of a
power series is holomorphic in the disc of its convergence.
b) The series
∞
P
0
an n+1
z
n+1
is converging in DR to a function F(z) such that F 0 (z) = f (z).
Proof. First observe that according to the Hadamard criterion the power series
and
∞
P
0
∞
P
0
an n+1
z
n+1
have the same radius of convergence. Indeed,
|an |
lim sup(n|an |) = lim sup |an | = lim sup
n+1
1
n
1
n
45
! 1n
.
an zn ,
∞
P
1
nan zn−1
Let us prove that
∞
P
1
nan zn−1 = f 0 (z). Recall an identity
un − vn
= un−1 + un−2 v + · · · + vn−1 .
u−v
Differentiating both sides with respect to v we get
−nvn−1 (u − v) + un − vn
= un−2 + 2un−3 v + · · · + (n − 1)vn−2 , or
(u − v)2
−nvn−1 (u − v) + un − vn
= (u − v)(un−2 + 2un−3 v + · · · + (n − 1)vn−2 ).
u−v
Suppose that |z| < R − and |h| < 2 . Let us plug in the above formula u = z + h, v = z:
(z + h)n − zn
− nzn−1 = h (z + h)n−2 + 2(z + h)n−3 z + · · · + (n − 1)zn−2 .
h
Observe that max(|z|, |z + h|) < R − 2 . Hence
!
(z + h)n − zn
n−2
n−2
n−1
n−1
− nz
≤ |h| R −
+2 R−
+ · · · + (n − 1) R −
h
2
2
2
n−2
|h|
n−2 n(n − 1) R − 2
= |h|(1 + 2 + · · · + (n − 1)) R −
=
.
2
2
Hence, we have
∞
f (z + h) − f (z) X
−
nan zn−1
h
1
!
∞
X
(z + h)n − zn
n−1
≤
− nz
an
h
2
n−2
∞
X
n(n − 1) R − 2
|h|
≤
|an |
.
2
1
n
P
|an | n(n−1)
xn−2 is the same as the radius of convergence
2
1
∞
∞
P
P
n−2
of the series an xn , i.e it is equal to R. Hence, |an | n(n−1)
R
−
≤ C() for a constant C()
2
2
But the radius of convergence of the series
1
1
which depends on , and therefore,
f (z+h)− f (z)
h
converges to
∞
P
1
46
nan zn−1 when h → 0.
To prove b) we apply a) to the series F(z) =
∞
P
0
an n+1
z .
n+1
5.3
Analytic vs holomorphic
A function f : U → C is called analytic if in a neighborhood of any point z0 ∈ U it can be
presented as a sum of a converging power series of z − z0 .
Lemma 5.6. Given an analytic function f : U → C, then the coefficients of its power expansion
are equal to its Taylor coefficients, i.e for any point z0 and a sufficiently small > 0 we have
∞
X f (n) (z0 )
f 00 (z0 ) 2
f (z0 + u) = f (z0 ) + f (z0 )u +
u + ··· +
un + . . . ,
2
n!
0
0
for |u| < .
Indeed, according to Proposition 5.5 a converging power series can be differentiated term-wise
∞
P
in the disc of its convergence. Hence, if f (z0 + u) = an un then
0
f (z0 ) = a0 ,
f 0 (z0 + u) =
∞
X
nan−1 un−1 , and hence
1
f 0 (z0 ) = a1 . Continuing this process we get
an =
f (n) (z0 )
.
n!
Theorem 5.7. The notions of a holomorphicity and analyticity are equivalent.
Proof. According to Proposition 5.5a) any analytic function is holomorphic. To see the converse,
take any point z0 ∈ U and choose r > 0 such that the closed disc D = {|z − z0 | ≤ r} of radius r
centered at z0 is contained in U. Changing the variable u := z − z0 we can express the function f (u)
in the open disc D = {|u| < r} by the Cauchy formula
Z
1
f (ζ)dζ
f (u) =
.
2πi
ζ−u
{|ζ|=r}
47
We have
f (ζ)
f (ζ) 1
=
ζ−u
ζ 1−
∞
u
ζ
=
f (ζ) X un
.
ζ 0 ζn
Let us prove that the power series in the right hand side can be integrated term-wise and thus we
get


Z

∞ 

X

f (ζ)dζ
f (ζ)dζ  n

=

u .
ζ−u
ζ n+1 

0
Z
|ζ|=r
(5.3.1)
|ζ|=r
Then for any u ∈ D, |u| < r we have
Mr
f (ζ)
≤ n+1 ,
n+1
ζ
r
where we denoted Mr := max | f (ζ)|. Hence,
|ζ|=r
Z
|u|
f (ζ)dζ n
u
≤
2πM
r
ζ n+1
r
!n
,
|ζ|=r
and thus the power series in the right-hand side absolutely converges in D.
Let us choose any ρ < r. Then for any u ∈ D, we have


Z
Z

N 

X

f (ζ)dζ  n
f (ζ)dζ


−

u
ζ−u
ζ n+1 

n=0
|ζ|=r
|ζ|=r
Z X
Z
Z X
N
∞
f (ζ)dζ n
f (ζ)dζ n
u −
u =
=
n+1
n+1
ζ
ζ
n=0
n=0
|ζ|=r
|ζ|=r
Z2π
=
0
∞
X
n=N+1
i f (reiθ )
un n inθ dθ
re
|ζ|=r
Z2π
≤
0
∞
X
f (ζ)dζ n
u
n+1
ζ
n=N+1
2π
Z X
∞
∞
X
Mr n
|u|n | f (reiθ )|
dθ ≤
|u| dθ
n
r
rn
n=N+1
n=N+1
0
!n
∞
X
|u|
≤ 2πMr
r
n=N+1
The series
∞ n
P
|u|
n=0
r
converging for |u| < r, and hence lim
∞ n
P
|u|
N→∞ n=N+1
r
= 0, which proves formula
(5.3.1).
48
Remark 5.8. The above argument also shows that if f : U → C is a holomorphic function and
for a ∈ U the disc Dr (a) = {|z − a| < r} is contained in U then the radius R of convergence of the
Taylor expansion of f at the point a satisfies the inequality R ≥ r.
49
50
Chapter 6
Properties of holomorphic functions
6.1
Exponential function and its relatives
So far the only examples of holomorphic functions we had were polynomials and rational functions
P(z)
,
Q(Z)
where P, Q are polynomials, in the domain where Q(z) , 0. The theorem equating holomor-
phic and analytic functions allows us to greatly extend the set of examples. We begin in this section
with exponential function and its close relatives.
As we already pointed out the exponential function is defined by the formula
∞
X
zn
f (z) =
.
n!
0
We also define
sin z =
∞
X
(−1)n
0
∞
X
z2n+1
,
(2n + 1)!
z2n
,
(2n)!
0
∞
ez − e−z X z2n+1
sinh z =
=
,
2
(2n
+
1)!
0
∞
z
−z
X
e +e
z2n
cosh z =
=
2
(2n)!
0
cos z =
(−1)n
51
The radius of convergence of all these series is ∞, and hence the above formulas define holomorphic function on the whole C.
Lemma 6.1.
1) ez1 +z2 = ez1 ez2 ;
2) (ez )0 = ez ;
3) eiz = cos z + i sin z;
4) cos z = cosh iz, sin z = −i sinh iz;
First two properties follows from the formulas of multiplication and differentiation of power of
series, see Exercise 5.4 and Proposition 5.5a). Formula 3) follows the comparison of series in the
left and right hand sides. Formula 4) follows from 3).
It is also interesting to observe that the exponential function is periodic with the imaginary
period 2πi, and that the identity
cos2 z + sin2 z = 1
holds for all z ∈ C and not only when z is real.
6.2
Entire functions
Holomorphic on the whole C functions are called entire. The functions ez , sin z, cos z, sinh z, cos hz
considered in Section 6.1 are examples of entire functions. The sum of any power series with
the infinite radius of convergence is an entire holomorphic function. Remark 5.8 implies that the
converse is also true:
If f : C → C is an entire function, then its Taylor expansion at any point has an infinite radius
of convergence.
Theorem 6.2 (Liouville’s theorem). If an entire function is bounded it is a constant.
52
Proof. This is a corollary of the Cauchy inequality 4.8. Indeed, suppose | f (z)| ≤ M, then the
inequality 4.8 implies that for any k we have
| f k (0)| ≤
Mk!
for any R > 0.
Rk
Hence f k (0) = 0 for k > 0. Therefore,
f (z) = f (0) + f 0 (0)z +
1 00
f (0)z2 + · · · = f (0).
2
The following so-called little Picard Theorem significantly strengthen Liouville’s theorem.
Theorem 6.3 (Little Picard theorem). If an entire function does not take two distinct values z1 , z2 ∈
C then it is a constant.
Note that the function exp z = ez takes all values except 0. Hence, the statement of the little
Picard Theorem is sharp. We will prove this theorem later in the notes.
Theorem 6.4 (Fundamental theorem of algebra). Any polynomial P(z) = a0 +a1 z+· · ·+an zn , an , 0,
of degree n > 0 has a root.
Proof. Suppose P(z) , 0 for all z ∈ C. Then g(z) :=
1
P(z)
is an entire holomorphic function. On the
other hand,
!
|an−1 | |an−2 |
|a0 |
|P(z)| ≥ |z | |an | −
−
− ··· − n .
|z|
|z|2
|z|
n
But
|an−2 |
|z|2
+ ··· +
|a0 |
→ 0,
|z|n |z|→∞
and therefore
|an−2 |
|a0 | |an |
+
·
·
·
+
≤
|z|2
|z|n
2
for |z| sufficiently large. But then
!
|an−1 | |an−2 |
|a0 |
|an ||z|n
|P(z)| ≥ |z | |an | −
−
− ··· − n ≥
,
|z|
|z|2
|z|
2
n
53
and therefore
|g(z)| ≤
2
→ 0.
|an ||z|n |z|→∞
This implies that the function g is bounded, and therefore by Liouville’s theorem it is constant. But
this contradicts to the assumption that the degree n is positive.
Theorem 6.4 implies that the polynomial P(z) of degree n with complex coefficients has n roots,
counted with multiplicities. Indeed, by Theorem 6.4 there exists at least one root z1 . Then P(z) can
be divided by (z − z1 ):
P(z) = (z − z1 )P1 (z),
where degP1 = degP − 1. If degree of P1 (z) is still positive one can continue the process and get
P(z) = (z − z1 )(z − z2 )P2 (z). Continuing the process we decompose P into a product of linear terms:
P(z) = a(z − z1 ) . . . (z − zn ).
6.3
Analytic continuation
Let us recall that a domain U is called connected if one cannot present it as the union U = U1 ∪ U2
of disjoint non-empty open sets. Equivalently, a disconnected domain is a domain which admits a
continuous function on U which takes exactly two values: 0 and 1.
There is a related notion of path-connectedness. A domain U ⊂ C is called path-connected if for
any two points A, B ∈ U there exists a continuous path φ : [0, 1] → U such that φ(0) = A, φ(1) = B.
The notions of connectedness and path connectedness coincide for open sets (for more general sets
path connectedness is a stronger notion).
Lemma 6.5. Let f : U → C be a holomorphic function on a connected domain U. Suppose that
there exists a sequence of distinct points zn ∈ C, n = 1, . . . ,, such that f (zn ) = 0, and lim zn = a ∈
n→∞
U. Then f ≡ 0 in U.
In other words, zeroes of a holomorphic function are always isolated.
54
Proof. Denote by A the set of all points a ∈ U which satisfy the conditions of the lemma, i.e. that
there exists a sequence of distinct points zn ∈ C, n = 1, . . . , ∞, such that f (zn ) = 0, and lim zn = a.
n→∞
Then by continuity we have f (a) = 0 for every a ∈ U. Let us prove that the set A is open. For every
a ∈ A the holomorphic function f can be expanded to a converging power series in a sufficietly
small disc centered at a:
f (a + u) = c1 u + c2 u2 + . . . .
If f is not identically 0 in a neighborhood of a then there is k > 0 such that ck , 0 and c j = 0 for
all j < k. Then
f (a + u) = ck uk (1 + g(u)), where g(u) =
ck+1
ck+2 2
u+
u + ....
ck
ck
The function g is holomorphic in a neighborhood of u = 0 and we have g(0) = 0. Hence, there
exists r > 0 such that |g(u)| <
1
2
for |u| < r. Therefore,
1
| f (a + u)| ≥ |ck ||u|k , for |u| < r.
2
But this implies that f (z) , 0 provided that z , a and |z−a| < r. But this contradicts the assumption
of existence of a sequence zn → a such that f (zn ) = 0. Hence, f is identically equal to 0 in a
neighborhood of a, i.e. A is open.
Suppose that U \ A , ∅. Then for any b ∈ U \ A the point b there is a neighborhood Ub ⊂ U
where there is no zeroes of f with a possible exception of b. But then Ub ⊂ U \ A, and hence U \ A
is open. But this contradicts the connectedness of U, and hence A = U, i.e. the function f is equal
to 0 identically on U.
Given domains U ⊂ V ⊂ C we say that a holomorphic function f : V → C is a holomorphic
extension of a holomorphic function g : U → C if f |U = g. Lemma 6.5 implies that any two
holomorphic extensions of f to a bigger domain coincide.
Example 6.6. 1) The radius of convergence of the series
provides a holomorphic extension of
∞
P
0
∞
P
0
zn is 1. However the function f (z) =
z from the unit disc {|z| < 1} to C \ 1.
n
55
1
1−z
2) The radius of convergence of the series
∞ n
P
z
n
1
is also equal to 1. But as we will see in the
next section this function extends as a holomorphic function to C \ {z ∈ C; Re z > 1, Im z = 0}. An
attempt to extend it further leads to the notion of a Riemann surface which will be discussed later
in the notes.
6.4
Complex logarithm
Consider a closed differential 1-form
dz
z
on C \ 0. While this form is not exact on C \ 0 it becomes
exact when restricted to any simply connected subdomain U ⊂ C \ 0. For instance, take U := C \ R,
where R is the ray R = {z ∈ C; Rez ≤ 0, Imz = 0}. The primitive of
dz
,
z
which is called logarithm (or
sometimes the principal branch of the logarithm and denoted by log z (or ln z), can be computed
by the formula
log z =
Z
Γz
dz
,
z
where Γz is any path connecting 1 with the point z. Then
d log z =
As we already computed above,
dz
z
log z =
dz
1
, or (log z)0 = .
z
z
= d ln r + idφ, and therefore
Z
Γz
dz
=
z
Z
d(ln r) + i
Γz
Z
dφ
Γz
= log r + iφ, φ ∈ (−π, π).
Thus the real part of the complex logarithm log z is equal to log |z|, while the imaginary part is
equal to arg z.
Lemma 6.7 (Properties of the logarithm).
2) log(1 + z) =
∞
P
1
1) elog z = z
n
(−1)n zn for |z| < 1.
56
Indeed, elog z = elog r+iφ = reiφ = z. To prove 2) we observe that d(log(1 + z)) =
1
But 1+z
=
∞
∞
P
P
n
(−1)n zn for |z| < 1. Hence, by integrating both parts of this equality we get log(1 + z) = (−1)n zn
dz
.
1+z
1
1
for |z| < 1.
When trying to extend log z to the whole punctured plane C \ 0 we get a multivalued function
defined up to a multiple of 2πi. In particular, the equality log z1 z2 = log z1 + log z2 holds only up to
a multiple of 2πi.
More about the logarithm
More generally, given any simply connected domain U = 0 we define a logarithm branch logU z in
U as follows. Choose a point z0 ∈ U and a path δ connecting 1 and z0 in C \ 0, and for every point
z ∈ U choose a path γz connecting z0 in z in U. Define
Z
dz
U
log z =
.
z
δ∪γz
Simply connectedness of U guarantees that the integral is independent of the choice of γz . However
it does depend on δ, and a different choice of δ changes the value of the logarithm by adding a
multiple of 2πi.
Lemma 6.8. Let U ⊂ C \ 0 be a simply-connected domain and U0 := C \ R = C \ {z ∈ C; Rez ≤
0, Imz = 0}. Let logU : U → C be a logarithm branch defined above, and ln : U0 → C the principal
logarithm branch. Then for any point z ∈ U ∩ U0 we have
logU z = ln z + 2kπi
for some integer k (which is locally independent on z, but globally may depend on z).
R
R
R
Indeed, logU z − ln z = dzz , where λ a loop, but dzz = i dφ = 2kπi.
λ
Corollary 6.9. elog
U
z
λ
λ
= z.
Indeed, if z ∈ U \ R = U ∩ U0 then according to Lemma 6.8 we have
U
elog
z
= eln z+2kπi = z.
57
If z0 ∈ R then the same holds by continuity.
In the situation when logU is defined we can also define a branch of the function za for any
complex number a by the formula
U
za = ea log z .
In particular, if a =
1
n
1
then a branch of z n =
√ n
√n
z defined this way satisfies n z = z. Indeed,
√n 1 U n
U
z = e n log z = elog z = z.
6.5
Schwarz reflection principle
The following result provides an interesting case of a holomorphic extension.
Theorem 6.10 (Schwarz reflection principle). Let U ⊂ {Im z > 0} ⊂ C be a domain in an upper
half plane. Suppose an interval I = (a, b) ⊂ R ⊂ C is contained in the boundary ∂U. Denote
b := U ∪ I ∪ U.
U
Let f : U → C be a holomorphic function which extends continuously to I and takes real values
b
on I. Then f holomorphically extends to U.
Proof. Define f (z) = f (z̄) for each z ∈ U and extend it by continuity to I. To show that f is
b The function f is holomorphic in U and
holomorphic consider any piecewise linear loop γ ⊂ U.
R
U and extends continuously to I from both sides. Hence, the integral f (z)dz is equal to 0 over
γ
b is split by the interval I into several loops, each
loops δ in U ∪ I and U ∪ I. But any loop γ in U
R
one is either in U ∪ I or U ∪ I. Hence, f (z)dz = 0, and by Proposition 4.11 the function f is
γ
b
holomorphic in U.
58
Chapter 7
Isolated singularities, residues and
meromorphic functions
7.1
Holomorphic functions with isolated singularities
Let U ⊂ C be an open domain. A subset Z ⊂ U is called discrete if for every point u ∈ Z there
exists an > 0 such that the disc D (z) = {z; |z − u| < } is contained in U and has no other points
of Z besides u. In other words, any point of u has a neighborhood which does not contain other
points of Z. The set Z could be finite, or countable, but in the latter case all its accumulation points
do not belong to U. If a discrete set is contained in a compact set then it is finite.
Given a discrete subset Z ⊂ U a holomorphic function f : U \ Z → C is sometimes called a
function on U with isolated singularities at the points of Z.
Some of these singularities could be fictitious, or removable, i.e. the function f can actually be
extended to this point as a holomorphic function.
Theorem 7.1. Let a be an isolated singularity of a holomorphic function given on U = {0 <
|z − a| < r}. Suppose that | f (z)| ≤ C|z − a|−σ for σ < 1. Then a is a removable singularity. In
particular if f is bounded near a then a is removable.
Proof. Consider the function g(z) = (z − a)2 f (z) and extend it to the point a by setting g(a) = 0.
59
Then g is differentiable at a and g0 (a) = 0. Indeed
|g0 (a)| = | lim
z→a
g(z)
| ≤ lim | f (z)|z − a|| ≤ |z − a|1−σ = 0,
z→a
z−a
if σ < 1. Hence, the function g is holomorphic in {|z − a| < r} and vanishes at a together with its
first derivative. Hence, g(z) = (z − a)2 h(z), where h is a holomorphic function. But then f (z) = h(z)
in U = {0 < |z − a| < r}, and hence h(z) is the required holomorphic extension of f to the disc
{|z − a| < r}.
Non-removable singularities are divided into two types, poles and essential singularities.
A point u is called a pole of order k if in a neighborhood of u the function f can be written as
f (z) =
g(z)
(z−u)k
where g is a holomorphic function such that g(u) , 0. A non-removable singularity
which is not a pole is called essential. A holomorphic function with isolated singularities which
are all non-essential is called meromorphic. For instance, any rational function
P(z)
.
Q(z)
i.e. the ratio of
two polynomials is a meromorphic function on C.
The following theorem characterizes poles among isolated singularities.
Proposition 7.2. An isolated singularity u of f is a pole if and only if | f (z)|z→u → ∞.
Thus for an essential singularity u the modulus | f (z)| is unbounded near u but there is no finite
or infinite limit lim | f (z)|.
z→u
Proof. If u is a pole then near u we can write f (z) =
g(z)
,
(z−u)k
where g is a holomorphic function and
g(u) , 0. Hence when z → u we have |g(z)| → |g(u)| , 0 and
Suppose now that | f (z)|z→u → ∞. Then h(z) :=
1
→ 0.
f (z) z→u
1
|z−u|k
→ ∞. Hence, lim | f (z)| = ∞.
z→u
Thus, u is a removable singularity for
h and u is its zero of some order k. Then we can write h(z) = (z − u)k h̃(z), where h̃(u) , 0. Hence,
1
1
h̃(z)
f (z) =
=
h(z) (z − u)k
has a pole of order k at u.
The following, so called great Picard Theorem is a far-going generalization of the little Picard
Theorem, see Theorem 6.3, as well as Proposition 7.2.
60
Theorem 7.3. If an analytic function f : U → C has an essential singularity at a point u ∈ U,
then for any neighborhood Ω 3 u, Ω ⊂ U, the restriction f |Ω\u : Ω \ u → C takes all values in C
with a possible exception of one point a ∈ C
Remark 7.4. It follows that the exceptional value a ∈ C is independent of a neighborhood Ω if
it is small enough, and that on any Ω \ u the function f takes each value in C \ a infinitely many
times.
The proof of the great Picard Theorem goes beyond this class. Interested students can read
it, e.g. in Chapter 4 of the book “Complex Analysis on One Variable” by R. Narasimhan and Y.
Nievergelt, Birkhäuser, 2001.
Exercise 7.5. Deduce the little Picard theorem from the great one.
7.2
Residues
Suppose that a holomorphic function f : U \ u → C has an isolated singularity at u. Take r > 0
such that Dr (u) = {z; |z − u| ≤ r} ⊂ U and define the residue of f at u by the formula
Z
1
Resu f =
f (z)dz.
2πi
(7.2.1)
|z−u|=r
In view of the Cauchy theorem the integral (7.2.1) is independent of the choice of r. If singularity
is removable, then again the Cauchy theorem implies that Resu f = 0.
Example 7.6.





1 
1
Res0 k = 


z


0,
k = 1;
k > 1.
Theorem 7.7 (Residue theorem). Let U be a domain with a piecewise smooth boundary Γ := ∂U
and compact closure. Suppose f : U \ Z → C be a holomorphic function with the set Z of isolated
singularities. Suppose f extends continuously to Γ. Then


Z
X

f (z)dz = 2πi  Resu f  .
u∈Z
Γ
61
Proof. This is just a reformulation of the Cauchy theorem. First, we observe that in view of
compactness of U there could be only finitely many of isolated singularities in U (why?): Z =
{u1 , . . . , uk }. There exist r1 , . . . rk > 0 such that the discs Du j (r j ) = {|z − u j | ≤ r j } are contained in U
and do not intersect each other. Hence, the Cauchy theorem yields:


Z
X Z
X

f (z)dz =
f (z)dz = 2πi  Resu f  .
Γ
u∈Z
∂Dr (u j )
Theorem 7.7 would provide a way of computing contour integrals if we could develop effective
methods for computing the residues. The next proposition explains how to compute residues for
poles.
Proposition 7.8. Suppose f has a pole of order n at a point u, i.e. f (z) =
g(z)
,
(z−u)n
where g is a
holomorphic function such that g(u) , 0. Then
Resu f =
1
g(n−1) (u), where g(z) = (z − u)n f (z).
(n − 1)!
(7.2.2)
In particular, if u is a pole of order 1 we have
Resu f = g(u) = lim(z − u) f (z).
z→u
(7.2.3)
Warning: Formula (7.2.3) can only be used if you already know that the pole is simple. In particular,
if the limit in (7.2.3) is 0 then this means that the pole is not simple.
Proof. Consider the Taylor expansion of g at the point u:
g(z) = g(u) + g0 (u)(z − u) + · · · +
g(n−1) (u)
g(n) (u)
(z − u)n−1 +
(z − u)n + . . . .
(n − 1)!
(n)!
Hence,
g(z)
g(u)
g0 (u)
g(n−1) (u)
=
+
+
·
·
·
+
+ h(z),
(z − u)n (z − u)n (z − u)n−1
(n − 1)!(z − u)
where h is a holomorphic function at a neighborhood of u. But then
!
Z
1
g(u)
g0 (u)
g(n−1) (u)
Resu f =
+
+ ··· +
+ h(z) dz
2πi
(z − u)n (z − u)n−1
(n − 1)!(z − u)
|z−u|=r
Z
1
g(n−1) (u)dz
1
=
=
g(n−1) (u).
2πi
(n − 1)!(z − u) (n − 1)!
f (z) =
|z−u|=r
62
We will later discuss computation of residues at essential singularities.
7.3
Application of the residue theorem to computation of integrals
The residue theorem yields computation of many definite integrals of functions of 1 real variables
which is difficult to compute using elementary methods. We consider here 3 examples.
R∞ dx
1. Compute 1+x
4.
0
R
1
Consider a meromorphic function f (z) = 1+z
f (z)dz, where
4 and compute
ΓR
ΓR = ∂{z; Imz ≥ 0, |z| ≤ R}.
Then
Z
f (z)dz =
ΓR
ZR
−R
dx
+
1 + x4
Z
f (z)dz,
S R+
where S R+ = {z; Imz ≥ 0, |z| = R}. We have
Zπ
Z
f (z)dz ≤
S R+
πR
Rdφ
= 4
→ 0.
4
R − 1 R − 1 R→∞
0
Therefore,
Z∞
Z
f (z)dz →
ΓR
R→∞
−∞
dx
.
1 + x4
The function f has simple poles at the points z1 = eiπ/4 , z2 = e3iπ/4 ∈ {z; Imz ≥ 0, |z| ≤ R}, provided
that R > 1. The residues of f at z j , j = 1, 2 are equal to
lim
z→z j
z − zj
1
1
=
=
.
1 + z4 (1 + z4 )0 |z=z j 4z3j
63
Hence,
√
!
e−3πi/4 + e−9πi/4
π π 2
f (z)dz = 2πi
= π sin =
.
4
4
2
Z
ΓR
Therefore,
Z∞
0
dx
1
= lim
4
1+x
2 R→∞
√
π 2
.
f (z)dz =
4
Z
ΓR
Remark 7.9. Similar techniques applies to integrals
R∞
R(x)dx, when this integral is converging,
−∞
where R(x) =
2. Compute
P(x)
Q(x)
is a rational function.
R∞ cos xdx
1+x2
.
0
Consider a function
eiz
.
1 + z2
f (z) =
Denote
UR,a := {z; 0 ≤ Im z ≤ a, |Re z| ≤ R}.
Then
Z
∂UR,a
+
eiz dz
=
1 + z2
Z0
a
ZR
−R
e x dx
+
1 + x2
e−iR e−y dy
+
1 + (−R + iy)2
Then we have
Za
|I2 | ≤
Z−R
R
Za
0
eiR e−y dy
1 + (R + iy)2
e−a eix dx
= I1 + I2 + I3 + I4 .
1 + (x + ia)2
e−y dy
1
≤ 2
2
R −1 R −1
0
Z∞
e−y dy ≤
0
Similarly,
I3 → 0.
R→∞
Furthermore,
|I4 | ≤
2Re−a
.
a2 − 1
64
R2
1
→ 0.
− 1 R→∞
Choose a = ln R then
|I4 | ≤
2
→ 0.
(ln R)2 − 1 R→∞
Hence,
Z∞
−∞
eix dx
= lim
1 + x2 R→∞
= 2πiResi
ZR
−R
eix dx
= lim
1 + x2 R→∞
Z
∂UR,ln R
eiz dz
1 + z2
e dz
π
=
.
1 + z2 e
iz
Therefore,
Z∞
0
3. Compute
R∞ sin x
x
cos xdx 1
=
1 + x2
2
Z∞
−∞
 ∞

Z ix 
cos xdx 1
e dx 
π

 = .
= Re 
2
2
 1 + x  2e
1+x
2
−∞
dx.
0
Let UR,ln R the domain defined in the previous example. Let D denote the disc {|z| < }. Set
UR,ln R, := UR,ln R \ D .
Then we have
0=
Z
∂UR,ln R,
Zln R
0
 
ZR ix 
Z ix
e dx 
e dz  e dx
 −
= 
+

z
x
x 
iz
−R
eiR e−y dy
+
R + iy
Z0
ln R
e−iR e−y dy 1
+
−R + iy
R
Z
∂D ∩{Im z≥0}
Z−R
R
eix dx
.
x + i ln R
As in Example 2 the last three terms converge to 0 when R → ∞.
We claim that
Z
∂D ∩{Im z≥0}
eiz dz
eiz
−→ πiRes0 = πi.
z →0
z
Indeed, denote S + := ∂D ∩ {Im z ≥ 0}. Then
Z
S +
dz
=i
z
Z
0π
65
dθ = πi.
eiz dz
+
z
On the other hand,
eiz dz
−
z
Z
S +
Zπ
Z
S +
Zπ
dz
≤
z
iθ
|eie − 1|dθ
0
|e − 1|dθ = π(e − 1) → = 0.
≤
→0
0
We used here an inequalty
|ez − 1| =
∞
X
zk
1
Finally,
Z
eix dx
+
x
ZR
−R
Hence,
Z∞
0
k!
∞
X
|z|k
≤
1
eix
dx = 2
x
ZR
= e|z| − 1.
k!
eix
dx −→
R→∞
x
Z∞
→0 0
eix
dx.
x
 ∞

Z ix 
sin x
1
 e
 π
dx = Im 
dx = .

x
2
x  2
0
R∞
Similar techniques applies to integrals of the form
eix R(x)dx, (and in particular to integrals
−∞
R∞
(cos x)R(x)dx and
−∞
R∞
(sin x)R(x)dx) when this integral is converging, where R(x) =
P(x)
Q(x)
−∞
rational function.
4. Compute the integral
I=
Zπ
0
Denote z = eiθ then cos θ =
z+z−1
2
dθ
, a > 1.
a + cos θ
, and hence a + cos θ =
z2 +2az+1
.
2z
Consider the integral
J=
=i
Z
∂D
Zπ
0
dz
1
=
2
z + 2az + 1 2
dθ
= iI,
a + cos θ
66
Z2π
0
idθ
a + cos θ
Denote D := {z; |z| < 1},
is a
i.e. I = −iJ.
On the other hand, we can apply the residue theorem to compute the integral J. the function
√
1
f (z) = z2 +2az+1
have simple poles at the points A± := −a ± a2 − 1 and we have |A+ | < 1 and
|A− | > 1. We have ResA+ f =
1
A+ −A−
=
√1
.
2 a2 −1
Hence,
I = −iJ = −i(2πi)ReszA+ = √
π
a2 − 1
Remark 7.10. The same method applies to integrals of the form
Z2π
R(cos t, sin t)dt,
.
0
where R is a rational function.
7.4
Complex projective line or Riemann sphere
Consider the space Cn of n-tuples (z1 , . . . , zn ) of complex numbers z j ∈ C. This is an example of a
complex vector space. One can add vectors and multiply them by complex numbers:
(z1 , . . . , zn ) + (z01 , . . . , z0n ) = (z1 + z01 , . . . , zn + z0n ),
λ(z1 , . . . , zn ) = (λz1 , . . . , λzn ), λ ∈ C
Similar to the real case one can projectivise Cn . The complex projective space of dimension n,
denoted CPn , is defined as the space of all complex lines through the origin. We will need for our
purposes mostly the 1-dimensional complex projective space, or as it is called complex projective
line. We analyze this notion below.
Any non-zero vector z = (z1 , z2 ) ∈ C2 generates the 1-dimensional complex subspace, or complex line
lz = Span(z) = {λz; λ ∈ C} ⊂ C2 .
The line lz can be viewed as a point of CP1 . Any proportional vector z̃ = µz, µ ∈ C, generates the
same line: lz̃ = lz . Hence, we equivalently can define CP1 as the space of points in C2 \ 0 up to a
complex proportionality.
67
Let us fix an affine line L1 = {z2 = 1} ⊂ C2 . Any line in C2 through the origin, except the line
l(1,0) = {z2 = 0}, intersects L1 at exactly one point. Namely, if l = lz for z = (z1 , z2 ) with z2 , 0
then it intersects L1 at the point (u =
z1
, 1).
z2
So one can view CP1 as C with one point added “at
infinity”.
On the other hand, there is nothing special in this point at infinity. If instead we take an affine
line L2 = {z1 = 1} ⊂ C2 then any line through the origin except the line l(0,1) {z1 = 0} intersects L2
in exactly one point. Namely, if l = lz for z = (z1 , z2 ) with z1 , 0 then it intersects L1 at the point
(v =
z2
, 1).
z2 1
So in CP1 \ (l(0,1) ∪ l(1,0) we have two coordinates u and v related by the formula u = 1v .
Thus, CP1 \ {l(1,0) } and CP1 \ {l(0,1) } can be identified with C, and we can say that CP1 is
obtained by gluing two copies of C along C \ 0 using the gluing map z 7→ 1z . It follows that CP1 is
diffeomorphic to the 2-sphere. To see this, let us take the unit sphere Σ = {x2 + y2 + z2 = 1} ⊂ R3
and view the coordinate plane (x, y) as C. Let N = (0, 0, 1) and S = (0, 0, −1) be the North and
South poles of Σ. Consider the stereographic projections StN : Σ \ S → C and StS : Σ \ N → C
from the North and South poles, respectively. Let us associate with any point p ∈ Σ \ N its complex
coordinate u = StN (p), and with any point p ∈ Σ \ S its complex coordinate v = StS (p), where the
bar denote the complex conjugation. Then one can check that for p ∈ Σ \ (S ∪ N) coordinates u
and v are related by u = 1v , exactly as we had seen above in CP1 .
This leads to the following interpretation of CP1 . We add to C one extra point ∞. Disc complements Ur := {|z| > r} form a system of neighborhoods of ∞. The function u =
1
z
can be viewed
as a coordinate at this neighborhood which is equal to 0 at infinity. Given a holomorphic function
g : Ur → C we say that it extends to ∞ if the function g( u1 ) extends as a holomorphic function to 0.
We say that g has a pole or zero of order m at infinity if so does the function g( u1 ). For instance, any
polynomial P(z) = zn + a1 zn−1 + · · · + an of degree n has a pole of order n at ∞. Indeed, replacing
z=
1
u
we observe that the function
!
1
1 + a1 u + a22 u2 + . . . an un
=
f (u) := P
u
un
has a pole of order n at 0. Similarly, the function
1
P(z)
68
has a zero of order n at infinity.
In view of the above interpretation the complex projective line CP1 is also sometimes called
the Riemann sphere, or extended complex plane C and denoted C.
We had already seen that if u ∈ C is a pole of a function f then | f (z)| → → ∞. But this means
z→u
that if we interpret C as a subset of CP1 = C = C ∪ {∞} then we can continuously extend f to the
point u by setting f (u) = ∞. Moreover, from the point of view of the coordinate u =
1
z
the point ∞
has a coordinate u = 0 and hence it is no different than any other point on CP1 . Thus we conclude
that meromorphic functions on a domain U ⊂ C are just CP1 -valued holomorphic functions.
If a meromorphic function on C has a pole at infinity then it extends to a meromorphic function
on CP1 , i.e. a holomorphic map CP1 → CP1 .
Theorem 7.11. Any rational function R(z) =
P(z)
Q(z)
is meromorphic on CP1 , and conversely any
meromorphic function on CP1 , i.e. a holomorphic map CP1 → CP1 is rational, i.e. the ratio of 2
polynomials.
Proof. Let R(z) =
P(z)
,
Q(z)
where P(z) = a0 zn + a1 zn−1 + · · · + an , Q(z) = b0 zm + b1 zn−1 + · · · + bm , where
a0 , b0 , 0. Then if n > m then R(z) has a pole of order n − m at ∞, if n < m it has a zero of order
m − n at ∞, and if n = m then ∞ is a removable singularity and not a zero.
Conversely, let f : CP1 → CP1 be a holomorphic function. In view of compactness of CP1
the function f : CP1 → CP1 has finitely many poles and zeroes. Let q1 , q2 , . . . , qk ∈ C ⊂ CP1
be zeroes or poles of f . Denote by r1 , . . . , rk the multiplicity of zeroes and poles assuming them
negative for poles. Also ∞ could be a pole or zero of multiplicity r∞ .
Consider the rational function
R(z) := (z − q1 )r1 . . . (z − qk )rk
The functions R(z) and f (z) have zeroes and poles of the same multiplicities at the same points in
C. Then h(z) :=
f (z)
R(z)
has no poles and zeroes in C. We argue that ∞ is also not a pole and hence h
is a non-zero constant C (according to Liouville’s theorem). Indeed, if ∞ is a pole then for
1
h
it is a
zero and then h1 has to be a non-zero constant which implies that h(∞) , 0. Therefore, f (z) = CR(z)
is rational.
69
Corollary 7.12. For any meromorphic function f : CP1 → CP1 the total multiplicity of all zeroes
is equal to the total multiplicities of all poles.
Proof. According to Theorem 7.11 f (z) =
P(z)
.
Q(z)
The total multiplicity of poles of f in the finite part
of C is equal to degree d(Q) of Q while the total multiplicity of zeroes of f in the finite part of C
is equal to the degree d(P) of P. As we already seen in the proof of Theorem 7.11, if d(P) > d(Q)
then f has a zero at ∞ of order d(P) − d(Q), while if d(Q) > d(P) then f has a pole at ∞ of order
d(Q) − d(P). If d(P) = d(Q) the ∞ is neither pole nor zero. In all cases the difference between the
total number of zeroes and poles, including ∞, is equal to 0.
7.5
Residue of meromorphic differential forms
Recall that we defined the residue of a meromorphic function f : U → CP1 at its pole a ∈ U by
the formula
1
Resa f =
2πi
Z
f (z)dz.
∂D (a)
In fact, it would be better to call Resa f the residue of the meromorphic differential 1-form f (z)dz
rather than the residue of the meromorphic function f (z). The reason for this is that the residue of
a meromorphic function depends on the choice of a holomorphic coordinate near the point a. For
1
instance, Res0 1z = 1, but if we make a change of coordinate z = 2u then Res0 2u
= 12 . On the other
hand, the next lemma shows that the residue of a meromorphic differential forms remains invariant
when we make a holomorphic change of coordinate.
Lemma 7.13. Consider a meromorphic form f (z)dz, where the function f defined in a neighborhood U 3 0 has a pole at 0. Consider a change of coordinate z = h(u), where h : U 0 → U where h
is a biholomorphism such that h(0) = 0. Then
Res0 ( f (z)dz) = Res0 ( f (h(u)dh = f (h(u))h0 (u)du).
Proof.
70
1
Res0 ( f (h(u))h (u)du =
2πi
Z
0
1
f (h(u))h (u)du =
2πi
Z
0
∂D
f (z)dz.
∂h(U)
But for a biholomorphism h which preserves 0 we have
Z
Z
1
1
f (z)dz =
f (z)dz = Res0 ( f (z)dz)).
2πi
2πi
∂h(U)
∂Dδ (0)
The invariance of residues for meromorphic differential 1-forms allows us to define the residue
of a meromorphic function f : CP1 → CP1 at ∞. Indeed, in a neighborhood of ∞ we can choose
u=
1
z
as a coordinate and define
!
Res∞ ( f (z)dz) := Res0 f
1
1
d
u
u
!!
 1 
 f u

= −Res0  2 du .
u
For instance,
Res∞
!
!
!
du
du
dz
= −Res0
= −1; and Res∞ (zdz) = −Res0 3 = 0.
z
u
u
It is interesting to observe that while the meromorphic function
1
z
has a 0 and infinity and z has a
pole at infinity the meromorphic differential form dzz has a simple pole at infinity, while zdz has at
and zdz = − du
. We also observe that that for
infinity a pole of order 3. Indeed, dzz = 1u d u1 = − du
u
u3
both meromorphic forms the sum of residues in all its poles is equal to 0. It turns out that this is a
general fact as the following exercise shows.
Exercise 7.14. Let f : CP1 → CP1 be a meromorphic function. Prove that the sum of residues of
all poles of the meromorphic differential 1-form f (z)dz is equal to 0.
7.6
Argument principle
Theorem 7.15. Let U be a domain with a piecewise smooth boundary and f : U → CP1 be
a meromorphic function which C 1 -extends to the boundary ∂U without poles and zeroes on ∂U.
71
Let q1 , . . . , qk ∈ U be the zeroes of f of multiplicities r1 , . . . , rk , and p1 , . . . , pl be poles of f of
multiplicities s1 , . . . , sl . Then
1
2πi
l
k
X
f 0 (z)dz X
si .
rj −
=
f (z)
1
1
Z
Γ
Proof. According to Cauchy theorem
k
f 0 (z)dz X
=
f (z)
1
Z
Γ
Z
|z−q j |=
l
f 0 (z)dz X
+
f (z)
1
Z
f 0 (z)dz
,
f (z)
|z−pi |=
where > 0 is chosen so small that the discs D (q j ) and D (pi ), j = 1, . . . , k, i = 1, . . . , l, are
pairwise disjoint and are contained in U. If is small enough then in D (q j ) we have
f (z) = (z − q j )r j g j (z),
where g j (z) , 0, and in D (pi ) we have
f (z) = (z − p j )−si hi (z),
where hi (z) , 0. Then
f 0 (z)dz
=
f (z)
Z
|z−q j |=
(z − q j )r j g j (z)
|z−q j |=
Z
=
r j (z − q j )r j −1 g j (z) + (z − q j )r j g0j (z))dz
Z
r j dz
+
z − qj
|z−q j |=
Z
g0j (z)
g j (z)
= 2πr j i.
|z−q j |=
The second integral is equal to 0 because the function
g0j
gj
is holomorphic in the whole disc D (q j ).
Similarly, we get
f 0 (z)dz
=
f (z)
Z
|z−p j |=
=
Z
|z−p j |=−
(−si (z − p j )−s j −1 h j (z) + (z − p j )−s j h0j (z))dz
Z
(z − p j )−s j h j (z)
|z−p j |=
si dz
+
z − pj
Z
h0j (z)
h j (z)
= −2πs j i.
|z−p j |=
72
Hence,
1
2πi
Z
Γ
k
l
X
f 0 (z)dz X
s j.
rj −
=
f (z)
1
1
Corollary 7.16 (Rouché’s theorem). Let U be a domain with a piecewise smooth boundary. Let
f, g : U → C be holomorphic functions which C 1 -extend to ∂U. Suppose that |g(z)| < | f (z)| for all
z ∈ ∂U. Then the holomorphic functions f and f + g have the same number of zeroes in U counted
with multiplicities.
Exercise 7.17. Prove that it is sufficient to assume that f, g continuously, and not necessarily C 1
extend to U.
Proof. Consider a 1-parametric family of functions ft := f + tg, t ∈ [0, 1]. By assumption on ∂U
we have
| ft | ≥ | f | − t|g| > 0.
Hence we can apply the argument principle to conclude that the total number nt of zeroes of the
function ft counted with multiplicities is given by the formula
Z 0
ft (z)
1
nt =
.
2πi
ft (z)
∂U
This integral takes only integer values, but at the same time it continuously depends on t. Hence it
is a constant, which implies that the number n0 of zeroes of f is equal to the number n1 of zeroes
of f + g.
Corollary 7.18 (Open image theorem). Let U be a connected open domain and f : U → C a
non-constant holomorphic map. Then the image f (U) ⊂ C is open.
Proof. Take u ∈ U. Without a loss of generality we can assume that u = 0 and f (u) = 0. Let us
write a Taylor expansion of f at u = 0:
f (z) = a1 z + a2 z2 + . . . .
73
Let ak be the first coefficient which is not 0. Then
f (z) = ak zk + ak+1 zk+1 + · · · = zk (ak + h(z)) ,
where ak , 0 and h(0) = 0. If r is small enough then |h(z)| ≤
|ak |
2
for |z| ≤ r. Choose ρ <
|ak |rk
.
4
We
claim that the disc Dρ = {|z| < ρ} is contained in f (U). Indeed, for any point v ∈ Dρ the equation
ak zk = v has exactly k solutions in Dr (u). In other words, the function g(z) = ak zk − v has k zeros in
Dr (0). Note that for z ∈ Dr (0) and v ∈ Dρ we have
|zk h(z)| = rk |h(z)| <
|ak |rk
,
2
while for |z| = r we have
|ak zk − v| ≥ |ak |rk −
|ak |rk
> |zk h(z)|.
4
Hence, according to Rouché’s theorem the function
f (z) − v = (ak zk − v) + zk h(z)
has also k ≥ 1 zeroes in Dr (u), i.e. the point v is in the image f (U).
Corollary 7.18 implies (why?)
Corollary 7.19 (Maximum modulus principle). A non-constant holomorphic function f : U → C
cannot attain the maximum of its modulus | f (z)| at an interior point of U.
7.7
Winding number
Consider a loop γ : [0, 1] → C \ 0, γ(0) = γ(1). It winding number w(γ) is defined as
1
w(γ) :=
2πi
Z
γ
dz
=
z
Z
dθ.
γ
The winding number depends on the orientation of the loop but not on its parameterization.
The following proposition is a reformulation of the argument principle.
74
Proposition 7.20. Let U ⊂ C be a domain with a piecewise smooth boundary and f : U → C is a
holomorphic function which extends as a C 1 -function to U. Let γ1 , . . . , γk : [0, 1] → ∂U parameterizing the boundary components of U according to their orientation as boundary components of
U. Denote
e
γ j ; = f ◦ γ j , j = 1, . . . k.
Then
k
X
w(e
γ j ) = n,
1
where n is the number of zeroes of the function f counted with multiplicities.
Proof. Using Theorem 7.15 we get
1
n=
2πi
Z
Γ
k
f 0 (z)dz X 1
=
f (z)
2πi
1
Z
γj
k
df X 1
=
f
2πi
1
Z
f ◦γ j
k
dz X
=
w(e
γ j ).
z
1
75
76
Chapter 8
Harmonic functions
8.1
Harmonic and holomorphic functions
The Laplace differential operator ∆ on functions on domains in C is defined as ∆ :=
∆f =
∂2
∂x2
+
∂2
,
∂y2
i.e.
∂2 f ∂2 f
+
∂x2 ∂y2
for any C 2 -function f . The Laplace operator can be rewritten in the complex notation as
∆=4
Indeed,
∂ ∂
∂2 f
, i.e. ∆ f = 4
.
∂z ∂z̄
∂z∂z̄
∂2
∂ ∂
1 ∂
∂
=
=
−i
∂z∂z̄ ∂z ∂z̄ 4 ∂x
∂y
!
!
!
∂
1 ∂2
∂
∂2
+i
=
+
.
∂x
∂y
4 ∂x2 ∂y2
because mixed derivative terms cancel.
A real- or complex-valued C 2 -smooth function f on a domain U ⊂ C is called harmonic if
∆ f = 0.
Example 8.1. Any (inhomogeneous) linear function u(x, y) = ax + by + c is harmonic. The function
ln |z| is harmonic on C \ 0.
The notions of a harmonic functions can be extended to much more general setup, in particular
to domains in higher dimensional Euclidean spaces. However in the (real) 2-dimensional case the
77
theory of harmonic and holomorphic functions are intertwined in a very special and interesting
way.
Theorem 8.2. If f = u + iv : U → C is a holomorphic function, then f , and therefore its real
and imaginary parts u and v are harmonic. Conversely if u : U → R is a harmonic function
and the domain U is simply connected, then there exists a unique up to an additive constant harmonic function v : U → C, called harmonic conjugate of u, such that the function f = u + iv is
holomorphic.
Proof. Suppose f = u + iv : U → C is a holomorphic function. Then
!
∂ ∂f
∆ f = ∆u + i∆v = 4
= 0,
∂z ∂z̄
because
∂f
∂z̄
= 0, and hence ∆ f = 0 and ∆u = ∆v = 0.
Conversely, suppose that u is harmonic in a simply connected domain U. Then g := u x − iuy
is holomorphic. To see this we verify the Cauchy-Riemann equations for g. We have ∆u = u xx +
uyy = 0, and hence (u x ) x = (−uy )y . We also have (u x )y = −(−uy ) x due to the equality of mixed
derivatives. In view of simply connectedness of U the holomorphic 1-form is exact, i.e. there
exists a holomorphic function f = e
u + ie
v : U → C such that d f = g(z)dz, or
!
1 ∂
∂
1
−i
(e
u + ie
v) = e
ux + e
vy + i(e
vx − e
uy ) = g(z) = u x − iuy .
2 ∂x
∂y
2
Taking into account the Cauchy-Riemann equations we get
1
(e
ux + e
vy ) = e
ux = ux ,
2
1
(e
uy − e
vx ) = e
uy = uy .
2
Hence, de
u = du and therefore e
u = u + C, and we can choose C = 0. Thus the function e
v is
a harmonic conjugate of u, Any two holomorphic functions with the same real part differ by a
constant, and hence the harmonic conjugate is defined up to adding a constant.
Remark 8.3. If U is not simply connected then the harmonic conjugate may not exist as a univalent
function v, while its differential dv is well defined as a closed 1-form. For instance, consider the
78
harmonic function u = ln r = ln
p
x2 + y2 = ln |z| on C \ 0, z = reiφ = x + iy. Then its harmonic
conjugate is φ = arg z, which is multivalued. At the same time the form dφ =
zdy−ydx
.is
x2 +y2
a well
defined closed 1-form.
8.2
Properties of harmonic functions
Theorem 8.2 implies that similarly to the case of holomorphic functions
Corollary 8.4. Two harmonic functions u, ũ : U → R on a connected domain U which coincide
on a subdomain U 0 ⊂ U coincide in U
Proof. Take a point a ∈ U 0 and any point b ∈ U. There exists a simply connected subdomain
V ⊂ U which contains the points a, b. Indeed, take any embedded path connecting a and b and
choose its neighborhood as V. Let v, ṽ be the harmonic congugate of u and e
u on V. Holomorphic
functions f := u + iv, e
f =e
u + ie
v have the same real parts near the point a, and hence its imaginary
parts near a differ by an additive constant. Hence, by adjusting this constant we can assume that
f = e
f near a. But then by uniqueness of the holomorphic continuation we have f = e
f on V, and in
particular f (a) = e
f (a) and hence u(a) = Re f (a) = Re e
f (a) = e
u(a).
An important fact is that the notion of a harmonic function is invariant with respect to a holomorphic change of coordinate.
e → U a holomorphic function. Then
Lemma 8.5. Let h : U → C be a C 2 -function and f : U
∆(h ◦ f )(z) = (∆h)( f (z))| f 0 (z)|2 .
In particular, if h is harmonic then so is h ◦ f .
Proof.
79
!
∂ ∂(h ◦ f )
(z)
∆(h ◦ f )(z) = 4
∂z̄
∂z


!
∂  ∂h
∂h ∂ f 
0

= 4 
( f (z)) f (z) +
∂z̄ ∂z
∂z̄ ∂z
!
!
∂ ∂h
0
( f (z)) f (z)
=4
∂z̄ ∂z
!
∂2 h
∂ f¯
∂2 h
∂f
0
0
=4
( f (z)) (z) f (z) +
( f (z)) (z) f (z)
∂z̄∂z
∂z̄
∂z∂z̄
∂z̄
= ∆h( f (z)) f 0 (z) f 0 (z) = ∆h( f (z))| f 0 (z)|2 ,
because
∂f
∂z̄
=
∂ f¯
∂z
= 0 and
∂ f¯
∂z̄
= f 0 (z).
Example 8.6. If f is a holomorphic function then h(z) = ln | f (z)| is harmonic. Indeed, the function
h(z) is the composition of a holomorphic function f with a harmonic function ln |z|.
Theorem 8.7 (Mean value theorem). For any harmonic function h : U → C, any point a ∈ U and
r > 0 such that Dr (a) = {|z − a| ≤ r} ⊂ U one has
1
h(a) =
2π
Z2π
h(a + reit )dt.
0
Proof. It is sufficient to assume that h takes real values (because we can prove the theorem separately for the real and imaginary parts. Take an open slightly larger disc Dρ (a) = {|z − a| < ρ} ⊂ U,
ρ > r. The disc Dρ (a) is simply connected. Hence, we can find a harmonic function g : Dρ (a) → C
which is harmonic conjugate to h, i.e. f := h + ig is a holomorphic function. Then by the Cauchy
integral formula we have
1
f (a) =
2πi
=
1
2π
Z2π
Z
∂Dr (a)
f (z)dz
1
=
z−a
2πi
Z2π
f (a + reit )ireit
dt
reit
0
f (a + reit )dt.
0
Taking the real part of this equality we get the required formula for the function h.
80
Corollary 8.8 (Maximum principle for harmonic functions). Let h : U → R be a non-constant
harmonic function on a connected domain U. Then it cannot achieve a local maximum at any
(interior) point of U.
Proof. Suppose that for a ∈ U we have u(a) ≥ u(a + z) for all |z| < . The function h cannot be
constant on D (a), because then it would be a constant by the uniqueness of harmonic continuation,
see Corollary 8.4. Therefore there exists |z0 | < such that h(z0 ) < h(a). But then
1
2π
Z2π
f (a + |z0 |eit ) < f (a),
0
which contradicts Theorem 8.7.
Remark 8.9. Maximum principle for harmonic functions can also be deduced from the open image
theorem for holomorphic functions.
One of the corollaries of the maximum principle is the uniqueness of an extension of a harmonic
function from a boundary of a domain.
Corollary 8.10. Let f, g : U → R be two harmonic functions which extend continuously to the
boundary ∂U. Suppose that
f |∂U = g|∂U .
Then f = g on U.
Proof. Suppose for a ∈ U we have f (a) > g(a). Then the maximum of the harmonic function f − g
is achieved in an interior point of U which is impossible.
81
82
Part II
Conformal mappings and Riemann
mapping theorem
83
Chapter 9
Conformal mappings and their properties
9.1
Biholomorphisms
Let f : U → C be a holomorphic function. Recall that the image V := f (U) is an open set. Suppose
that f is injective, i.e. f (z1 ) , f (z2 ) for any z1 , z2 ∈ U, z1 , z2 . Then f can be viewed as a 1 − 1,
i.e. bijective map U → V.
Lemma 9.1. If f : U → V is bijective and holomorphic, then the derivative f 0 never vanishes,
and the inverse map f −1 : V → U is also holomorphic.
Proof. Suppose f 0 (a) = 0 for a ∈ U. Then expanding f to a Taylor series near the point a we have
f (z) − f (a) =
f 00 (a)
(z − a)2 + · · · = (z − a)k (c + g(z)),
2
where k ≥ 2, c , 0 and g(z) is a holomorphic function such that g(a) = 0. Hence, there exists > 0
such that if |z − a| ≤ then |g(z)| <
|c|
.
2
Take any b , 0, |b| <
|c| k
.
2
Note that the equation c(z − a)k = b
has exactly k solutions in the disk D (a) = {|z − a| < }. For any z ∈ ∂D (a) we have
|c(z − a)k − b| >
|c| k
> k |g(z)| = |(z − a)k g(z)|.
2
Note that
f (z) − ( f (a) + b) = (c(z − a)k − b) + (z − a)k g(z).
85
Hence, Rouché’s theorem implies that the equation f (z) = f (a) + b has the same number of solutions as the equation c(z − a)k = b, which is k > 1, which contradicts to the injectivity of f . This
proves that f 0 (z) , 0 for all z ∈ D (a).
But then the chain rule implies the the inverse map h = f −1 : V → U is also hoomorphic and
h0 ( f (z)) =
1
.
f 0 (z)
A bijective holomorphic map f : U → V is called a biholomorphism. Thus the map f −1 : V →
U, inverse to a biholomorphism is itself a biholomorphism.
9.2
Conformal mappings
Let us assume that C = R2 is endowed with the standard Euclidean metric. Any orientation preserving orthogonal transformation of R2 is a rotation, i.e in complex notation is given by z 7→ eiθ z.
Note that any complex linear map C → C has the form z 7→ cz, i.e. z 7→ reiθ z, where c = reiθ . Geometrically this map is characterized by two properties: it preserves the orientation and it preserves
all angles. Maps with this properties are called linear conformal. Any linear conformal map is a
composition of a rotation with a scaling (homothety) z 7→ rz, and therefore linear conformal maps
R2 → R2 are exactly the same as linear complex maps C → C.
A differentiable in the real sense bijective map f : U → V is called conformal if its differential
dz f : Cz → C f (z) is linear conformal for any z ∈ U.
The above discussion implies that conformal maps U → V coincide with biholomorphisms
U → V.
Remark 9.2. A not necessarily linear map f : U → R2 is called an isometry if its differential
at every point is orthogonal, i.e. preserves the Euclidean metric. However, one can show that any
isometry U → R2 has to be an affine map: it is a composition of a rotation with a parallel translation. It is a remarkable fact that by relaxing the isometry condition to the conformality condition
one greatly enlarges the class of maps.
If there exists a conformal map (or a biholomorphism) f : U → V, then U and V are called
86
biholomorphic or conformally equivalent.
9.3
9.3.1
Examples of conformal mappings
Unit disc and the upper-half plane
We begin with exploring the inversion operation inv : C \ 0 → C \ 0, which in polar coordinates is
given by the formula r 7→ 1r . In complex notations we have inv(z) = (z)−1 .
Lemma 9.3. Image inv(l) of the line l = {Im z = d} is the circle {|z −
Proof. We have inv(x + id) =
1
x−id
=
x+id
.
d2 +x2
i
|
2d
=
1
}.
2d
Therefore,
x + id
i
i
= 2
−
2d
d + x2 2d
√
d4 + x4 − 2d2 x2 + 4d2 x2
1
2id2 + 2dx − id2 − ix2
=
=
=
2
2
2
2
2d(d + x )
2d(d + x )
2d
inv(x + id) −
Therefore, the map z 7→
1
z
maps the half-plane {Im z > d} onto the open disc {|z +
i
|
2d
<
1
}.
2d
Consequently,
Proposition 9.4. The map
z 7→
2
iz + 1
+i=
z+i
z+i
is a biholomorphism between the upper half plane H = {Im z > 0} and the unit disc D = {|z| < 1}.
Exercise 9.5. Show that
z 7→
i−z
i+z
defines another conformal equivalence H → D, and that the inverse map D → H is given by
z 7→ i
1−z
.
1+z
87
Exercise 9.6. Show that the map z 7→ z +
1
z
is a conformal isomorphism of C \ D onto
C \ {−2 < Re z < 2; Im z = 0}.
It is called Joukovsky’s map.
9.3.2
Strips and sectors
The map z 7→ ez conformally maps the infinite strip Pa := {−a < Im z < a}, a < π onto the sector
S a := {−a < arg z < a}. The strip Pπ := {−π < Im z < π} is mapped by the exponential map onto
the domain C \ {Re z ≤ 0, Im z = 0}, the complement of the negative real ray in C.
The map z 7→ z2 establishes a biholomorphism between the upper-half plane H = {Im z > 0}
and the complement C \ {Re z ≥ 0, Im z = 0} of the positive real ray. The map z 7→ zα for 0 < α < 1
establishes a biholomorphism between the the upper-half plane H and the sector {0 < arg z < απ}.
Consequently, taking compositions of the above biholomorphisms we can establish more conformal equivalences. For instance, the composition log(−z2 ) of the maps z 7→ z2 , z 7→ −z and
z 7→ log z maps the upper-half plane H onto the strip Pπ := {−π < Im z < π}.
9.4
Schwarz lemma
The following statement known as the “Schwarz lemma” will be useful for our further study of
conformal mappings.
Theorem 9.7 (Schwarz lemma). Let f : D → D be a holomorphic map with f (0) = 0 (D denotes
the unit disc). Then
(i) | f (z)| ≤ |z|;
(ii) If for some z0 ∈ D, z0 , 0, we have | f (z0 )| = |z0 | then f is a rotation;
(iii) | f 0 (0)| ≤ 1, and if | f 0 (0)| = 1 then f is a rotation.
88
Proof. The equality f (0) = 0 implies that g(z) :=
f (z)
z
is holomorphic. If |z| = r then
f (z)
| f (z)| 1
=
≤ .
z
r
r
Hence, applying the maximum modulus principle we conclude that this true for all |z| < r, and
therefore passing to the limit r → 1 we get | f (z)| ≤ |z| for all z ∈ D. If | f (z0 )| = |z0 | then z0 is an
interior maximum point of the function g(z) =
f (z)
,
z
and hence f (z) = cz and |c| =
f (z0 )
z0
= 1, i.e. f
is a rotation which proves (ii). Finally the inequality | f 0 (0)| ≤ 1 follows from the Cauchy formula.
We also notice that f 0 (0) = lim
z→0
f (z)
z
= g(0), and and if | f 0 (0)| = |g(0)| = 1, then again the maximum
modulus principle implies that g(z) = c, with |c| = 1, and hence f (z) = cz is a rotation.
Corollary 9.8. Let f : D → D be a conformal equivalence such that f (0) = 0. Then f is a rotation,
i.e f (z) = eiθ z, θ ∈ R.
Proof. By 9.7(iii) we have | f 0 (0)| ≤ 1, and applying 9.7(iii) to f −1 we get |( f −1 )0 (0)| =
Hence, | f 0 (0)| = 1, and applying again 9.7(iii) we conclude that f is a rotation.
9.5
1
| f 0 (0)|
≥ 1.
Automorphisms of the Riemann sphere, C, the unit disc and
the upper-half plane
Given a domain U its self-biholomorphisms U → U are called automorphisms. Composition of
automorphisms are automorphisms and inverse automorphisms are automorphisms as well. Hence,
automorphisms of a given domain form a group.
9.5.1 GL(n, C), GL(n, R), PGL(n, C), PGL(n, R) and PGL+ (n, R) = PS L(n, R)
The notation GL(n, C) and GL(n, R) stand for the group of complex and real linear transformations
of Cn and Rn , respectively, or equivalently the groups of n × n complex and real non-degenerate
matrices. These groups are called the general groups of complex and real linear transformations,
respectively.
89
Note that a linear transformation A : Cn → Cn defines also a transformation of the projective
space CPn−1 . Indeed, the transformation A maps lines to lines. Such transformations of CPn−1 are
e : Cn → Cn define
called complex projective transformations. Note that two transformations A, A
e = cA, c ∈ C. Projective
the same transformation of CPn−1 if and only if they are proportional: A
transformations of CPn−1 form the complex projective linear group PGL(n, C). Thus an element of
PGL(n, C) is an n × n complex matrix up to a complex scalar factor.
Similarly, one defines the real projective group PGL(n, R) of projective transformations of
RPn−1 . An element of this group can be viewed as an n × n real matrix up to a real scalar factor. If n is even, the group PGL(n, R) consists of two connected components PGL+ (n, R) and
PGL− (n, R), of orientation preserving and reversing orientations. Note that PGL+ (n, R) is a subgroup of PGL(n, R) while PGL− (n, R) is not. If n is odd then multiplying a matrix by −1 one
changes the sign of its determinant and hence PGL(n, R) is connected.
. The group PGL+ (n, R) is also denoted PS L(n, R). The notation S L(n, R) stands for the special
linear group, i.e. the group of n × n matrices with determinant 1. The projective special group
PS L(n, R) is obtained from S L(n, R) by identifying matrices A and −A. Clearly we get the same
thing by identifying matrices A and −A in S L(n, R), or by identifying all proportional matrices in
PGL+ (n, R). Hence, PGL+ (n, R) = PS L(n, R).
It turns out that the groups PGL(2, C) and PS L(2, R) can be also interpreted as groups of
conformal automorphisms of some special domains. Namely, we will see below that elements of
PGL(2, C) serve as automorphisms of the Riemann sphere CP1 , while elements of PS L(2, R) act
as automorphisms of upper-half plane H.
9.5.2
Automorphisms of CP1 and C


a b


Take an element A ∈ PGL(2, C), i.e. a complex matrix 
up to a complex scalar factor and
c d
associate with it a fractional linear transformation
fA
z 7→
az + b
.
cz + d
90
Lemma 9.9. The function z 7→ fA (z) =
az+b
cz+d
is a conformal automorphism CP1 → CP1 .
Proof. The function f is meromorphic, and hence a holomorphic map CP1 → CP1 . So we only
need to check that it is bijective. But this follows from the fact that for any w ∈ C ⊂ CP1 we can
uniquely solve the equation
az+b
cz+d
= w:
z=
dw − b
,
(ad − bc)(−cw + a)
d
and if w = ∞ then z = − c(ad−bc)
if c , 0, and z = ∞ otherwise.
Lemma 9.10.
fAB = fA ◦ fB .
Proof. The above property can be easily verified by the direct computation. However, we present
here a more conceptual proof.
Recall that CP1 is the space of complex lines in C2 , or equivalently the space of pairs (z1 , z2 ) ,
(0, 0) of complex numbers up to proportionality (z1 , z2 ) ∼ (λz1 , λz2 ), λ ∈ C. CP1 can be covered by
two coordinate charts, with a coordinate u = zz12 , where z2 , 0 and v =


a b


A matrix A = 
∈ GL(2, C) acts on C2 by
c d
1
u
=
z2
,
z1
where z1 , 0.


az + bz 
1
2


z 7→ Az = 
.
cz + dz 
1
2
This action defines the action on CP1 , which in terms of the coordinate u has the form
u=
az1 + bz2 au + b
z1
7→
=
,
z2
cz1 + bz2 cu + d
i.e. it acts exactly by fractional linear transformations. But on C2 the composition of transformation
corresponds to multiplication of matrices, and hence so does the composition of fractional linear
transformations.
It turns out that
91
Proposition 9.11. Any conformal automorphism CP1 → CP1 is a fractional linear transformation.
Proof. Holomorphic maps CP1 → CP1 , i.e. meromorphic functions, are, according to Theorem
7.11 are rational. Let us analyze when a rational function f (z) =
P(z)
Q(z)
defines an bijective map
CP1 → CP1 . We can assume that the polynomial P(z) and Q(z) have no common divisors, which
is equivalent to the fact that they do not have any common zeroes. If the degree of the polynomial
P is > 1 then the function f (z) has at least two distinct zeroes, or one of the zeroes has multiplicity
> 1. In both cases this implies that f is not 1-1. Indeed, if there are two distinct zeroes z1 , z2 then
f (z1 ) = f (z2 ) = 0 of za is a multiple zero then the function f in a neighborhood of z1 can be written
as c(z − z1 )k (1 + h(z)), where k > 1, c , 0 and h(0) = 0. Then arguing as in the proof of Lemma 9.1
we conclude that f is not 1-1 on this neighborhood. Applying the same argument to the function
1
f
we show that the degree of Q is also ≤ 1, and hence f is fractional linear.
Combining Proposition 9.11 and Lemma 9.9 we get
Theorem 9.12. The group Aut(CP1 ) of conformal automorphisms of the Riemann sphere is isomorphic to PGL(2, C). The elements of PGL(2, C) act on CP1 by fractional linear transformations.
Proposition 9.13. For any 3 points z0 , z1 , z2 of CP1 there exists a unique automorphism f ∈
PGL(2, C) such that f (0) = z0 , f (1) = z1 , f (∞) = z2 .
Proof. Let f (z) =
az+b
.
cz+d
The required conditions amount to the system of equations on the coeffi-
cients of f (which are given up to a proportionality factor):
b = dz0 ,
a + b = z1 (c + d),
a = cz2 .
We can set d = 1 and then get b = z0 and
cz2 + z0 = cz1 + c,
a = cz2 .
92
Thus, c =
c−z0
,
z2 −z1
a=
(c−z0 )z2
.
z2 −z1
Given 4 points z1 , z2 , z3 , z4 their cross ratio is defined as
(z1 , z2 ; z3 , z4 ) =
(z3 − z1 )(z4 − z2 )
.
(z3 − z2 )(z4 − z1 )
For instance, we have
(1, z; ∞, 0) =
(∞ − 1)(0 − z)
= z.
(∞ − z)(0 − 1)
Exercise 9.14. Prove that in order that two 4-tuples of points were equivalent under a conformal
equivalence of CP1 it is necessary and sufficient that they had the same cross-ratio.
Any automorphism f : C → C extends, by the removal of singularities lemma to CP1 (why?).
Hence, f (z) =
b
.
d
az+b
,
cz+d
and f (∞) = ∞ implies that c = 0. Therefore, f (z) = Az + B, where A = da , B =
In other words,
Proposition 9.15. Any automorphism of C is a composition of a rotation, a scaling (homothety)
and a parallel transport.
9.5.3
Automorphisms of H and D
It turns out that any automorphism of H or D extends to an automorphism of CP1 and hence Aut(H)
and Aut(D) are subgroups of Aut(CP1 ) consisting of fractional linear transformations of CP1 which
map H and D onto themselves, or as one says, leave them invariant.
Lemma 9.16. A fractional linear transformation f (z) =
az+b
cz+d
leaves the upper-half plane H invari

a b
 is proportional to a
ant (i.e. f (H) = H) if and only if f ∈ PS L(2, R), i.e. when the matrix 
c d
real matrix with a positive determinant.
Proof. Indeed, if a, b, c, d are real and det A = ad − bc > 0 then for each z = x + iy with y > 0 we
have
az + b ax + b + iay (ax + b + iay)(cx + d − icy)
=
=
,
cz + d cx + d + icy
(cx + d)2 + c2 y2
93
and
!
az + b
ax + b + iay ay(cx + d) − cy(ax + b)
y(ad − bc)
Im
=
=
=
> 0.
2
2
2
cz + d
cx + d + icy
(cx + d) + c y
(cx + d)2 + c2 y2
We leave it as an exercise to prove the converse, that is if for a fractional linear transformation
a b


f we have f (H) = H then f ∈ PS L(2, R), i.e. the matrix 
is proportional to a real matrix
c d
with a positive determinant.
Corollary 9.17. For any points α, β ∈ H there exists an automorphism f : H → H such that
f (α) = β
Proof. It is sufficient to consider the case α = i. Let β = p + iq. We have q > 0. We need to find
real a, b, c, d solving the equation
ai + b
= p + iq,
ci + d
or
ai + b = d p − cq + i(cp + dq).
This yields the linear system
pd − qc = b
qd + pc = a.
The determinant p2 + q2 > 0. Hence, for any a, b we can solve the system with respect to c, d. For
instance, taking b = 1, a = 0 we find
p
+ q2
−q
c= 2
p + q2
d=
p2
Note that
0
−q
p2 +q2
1
p
p2 +q2
=
p2
q
> 0,
+ q2
and hence the fractional linear transformation belongs to PS L(2, R).
94
Remark 9.18. The statement of the corollary means that PS L(2, R) acts on H transitively. One says
that a group G acts transitively on a set X if for any two points x0 , x1 ∈ G there is a transformation
from G which moves x0 to x1 . As a corollary we get that fractional linear transformations act
transitively on D as well.
Theorem 9.19. Any conformal automorphism of D and H is fractional linear, and hence in the
case of H is given by a matrix from PS L(2, R).
Proof. Let f : D → D be a conformal automorphism. By composing f with a fractional linear
transformation g : D → D we can arrange g( f (0)) = 0, because fractional linear automorphisms
of D act transitively. Applying Corollary 9.8 we then conclude that g ◦ f is a multiplication by eiθ ,
and hence f = eiθ g−1 is fractional linear. But H and D are conformally equivalent via a fractional
linear transformation. Hence, any automorphism of H is fractional linear as well, and by Lemma
9.16 it belongs to PS L(2, R).
Exercise 9.20. Prove that a fractional linear transformation f (z) =
az+b
cz+d
leaves the unit disc D
invariant (i.e. f (D) = D) if and only if f has the form
f (z) = eiθ
α−z
,
1 − ᾱz
where α ∈ D and θ ∈ [0, 2π).
Exercise 9.21. Prove that C is not conformally equivalent to H (or D).
Hint: Apply the Liouville theorem for a holomorphic map C → D.
9.6
Summary of useful conformal maps
D → CP1 \ D
1
z 7→ .
z
95
(9.5.1)
H→D
i−z
iz + 1
or z 7→
.
z+i
i+z
z−a
General form : z 7→ eiθ
, a ∈ H, θ ∈ [0, 2π).
z−a
z 7→
This conformal equivalence sends the point a to the center of the disc.
D→H
z 7→ i
1−z
.
1+z
D→D
z 7→ eiθ
α−z
, α ∈ D, θ ∈ [0, 2π).
1 − ᾱz
This automorphism sends the point α to the center of the disc.
Strip to sector
z → ez maps a strip {0 < Im z < α} onto the sector 0 < arg z < α, α ∈ (0, 2π).
In particular, for α = π it maps the strip {0 < Im z < π} onto H.
H→C\R
z → z2 , where R = {Im z = 0, Re z > 0}.
Sector to H
z → zα maps the sector 0 < arg z <
π
1
onto H, α >
α
2
Complement of an interval onto the complement of a disc
1
z 7→ z + ,
z
see Exercise 9.6.
96
9.7
Examples
In this section we consider a few examples of using conformal maps from Section 9.6.
1. Consider the domain U = H \ {|z − i| ≤ a}, 0 < a < 1. Fund r ∈ (0, 1) such that there is a
conformal equivalence U → D \ {|z| ≤ r} and find this equivalence.
It is easier to go from the other end. The map
f (z) = i
1−z
1+z
maps D onto H. Let us fund the image f (Dr = {|z| ≤ r}). We know that this image is a circle. We
¯ The operation z 7→ −z̄ is the reflection with respect to the imaginary axis.
note that f (z̄) = − f (z).
This implies that the circle f (Dr ) is symmetric with respect to the imaginary axes, and hence, its
center is on the imaginary axes. We also note that the map f sends the diameter Dr ∩ {Im z = 0}
onto an interval (mi, Mi) ⊂ ıR. Therefore the diameter d(r) of the circle f (Dr ) is equal M − m. The
points mi, Mi are images of the points ±r ∈ D, i.e.
m=i
1−r
1+r
,M =i
.
1+r
1−r
Now recall that we want to map the disc Dr onto the disc {|z − i| ≤ a}. In particular, we would
like to have the lower point of the image of Dr to coincide with the point 1 − a. To achieve that let
us compose the map f with the linear map z 7→ cz, where we choose the real number c to satisfy
cm = 1 − a, i.e. c =
1−a
m
=
(1−a)(1+r)
.
1−r
Thus the function
(1 − a)(1 + r)i(1 − z)
b
f (z) =
(1 − r)(1 + z)
maps the disc Dr onto the disc with the lower point (1 − a)i, as required. However, the upper point
of the disc b
f (Dr ) is at the point cMi =
i(1−a)(1+r)2
.
(1−r)2
1+r
1−r
q
Denoting b :=
1+a
,
1−a
We need to have cM = 1 + a, i.e.
!2
=
1+a
.
1−a
and solving the above equation with respect to r we get
r=
b−1
.
b+1
97
(9.7.1)
The required conformal equivalence U → D \ {|z| ≤ r} for r given by formula (9.7.1) is the
inverse of the map b
f.
2. Find a conformal equivalence between the domains U := D ∩ {z = x + iy; x, y > 0} and
V := C \ (z ∈ C; |Re z| ≥ 1, Im z = 0}.
Recall that the function
z 7→ z2
maps U onto H.
The function
z 7→
i−z
i+z
maps H onto D.
The function
z 7→
1
z
maps D onto CP1 \ D.
The Jukovsky function
1
1
z 7→
z+
2
z
!
maps C \ D onto C \ {|Re z| ≤ 1, Im z = 0}. One can also observe that the function extends to ∞ and
maps ∞ to ∞. Hence, it defines a biholomorphism CP1 \ D onto CP1 \ {|Re z| ≤ 1, Im z = 0}.
Finally the map
z 7→
1
z
maps CP1 \ {|Re z| ≤ 1, Im z = 0} onto C \ (z ∈ C; |Re z| ≥ 1, Im z = 0} = V.
The required map is the composition of all of the above:
i + z2 i − z2
+
f (z) = 2
i − z2 i + z2
98
!−1
=
1 + z4
.
1 − z4
Chapter 10
Riemann mapping theorem
Theorem 10.1 (B. Riemann–P. Koebe). Any simply connected domain U ⊂ C, U , C is conformally equivalent to D. In other words, any two simply connected domains in C which are different
from C are conformally equivalent.
This theorem is one of the crown achievements of Mathematics of 19th century (except that the
first proper proof was given only in 20th century by Koebe).
Plan of the proof
Assume that U , C.
Step 1: Constructing of an injective map f : U → D. See Proposion 10.9. Note that by assumption
there is at least one point a < U. If we manage to conformally map U onto a domain missing D (a)
this would prove the claim (why?).
Step 2: Proving that the limit of a uniformly on compact set converging sequence of injective
holomorphic maps is injective. See Lemma 10.11.
Step 3: Finding a converging sequence of injective maps fn : U → D with fn (z0 ) = 0 maximizing
the modulus of the derivative | fn0 (z0 )|.
See Proposition 10.10. Note that Step 2 implies that f is injective.
99
Step 4: Prove that f constructed in Proposition 10.10 is surjective. See Proposition 10.12. This
concludes the proof of Theorem 10.1.
Before proving theorem we need to develop some additional tools.
10.1
Functional analytic background
10.1.1
Arzelá-Ascoli theorem
The Arzelá-Ascoli theorem is a general fact, not specific to the context of holomorphic functions.
Let K ⊂ Rn be a compact set, and fn : K → Rm a sequence of continuous maps.
We say that fn uniformly converges to a map f : K → Rm if for any > 0 there exists an integer
N such that | fn (x) − f (x)| < for all n ≥ N and all x ∈ K.
Lemma 10.2. The uniform limit of a sequence of continuous maps is continuous.
Proof. For any points a, x ∈ K we have
| f (x) − f (a)| = |( f (x) − fn (x)) + ( fn (x) − fn (a)) + ( fn (a) − f (a))|
(10.1.1)
≤ | f (x) − fn (x)| + | fn (x) − fn (a)| + | fn (a) − f (a)|.
(10.1.2)
Uniform convergence implies that for any there exists N such that | f (x) − fn (x)|, | fn (a) − f (a)| <
4
for n ≥ N. Here N is independent of a and x. On the other hand, fN is continuous and hence there
exists δ > 0 such that | fN (x) − fN (a)| <
4
provided that |x − a| < δ. Hence, (10.1.1) implies that
| f (x) − f (a)| < if |x − a| < δ, i.e. f is continuous at a.
It is straightforward to prove that a uniform convergence is equivalent to the uniform Cauchy
property: for any there exists n such that | fn (x) − fm (x)| < for all x ∈ K and all m, n ≥ N. Thus,
according to Lemma 10.2 a uniformly Cauchy sequence of continuous functions always converges
to a continuous function.
We say that fn is uniformly bounded on a compact set K if there exists a constant C > 0 such
that | fn (x)| < C for all n and all x ∈ K.
100
We say that fn is equicontinuous if for any > 0 there exists δ > 0 such that | fn (x) − fn (y)| < for all n and any x, y ∈ K such that |x − y| < δ.
Lemma 10.3. Given a domain U ⊂ C, consider a sequence of holomorphic functions fn : U → C.
Suppose the sequence of their derivatives fn0 (z) is uniformly bounded on every compact subset
K ⊂ U. Then fn is equicontinuous on every compact subset.
Proof. Let K ⊂ U be an open set. Assuming U , C consider the distance function from a point in
K to the boundary of U:
dist∂U (z) = inf(|z − w|, w ∈ ∂U).
This function is positive and continuous. Hence it achieves the minimum positive value σ0 > 0.
S
Take a positive σ < σ0 (if U = C take any σ > 0), and denote Uσ (K) =
Dσ (z). The closure
z∈K
Uσ (K) is compact and contained in U. Hence, by assumption there exists a constant C > 0 such
that | fn0 (z)| < C for all n and all z ∈ Uσ . Note that for any z, z0 ∈ K such that |z − z0 | < σ the interval
Iz,z0 = {z + t(z0 − z); t ∈ [0, 1]} connecting the points z and z0 is contained in Uσ (K). Let us write
fn = un + ivn and consider the functions e
un (t) = un (z + t(z0 − z)),e
vn (t) = vn (z + t(z0 − z)). Then, using
the intermediate value theorem we conclude
un (z0 ) − un (z) = e
un (1) − e
un (0) = e
u0n (θ1 ),
(10.1.3)
vn (z0 ) − vn (z) = e
vn (1) − e
vn (0) = e
v0 (θ2 ),
(10.1.4)
for some values θ1 , θ2 ∈ [0, 1]. Note that
d fn (z + t(z0 − z))
= f 0 (z + t(z0 − z))(z0 − z),
dt
and hence,
d fn (z + t(z0 − z))
= | f 0 (z + t(z0 − z))| |z0 − z| ≤ C|z0 − z|
dt
for all t ∈ [0, 1] and all z0 , z ∈ K such that |z0 − z| < σ. On the other hand,
and hence we also have
|e
u0n (t)|, |e
v0n (t)| ≤ C|z0 − z|
101
d fn (z+t(z0 −z))
dt
=e
u0n (t) + ie
v0n (t),
for all t ∈ [0, 1]. Combining with (10.1.3) we get
|un (z0 ) − un (z)| < C|z0 − z|, |vn (z0 ) − vn (z)| < C|z0 − z|
Finally, we have
| fn (z) − fn (z0 )| =
p
√
(un (z0 ) − un (z))2 + (vn (z0 ) − vn (z))2 ≤ C 2|z0 − z|,
which implies the equicontinuity of the sequence fn (z) on K.
Theorem 10.4 (Arzelá-Ascoli). Let K ⊂ Rn be a compact set If the sequence f j : K → Rm is
equicontinuous and uniformly bounded, that there is a subsequence f jk which uniformly converges
to a continuous function f : K → Rm .
Proof. First we choose a sequence of points x j ∈ K which form an everywhere dense set in K, i.e.
for any point z ∈ K there is a subsequnce xnk converging to z. To do that we enclose K (which is by
assumption bounded) into a large cube I := {|xi | < C, i = 1, . . . , n} with the side 2C. Take any point
x1 ∈ K. Divide the cube into 2n smaller cubes with the side C, and pick in each of them a different
from x1 point from K (if the intersection of this sub-cube with K is non-empty). We further divide
each of the sub-cubes into 2n cubes with the side C2 , and choose again a point from K, if it exists in
each of them. Continuing this process we define the required sequence x1 , x2 , . . . .
The sequence fk (x1 ) is bounded. Hence by Bolzano-Weierstrass we can find a converging subsequence fk1 (x1 ). Repeating the argument for the sequence fk1 (x2 ) we find its converging subsequence fk2 (x2 ). Continuing inductively we find subsequences
{ fk1 } ⊃ { fk2 } ⊃ . . .
such that the subsequence fkm converges at points x1 , . . . , xm . Let us prove that the diagonal sequence fkk converges on the whole compact set K. First of all, by construction the diagonal subsequence fkk converges at points x j , j = 1, . . . from the chosen everywhere dense sequence. This
means for any > 0 and any m there exists N = Nm such that | fnn (xm ) − fkk (xm )| < for all n, k ≥ N.
The equi-continuity of fn implies that there exists δ > 0 such that | fnn (x) − fnn (y)| < for any n
102
and any x, y ∈ K such that |x − y| < δ But the set xk is everywhere dense. Hence, there exists a
subsequence xm j → x. In particular there exists J such that for j > J we have |xm j − x| < δ. Thus,
| fnn (x) − fkk (x)| ≤ | fnn (x) − fnn (xm j )| + | fnn (xm j ) − fkk (xm j )| + | fkk (x)) − fkk (xm j )| ≤ 3.
for all sufficiently large n and k. This means that fnn satisfies a uniform Cauchy property and hence
it uniformly converges to a continuous function.
Corollary 10.5. Let U ⊂ Rn be an open domain. If the sequence fn : U → X is equicontinuous
and uniformly bounded on every compact subset K ⊂ U, then there is a subsequence fnk which
converges, uniformly on all compact sets, to a continuous function f : U → X.
Proof. One can exhaust U by a sequence of compact subsets K1 ⊂ K2 ⊂ . . . U,
S
j
K j = U, and
then apply Arzelá-Ascoli to choose a subsequence converging on K1 , from it choose a subsequence
converging on K2 , etc. Finally choose a diagonal subsequence.
10.1.2
Montel’s theorem
Theorem 10.6. Let fn : U → C be a family of holomorphic functions which uniformly on compact
sets converges to a function f : U → C. Then f is holomorphic.
Proof. According to Lemma 10.2 the function f is continuous. Let us show that f is holomorphic.
Take any point u ∈ U and consider a disk D (u) = {|z − u| < }. Then holomorphic forms fn (z)dz
R
are exact in D (u), and hence for any loop γ in D (u) and any k we have fk (z)dz = 0. The uniform
γ
convergence fk → f allows us to pass to the limit under the integral. Thus, we get
Z
f (z)dz = 0
γ
for any loop γ. Therefore, we can apply Theorem 4.9 to conclude that f is holomorphic in D (u),
and hence, everywhere in U.
103
Theorem 10.7 (Montel’s theorem). Suppose that a family F of holomorphic functions U → C is
uniformly bounded on all compact subsets K ⊂ U. Then one can find a subsequence uniformly
converging on all compact sets K ⊂ U to a holomorphic function f : U → C.
Proof. A family of uniformly bounded on compact sets holomorphic functions is uniformly continuous on compact sets. Indeed, the Cauchy inequality guarantees that the family of derivatives
{ f 0 ; f ∈ F } is also uniformly bounded on compact sets. Indeed, take a compact set K ⊂ U. By
assumption, there exists a constant M > 0 such that | f (z)| < M for all f ∈ F and all z ∈ K. Then,
arguing as in the proof of Lemma 10.3 we can find σ > 0 such that for any point u ∈ K we have
Dσ (u) = {z; |z − u| ≤ σ} ⊂ U.. Hence, Cauchy inequality implies that | f 0 (u)| <
M
σ
for all u ∈ K and
all f ∈ F . According to Lemma 10.3 this implies the equicontinuity.
Hence, we can apply Arzelá-Ascoli to extract a sequence fk ∈ F uniformly converging on compact sets to a continuous function f : U → C. But then Theorem ?? implies that f is holomorphic.
Remark 10.8. A family F of holomorphic functions f : U → C is called normal, if every sequence in F contains a subsequence which uniformly converging on all compact subsets of U (but
not necessarily to a function from F ). Hence, Montel’s theorem says that a uniformly bounded on
compact set family is normal.
10.2
Proof of the Riemann mapping theorem
10.2.1
Embedding into D
Proposition 10.9. Let U ⊂ C be a simply connected open subset of C, not equal to C. Then there
exists an injective holomorphic map f : U → D.
Proof. Without loss of generality we can assume 0 < U. Choose a point z0 ∈ U and define the
logarithm branch logU z. We have elog
U
(z)
= z, and in particular, the map logU is injective.
104
We claim that there exists > 0 such that the disc D (logU (z0 )+2πi) is contained in C\logU (U).
Indeed, if this is not true then there exists a sequence logU (zn ) ∈ logU (U) ∩ D 1n (logU (z0 ) + 2πi),
U
n = 1, 2, . . . , and hence lim logU (zn ) = logU (z0 ) + 2πi. Therefore, elog
n→∞
elog
U
(zn )
(zn )
→ e2πi+log
U
(z0 )
= z0 . But
= zn , and therefore lim zn = z0 . But then logU (zn ) → logU (z0 ) , logU (z0 ) + 2πi, which is a
n→∞
n→∞
contradiction.
Hence the holomorphic function f injectively maps U into the complement of the disc D (2πi).
On the other hand, the map g(z) =
z−2πi
conformally maps the complement of D (2πi) in the Rie-
mann sphere onto D. Hence, composing logU and g we get the required injective holomorphicmap
f (z) =
log (z) − 2πi
U
of U into D.
10.2.2
Maximizing the derivative
In Proposition 10.9 we constructed an injective holomorphic map f : U → D such that f (z0 ) = 0.
Let us denote by F the set of all holomorphic maps with this property. Note that by Cauchy
inequality | f 0 (z0 )| is uniformly bounded by some constant C (which depends on the distance of z0
to ∂U) for all f ∈ F . Let Cmax := sup | f 0 (z0 )|. Thus Cmax ≤ C.
f ∈F
Proposition 10.10. There exists f ∈ F such that | f 0 (z0 )| = Cmax .
Proof. Take a sequence { fn } of functions from F with | fn0 (z0 )| → Cmax . The set { fn } is uniformly
bounded on compact subsets (because of Cauchy inequality, see the argument in the proof of
Proposition 10.9), and hence it is normal). Therefore, there is a subsequence uniformly on compact
sets converging to a function f which satisfies | f 0 (0| = Cmax . We also have | f (z)| ≤ 1, z ∈ U, and by
maximum modulus principle we conclude that | f (z)| < 1. In other words, f (U) ⊂ D. Taking into
account that it is non-constant (because its derivative at z0 is , 0 we conclude from Lemma 10.11,
which we prove below, that f is injective, and thus belongs to F .
105
10.2.3
Preservation of injectivity
Lemma 10.11. Let U ⊂ C be a connected domain and fn : U → C a sequence of injective
holomorphic functions. Suppose that fn → f uniformly on compact sets. Then if f is not constant
then it is injective as well.
Proof. Suppose that f is not injective, i.e. there are points z1 , z2 ∈ U, z1 , z2 , such that f (z1 ) =
f (z2 ) = w. Consider the map g(z) := f (z) − w and gn (z) = fn (z) − fn (z1 ). Then z1 are z2 are zeroes of
g. By assumption g is not a constant, and hence the connectivity of U implies that zeroes z1 and z2
are isolated. Take > 0 small enough such that D (z2 ) = {|z − z2 | ≤ } ⊂ U and inside D (z2 ) there
are no other zeroes of g, and in particular z1 < D (z2 ). Hence, according to the argument principle
we have
1
lim
n→∞ 2πi
Z
∂D
g0n (z)
1
dz =
gn (z)
2πi
Z
∂D (z2 )
g0 (z)
dz ≥ 1.
g(z)
Then this implies that for a sufficiently large n the function gn has a zero in the disc D (z2 ) = z1 .
But this contradicts to the fact that gn has a unique zero z1 because fn is injective.
Thus the map f constructed in Proposition 10.10 is injective. In the next section we will show
that the constructed map f is also surjective, i.e. it is the required biholomorphism U → D.
10.2.4
Surjectivity
The following proposition completes the proof of the Riemann mapping theorem 10.1. Recall that
we denoted by F the set of injective holomorphic maps f : U → D such that f (z0 ) = 0.
Proposition 10.12. Let f : U → D be a map from F which satisfies | f 0 (z0 )| = Cmax . Then f (U) =
D.
Proof. Suppose f is not surjective, i.e. there exists a ∈ D such that a , f (z) for any z ∈ U. Of
course, a , 0, because 0 = f (z0 ). We will use the point a to construct a holomorphic map e
f ∈F
with | e
f 0 (z0 )| > | f 0 (z0 )|, which would contradicts our assumption that | f 0 (z0 )| = Cmax . Consider a
106
conformal automorphism ψa : D → D given by the formula
ψa (z) =
a−z
.
1 − āz
It interchanges points 0 and a, i.e. ψa (0) = a and ψa (a) = 0. Hence, the inverse fractional linear
transformation ψ−1
a also interchanges the points 0 and a.
Consider the domain U 0 = ψa ( f (U)). Then 0 < U 0 , but a ∈ U 0 . The domain U 0 is simplyconnected because it is biholomorphic to a simply connected domain U. Consider a logarithm
√
0
branch logU : U 0 → C in U 0 and define the z branch
1
s(z) := e 2 log
U0
(z)
.
It satisfies (s(z))2 = z. Consider the function
e
f (z) = ψ s(a) ◦ s ◦ ψa ◦ f.
We claim that e
f ∈ F and | e
f 0 (z0 )| > Cmax . Indeed, we have
e
f (z0 ) = ψ s(a) (s(ψa ( f (z0 )))) = ψ s(a) (s(ψa (0))) = ψ s(a) (s(a)) = 0.
To verify the injectivity of the function e
f we observe that it is a composition of injective functions
f , ψa , s and ψ √a .
To estimate | e
f 0 (z0 )| consider the diagram
f
z
f
}|
ψa
0 s
ψ s(a)
ψ−1
s(a)
{
z7→z2
ψ−1
a
U → f (U) → U → D → D |→ D {z
→ D→
} D.
|
{z
}
h
e
f
2
−1
e
Taking into account that ψ−1
s(a) ◦ ψ s(a) = Id, (s(z)) = z and ψa ◦ ψa = Id we conclude that f = h ◦ f .
The function h : D → D is defined by the formula
2 −1
h(z) = ψ−1
ψ
(z)
.
a
s(a)
This function satisfies
−1
2
h(0) = ψ−1
a ((ψ s(a) (0)) ) = ψa (a) = 0,
107
but is not injective, because the function z 7→ z2 is not injective. Hence, the Schwarz lemma implies
that |h0 (0)| < 1.
On the other hand, we have f = h ◦ e
f . Using the chain rule we compute f 0 (z0 ) = h0 ( f (z0 )) ·
e
f 0 (z0 ) = h0 (0) · e
f 0 (z0 ) and hence
|e
f 0 (z0 )| =
| f 0 (z0 )|
> | f 0 (z0 )| = Cmax
|h0 (0)|
because |h0 (0)| < 1. This contradiction concludes the proof of Proposition 10.12, and with it the
proof of the Riemann mapping theorem.
10.2.5
Discussion: boundary regularity
Given a simply connected domain U one cannot expect, in general, any control of the boundary
behavior of a conformal map f : D → U provided by the Riemann mapping theorem, because the
boundary of a general domain could be quite terrible. However, it turned out that if boundary is
reasonable then the boundary behavior of the conformal map f is reasonable as well.
Let us recall that a curve Γ ⊂ C is called a C k -submanifold if for any a ∈ Γ there exists > 0
and a C k -diffeomorphism (i.e. a bijective C k -map h with a C k -inverse) of D onto a neighborhood
U 3 0, such that h(Γ) = h(U) ∩ {y = 0}.
Note that any 2 conformal equivalences f, f˜ : D → U differs by an automorphism of D, which
is smooth (and even real analytic) on the boundary. Hence the boundary behavior properties are
the same for f and e
f.
Theorem 10.13. If the boundary ∂U of a simply connected domain U is a C k -submanifold of C
and f : D → U a conformal equivalence. Then f extends to a C k−1 -diffeomorphism f between the
closures: f : D → U. If ∂U is a C 0 -submanifold, then f extends to D as a homeomorphism.
The proof of this theorem (which was first proven in a weaker form by P. Painlevé) goes beyond
this course.
108
10.3
Annuli
10.3.1
Conformal classification of annuli
Conformal classification of not simply connected domains is less boring. As an example, we consider this problem for annuli.
Given r, R > 0, r < R The domain A(r, R) = {r < |z| < R} is called an annulus.
Lemma 10.14. There exists a biholomorphism h : A(r, R) → A(r, R) which switches the boundary
circles, i.e. h({|z| = r}) = {|z| = R}.
Proof. This is done by the map z 7→
rR
.
z
Clearly, any two annuli A(r, R) and A(r0 , R0 ) with
R
r
=
R0
r0
are conformally equivalent. Indeed,
the required conformal equivalence A(r, R) → A(r0 , R0 ) is the linear map z 7→
R0
z.
R
It turns that this
sufficient condition together with the one arising from Lemma 10.14 is also necessary.
One can also allow in the definition of an annulus to allow r to be 0 and/or R = ∞. Thus,
A(0, 1) = D \ 0, A(0, ∞) = C \ 0, A(1, ∞) = C \ D.
Note that the map z 7→
1
z
establishes a conformal equivalence of A(0, 1) and A(1, ∞). However,
Lemma 10.15. A(0, 1) and A(0, ∞) are not conformally equivalent.
Proof. Indeed, suppose f : A(0, 1) → A(0, ∞) is a conformal equivalence. Then either lim f (z) = 0,
z→0
or lim f (z) = ∞ (why?). In the former case the removal of singularities theorem allows us to extend
z→0
f to 0 and hence we get a conformal equivalence D → C, which is impossible. In the latter case
we first compose f with the automorphism C \ 0 → C \ 0 given by the function z 7→
1
z
and then
repeat the previous argument.
Theorem 10.16. Suppose that r , 0 and R , ∞. Then two annuli A(r, R) and A(r0 , R0 ) are conformally equivalent if and only if
R
R0
ln = ln 0 .
r
r
109
The quantity ln Rr is called the conformal modulus of the annulus A(r, R) and will be denoted
by m (A(r, R)).
Before proving this theorem we first we need to discuss Laurent series.
10.3.2
Laurent series
When studying meromorphic functions we already encountered series containing negative powers.
For instance,
∞
zn
ez X
, z , 0.
=
z2 n=−2 (n + 2)!
Sometimes we have to deal with series containing negative powers all the way up to −∞. For
instance, we have
1
z
e =
∞
X
z−n
n!
0
A series of the form
S (z) =
, z , 0.
∞
X
ak zk
−∞
is called Laurent series. The series is a sum of two series S − (z) =
∞
P
1
a−k z−k and S + (z) =
∞
P
ak zk ,
0
and convergence of S (z) means convergence of both S ± (z). The series S + (z) is the standard power
series and it converges in its disc of convergence DR = {|z| < R}, while S − (z) is a power series in the
variable u =
1
z
and it converges in the disc |u| <
1
r
around ∞, or equivalently in the complement of
the disc Dr = {|z| ≤ r}. Thus if r > R then the Laurent series S (z) does not converges anywhere, and
if r < R it (absolutely) converges in the annulus A(r, R) = {r < |z| < R} and defines a holomorphic
function S : A(r, R) → C. It turns out that
Proposition 10.17. Any holomorphic function S : A(r, R) → C can be presented as the sum of an
∞
P
absolutely converging in A(r, R) Laurent series, S (z) = ak zk .
−∞
Proof. This follows from the Cauchy formula. Let us take a slightly smaller annulus A(r0 , R0 ) ⊂
A(r, R). Then for any z ∈ A(r0 , R0 ) we have
Z
Z
f (ζ)dζ
=
f (z) =
ζ−z
∂A(r0 ,R0 )
∂DR0
110
f (ζ)dζ
−
ζ−z
Z
∂Dr0
f (ζ)dζ
.
ζ−z
Changing the variable ζ =
1
u
and z =
Z
∂Dr0
1
v
in the second integral we get
v f ( 1u )du
.
u(u − v)
Z
f (ζ)dζ
=−
ζ−z
∂{|u|< r10 }
Therefore,
f (z) =
Z
∂DR0
=
Z
∂DR0
f (ζ)dζ
−
ζ−z
Z
∂Dr0
f (ζ)dζ
+
ζ−z
f (ζ)dζ
ζ−z
v f ( 1u )du
.
u(u − v)
Z
∂{|u|< r10 }
Arguing as in Theorem 5.7 we can expand the first integral in a power series in z converging for
|z| < R and expand the second integral in a power series in v converging for |v| < 1r . Changing back
v 7→ z =
1
v
we get the required Laurent expansion in z variable.
We will also need the following formula (due to Green)
Lemma 10.18. Let f (z) =
∞
P
an zn for z ∈ A(r, R). Suppose that the map f is injective. Denote by
−∞
A(ρ), ρ ∈ (r, R) the area of the domain in C bounded by the curve f ({|z| = ρ}). Suppose that f
sends the circle {|z| = ρ} oriented as the boundary of the disc {|z| ≤ ρ} to f ({|z| = ρ}) oriented as
the boundary of C.
Then
A(ρ) = π
∞
X
n|an |2 ρ2n .
−∞
Proof. Using Proposition 4.3 we have
i
A(ρ) = −
2
Z
i
f (z) f 0 (z)dz = −
2
|z|=ρ

 ∞
Z X
∞
X






 an z̄n   mam zm−1 
|z|=ρ
−∞
−∞
Z
Z2π
∞
∞
∞
X
i X
1 X
m−1 n
=−
mam an
z z̄ dz =
mam an ei(m−n)φ ρm+n dφ = π
n|an |2 ρ2n ,
2 m,n=−∞
2 m,n=−∞
n=−∞
0
|z|=ρ
because
R2π
eikφ dφ = 0 unless k = 0.
0
111
10.3.3
Proof of Theorem 10.16
Theorem 10.16 follows from the following stronger result.
Proposition 10.19. Suppose there exists an injective holomorphic map
f : A(r0 , R0 ) → A(r, R)
such that for ρ ∈ (r0 , R0 ) the circle S ρ := {|z| = ρ} bounds a domain in C which contains the disc
Dr (0) = {|z| < r}. Then
m(A(r0 , R0 )) ≤ m(A(r, R))
Proof. Without loss of generality we can assume that r0 = r = 1. We will view the annuli A(1, R)
and A(1, R0 ) as subdomains of the discs DR = {|z| < R} and DR0 = {|z| < R0 }. Let Vρ denote
the closed subdomain of DR bounded by f (S ρ ) for ρ ∈ (1, R0 ). By assumption Vρ ⊃ Dr (0). We
can assume that Vρ ⊂ Veρ if ρ < e
ρ. Otherwise, we can use Lemma 10.14 to switch the boundary
components of the annulus A(1, R0 ). This ensures that the the map f |S ρ : S ρ → f (S ρ ) preserves
orientations of S ρ and f (S ρ ) as boundaries of Dρ and Vρ . Hence we can apply formula 10.18 to
compute the area A(ρ) := Area(Vρ ):
A(ρ) = π
∞
X
n|an |2 ρ2n .
−∞
Passing to the limits when ρ → 1 and ρ → R0 and taking into account that
D1 ⊂ Vρ ⊂ DR
for any ρ ∈ (1, R0 ) we get
π≤π
∞
X
n|an |2 ≤ π
−∞
∞
X
n|an |2 (R0 )2n ≤ πR2 .
−∞
Let us multiply the first inequality by R2 :
πR2 ≤ π
∞
X
n|an |2 R2 .
−∞
112
We also have
π
∞
X
n|an |2 R2 ≤ π
∞
X
−∞
2
n|an |2 R2n .
−∞
2n
Indeed, for R ≥ 1 we have nR ≤ nR for any integer n regardless its sign. Hence,
π
∞
X
n|an | (R ) ≤ πR ≤ π
2
0 2n
2
∞
X
−∞
n|an |2 R2n .
−∞
But the function A(ρ) is strictly increasing. Hence the inequality
π
∞
X
n|an |2 (R0 )2n ≤ π
−∞
∞
X
n|an |2 R2n
−∞
0
implies R ≤ R, and hence
m(A(r0 , R0 )) = ln R0 ≤ ln R ≤ m(A(r, R)).
Thus, there is a unique annulus of each finite conformal modulus and exactly two annuli of
infinite modulus.
10.4
Dirichlet problem
One of important applications of conformal mappings is for the solution of the following
Dirichlet problem for harmonic functions. Given a domain U and a continuous function φ :
∂U → R find a harmonic function u : U → R which continuously extends to ∂U and uU = φ.
One can make sense of Dirichlet problem even for discontinuous but integrable functions,
where one requires boundary convergence at the points of continuity.
Note that the maximum principle for harmonic functions guarantees that the Dirichlet problem
for a bounded domain U has a unique solution. Indeed, if ∆u = ∆e
u = 0 and u|∂U = e
u|∂U then u − e
u
is harmonic and (e
u − u)|∂U = 0. Hence, Corollary 8.8 implies that e
u = u.
Thanks to the Riemann mapping theorem, solving Dirichlet problem for simply connected
domains can be reduced to solving it for D and understanding the behavior of the conformal equivalence h : U → D. Indeed, suppose a conformal equivalence h : D → U continuously extends
113
to ∂U (comp. Theorem 10.13), then if g : D → R is a solution of the Dirichlet problem for D
for the boundary value ψ ◦ g : ∂D → R, then, according to Lemma 8.5 the function h = g ◦ h is
harmonic and solves the Dirichlet problem for U with the boundary data ψ. Hence, it is important
to solve the Dirichlet problem for the disc D. This is done below via an explicit formula proven by
Schwarz, but first written by Poisson.
10.4.1
Poisson integral and Schwarz formula
Proposition 10.20 (Schwarz’s formula). Let u : D → R be a harmonic function on a closure D of
the unit disc in C. Then for any a ∈ D we have


Z


ζ+a
dζ 
 1
u(a) = Re 
u(ζ)  .
ζ−a
ζ 
 2πi
(10.4.1)
|ζ|=1
Remark 10.21. Note that Schwarz’s formula (10.4.1) gives an explicit expression for the holomorphic function f (z) whose real part is the harmonic function u(z). Indeed the function
Z
ζ+z
dζ
1
u(z)
f (z) =
2πi
ζ−z
ζ
|ζ|=1
is holomorphic and according to Schwarz’s formula, u(z) = Re f (z).
Proof of Proposition 10.20. Consider a fractional linear change of coordinates
ζ = R(v) =
v+a
, a ∈ D.
1 + av
Note that R is an automorphism of D and R(0) = a. The function u(R(v)) is harmonic in D and
hence the mean value theorem (see Theorem 8.8) for this function asserts:
1
u(a) = u(R(0)) =
2π
Z2π
u(R(eiθ ))dθ.
0
The integral in the right-hand side can be rewritten as the complex contour integral
Z
1
dv
u(a) =
u(R(v)) .
2πi
v
|v|=1
114
Let us change back the variable v to ζ,
v = R−1 (ζ) =
ζ−a
.
1 − aζ
We have
!
!
dv
1
a
ζ
aζ
dζ
=
+
dζ =
+
.
v
ζ − a 1 − aζ
ζ − a 1 − aζ ζ
Therefore,
1
u(a) =
2πi
Z
Z
dv
1
u(R(v)) =
v
2πi
|v|=1
!
ζ
dζ
aζ
u(ζ)
+
ζ − a 1 − aζ ζ
|ζ|=1
But for |ζ| = 1 we can write 1 = ζζ, and therefore
ζ
aζ
ζ
aζ
ζ
a
+
=
+
=
+
ζ − a 1 − aζ ζ − a ζζ − aζ ζ − a ζ − a
1 − |a2 |
=
.
|ζ − a|2
Hence, the expression
A :=
a
ζ
+
ζ−a ζ−a
is real, and thus


1  ζ

1
a
ζ
a

A=
A + A = 
+
+
+
2
2 ζ−a ζ−a ζ−a ζ−a


1  ζ + a ζ + a 
ζ+a
 = Re
= 
+
.
2 ζ−a ζ−a
ζ−a
Combining all the formulas, we get
!
!
Z
Z
1
ζ
aζ
dζ
1
ζ+a
dζ
u(a) =
u(ζ)
+
=
Re
u(ζ) .
2πi
ζ − a 1 − aζ ζ
2πi
ζ−a
ζ
|ζ|=1
We note that
1 dζ
i ζ
|ζ|=1
= dθ is real valued, and hence
1
2πi
Z
|ζ|=1


Z


ζ+a
ζ+a
dζ
dζ 
 1
Re
u(ζ) = Re 
u(ζ)  ,
ζ−a
ζ
ζ−a
ζ 
 2πi
!
|ζ|=1
and we get the required formula (10.4.1).
It is useful to rewrite formula (10.4.1) in some different forms.
115
Proposition 10.22. Let u : D → R be a harmonic function on a closure D of the unit disc in C.
Then for any a ∈ D we have
1
u(a) =
2πi
!
ζ+a
dζ
Re
u(ζ)
ζ−a
ζ
Z
|ζ|=1
1
=
2π
=
1
2π
Z2π
0
Z2π
!
eiθ + a
u(eiθ )dθ.
Re iθ
e −a
(10.4.2)
1 − |a|2
u(eiθ )dθ
|eiθ − a|2
0
Equivalently, for a = reiφ we have
1
u(re ) =
2π
Z2π
iφ
0
1 − r2
u(eiθ )dθ.
1 − 2r cos(θ − φ) + r2
(10.4.3)
The proof is by a straightforward computation using formula (10.4.1). The latter integral called
Poisson integral.
Applying formula (10.4.2) to u = 1 we get
Corollary 10.23.
Z2π
1 − |a|2
dθ = 2π
|1 − aeiθ |2
0
for any a ∈ D.
10.4.2
Solution of the Dirichlet problem for the unit disc
We will now use Schwarz formula and Poisson integral for solving Dirichlet problem for the unit
disc. Let ψ : ∂D → R be a piece-wise continuous integrable function. Denote
1
Pψ (z) :=
2πi
Z
∂D
1 − |z|2
dζ
1
ψ(ζ) =
2
|ζ − z|
ζ
2π
Z2π
0
.
116
1 − |z|2 iθ ψ e dθ
|eiθ − z|2
(10.4.4)
Theorem 10.24 (H.A. Schwarz). For any piece-wise continuous integrable function ψ : ∂D → R
we have
a) Pψ (z) is a harmonic function in C \ ∂D; moreover, if the function ψ is equal to 0 in a neighborhood of a point eiθ ∈ ∂D then the function Pψ (z) is harmonic in a neighborhood of this
point.
b) If ψ is continuous at a point eiθ ∈ ∂D then
lim Pψ (z) = ψ(eiθ ).
z→eiθ ,z∈D
In particular if ψ is continuous on ∂D then Pψ (z) extends to a continuous function on D
which is equal to ψ on ∂D.
Thus Pψ (z) solves the Dirichlet problem for the boundary data ψ.
Before proving Theorem 10.24 let us list some elementary properties of the integral Pψ
Lemma 10.25.
(1) The operator ψ 7→ Pψ is linear, i.e. given two piecewise continuous functions
ψ1 , ψ2 : ∂D → R and complex numbers a1 , a2 ∈ C we have
Pc1 ψ1 +c2 ψ2 = c1 Pψ1 + c2 Pψ2 ;
(2) if ψ ≥ 0 then Pψ ≥ 0;
(3) if ψ = c is a constant them Pψ = ψ = c,
(4) if c < ψ < C for constants c and C then c < Pψ < C.
Proof. (1) is straightforward, (2) follows from the fact that
1−|z|2
|eiθ −z|2
10.23, and (4) is a corollary of (2) and (3).
Proof of Theorem 10.24.
a) First note that the function
1
f (z) =
2πi
> 0, (3) follows from Corollary
Z
∂D
ζ+z
dζ
u(ζ) .
ζ−z
ζ
117
is holomorphic for all z ∈ C \ ∂D. Indeed, the condition z < D implies that ζ − z , 0 for all ζ ∈ ∂D.
Henece, the function under the integral is holomorphic in z, and hence, the claim follows from the
fact that the integral can be differentiated with respect to the parameter z under the integral sign.
Moreover, if the function ψ is equal to 0 in a neighborhood of a point eiθ ∈ ∂D then the function
f (z) is holomorphic in a neighborhood of this point. Indeed, suppose ψ(ζ) = 0 in a neighborhood
U ⊂ ∂D. Denote A := ∂D \ A. Then
1
f (z) =
2πi
Z
∂D
ζ+z
dζ
1
u(ζ) =
ζ−z
ζ
2πi
Z
ζ+z
dζ
u(ζ) .
ζ−z
ζ
A
But for z ∈ U we have ζ − z , 0 for ζ ∈ A, and hence the function under the integral is holomorphic at the point z, and the same argument about differentiation of the integral with respect to the
parameter z applies.
Next, we observe that
!
1 − |z|2
ζ+z
= Re
,
|ζ − z|2
ζ−z
and hence Pψ (z) is the real part of the function f (z), and hence, Pψ (z) is harmonic everywhere,
where f (z) is holomorphic.
b) Suppose that the function ψ is continuous at a point eiθ0 ∈ ∂D. Without loss of generality we
can assume that ψ(eiθ0 ) = 0. Indeed, otherwise we can replace ψ by ψ − ψ(eiθ0 ), thanks to Lemma
10.25(3) which implies that the theorem holds for constant functions ψ. Given any > 0 choose
δ > 0 so small that |ψ(eiθ )| < if |θ − θ0 | < δ. Choose two complementary arcs A1 := {|θ − θ0 | < δ}
and A2 := ∂D \ A1 .
Let us decompose the function ψ as ψ1 +ψ2 where ψ1 |A2 = 0 and ψ2 |A1 = 0. Then Pψ = Pψ1 +Pψ2 .
The functions Pψ1 and Pψ2 have the following properties:
• (i) Pψ1 is harmonic and hence continuous at interior points of A2 and Pψ2 is continuous on
A1 . This follows from part a).
• (ii |Pψ1 (z)| < for z ∈ D. This is a corollary of Lemma 10.25(4).
118
• (iii) Pψ2 |A1 = 0. Indeed, for z ∈ A1 we have
Z
1
Pψ2 (z) =
2π
θ<(θ0 −δ, θ0 +δ)
1 − |z|2
ψ2 (eiθ )dθ.
|eiθ − z|2
But |z| = 1 on ∂D, and eiθ − z , 0 when θ < (θ0 − δ, θ0 + δ), and hence the integrand is equal
to 0.
Properties (i) and (iii) imply that there exists σ > 0 such that if |z − eiθ0 | < σ and z ∈ D then
|Pψ2 (z) − Pψ2 (eiθ0 | = |Pψ2 (z)| < . Thus, for |z − eiθ0 | < σ and z ∈ D we have
|Pψ (z) − ψ(eiθ )| = |Pψ (z)| ≤ |Pψ1 (z)| + |Pψ2 (z)| < 2.
Hence,
lim
Pψ (z) = ψ(eiθ ).
iθ
z→e ,z∈D
10.4.3
Solving the Dirichlet problem for other domains
As it was already mentioned at the beginning of Section 10.4 the Riemann mapping theorem allows
us to transform solutions of the Dirichlet problem from the unit disc to other simply connected
domains.
Indeed, consider a simply connected domain U with a piecewise smooth (non-empty!) boundary. The domain U need not be even compact. Let ψ : ∂U → R be a piecewise continuous integrable function. Choose a conformal equivalence f : U → D. According to Theorem 10.13
e :=
it extends continuously to the smooth part of the boundary ∂U and consider the function ψ
e is integrable on ∂D (as we will see below it is not necessarily the case),
ψ ◦ f −1 : ∂D → R. If ψ
then we can consider the solution u := Pψe of the Dirichlet problem for D with the boundary value
e. In other words, u is harmonic in D, it extends, continuously to the points of ∂U, where ψ
e is
ψ
e. Then b
continuous, and coincides there with ψ
u := u ◦ f is the solution of the Dirichlet problem for
U with the boundary data ψ.
As an example, let us consider the Dirichlet problem for a strip
Ω := {z ∈ C; 0 < Im z < π}.
119
Problem 10.26. Find a harmonic function u : Ω → R which satisfies the boundary conditions
u(x + iπ) = f1 (x), u(x) = f0 (x), x ∈ R,
where f0 , f1 are bounded continuous functions.
Solution. The exponential function exp(z) = ez maps Ω conformally onto H, and composing it
with a conformal equivalence g : H → D, g(z) =
i−z
,
i+z
v = h(z) =
we get the required biholomorphism
i − ez
i + ez
of Ω onto D. We have
z = f −1 (v) = log
i − iv .
1+v
Transporting the function u : ∂Ω → R to ∂D, we get a function
i − iv e
u(v) = u log
.
1+v
Let us apply Schwarz extension formula (10.4.1) to extend e
u to a harmonic function on D:


Z


ζ+v
dζ 
 1
e
e
u(v) = Re 
u(ζ)  .
ζ−v
ζ 
 2πi
{ζ=1}
Now we transport the solution back to Ω, i.e. we change variable
i − ez
, z ∈ Ω,
v = h(z) =
i + ez
and also change the variable of integration:
ζ=
i − eη
, η ∈ ∂Ω.
i + eη
We have
ζ + v (i − eη )(i + ez ) + (i − ez )(i + eη )
=
ζ − v (i − eη )(i + ez ) − (i − ez )(i + eη )
1 + eη+z
=i z
e − eη
!
dζ
eη
eη
η
η
= d log ζ = d log(i − e ) − d log(i + e ) = −
+
dη
ζ
i − eη i + eη
2ieη
= 2η
dη.
e +1
120
Thus,


Z


ζ
+
v
dζ
1


e
u(ζ) 
u(z) := e
u(h(z)) = Re 
 2πi
ζ−v
ζ 
∂D



 Z
η
η+z
e
1+e

 i
u(η) 2η
dη
= Re 
z
η
π
e −e
e +1 
∂Ω

 ∞
Z∞

Z
t
t+iπ
t+iπ+z
t+z
e
e
1+e
1  1 + e


f0 (t) 2t
dt −
f1 (t) 2(t+iπ)
dt
= −Im 
z
t
z
t+iπ

π
e −e
e +1
e −e
e
+1 
−∞
−∞

 ∞
Z∞

Z
t+z
t
t
t+z
1−e
e
e
1  1 + e

dt +
dt
f0 (t) 2t
f1 (t) 2t
= −Im 
z
t
z
t
π
e −e
e +1
e +e
e +1 
−∞
−∞
Setting z = x + iy we get
1 + et+z 1 + et+x+iy (1 + et+x+iy )(e x−iy − et )
= x+iy
=
ez − et
e
− et
|e x+iy − et |2
x−iy
t
t+2x
2t+x+iy
e
−e +e
−e
;
=
2x
2t
x+t
e + e − 2e cos y
1 − et+z 1 − et+x+iy (1 − et+x+iy )(e x−iy + et )
= x+iy
=
ez + et
e
+ et
|e x+iy + et |2
x−iy
t
t+2x
2t+x+iy
e
+e −e
−e
.
=
e2x + e2t + 2e x+t cos y
Hence,
!
1 + et+z
sin y(e x + e2t+x )
Im z
= − 2x
e − et
e + e2t − 2e x+t cos y
!
sin y(e x + e2t+x )
1 − et+z
Im z
=
−
.
e + et
e2x + e2t + 2e x+t cos y
and
 ∞

Z∞
Z

x
2t+x
x
2t+x
t
t
e
e
1 
sin y(e + e )
sin y(e + e )

u(x + iy) = − 
f0 (t) 2t
dt +
f1 (t) 2t
dt .
2x
2t
x+t
2x
2t
x+t
π  e + e − 2e cos y
e +1
e + e + 2e cos y
e +1 
−∞
−∞
121
Furthermore, we have
(e x + e2t+x )et
e x+t (e x−t + et−x − 2 cos y)(e2t + 1)
e−t + et
1
= x−t
=
;
t−x
t
−t
(e + e − 2 cos y)(e + e ) 2(cosh(x − t) − cos y)
(e x + e2t+x )et
1
= x+t x−t
.
e (e + et−x + 2 cos y)(e2t + 1) 2(cosh(x − t) + cos y)
Hence,
 ∞

Z∞
Z

sin y f1 (t)dt
1 
sin y f0 (t)dt


+
u(x + iy) =
.
2π  cosh(x + t) − cos y
cosh(x + t) + cos y 
−∞
−∞
Changing the variable of integration to τ = x + t we get
 ∞

Z∞
Z

f1 (τ − x)dt 
f0 (τ − x))dt
sin y 
 .
+
u(x + iy) =

2π  cosh τ − cos y
cosh τ + cos y 
−∞
−∞
122
Chapter 11
Riemann surfaces
11.1
Definitions
A Riemann surface is a 1-dimensional complex manifold. For those, familiar with a notion of a
smooth 2-dimensional real manifold the difference is that instead of pair of local coordinates in
a coordinate chart, one has 1 complex coordinate and transition maps between two overlapping
coordinate chart is required to be holomorphic.
More precisely, a set S is called a Riemann surface, or a 1-dimensional complex manifold if
there exist subsets Uλ ⊂ S , λ ∈ Λ, where Λ is a finite or countable set of indices, and for every
λ ∈ Λ a map Φλ : Uλ → C such that
RS1. S =
S
λ∈Λ
Uλ .
RS2. The image Gλ = Φλ (Uλ ) is an open set in C.
RS3. The map Φλ viewed as a map Uλ → Gλ is one-to-one.
RS4. For any two sets Uλ , Uµ , λ, µ ∈ Λ the images Φλ (Uλ ∩ Uµ ), Ψµ (Uλ ∩ Uµ ) ⊂ C are open and
the map
n
hλµ := Φµ ◦ Φ−1
λ : Φλ (U λ ∩ U µ ) → Φµ (U λ ∩ U µ ) ⊂ R
123
is a biholomorphism.
Sets Uλ are called coordinate neighborhoods and maps Φλ : Uλ → C are called coordinate
maps. The pairs (Uλ , Φλ ) are also called local coordinate charts. The maps hλµ are called transiton
maps between different coordinate charts. The inverse maps Ψλ = Φ−1
λ : G λ → U λ are called
(local) parameterization maps. An atlas is a collection A = {Uλ , Φλ }λ∈Λ of all coordinate charts.
One says that two atlases A = {Uλ , Φλ }λ∈Λ and A0 = {Uγ0 , Φ0γ }γ∈Γ on the same Riemann surface
S are equivalent, or that they define the same conformal structure on S if their union A ∪ A0 =
{(Uλ , Φλ ), (Uγ0 , Φ0γ )}λ∈Λ,γ∈Γ is again an atlas on X. In other words, two atlases define the same smooth
structure if transition maps from local coordinates in one of the atlases to the local coordinates in
the other one are given by smooth functions.
A subset G ⊂ S is called open if for every λ ∈ Λ the image Φλ (G ∩ Uλ ) ⊂ Rn is open. In
particular, coordinate charts Uλ themselves are open, and we can equivalently say that a set G is
open if its intersection with every coordinate chart is open. By a neighborhood of a point a ∈ S we
will mean any open subset U ⊂ S such that a ∈ U.
Usually (but not always) it is required that a Riemann surface S admits a countable atlas and
satisfy the following additional axiom, called Hausdorff property:
RS5. Any two distinct points x, y ∈ S have non-intersecting neighborhoods U 3 x, G 3 y.
Given two smooth Riemann surfaces S and Se a map f : S → Se is called holomorphic if if
e λ ) in Se, such that
eλ , Φ
for every point a ∈ S there exist local coordinate charts (Uλ , Φλ ) in S and (U
eλ and the composition map
a ∈ Uλ , f (Uλ ) ⊂ U
Ψλ
f
eλ
Φ
eλ → Rn
Gλ = Φλ (Uλ ) → Uλ → U
is holomorphic. In other words, a map is holomorphic, if it is holomorphic when expressed in local
coordinates.
A map f : S → S 0 is called a biholomorphism if it is holomorphic, invertible, and the inverse
is also a holomorphic.
124
Example 11.1. 1. Any open subset of C is a Riemann surface.
2. The Riemann sphere CP1 is a Riemann surface. It can be covered by two coordinate charts with
coordinates z, ζ ∈ C which are on the overlap C \ 0 are related by z = 1ζ .
11.2
Uniformization theorem (or strong Riemann mapping theorem)
It turns out that the Riemann mapping theorem holds in a stronger form, called uniformization
theorem.
Theorem 11.2 (Uniformization theorem). Any simply connected Riemann surface is conformally
equivalent to either to CP1 , or C, or H ( D).
The proof of this theorem goes beyond this course. The first rigorous proofs were given by
H. Poincaré and P. Koebe in 1907. Since that time many proofs based on different ideas were
found.
The significance of the uniformization theorem will become clear in our discussion below. In
some sense it will allow us to classify all Riemann surfaces (see Theorem 11.11).
11.3
Quotient construction
An important class of examples is given by the following construction. Let S be a Riemann surface
and Aut(S ) its group of biholomorphisms S → S . A discrete group of biholomorphisms of S is a
finite or countable subgroup G ⊂ Aut(S ) such that
• The trajectory (also called an orbit) Gx = {g(x); g ∈ G} of any point x ∈ S is a discrete set.
We also spell out the definition of a subgroup:
• For any f, g ∈ G, f ◦ g is in G;
125
• Id ∈ G and for any g ∈ G we have g−1 ∈ G;
We say that G acts freely on S if each element g ∈ G, g , Id, acts on S without fixed points,
i.e. g(x) , x for any x ∈ S , g ∈ G, g , Id.
Example 11.3. a). Consider the group G = Z2 = Z ⊕ Z of pairs of integer numbers. Consider the
action of G on C2 by translations:
z = x + iy 7→ g(z) = (x + m) + i(y + n), for g = (m, n) ∈ G.
Then G is a discrete group acting freely on C.
b) Let G=Z/p, the finite group of p elements acting on C \ 0 by the formula
2πi
z 7→ e p z.
Then this action is discrete and free. The same action on C is discrete, but not free (why?).
Given a discrete group of transformations acting freely on a Riemann surface S we can define
a new surface S /G, called the quotient of S by G whose points are orbits Gx, x ∈ S , of points of
S . There is an obvious projection π : S → S /G (which we will call tautological) which sends a
point x ∈ S to its orbit Gx. Any point x ∈ S has a coordinate neighborhood U x , such that for any
g ∈ G, g , Id, we have g(x) < U x . That means that the projection π|U x is injective and we call π(U x )
the coordinate neighborhood of the point X = π(x) ∈ S /X.
Example 11.4. 1. Consider the group Z acting on C by the formula z =7→ z + 2kπi for k ∈ Z. Then
the points of the quotient C/Z are complex numbers up to addition of a multiple of 2πi.
Lemma 11.5. The quotient C/Z is conformally equivalent to C \ 0.
Indeed, the exponential map exp : z 7→ ez is the required biholomorphism C/Z → C \ 0 . The
inverse map Log : C \ 0 → C/Z is well defined (unlike the map to C which is multiple valued).
2. Consider the same action restricted to to the half-plane U = {Re z > 0}. Then U/Z is conformally
equivalent to the semi-infinite annulus A(1, ∞) = {1 < |z| < ∞}. The holomorphic function exp |U
induces a biholomorphism U/Z → A(1, ∞).
126
3. Consider another action of Z, this time on C \ 0:
z 7→ 2k z, k ∈ Z
Then the quotient (C \ 0)/Z is a 2-torus T .
4. Choose two complex numbers ω1 , ω2 ∈ C \ 0 such that
z1
z2
< R. Consider an action of Z2 = Z ⊕ Z
on C given by the formula
z 7→ kω1 + `ω2 for (k, `) ∈ Z ⊕ Z.
The quotient T (ω1 , ω2 ) = C/Z ⊕ Z is again a torus. We will see that its conformal structure of
T (ω1 , ω2 ) depends on the choice of w1 , w2 . These quotients T (ω1 , ω2 ) are also called elliptic curves.
11.4
Covering maps
Given two Riemann surfaces U, V a holomorphic map p : U → V is called a covering map if
for any point x ∈ V there exists a neighborhood V x 3 x such that the preimage p−1 (V x ) can be
presented as a union of finitely or countably many disjoint open sets:
p−1 (V x ) = U1 ∪ U2 ∪ . . . ,
such that p|U j : U j → V x is a biholomorphism for each j = 1, 2, . . . . If p : U → V is a covering
map than U is called a cover of V 1
Exercise 11.6. Prove that if V is connected (and we will always assume that), then each point
x ∈ V has the same number (which can be infinite) of pre-mages.
This number is called the order of the covering. Sometimes if p : U → V is a covering of order
k, then one calls it a k-sheeted covering. A covering of order 1 is a biholomorphism.
1
Be aware that the word cover or covering is used in Mathematics also in a different sense, as a covering of a space
by overlapping open sets.
127
Example 11.7. 1.The map exp : C → C \ 0, exp(z) = ez , is a covering map (of infinite order).
Indeed, we have
C \ 0 = V+ ∪ V− , V+ := C \ {z; Im z = 0, Re z ≤ 0}, V− := C \ {z; Im z = 0, Re z ≥ 0}.
We have
exp (V+ ) =
−1
exp−1 (V− ) =
∞
[
k=−∞
∞
[
U +j , U +j := {z = x + iy; y ∈ ((2k − 1)π, 2kπ)};
U +j , U +j := {z = x + iy; y ∈ (2kπ, (2k + 1)π)}.
k=−∞
The restrictions exp |Uk± : Uk± → V± are biholomorphisms. The inverse maps are given by branches
of the logarithm.
2. Similarly, the maps πk : C \ 0 → C \ 0 given by the formula
πk (z) = zk , k = ±1, ±2, . . .
are covering maps. In this case every point of C \ 0 has exactly k-pre-images, i.e. πk : C \ 0 → C \ 0
is a k-sheeted covering.
11.5
Quotient construction and covering maps
The previous example can be generalized as follows. Let S be a Riemann surface and G ⊂ Aut(S )
be a discrete group of its automorphisms. Consider the tautological projection πG : S → S /G.
Lemma 11.8. The tautological projection πG : S → S /G is a covering map.
Proof. Take any x ∈ S and consider its orbit Gx = {g(x); g ∈ G. Then by definition of a discrete
group and its free action there exists a neighborhood U x 3 x such that g(U x ) ∩ e
g(U x ) = ∅ for
e := G(U x ). U
e is by
g ,e
g. The tautological projection πG : X → X/G sends U x onto its trajectory U
ex ) is the union
definition an open set containing the point Gx ∈ S /G and the preimage πG−1 (U
ex ) =
πG−1 (U
[
g∈G
128
g(U x )
of disjoint open sets g(U x ), and the restriction of the projection πG to each of these sets, πG |g(U x ) :
e is a biholomorphism. Hence, πG : S → S /G is a covering map.
g(U x ) → U
Given a covering p : U → V, a biholomorphism f : U → U is called a deck transformation
if p ◦ f = p, i.e. if f preserves fibers, i.e. it maps p−1 (z), z ∈ V, onto itself for any z ∈ V. Deck
transformations form a discrete subgroup of the group of biholomorphisms Aut(U) of U, which is
called the group of deck transformations.
Note that if πG : S → S /G for a discrete subgroup G ⊂ Aut(S ) then G is the group of
deck transformation of this covering. Covering maps p : U → V of this type can be equivalently
characterized by the property that the deck transformation group acts transitively on fibers, which
means that for any v ∈ V and any two points u1 , u2 ∈ p−1 (v) there exists a deck transformation
g ∈ Aut(U) such that g(u1 ) = u2 . Covering maps which satisfy this condition are called regular, or
sometimes also called Galois cover, and the group G of deck transformations of a Galois cover is
called the Galois group.
Lemma 11.9. Let p : U → V be a regular cover, and G its Galois group. Then V is biholomorphically equivalent to U/G.
Proof. Let πG : U → U/G be the tautological projection. We claim that there exists a holomorphic
map p : U/G → V such that p = p ◦ πG . Indeed, by definition, an orbit Gy of a point y ∈ U
is contained in the fiber p−1 (p(y)) over the projection p(y) of the point y. Hence, we can define
p(Gy) = p(y). Moreover, by regularity of the cover Gy = p−1 (p(y)), and hence the map p is 1-1.
The fact that it is holomorphic follows from the definition of the conformal structure on U/G.
11.6
Universal cover
Given a connected Riemann surface V, its cover p : U → V is called universal, if U is simply
connected (recall that simply connectedness also assumes connectedness).
Example 11.10. Consider covering maps C → C \ 0 given by the formulas z 7→ ez and z 7→ zn .
The first one is universal while the second one is not.
129
The following theorem explains the origin of the term universal.
Theorem 11.11.
1. For any Riemann surface V there exists a universal cover p : U → V.
2. Let V1 and V2 be two Riemann surfaces and p1 : U1 → V1 and p2 : U2 → V2 its universal
covers. Then for any holomorphic map f : V1 → V2 there exists a holomorphic map F :
U1 → U2 such that the following diagram commutes:
U1
p1
F
/
U2
p2
V1
f
/
V2
,
i.e. f ◦ p1 = p2 ◦ F.
While the proof of this theorem is not difficult we do not have time to discuss it in this course.
As a corollary we see that the the universal cover is unique in the following sense.
Corollary 11.12. Let p1 : U1 → V and p2 : U2 → V be two universal covers. Then there exists a
biholomorophism f : U1 → U2 such that the following diagram commutes:
/ U2
F
U1
p1
p2
V
~
i.e. p1 = p2 ◦ F.
Indeed, we can apply Theorem 11.11(ii) to the identity map V → V as f .
If U = CP1 then it is simply connected itself, and hence it serves as its own universal (onesheeted) cover.
Exercise 11.13. Prove that the universal cover is always regular.
130
Theorem 11.14 (Uniformization theorem revisited). The universal cover of any Riemann surface
is conformally equivalent to H, C or CP1 . For any Riemann surface other than CP1 there exists a
discrete subgroup of Aut(C), resp. Aut(H), which freely acts on C, resp. H, such that U is conformally equivalent to C/G, resp. H/G.
Thus Theorem 11.14 reduces the conformal classification of Riemann surfaces to the classification of discrete subgroups of the groups of automorphisms of C and H which acts without fixed
points. The main variety of such subgroups is in Aut(H). But the Riemann surfaces covered by C
are also very important, They turned out to be only tori (elliptic curves) and C \ 0. We will discuss
them in Section 12.
Proof. Let V be a Riemann surface and U its universal cover. Let G be the group of its deck
transformations. According to Exercise 11.13 the universal cover is regular, and hence Lemma
11.9 implies that U/G is biholomorphic to V. By assumption, V, and hence U, is not CP1 , and
hence according to Theorem 11.11 U it is conformally equivalent to either H or C.
11.7
Lattices
A lattice Λ ⊂ C = R2 is any discrete subgroup. The rank of the lattice is the real dimension of a
real subspace SpanR (Λ). Because dim R2 = 2, any non-trivial (i.e. not consisting of only 0) lattice
could be either of rank 1 or of rank 2.
Lemma 11.15. Any rank 1 lattice has the form Λ = {kω} for ω ∈ C, ω , 0, k ∈ Z. Any rank two
lattice has the form Λ = {kω1 + `ω2 } where k, ` ∈ Z, ω1 , ω2 ∈ C, ω1 , ω2 , 0 and
ω2
ω1
< R.
Proof. We consider here only the case of a rank 2 lattice and leave the rank 1 case as an exercise
to the reader. The discreteness of the lattice guarantees that there exists ω1 ∈ Λ \ 0 such that
|ω1 | ≤ |ω| for all ω ∈ Λ \ 0. Denote L = SpanR (ω1 ). We claim that there exists ω2 ∈ Λ \ L such that
dist(ω2 , L) ≤ dist(ω, L) for any ω ∈ Λ \ L. Indeed suppose there exists a sequence ω j ∈ Λ \ L such
131
that dist(ω j , L) < dist(ω j+1 , L), for all j = 2, 3, . . . . that for any ω ∈ Λ and any k ∈ Z we have
dist(ω, L) = dist(ω + kω1 , L).
(Recall that ω1 ∈ L). Denote d := dist(ω2 , L). Then by adding to ω j an appropriate multiple of kω1
p
of ω1 we can arrange that |ω − kω1 | ≤ |ω1 |2 + d2 but then we get an infinite set of points of Λ in
a bounded domain which contradicts the fact that Λ is discrete. Take now the lattice
e := {kω1 + `ω2 ; k, ` ∈ Z}
Λ
e ⊂ Λ. We claim the Λ
e = Λ. Indeed, if there exists ω ∈ Λ \ Λ,
e
generated by ω1 and ω2 . Clearly, Λ
then by adding to ω a multiple of ω2 we can bring it to a distance < d to L:
dist(ω + kω2 , L) < d,
but by the choice of ω2 it means that
dist(ω + kω2 , L) = 0,
i.e. ω+kω2 ∈ L. But then, adding an appropriate multiple of ω1 we can arrange that |ω+kω2 +`ω1 | <
e
|ω1 |, but then ω + kω2 + `ω1 = 0, i.e. ω ∈ Λ.
A lattice Λ can be interpreted as a subgroup of Aut(C) acting on C by translations. It turns out
that converse is also true:
Lemma 11.16. Any discrete subgroup G of Aut(C) which acts on C freely is a lattice.
Proof. Any conformal automorphism g : C → C has the form g(z) = az + b. We claim that if a , 1
g has a fixed point. Indeed in that case the equation az + b = z has a solution z0 =
b
.
1−a
Hence,
g(z) = z + b, i.e. G is a subgroup of the subgroup C ⊂ Aut(C) which consists of translations.
Therefore it is a lattice.
Hence we can form a quotient torus T (ω1 , ω2 ) = C/Λ. Clearly, some of these tori are conformally equivalent, which allows us to restrict the class of lattices we want to study. Here are some
of the operations which allows us to do that.
132
1. Action z 7→ cz, c ∈ C. The tori T (ω1 , ω2 ) and T (cω1 , cω2 ) for any complex number c ∈ C \ 0
are conformally equivalent. Indeed, the linear map z 7→ cz sends the lattice Λ(ω1 , ω2 ) generated by
ω1 , ω2 to the lattice Λ(cω1 , cω2 ) generated by cω1 , cω2 . Hence, orbits of the former lattice are sent
by this map to the orbits of the latter one, and this map is clearly invertible.
Thus T (ω1 , ω2 ) T (1, τ =
ω2
),
ω1
2. Automorphisms of a lattice. The lattice Λ(ω1 , ω2 ) is preserved by interchanging ω1 and ω2 .
Hence, we can always assume that ω1 and ω2 define the standard orientation of R2 = C (i.e. the
same orientation as defined by 1 and i. This is equivalent to the requirement that τ =
ω2
ω1
∈ H.
In particular any lattice is conformally equivalent to a lattice Λ(1, τ) where τ ∈ H via an
automorphism C → C, and hence every torus T (ω1 , ω2 ) is conformally equivalent to the torus
T (1, τ) with τ ∈ H.
Let us denote by PS L(2, Z) the subgroup of Aut(H) = PS L(2, R) which consists of matrices
with integer entries (and determinant 1). It is called the modular group.
e2 ) are conformally
Theorem 11.17 (Conformal classification of tori). Tori T (ω1 , ω2 ) and T (e
ω1 , ω
ω2
ω1
and e
τ = ωωee12 are related by the action of an element of the modular


n m


group, i.e. there exists a matrix 
with integer values and det = kn − m` = 1 such that
 ` k 
equivalent if and only if τ =
nτ + m
.
`τ + k
e
τ=
e = Λ(e
e be a conformal equivalence.
e2 ). Let f : C/Λ → C/Λ
Proof. Denote Λ := Λ(ω1 , ω2 ), Λ
ω1 , ω
Then according to the uniqueness of universal cover theorem 11.11 there exists a biholomorphism
F : C → C such that the following diagram commutes:
C
πΛ
F
/C
πΛ
e
C/Λ
,
133
f
/ C/Λ
e
e are tautological projections. We have
i.e. f ◦ πΛ = πΛe ◦ F, where πΛ : C → C/Λ and πΛe : C → C/Λ
e and hence F(z + 1) = F(z) + c1 , F(z + τ) = F(z) + c2 , where c1 , c2 ∈ Λ.
e Hence,
F(z + Λ) = F(z) + Λ,
F 0 (z) is a bi-periodic function with periods 1 and τ, and hence a constant, but then F(z) = A + Bz,
e differ by a rotation, scaling and translation. After
where A, B ∈ C. and the two lattices Λ and Λ
e we see that (ω1 , ω2 )
we used the transformation F(z) = A + Bz to identify the lattices Λ and Λ
e2 ) are two different bases of the same lattice Λ, which in addition by assumption
and (e
ω1 , ω
define


 k ` 

 such that
the same orientation of C. Hence, there exists and integer valued matrix A = 
m n
e1 = kω1 + `ω2 , ω
e2 = mω1 + nω2 . The matrix is invertible (as an integer valued matrix) and it is
ω
orientation preserving. Hence, det A = kn − m` = 1.
e
τ=
e2 nτ + m
ω
.
=
e1
ω
`τ + k
The action of the modular group on H is discrete but not free. Hence, the quotient H/PS L(2, Z)
is a what is called singular Riemann surfaces.
Theorem 11.18. Denote
(
1
1
U := τ ∈ H; |z| ≥ 1, − ≤ Re τ ≤ .
2
2
)
Every torus T (ω1 , ω2 ) is conformally equivalent to a torus T (1, τ) with τ ∈ U.
Proof. As it is explained in the proof of Theorem 11.17 given a lattice Λ = Λ(ω1 , ω2 ) automorphisms of this lattice are real linear transformation of C = R2 which map a basis ω1 , ω2 of Λ to
e1 , ω
e2 of Λ, i.e. ω
e1 = kω1 + `ω2 , ω
e2 = mω1 + nω2 , where k, `, m, n are integers. The
another basis ω
condition that this is an automorphism means that the matrix is invertible, and the inverse is also
integer valued, which means that det = kn − m` = ±1, and taking into account that we consider
only orientation preserving automorphisms, we should have kn − m` = 1. On the other hand, as
we discussed above, by a complex linear isomorphism C → C given by z 7→
lattices Λ(ω1 , ω2 ) and Λ(1, τ =
ω2
),
ω1
1
z,
ω1
we identify the
where τ ∈ H (due to our orientation convention). The change
134
e2 ) amounts to changing
of a basis (ω1 , ω2 ) 7→ (e
ω1 , ω
τ 7→ e
τ=
e2 nτ + m
ω
=
.
e1
ω
`τ + k
In other words, the action of S L(2, Z) on H via fractional linear transformations is equivalent to
the action of S L(2, Z) as the automorphism group of the lattice Λ = Λ(ω1 , ω2 ).
e2 )
Hence, our problem can be interpreted as the problem of finding an appropriate basis (e
ω1 , ω
e1 , ω
e2 . The condition
of the lattice Λ(ω1 , ω2 ). First, let us interpret the conditions on e
τ in terms of ω
|e
τ| ≥ 1 means |e
ω2 | > |e
ω1 |. To clarify geometric meaning of the condition |Ree
τ| <
1
2
let us write
e
τ = x + iy. Then
e2 = e
e1 = (x + iy)e
ω
τω
ω1 = xe
ω1 + y(ie
ω1 )
e1 and ie
e2 | =
The vectors ω
ω1 are orthogonal; and have the same length. So the condition |Re ω
|x| ≤
1
2
e2 onto the line L := Span(e
means that the orthogonal projection of ω
ω1 ) lies in the interval
[− ωe21 , ωe21 ] ⊂ L.
e2 ) should be the following (comp.
Hence, the strategy for finding the appropriate basis (e
ω1 , ω
e1 the shortest vector of Λ. This guarantees that whatever
the proof of Lemma 11.15). Choose as ω
e2 the condition e
vector we choose as ω
τ=
e2
ω
e1
ω
≥ 1 would be automatically satisfied. Note that all
points of Λ are located on affine lines L + λ, λ ∈ Λ, parallel to L. Let L1 be the line L + λ closest to
e1 , λ define the complex (i.e. counterL and chosen on the appropriate side of L, i.e. to ensure that ω
clockwise) orientation of C. Points of Λ on L1 are spaced at a distance |e
ω1 |. Hence, every closed
interval of length |e
ω1 | contains at least one point of Λ, and therefore there exists λ ∈ L such that
the orthogonal projection of λ to L lies in the interval [− ωe21 , ωe21 ] ⊂ L. Set ω2 := λ. Then (ω1 , ω2 )
form a basis of Λ with the required properties.
Remark 11.19. Theorem 11.18 can be strengthen as follows:
Every torus T (ω1 , ω2 ) is conformally equivalent to a unique torus T (1, τ) with
π π iθ
τ ∈ Int U ∪ τ = e , θ ∈ ,
.
3 2
135
11.8
Branched covers
We begin with a lemma describing the behavior of a holomorphic function near its critical point.
Lemma 11.20. Let f : Ω → C be a holomorphic function defined on a domain Ω 3 0. Suppose
f (0) = f 0 (0) = · · · = f k−1 (0) = 0 and f k (0) , 0 for some k > 1. Then there exists a local coordinate
u near 0 such that the map f is given by u 7→ uk . In other words, there exists a biholomorphism
e ⊂ C defined on a neighborhood U 3 0, U ⊂ Ω such that h(0) = 0 and the following
h:U →U
diagram commutes:
u7→uk
e_
U
/
?C
.
f
h
U
Proof. We can write f (z) = zk g(z) where g(z) = a , 0. Let log z be a branch of the logarithm which
is defined in a neighborhood V 3 a = g(0) , 0, and let U be a neighborhood of 0, so small that
g(U) ⊂ V. Define a function h(z) on U by the formula
h(z) = ze
log g(z)
k
.
We have h(0) = 0 and h0 (0) , 0. Hence, decreasing, if necessary, the neighborhood U we can
e = h(U) 3 0. But we have f (z) = zk g(z) =
assume that h is a biholomorphism of U onto its image U
hk (z).
A holomorphic map f : S 1 → S 2 is called a branched cover if there is a discrete set B ⊂ S 2
such that
- the restriction
f |S 1 \ f −1 (B) : S 1 \ f −1 (B) → S 1 \ B
is a covering map;
- each point b ∈ B has a neighborhood U such that f −1 (U) can be presented as a disjoint union
U1 ∪ U2 ∪ . . . , and there exist local coordinates w on U and z j on U j , j = 1, 2, . . . such that
136
f |U j : U j → U can be witten in these coordinates as w = zk j , where k j are positive integer
numbers.
Remark 11.21. If all k j are equal to 1 then the branched cover is just the usual cover. Hence, it is
usually assumed that for each branching point, at least for one of the exponents we have k j > 1.
The set B is called the branching, or branch locus and the points b ∈ B are called branch points
of f . Note that if S 2 is connected then S 2 \ B is connected as well. Hence the order of the covering
f |S 1 \ f −1 (B) : S 1 \ f −1 (B) → S 1 \ B is well defined. This number is usually referred to as the order
of the branched covering f . Thus for a branched cover f : S 1 → S 2 when S 2 is connected, the
number of pre-images of any point z \ B is the same, and it is equal to the order of the branched
cover.
Lemma 11.22. Let f : S 2 → S 1 be a branched map of order d. Suppose that a branching point
p
P
b ∈ B has p pre-images and (k1 , . . . , k p ) are exponents of its pre-images Then k j = d. In other
j=1
words, a branching point has d pre-images if one counts them with multiplicities given by the
exponents k j .
Proof. By definition the point b ∈ B has a neighborhood U such that f −1 (U) can be presented as a
disjoint union U1 ∪ U2 ∪ . . . , and there exist local coordinates w on U and z j on U j , j = 1, 2, . . .
k
such that f |U j : U j → U can be witten in these coordinates as w = z j j . Hence, for a point b0 ∈ U \ b
the preimage f −1 (b0 ) has exactly k j pre-images in U j , so the total number of pre-images, which is
p
P
by definition is the order d of the branched cover f , is equal to k j .
j=1
The set of exponents (k1 , . . . , k p ) is called the type of the branching point b. Hence, comparing
the number of preimages # f −1 (b) of a branching point b and a number of preimages # f −1 (b0 ) of a
p
P
nearby point p0 differ by (k j − 1) = d − p. Sometimes one refers to exponents k j (when k j > 1 as
1
the order of the corresponding pre-images of the branching points.
Example 11.23. The map z → z p is a branched cover C → C of order p with the unique branching
point 0. Viewed as a holomorphic map CP1 → CP1 the map z 7→ z p is a branched cover of order p
with branching points 0, ∞ ∈ CP1 .
137
It turns out that any non-constant holomorphic map between two compact Riemann surfaces is
a branched cover.
Theorem 11.24. Let S 1 , S 2 be compact Riemann surfaces and f : S 1 → S 2 a non-constant
holomorphic map. Suppose S 2 is compact. Then f : S 1 → S 2 is a branched cover.
Proof. First we observe that the map f is surjective, i.e f (S 1 ) = S 2 . Indeed, by the open image
theorem f (S 1 ) is an open set. On the other hand, the image of a compact set is compact, and hence
f (S 1 ) is also a closed subset of S 2 . Hence, the connectedness of S 2 implies that f (S 1 ) = S 2 . As f
is not constant there are only finite set C of its critical points, i.e. points c where the differential d f
vanishes. Denote B := f (C) ⊂ S 2 . We claim that f is a branched cover with the branching locus
B. Take a point b ∈ B denote by a1 , . . . , a p its pre-images. For a sufficiently small neighborhood
U 3 b in S 2 its pre-image f −1 (Ub ) can be presented as a disjoint union of neighborhoods V j 3 a j .
By making Ub smaller we can arrange that there exist a holomorphic coordinate u in Ub and
a holomorphic coordinate v j in V j for each j = 1, . . . , p. We can assume, in addition that the
coordinate u is centered at b, and v j is centered at a j , i.e. v j (a j ) = 0, u(b) = 0. Applying Lemma
11.20 (and possibly choosing U even smaller and further adjusting coordinates v j ) we can arrange
that the map f |V j : V j → U can be written as
k
u = f (v j ) = v j j , j = 1, . . . , p,
(11.8.1)
Here k j is the order of the first non-zero derivative at the point a j .
Consider now a point z ∈ S 2 \ B. Then for each point a ∈ f −1 (z) we have da f , 0. Hence,
there exists a neighborhood Va 3 a such that f |Va is a biholomorphism of Va onto a neighborhood
of z. Hence, f −1 (z) is a discrete set, and hence due to compactness of S 1 consists of finitely many
S
points. Choosing a neighborhood U 3 a in
f (Va ) we conclude that
a∈ f −1 (z)
f −1 (U) =
[
ea = Va ∩ f −1 (U))
(V
a∈ f −1 z
and
ea → U
f |Vea : V
138
is a biholomorphism. This proves that
f |S 1 \ f −1 (B) : S 1 \ f −1 (B) → S 2 \ B
is a covering map, and together with (11.8.1) this implies that f : S 1 → S 2 is a branched cover. Corollary 11.25. For any compact Riemann surface S , any non-constant meromorphic function
f : S → CP1
is a branched cover.
The branching point of f are critical values of f (i.e. points f (a) ∈ CP1 if da f = 0) and possibly
∞ in the case when f has non-simple poles.
11.9
Riemann surfaces as submanifolds
11.9.1
Affine case
Consider the space C2 with complex coordinates (z1 , z2 ). A function F : C2 → C is called holomorphic if its differential at each point is a complex linear function. This turns out to be equivalent
that F is holomorphic with respect to each of the variables z1 and z2 . However we will not need
this fact and will not use it.2
Lemma 11.26. Let F : C2 → C be a holomorphic function. Denote S := {F(z1 , z2 ) = 0}. Suppose
that for each point p ∈ S at least one of partial derivatives,
∂F
(p)
∂z1
and
∂F
(p)
∂z2
is not equal to 0.
Then S is a Riemann surface.
Proof. The holomorphic version of the implicit function theorem (which is proved in an exactly the
same way as its real version) asserts that if p = (a1 , a2 )
2
∂F
(p)
∂z1
, 0 then there exists a neighborhood
This fact is in a striking contrast with the real case, where for a differentiability of a function of 2 variables is not
sufficent to be differentiable in each of the variable separately.
139
U 3 a1 in C, a neighborhood V of the point p in C2 and a holomorphic function h : U1 → C such
that h(a1 ) = a2 and
V ∩ S = {z2 = h(z1 ); z1 ∈ U}.
Similarly, if
∂F
(p)
∂z2
, 0 then S near the point p can be presented as the graph of a function g given
on a neighborhood of a2 ∈ C. If both partial derivatives are not 0 then the functions h, g are inverse
of each other, and they themselves provide transition maps between the coordinates.
We call a Riemann surface as in Lemma 11.26 a 1-dimensional complex submanifold, or a
smooth (affine) holomorphic curve in C2 . If the defining function F(z1 , z2 ) is a polynomial, the the
holomorphic curve S is called algebraic.
Lemma 11.27. Any holomorphic curve in C2 is non-compact.
Proof. The function f := z1 |S : S → C is holomorphic and S is compact, then | f | achieves its
maximum at a point p ∈ S , but this contradicts to the maximum modulus principle applied to a
neighborhood of this point.
Remark 11.28. It is an open problem whether every connected non-compact Riemann surface is
conformally equivalent to a holomorphic curve in C2 .
11.9.2
Projective case
The projective plane CP2 is the space of all complex 1-dimensional subspaces in C3 , i.e. the space
of all complex lines in C3 through the origin. Equivalently, a point in CP2 can be viewed as a point
z = (z1 , z2 , z3 ) ∈ C3 \ 0 up to a complex proportionality coefficient:
(z1 , z2 , z3 ) ∼ (λc1 , λc2 , λc3 ), λ ∈ C \ 0.
z1 , z2 , z3 are called homogeneous coordinates in CP2 and usually denoted z1 : z2 : z3 .
b j in CP2 . Indeed, a
The subsets U j := {z j , 0} ⊂ C3 give rise to affine coordinate charts U
point z = (z1 , z2 , z3 ) ∈ U3 , viewed as a point in CP2 , is equivalent to ( zz31 , zz32 , 1), and hence (z13 =
140
z1
,z
z3 23
=
z2
)
z3
b1 we can introduce
can be viewed as coordinates in this neighborhood. Similarly, in U
coordinates (z21 =
z2
,z
z1 31
=
z3
),
z1
b2 coordinates (z12 =
and in U
z1
,z
z2 32
=
z3
).
z2
b3 with the affine
We will view C2 as a subset of CP2 by identifying it with the affine chart U
coordinates (z13 , z23 ). Hence CP2 \ C2 = {z1 : z2 : 0} is just the projective line, or the Riemann
sphere. In other words, we get CP2 by adding to C2 a projective line. We recall that we get CP1
from C by adding one point, which is of course can be viewed as the 0-dimensional projective
space.
Let us recall that a function F(z1 , z2 , z3 ) is called homogeneous of degree d if
F(λz1 , λz2 , λz3 ) = λd F(z1 , z2 , z3 )
for any λ ∈ C. A useful fact about homogeneous function is
Lemma 11.29 (Euler identity). Let F(z1 , z2 , z3 ) is homogeneous of degree k. Then
z1
∂F
∂F
∂F
+ z2
+ z3
= kF(z1 , z2 , z3 ).
∂z1
∂z2
∂z3
Proof. Differentiate the identity
F(λz1 , λz2 , λz3 ) = λk F(z1 , z2 , z3 )
with respect to λ and set λ = 1.
Suppose that F is a polynomial in variables z1 , z2 , z3 . Then it is homogeneous of degree d if all
its monomial have degree d:
F(z1 , z2 , z3 ) =
X
akmn zk1 zm2 zd3 , akmn ∈ C.
k+m+n=d
For instance, z3 z22 − z31 + z23 z1 is a homogeneous polynomial of degree 3.
Lemma 11.30. If F(z1 , z2 , z3 ) is a homogeneous function of any degree d, then the equation F(z1 , z2 , z3 ) =
0 defines a subset S in CP3 . If at every point p ∈ S we have
∂F
(p)
∂z j
, 0 for at least one of j = 1, 2, 3,
e j is an affine holomorphic curve.
then S is compact Riemann surface. The intersection S j := S ∩ U
141
Proof. We have
S 1 := {F(1, z21 , z31 ) = 0}, S 2 := {F(z12 , 1, z32 ) = 0}, S 3 := {F(z13 , z23 , 1) = 0}.
The condition of non-vanishing partial derivatives implies the same condition for S 1 , S 2 , S 3 , because if the non-vanishing derivative of F(z1 , z2 , z3 ), say at a point (z1 , z2 , 1), is with respect to z3
then the Euler identity gives
∂F
∂F
∂F
+ z2
=−
, 0,
∂z1
∂z2
∂z3
and hence one of two other partial derivatives have to not vanish as well. Hence the intersection
z1
of S with affine coordinate charts are holomorphic submanifolds, and so does S . The surface S is
compact because it is a closed subset of CP2 which is compact (why?).
The surface S is called a smooth projective holomorphic curve.
11.9.3
Projectivization
Given an affine algebraic curve C = {P(z1 , z2 ) = 0, where P is a polynomial, it can be extended to
a compact (possibly singular) holomorphic curve in CP2 by the following procedure.
The polynomial P is a combination of monomialsm zk1 z`2 . Let d = k+` be the highest total degree
of monomials which form P. Let us complete each monomial of zm1 zn2 of the degree m + n < d to a
monomial zm1 zn2 z3d−m−n , thus obtaining a homogeneous polynomial P(z1 , z2 , z3 ) of 3 variables.
The equation
P(z1 , z2 , z3 ) = 0
defines a projective curve C ⊂ CP2 which is called the projectivization of the affine curve C.
Example 11.31. 1. The equation z1 = 0 is already homogeneous, so the projectivization of the
affine line {z1 = 0} ⊂ C2 which is biholomorphic to C is the projective line {z1 = 0} ⊂ CP2 which
is biholomorphic to the sphere CP1 .
2. Consider a polynomial P(z1 , z2 ) = z21 + z22 + 1 = 0. Making it a homogeneous polynomial of
3 variables, we get P(z1 , z2 , z3 ) = z21 + z22 + z23 . Thus the projectivization of the affine curve
C = {z21 + z22 + 1 = 0} ⊂ C2
142
which is biholomorphic to C \ 0 is a projective curve
C = {z21 + z22 + z23 = 0} ⊂ CP2 ,
which is biholomorphic to CP1 , see the next section.
Note that the projectivization of a smooth curve C ⊂ C2 could be a singular curve. Consider
the following example.
Consider an affine holomorphic curve C = {z2 = z31 }. Its projectivization C is given by the
b2 = {z2 , 0 C is given by the equation
equation z2 z23 − z31 = 0. Note that the in the coordinate chart U
z232 = z312 = 0, and we see that the point with coordinates z32 = z12 = 0, i.e. the point with projective
coordinates (0 : 1 : 0) is singular.
11.10
Linear projective transformations
b : CPn → CPn . Indeed, A(λz) = λz, λ ∈
Any linear map A : Cn+1 → Cn+1 induces a map A
C, z ∈ Cn+1 . In other words, proportional vectors are mapped to proportional vectors. If A is nonb is called a linear projective transformation of CPn , or just a projective transfordegenerate then A
mation. We already had seen that for n = 1 linear projective transformations of CP1 , viewed as the
Riemann sphere, are just fractional linear transformation, or as they are called, Möbius transformations. In this section we will concentrate on the case n = 2.
Points in CP2 correspond to complex lines in C3 , while lines in CP2 are by definition the
images of planes in C2 . Hence each line in CP2 is itself biholomorphic to the projective line CP1 .
Any projective line L can be given by a homogeneous equation
`(z) = a1 z1 + a2 z2 + a3 z3 = 0.
The complement U := CP2 \ L can be viewed as an affine chart in CP2 . In order to define affine coordinates we pick two other linear functions `1 (z), `2 (z) such that `, `1 , `2 are linearly independent.
Then
u1 :=
`1
`2
, u2 :=
`
`
143
are affine complex coordinates in the affine chart U. Every projective line in CP2 , different from L
intersects U in an affine line
A1 u1 + A2 u2 + A3 = 0.
Projective pencils. A pencil in CP2 is by definition the set of all lines passing through a given point
a ∈ CP2 . We will denote a pencil of lines through a by Πa . Note that if L is a projective line and
a ∈ L then any two lines in L1 , L2 from Πa , different from L, intersect the affine chart U = CP2 \ L
along parallel affine lines. Note that given two distinct lines L1 = {`1 = 0} and L2 = {`2 = 0} in
a pencil Π any other lines from this pencil can be given by an equation a`1 + b`2 = 0 for some
a, b ∈ C.
Given any 3 lines
L1 = {`1 = 0}, L2 = {`2 = 0}, L3 = {`3 = 0}
such that `1 , `2 , `3 are linearly independent (or equivalently if the lines L1 , L2 , L3 do not belong to
one pencil) we say that L1 , L2 and L3 are in general position.
Lemma 11.32. Given 2 triples of lines in general position, there exists a linear projective transformation which sends one triple to another. In particular, any triple is projectively equivalent to
the triple {z1 = 0}, {z2 = 0}, {z3 = 0} of projective coordinate lines.
Proof. In C3 this is equivalent to mapping one triple of transverse planes P1 , P2 , P3 to another one
P01 , P02 , P03 by a linear transformation. Let us choose the bases v1 , v2 , v3 and v01 , v02 v03 such that
v1 ∈ P1 ∩ P2 , v2 ∈ P2 ∩ P3 , v3 ∈ P3 ∩ P1 , v01 ∈ P01 ∩ P02 , v02 ∈ P02 ∩ P03 , v03 ∈ P03 ∩ P01 .
Then the linear transformation which sends the basis v1 , v2 , v3 to the basis v01 , v02 v03 sends P1 to P01 ,
P2 to P03 and P3 to P03 .
Quadrics. A quadric Q in CP2 is a complex curve of degree two, i.e. given by a degree two
equation
Q = {F(z) = A11 z21 + A22 z22 + A33 z23 + 2A12 z1 z2 + 2A13 z1 z3 + 2A23 z2 z3 = 0}
144
in the homogeneous coordinates (z1 : z2 : z3 ). In an affine chart a quadric is given by a quadratic
equation in 2 variables with linear and constant terms. According to the implicit function theorem
the quadric is a smooth, or it is a submanifold of CP2 , if
!
∂F
∂F
∂F
(z),
(z),
(z) , 0
∂z1
∂z2
∂z3
for each z ∈ Q. We can easily compute that


 
 ∂F (z)
z1 
 ∂z1 
 
 
 ∂F 
 ∂z (z) = A z2  ,
 2 
 
 ∂F 
 
(z)
z3
∂z3
where


A11 A12 A13 




A = A12 A22 A23 




A13 A23 A33
is the matrix of the quadratic form F. Hence, the smoothness condition is equivalent to the condition det A , 0.
Theorem 11.33. All smooth quadrics are projectively equivalent.
Proof. Choose a point a ∈ Q. Consider the pencil Πa of lines through the point a. One of these
lines is tangent to Q at the point a. We denote it by L. Choose any other line L1 ∈ Πa and any
line L2 < Πa , i.e. L2 does not pass through the point a. According to Lemma 11.32 there exists a
projective linear transformation T : CP2 → CP2 such that T (L) = {z3 = 0}, T (L1 ) = {z1 = 0} and
T (L2 ) = {z2 = 0}.
Consider the affine chart U3 = {z3 , 0}(= T (CP2 \ L)). To simplify the notation denote the
affine coordinates there by
u1 := z13 =
z2
z1
, u2 := z23 = .
z3
z3
e := T (Q) ∩ U3 = T (Q \ L) in U3 can be written in the form
The equation of the affine part Q
αu32 + (bu1 + c)u2 + (ku21 + lu1 + m) = 0.
145
(11.10.1)
First, we observe that α = 0. Indeed, otherwise we would have 2 points on every line {u1 = const}
which would mean that the intersection of Q with a line of the pencil P consists of 3 points, which
is impossible because a quadric cannot intersect a line in more than 2 points.
Second, we argue that b = 0. Indeed, suppose b , 0 and set u1 = − bc . Then (11.10.1) implies
that ku21 + lu1 + m = 0, i.e. − bc is a rood of this quadratic polynomial, which means that
ku21 + lu1 + m = (bu1 + c)(pu1 + q),
and therefore, (11.10.1) takes the form
(bu1 + c)(u2 + pu1 + qu2 ) = 0.
(11.10.2)
But this equation defines two lines
c
{u1 = − } ∪ {u2 = −pu1 − q}.
b
But the two lines intersect at the point (u1 = − bc , u2 =
pc−qb
),
b
which contradicts to our assumption
that the curve Q is smooth. Therefore, (11.10.1) further reduces to
cu2 + ku21 + lu1 + m = 0,
(11.10.3)
where c , 0 (otherwise the curve would reduce to two lines parallel to the u2 -axis, and hence
intersecting at a; this would again contradicts the smoothness assumption). Hence, dividing the
equation by c and renaming the constant coefficients we get an equation
u2 = αu21 + βu1 + γ = 0,
(11.10.4)
We have α , 0. Indeed, otherwise, the equation would be linear, and hence define a line and not a
quadric. Finally by making successive linear changes of variables
• u1 7→ u1 +
β
2
we first kill the coefficient with the linear term u1 , thus getting an equation of
the form, u2 = αu21 + γ̃;
• u2 7→ u2 − e
γ getting u2 = αu21 , and finally
146
• u1 7→
√
α obtaining an equation u2 = u21 .
Thus, any smooth quadric is projectively equivalent to the projectivization of the complex parabola
u2 = u21 .
Exercise 11.34. Following the scheme of the proof of Theorem 11.33 show that any smooth cubic
(elliptic) curve C ⊂ CP2 , i.e. a curve given by a homogeneous equation of degree 3, is projectively
equivalent to the projectivization of an affine curve given by the equation u22 = u31 + pu1 + q.
11.11
Examples
1. Let C = {z1 z2 = 1} ⊂ C2 . Show that C is a Riemann surface conformally equivalent to C \ 0..
Note that if z1 , z2 ∈ C, then both z1 , z2 , 0. Hence we can write the equation as z2 =
just the graph of function
1
.
z1
1
,
z1
i.e. C is
Hence, z1 can be viewed as a coordinate on C. The map
!
1
u 7→ u, , u ∈ C \ 0
u
is the required biholommorphism. Indeed, this map is holomorphic, and 1-1. The inverse map is
given by the projection (z1 , z2 ) 7→ z1 .
2. Let the group Z acts on C as z 7→ z + k, k ∈ C. Find a biholomorphic C/Z → C \ 0.
Consider a map
z 7→ e2πiz .
It is a surjective map onto C\0. Moreover, this is a covering map, and the fiber over a point u ∈ C\0
consists of values of
1
2πi
log u, which is an orbit of the action of the group Z. Hence, we can define
a holomorphic map C/Z → C \ 0 which is 1:1 and hence a biholomorphism.
147
148
Part III
A few important meromorphic functions
149
In this part we introduce and study some famous meromorphic functions. First, in Chapter
12 we discuss the theory of elliptic functions, which are doubly periodic meromorphic functions,
or equivalently meromorphic functions on tori. The main role in this story plays the Weierstrass
function ℘(z). In Chapter 13 we introduce and discuss properties of the Euler Gamma function
Γ(z), and finally in Chapter 14 we define and study some elementary properties of the Riemann
zeta-function ζ(z) which plays a major role in Number Theory. Of course, we cannot discuss the
number-theoretic applications in the framework of this course.
151
152
Chapter 12
Elliptic functions
12.1
Elliptic integrals
12.1.1
Motivation
Several geometric and physical problems leads to computations of integrals of the form
Zx
Zx
p
dt
, or more generally
R(t, (P(t))dt,
√
P(t)
a
0
where P(t) is a polynomial and R(t, u) is a rational function of 2 variables. When deg P ≤ 2 this
integral is not hard to compute in terms of elementary functions. However, if deg P ≥ 3 then in
general this integral cannot be expressed through elementary functions. The integrals of this kind
when deg P = 3 or 4 are called elliptic. The term is motivated by Example 12.1.1 below.
Example 12.1. 1. Computing the arc length of an arc of an ellipse.
Consider an ellipse
)
x2 y2
E(a, b) = 2 + 2 = 1 ,
a
b
(
or parametrically given by
x = a cos t,
y = b sin t.
153
Integrating the length of the vector v = ( ẋ, ẏ) = (−a sin t, b cos t) we compute the arc length s(τ)
between the points (x(0), y(0)) = (a, 0) and (a cos τ, b sin τ), τ ∈ (0, π2 ), by the formula
Zτ p
s(τ) =
a2 sin2 t + b2 cos2 tdt.
0
Changing the variable sin t = u we get
Zsin τ p 2
Zsin τ r
(a − b2 )u2 + b2
1 + k 2 u2
du,
s(τ) =
du = b
√
1 − u2
1 − u2
0
where k := 1 +
a2
.
b2
0
Denote v := 1 − u2 . Then we get 1 + k2 u2 = (k2 + 1) − k2 v and dv = −2udu, or
du = − 2 √dv1−v . Therefore,
1
s(τ) = −
2
1−sin
Z 2τ
1
=
1
2
Z1 p
√
k2 + 1 − k2 v
dv
√
v(1 − v)
v(1 − v)(k2 + 1 − k2 v)
dv.
v(v − 1)
cos2 τ
2. Pendulum in the constant gravity field. The next example comes from Mechanics. In appropriate
units its equation of motion can be written as
θ̈ = − sin θ
and the law of conservation of energy gives
1 2
(θ̇) − cos θ = E.
2
Hence,
dθ p
= 2(E + cos θ),
dt
and separating variables we get
t=
Zθ
θ0
dτ
,
√
2(E + cos τ)
154
(12.1.1)
where θ0 := θ(0). Changing the variable u = cos τ we get
1
t= √
2
12.1.2
arccos
Z θ
du
.
p
(E + u)(1 − u2 )
arccos θ0
Elliptic curves
To understand the proper meaning and geometry related to elliptic integrals we take a closer look
√
at functions P(u) when P is polynomial of degree 3. In fact, for determinacy we will consider
just the polynomial P(u) = u3 − 1. It will be crucially important to allow the variable u to take
complex values, and hence to visualize the graph of this function we consider the algebraic curve
S = {z22 = P(z1 )}, where P(z1 ) = z31 − 1.
To verify that S is a smooth algebraic curve, and hence a Riemann surface we just need to
check, according to Lemma 11.26 that both partial derivatives never vanish at the same point
of S . Indeed, if
∂F
∂z1
∂F
∂z1
=
∂F
∂z2
= 0 then z2 = P(z1 ) = 0. But this implies that z1 = ζk and hence
= −P0 (ζk ) = −3ζk2 , 0 for ζk = 0, 1, 2.
Consider a holomorphic differential 1-form
dz1
z2
on C2 \ {z2 = 0}. The restriction α of
dz1
z2
to the
surface S is defined everywhere except points (ζk , 0) ∈ S .
Lemma 12.2. The form α extends holomorphicallly to the whole S as a nowhere vanishing holomorphic 1-form.
Proof. To extend α to points (ζk , 0) ∈ S let us observe that the equation z22 = P(z1 ) which holds on
S implies that 2z2 dz2 = P0 (z1 )dz1 , i.e. for z2 , 0 we have
dz1
dz2
=2 0
.
z2
P (z1 )
But P0 (ζk ) , 0, and hence the form
dz1
z2
(12.1.2)
can be defined as a holomorphic 1-form on a neighborhood
2
of (ζk , 0) as equal to 2 Pdz
0 (z ) , 0.
1
Given a path γ connecting in S two points (a0 , b0 ), (a1 , b1 ) ∈ S the integral
R
γ
α depends only on
the homotopy class of the pass γ, i.e. it remains unchanged if we continuously deform γ keeping
155
its end-points fixed. This integral gives the exact meaning to the integral
Za1
√
a0
du
u3 − 1
and we will study it below in Section 12.6.
12.1.3
Projectivization
The Riemann surface S is not compact and we would like to compactify it in a way similar to our
compactification of C into the Riemann sphere CP1 .
This is done by a general procedure, called projectivization which we already discussed in
Section 11.9.2 above.
To projectivize the affine algebraic curve
S = {z22 − z31 − 1 = 0}
we complete each monomial to degree3 by multiplying by an appropriate power zk3 , i.e. consider
the projective curve
S = {z3 z22 − z31 + z33 = 0}.
(12.1.3)
b1 , U
b2 , U
b3 . These equations are
Let us rewrite this equation in affine coordinates in the charts U
obtained by dividing (12.1.3) by z31 , z32 and z33 , respectively.
(i)
b1 = {z31 z221 − 1 + z331 = 0},
S ∩U
b2 = {z32 − z312 + z332 = 0},
(ii) S ∩ U
b3 = {z223 − z313 + 1 = 0}.
(iii) S ∩ U
In equation (iii) we recognize the equation for the Riemann surface S ⊂ C2 , written in coordinates (z13 , z23 ) instead of (z1 , z2 ).
b2 by a unique point
Lemma 12.3. S ⊂ CP2 is a compact Riemann surface. It differs from S ⊂ U
with projective coordinates p = (0, 1, 0).
156
b3 ⊂ U
b2 . Indeed, if z ∈ S \(U
b3 ∪ U
b2 ) then z2 = z3 = 0 which in view of equation
Proof. Note that S \ U
b3 = {p = (0, 1, 0)}.
(12.1.3) then implies that z1 = 0, which is impossible. On the other hand, S \ U
b2 about p the equation takes the form (ii):
In the coordinate chart U
G(z32 , z12 ) = z32 − z312 + z332 = 0.
We note
∂G
(p)
∂z32
= 1 , 0, and hence the implicit function theorem implies that S is a compact
Riemann surface, which is called the projectivization of the Riemann surface S , see Lemma 11.30.
b3 in Section 12.1.2 above.
Let α be the differential 1-form which we defined on S ⊂ U
Lemma 12.4. The form α extends to S as a holomorphic non-vanishing form.
Proof. Recall that the form α has two equivalent expressions (see (12.1.2),
α=
2dz23
dz23
dz13
= 0
=2 2 .
z23
P (z13 )
3z13
Both expression holds where P(z13 ) = z313 − 1 , 0 and P0 (z13 ) = 3z213 , 0. Both inequalities hold
when |z13 | > 1. Rewriting the latter expression for α in coordinates z12 and z32 (taking into account
that z13 =
z12
, z23
z32
=
1
z23
we get
α=−
2 dz32
.
3 z212
(12.1.4)
Recall that on S we have
z32 + z332 = z312 .
Differentiating we get
(1 + 3z232 )dz32 = 3z212 dz12 .
Plugging this expression into (12.1.4) we get
α = −2
dz12
.
1 + 3z232
(12.1.5)
This expression extends α to the point p with coordinates z12 = z32 = 0 as a non-vanishing
holomorphic form.
157
Lemma 12.5. The affine coordinate functions z13 |S =S ∩Ue3 : S → C and z23 |S : S → C extend as
meromorphic functions to S → CP1 , denoted z and w, respectively. These functions are related by
the equation
w2 = P(z) = z3 − z.
The holomorphic differential form α is equal to
α :=
dz
.
w
Proof. This follows from the removal of singularities theorem.
12.1.4
From a cubic curve to a torus T (ω1 , ω2 )
Proposition 12.6. Let S be a compact Riemann surfaces which admits a non-vanishing holomorphic 1-form α. Then there exists a lattice Λ = Λ(ω1 , ω2 ) ⊂ C and a biholomorphism
h : S → T (ω1 , ω2 ) = C/Λ(ω1 , ω2 ).
Moreover, the map (h ◦ πΛ ) : C → S pulls back the form α to du, where u is the coordinate in C:
(h ◦ πΛ )∗ α = du.
Here π|Λ : C → C/Λ is the tautological projection.
Proof. Let p : b
S → S be the universal cover of S . One can pull-back the form α to b
S . The
holomorphic differential 1-form β := f ∗ α is defined by the formula
βz (T ) = α p(z) (dz p(T )) ,
where z ∈ U, T is a tangent vector to U at the point z and d pa is the differential of p at the point z.
Similar to the form α the form β is not vanishing (because the differential dz p is not zero for every
z by the definition of a covering map. Being a holomorphic, the differential 1-form β is closed in b
S,
but then it is exact because U is simply connected. Therefore there exists a holomorphic function
f :b
S → C such that β = d f .
To complete the proof we need the following
158
Lemma 12.7. f : b
S → C is a biholomorphism.
Postponing the proof of the lemma we finish first the proof of Proposition 12.6.
According to Lemma 12.7
p ◦ f −1 : C → S
is a covering map. But then according to Theorem ?? and Lemma 11.16 there exists a lattice Λ ⊂ C
and a biholomorphism h : S → C/Λ such that p ◦ f −1 = h ◦ πΛ . The rank of the lattice has to be
equal 2, because if it is equal to 1 then the quotient C/Λ would be a cylinder which is not compact.
Hence, Λ = Λ(ω1 , ω2 ) and therefore S is conformally equivalent to the torus T (ω1 , ω2 ). Recall that
by construction d f = p∗ α. On the other hand we have d f = f ∗ du, where u is the coordinate in C.
Hence,
∗
du = ( f −1 )∗ (p∗ α) = p ◦ f −1 α = (h ◦ πΛ )∗ α.
Proof of Lemma 12.7
We split the proof into several steps.
Lemma 12.8 (Step 1). There exists r > 0 such that every point a ∈ b
S has a neighborhood ∆a such
that f (∆a ) = Dr ( f (a)) and f |∆a : ∆a → Dr ( f (a)) is a biholomorphism. If for a, b ∈ b
S , a , b we
have f (a) = f (b) then ∆a ∩ ∆b = ∅.
Proof. Take a point a ∈ U and denote a = p(a) ∈ S . By the definition of a covering there exists a
neighborhood U 3 a such that its pre-image p−1 (Ua ) can be presented as the union U1 ∪ U2 ∪ . . .
such that
• Ui ∩ U j = ∅ if i , j, and
• p|U j : U j → U) is a biholomorphism for j = 1, . . . .
159
e := U j .
In particular, there exists j such that U j 3 a. We denote U
Let ζ be a local holomorphic coordinate in U (it exists if U is chosen small enough). Then
e
e The non-vanishing holomorphic form
ζ := ζ ◦ p can be chosen as a local coordinate on each U.
α|U(a) can be written as α = g(ζ)dζ, where g(ζ) , 0. Hence the form β = p∗ α is equal g(e
ζ)de
ζ on
e The form β = p∗ α is exact, β = d f = f 0 (e
each U.
ζ)de
ζ. Hence, on f 0 (e
ζ) = g(e
ζ) , 0. Hence the map
e → C is injective if the neighborhood U is chosen small enough. The image V := f (u) is
f |Ue : U
open. Hence there exists r > 0 such that Dr ( f (e
a)) ⊂ V. Denote
e e
e
∆a = f −1 (D 2r ( f (a))) ∩ U,
∆a := f −1 (Dr ( f (a))) ∩ U.
Note that the choice of r depends only on the the point a ∈ S ,and not on a. Hence, in view of
compactness of S one can choose r > 0 which works for all points a ∈ S .
Suppose now that there exist a, b ∈ b
S such that a , b but f (a) = f (b). We claim that ∆a , ∆b .
Indeed, suppose there exists c ∈ ∆a ∩ ∆b . Then b ∈ ∆c and we have
f (∆c ) = D 2r ( f (c)) ∩ f (∆a ) = D 2r ( f (a)) .
Therefore, D 2r ( f (c)) ⊂ Dr ( f (a)). But this implies that
e
∆c ⊂ e
∆a := f −1 (Dr ( f (a))) ∩ U.
Hence, b ∈ e
∆a , but this contradicts to the fact that f |e∆a is injective.
Lemma 12.9 (Step 2). The map f : b
S → C is a covering map.
Proof. It remains to show that f is surjective. Indeed, in by Lemma 12.8 this would imply that for
each z ∈ C the pre-image f −1 (D 2r )(z) is a disjoint union of neighborhoods which are biholomorphically mapped by the map f onto D 2r )(z), which is the definition of a covering map. By the open
image theorem we know that f (U) is open. If f (U) , C then there exists a point z ∈ C \ f (U)
which is a boundary point of f (U). Hence, there exists a ∈ b
S such that | f (a) − z| < 2r . But then
z ∈ D 2r ( f (a)) = f (∆a ) ⊂ f (U), which is a contradiction.
Lemma 12.10 (Step 3). The map f : b
S → C is a biholomorphism.
160
Proof. This follows from simply connectedness of C and Theorem 11.11. Because C is simply
connected, the identity map Id : C → C is also a universal cover, but then according to Theorem
11.11.2 the universal cover is unique, i.e there exists a biholomorphism F : C → b
S such that
f ◦ F = Id. But then f = F −1 is a biholomorphism as well.
This concludes the proof of Lemma 12.7.
12.1.5
Summary of the construction
Let us summarize what we achieved in this section.
1. For a degree 3 polynomial P(u) = u3 + a1 u2 + a2 u + a3 without multiple roots we associated first
a Riemann surface
S = {w2 = P(z)}
in C2 and then compactified it to a Riemann surface S in CP2 ⊃ C2 , called projectivization of S .
2. We showed that the coordinate functions z|S , w| s extend to S as meromorphic functions satisfying
the equation w2 = P(z), and the differential form α =
dz
w
is defined on the whole S as a non-
vanishing holomorphic form.
3. We found a lattice Λ = Λ(ω1 , ω2 ) ⊂ C and a biholomorphism
f : T (ω1 , ω2 ) = C/Λ(ω1 , ω2 ) → S
such that
(πΛ ◦ f )∗ α = du
where u is the coordinate in C and πΛ : C → C/Λ is the tautological projection. In other words, if
we denote
e
e(u) = w( f (πΛ (u)))
z(u) = z( f (πΛ (u))), w
then
de
z(u)
de
z(u)
e(u).
= du, or,
=w
e(u)
w
du
161
Hence,
de
z(u)
du
!2
e(u)2 = P(u),
=w
i.e. the function e
z(u) is a solution of the differential equation
de
z(u)
du
!2
= P(u).
(12.1.6)
In the next section we will find this solution explicitly.
12.2
The Weierstrass ℘-function
12.2.1
The definition
A holomorphic or meromorphic function on T (ω1 , ω2 ) is exactly the same as a doubly periodic
function on C:
f (z + ω1 ) = f (u), f (u + ω2 ) = f (z).
There is no interesting holomorphic functions with this properties. Indeed any such function is
bounded, and by Liouville’s theorem has to be constant. Hence, we will be studying meromorphic
doubly periodic functions, i.e. holomorphic maps
T (ω1 , ω2 ) → CP1 .
Such functions are called elliptic. We begin with the famous example of an elliptic function, called
Weierstrass ℘-function. Suppose we are given a lattice Λ = Λ(ω1 , ω2 )). The Weierstrass ℘-function
is defined by the formula
℘(u) =
X
λ∈Λ\0
!
1
1
1
− 2 + 2.
2
(u − λ)
λ
u
(12.2.1)
Lemma 12.11. The series in formula (12.2.1) absolutely uniformly converges on compact sets in
C\Λ. The function is doubly periodic with periods ω1 , ω2 and hence define a meromorphic function
T (ω1 , ω2 ) → CP1 with a unique pole of order 2.
162
Proof. Denote = {inf |u − λ|; λ ∈ Λ}. then we have
1
2|λ||u| + |u|2
1
C
=
−
≤
,
(u − λ)2 λ2
|λ|2 |u − λ|2
|λ|3
where C depends on and R := |u|. Hence,
X
1
λ∈Λ\0
(u − λ)2
−
X 1
1
≤
C
.
λ2
|λ|3
λ∈Λ\0
But the series in the right hand side converges, as it is clear by comparison with the converging
integral
Z
dxdy
=
|u|3
Z∞ Z2π
1
C\{|u|≤1}
drdφ
= 2π.
r2
0
This proves that the function ℘(u) is meromorphic on C, To check that it is doubly periodic, let us
compute the derivative by term-wise differentiation:
X
℘0 (u) = −2
λ∈Λ
1
.
(u − λ)3
(12.2.2)
Clearly ℘0 (u) is doubly periodic with periods ω1 and ω2 , because addition to λ multiples of ω1 and
ω2 leaves the sum unchanged. Hence
d ℘(u + ω j ) − ℘(u) = ℘0 (u + ω j ) − ℘0 (u) = 0, j = 1, 1, 2
du
Hence, ℘(u + ω j ) − ℘(u) = c j , j = 1, 2 for some constants c1 and c2 . But ℘(u) is an even function:
℘(−u) = ℘(u). Hence
ω j
cj = ℘
−℘ −
= 0,
2
2
ω j
and therefore ℘(u) is doubly periodic and thus defines a meromorphic function T (ω1 , ω2 ) → CP1
with a unique pole of order to in the image of the origin 0 under the tautological projection C →
T (ω1 , ω2 ).
12.2.2
Differential equation for ℘(u)
Consider the Laurent expansion of ℘(u) about 0. We have
℘(u) =
1
+ au2 + bu4 + . . . .
u2
163
The absense of odd terms follows from the fact that ℘ is even (why?). The vanishing of the constant
term is clear from the fact that
℘(u) −
1 X
1
1
=
− 2
2
2
u
(u − λ)
λ
λ∈Λ
vanishes at 0.
Differentiating twice we get
℘00 (u) =
6
+ 2a + . . . .
u4
Therefore
℘00 (u) − 6℘2 (u) = −10a + . . .
is a holomorphic doubly periodic function, and hence constant, i.e.
℘00 (u) = 6℘2 (u) − 10a.
Thus we have
d 0
d 3
(℘ (u))2 = 2℘0 (u)℘00 (u) = 12℘0 (u)℘2 (u) − 20a℘0 (u) =
4℘ (u) − 20a℘(u) ,
du
du
i.e.
(℘0 (u))2 = 4℘3 (u) − 20a℘(u) + b,
where a, b are some constants which depend on the lattice Λ.
In fact, traditionally one uses different choice of constants and write this differential equation
in the form
(℘0 (u))2 = 4℘3 (u) − g2 ℘(u) − g3 .
(12.2.3)
To explain a somewhat strange notation g2 and g3 for the coefficients we need a deeper theory of
elliptic functions which we cannot discuss in this course.
Let us denote by e1 , e2 , e3 zeroes of the cubic polynomial 4w3 − g2 w − g3 , so that we have
4w3 − g2 w − g3 = 4(w − e1 )(w − e2 )(w − e3 ).
164
By Vieta’s theorem we conclude that
e1 + e2 + e3 = 0,
e1 e2 + e2 e3 + e3 e1 = −
g2
,
4
g3
.
4
e1 e2 e3 =
Let us observe from (12.2.2) that the derivative ℘(z) is an odd function ℘0 (−u) = −℘0 (u). It is
also bi-periodic with periods ω1 , ω2 , and hence
℘0 (ω1 − u) = −℘0 (u), ℘0 (ω2 − u) = −℘0 (u), and ℘0 (ω1 + ω2 − u) = −℘0 (u).
Hence,
℘0
ω 1
2
= 0, ℘0
ω 2
2
= 0, ℘0
ω + ω 1
2
= 0.
2
Comparing this equation with (12.2.3) we conclude that the value of the function ℘(u) at zeroes of
its derivative ℘0 (u) must coincide with e1 , e2 , e3 , i.e.
℘
ω 1
2
= e1 , ℘
ω 2
2
ω + ω 1
2
= e2 , ℘
= e3 .
2
The following theorem gives a necessary and sufficient condition for a polynomial w3 + aw + b
to be proportional to the polynomial 4w3 −g2 w−g3 which appears in the right hand side of equation
(12.2.3).
Theorem 12.12. For any polynomial P(w) = w3 + aw + b with 3 distinct roots e1 , e2 , e3 there
exists a lattice Λ(ω1 , ω2 ) such that the corresponding Weierstrass function satisfies (12.2.3) with
the polynomial 4P(w) in the right hand side, i.e.
(℘0 (u))2 = 4P(℘(u)).
Though the proof of this theorem is not difficult we will not discuss it in this course. Interested
students can read it, e.g in the book “Introduction to Complex Analysis” by Michael Taylor, AMS
GSM series. See there Proposition 6.4.1.
165
12.2.3
Identifying Weierstrass ℘-function with the solution of the pendulum
equation
As we had seen in Section 12.1,
1
t= √
2
Zu
u0
du
,
p
(E + u)(1 − u2 )
where u = arccos θ is the solution of the pendulum differential equation (12.1.1). We will see that
this integral can be computed in terms of the Weierstrass ℘-function.
First of all we note that by the change of variable v = u −
E
3
we kill the coefficient with the
square term in the cubic polynomial
E3
2E
E2
E
E2
+ Ev2 −
v+
−v+ −E
(E + u)(1 − u ) = −(u + Eu − u − E) = − v − Ev + v −
3
27
3
9
3
!!
2
2
3
E
2E
E − 2E
E
1
= − v3 +
v+ − +
−
:= − (4v3 − g2 v − g3 ),
3
27
9
3
4
2
3
2
3
2
and we get
i
τ := − √ t =
2
Zv
v0
dv
,
p
3
4v − g2 v − g3
or
1
dτ
= p
,
dv
4v3 − g2 v − g3
and hence for the inverse function function v(τ) we get a differential equation
dv p 3
= 4v − g2 v − g3
dτ
or
dv
dτ
(12.2.4)
!2
= 4v3 − g2 v − g3 ,
According to Theorem 12.12 there exists a lattice Λ(ω1 , ω2 ) such that the corresponding Weierstrass function ℘ = ℘Λ is a solution of this differential equation:
(℘0 (τ))2 = 4℘(τ) − g2 ℘(τ) − g3 .
166
!
By the uniqueness of a solution of a differential equation with the given initial data we conclude
that the solution of (12.2.4)with the initial condition v(0) = v0 is equal to v(τ) = ℘(τ + τ0 ), where
τ0 is chosen to satisfy the equality ℘(τ0 ) = v0 . Recalling that v = u +
E
3
= cos θ +
E
3
and τ = − √i 2 t
we conclude that the function
!
!
it
E
θ(t) = arccos ℘ − √ −
3
2
is the solution of the pendulum equation with the initial condition θ(0) = θ0 and where we set
v0 = cos θ0 + E3 .
12.2.4
More about geometry of the Weierstrass ℘-function
The meromorphic function ℘ : C/Λ(ω1 , ω2 ) = T (ω1 , ω2 ) → CP1 is a branch cover of CP1 . Recall
that the derivative ℘0 vanishes at the points
ω1 ω2
, 2
2
and
ω1 +ω2
.
2
The corresponding critical values
e1 , e2 , e3 which are the roots of the cubic polynomial P(w) = 4w3 − g2 w − g3 are branching points
167
of the branch cover ℘ : T (ω1 , ω2 ) → CP1 . The second derivatives do not vanish because the
polynomial P has no multiple roots. The function ℘ also has a unique pole of order 2. Hence, the
holomorphic map ℘ : T (ω1 , ω2 ) → CP1 has 4 branching points of order 2. The following picture
(called ”C.S. Peirce quincuncial projection”) illustrates this map (viewed as a doubly periodic map
from C).
168
Chapter 13
The Gamma function
13.1
Product development
The infinite product of complex numbers
∞
Q
j=1
Πn =
n
Q
p j is called convergent if there exists lim Πn , where
n→∞
p j , and this limit is not 0. Sometimes one slightly relaxes the latter condition by allowing
j=1
a finite number of terms in the products to be equal to 0, while the rest of the product is required
to converge to a non-zero number.
Lemma 13.1. The common term pn of a convergent product tends to 1 when n → ∞.
Indeed, pn =
Πn
Πn−1
→ 1.
In view of this lemma we can define (the principal branch of) log pn for all but finite number of
terms of the product.
Theorem 13.2. The product
∞
Q
p j converges if and only if the series
j=1
Proof. It is clear that if
∞
P
∞
P
log p j converges.
j=1
log p j converges then
j=1
∞
Q
p j converges. Indeed,
j=1
n
P
e
j=1
log p j
= Πn =
n
Y
j=1
169
p j,
so the convergence of partial sums S n :=
n
P
log p j implies the convergence of partial products to a
j=1
non-zero number.
The converse is slightly more tricky. Indeed, for the main branch of log it its not true in general
that log(ab) = log a + log b. Suppose there exists a non-zero limit lim Πn = Π. Then
Πn
Π
→ 1 and
pn → 1. Hence, for large n we can write pn = rn eiφn , where |φn | < π.
n
P
Let Πn = Rn eiΘn , Π = ReiΘ , where Θn =
φk . Then
k=1
Rn → R, and ln Rn → ln R,
and there exists a sequence of integer numbers kn such that
Θn − Θ − 2πkn = θn → 0.
Choose n large enough, so that |θn | <
π
2
and |φn | < π. Then
φn = Θn − Θn−1 = 2π(kn − kn−1 ) + θn − θn−1 ,
i.e.
φn − θn + θn−1 = 2π(kn − kn−1 ).
On the other hand,
|φn − θn + θn−1 | < 2π.
Therefore, the sequence kn stabilizes, i.e.
kn = kn−1 = K
Thus, Θn =
n
P
j=1
if n is large enough.
φ j → Θ + 2Kπ. Thus
n
X
j=1
log p j =
n
X
j=1
ln r j + i
n
X
φ j → log R + iΘ + 2πKi.
j=1
n→∞
The product
∞
Q
p j is called absolutely convergent if the series
j=1
1
∞
P
log p j absolutely converges. 1
j=1
Warning: the absolutely convergence implies but not equivalent to the convergence of the product
∞
Q
j=1
170
|p j | (why?)
Lemma 13.3. Denote pn = 1 + an . The product
∞
P
∞
Q
pn absolutely converges if and only if the series
n=1
|an | converges.
n=1
Proof. We have
| log (1+an )|
|an |
→ 1 if |an | → 0. But both (if and only if) assumptions imply that an → 0.
Hence, for large n we have
1 | log (1 + an )|
<
< 2, or
2
|an |
|an |
< | log(1 + an )| < 2|an |.
2
Hence both series are simultaneously converging or not.
Exercise 13.4. 1. Prove that if |z| < 1 then
(1 + z)(1 + z2 )(1 + z4 )(1 + z8 ) · · · =
1
.
1−z
Hint: First verify using Lemma 13.3 that the product absolutely converges, and then use the fact
that every integer has a unique binary presentation.
2. Show that the product
Y
p prime
1
1−
p
!−1
=
Y
p prime
1
1
1 + + 2 + ...
p p
!
diverges.
Hint: Any integer can be uniquely factored as a product of primes and the series
∞
P
n=1
1
n
diverges.
3. Prove that the series
X 1
p
p prime
diverges.
4. Let π(n) denote the number of primes ≤ n. Use Exercise 3 to show that there is no C, > 0 such
that
π(n) < Cn1− .
171
Remark 13.5. The actual rate of growth of the function π(n) is given by the formula
π(n) ∼
n
π(n)
, i.e. lim n = 1.
n→∞
ln n
ln n
This statement known as the prime number theorem has a long history, culminated in the proof in
1896 by J. Hadamard and C.J. de la Vallée Poussin, developing ideas of B. Riemann.
13.2
Meromorphic functions with prescribed poles and zeroes
Let us begin with the following simple lemma.
Lemma 13.6. Let f : C → C be an entire function without zeroes. Then there exists an entire
function g : C → C such that f (z) = eg(z) .
Proof. The function
f 0 (z)
f (z)
is holomorphic. Hence, the differential form
f 0 (z)dz
α :=
f (z)
is closed on C, and hence exact. Therefore, there exists a holomorphic function g(z) such that
dg = g0 (z)dz = α =
Thus g0 (z) =
f 0 (z)
f (z)
f 0 (z)dz
.
f (z)
and
!
d f 0 (z)
−g(z)
0
0
−g(z)
0
f (z)e
= ( f (z) − f (z)g (z))e
= f (z)
− g (z) e−g(z) = 0.
dz
f (z)
Therefore,
f (z) = Ceg(z) = eg(z)+log C , C , 0.
Corollary 13.7. Suppose an entire function f : C → C has finitely many zeroes. Denote zeroes not
at the origin by a1 , . . . , an (where multiple zeroes are repeated), and let m be the multiplicity of the
zero at the origin (possibly, m = 0). Then there exists an entire function g : C → C such that
!
n
Y
z
m g(z)
.
f (z) = z e
1−
aj
j=1
172
Proof. Apply Lemma 13.6 to the entire function
zm
f (z)
1−
n
Q
z
aj
j=1
which has no zeroes.
This corollary can be generalized to the case of infinitely many zeroes. But we will do it only
in some special cases.
13.3
Some product and series developments for trigonometric
functions
Lemma 13.8.
∞
X
1
π2
=
, z < Z.
sin2 πz −∞ (z − n)2
Proof. We first note that the series in the right hand side is uniformly converging on compact sets
in C \ Z, and hence the sum is a meromorphic function with double poles at points of Z. But so
is the left-hand side. The coefficients with
the same, so the difference
1
(z−n)2
in the Laurent expansion about n in both sides are
∞
h(z) =
X
π2
1
−
2
sin πz −∞ (z − n)2
is an entire holomorphic function. Note that
1
| sin(x + iy)| = |eix e−y − e−ix ey | ≥ e|y| − e−|y| → ∞,
|y|→∞
2
and hence
π2
→ = 0.
sin2 π(x + iy) |y|→∞
We also note
∞
X
−∞
∞
X
1
1
=
2
|x + iy − n|
(x − n)2 + y2
−∞
173
(13.3.1)
is monotone decreasing in |y|. On the other hand, h(x + iy) is 1-periodic in x. Hence, |h(z)| is
bounded, and therefore, in view of Liouville’s theorem,
h(z) = const.
Moreover, for x =
1
2
we have
(13.3.2)
2
2
!2
1
−n
−
n
1
1
2
=
−n
.
≤
2
2
2
y
2
1
+
4y
1
2
1
+
−n +y
2
( 21 −n)2
1
2
Hence,
∞
X
−∞
∞
1 X
1
2 ≤
→ 0.
1 + 4y2 −∞ 1 − n 2 |y|→∞
1
+
iy
−
n
2
2
1
(13.3.3)
Combining (13.3.1) and (13.3.3) we conclude that
!
1
h + iy → 0,
|y|→∞
2
and therefore, in view of (13.3.2) we get h(z) = 0.
Lemma 13.9.
!
1 X 1
1
π cot πz = +
+ .
z n,0 z − n n
Proof. First, note that the series in the right-hand side uniformly converges on compact sets in
C \ Z, and hence its sum is a meromorphic function with simple poles at points of Z ⊂ C. Hence,
we can differentiate the series term-wise to get

!
∞
X
d  1 X 1
1 
1
+  = −
.
 +
dz z n,0 z − n n
(z − n)2
−∞
On the other hand, we have
∞
(π cot πz)0 = −
X
π2
1
=
−
.
2
(sin πz)
(z − n)2
−∞
Thus,
!
1 X 1
1
− .
π cot πz = C + +
z n,0 z − n n
P 1
1
But both functions, π cot πz and 1z +
−
are odd (i.e. changing signs when z 7→ −z), and
z−n
n
therefore, C = 0.
n,0
174
Lemma 13.10.
!
∞
Y
Y
z nz
z2
sin πz = πz
1 − e = πz
1− 2 .
n
n
n,0
n=1
Proof. First note that the product πz
z
∞
Q
P
1 − nz e n absolutely converges because the series n12
1
n,0
absolutely converges, see Lemma 13.3. Moreover, it converges uniformly on every compact set in
z
Q
C. The function πz
1 − nz e n has simple zeroes at all real integer points 0, ±1, ±2, . . . . But so
n,0
does sin πz. Hence, the entire function
h(z) =
sin πz
z
Q
πz
1 − nz e n
n,0
has no zeroes, and hence by applying Lemma 13.6 we get:
sin πz = eg(z) πz
Y
z z
1 − en .
n
n,0
(13.3.4)
Let us prove that g(z) is a constant. To do this we compute the logarithmic derivative (i.e.
f0
)
f
of both parts of the equation (13.3.4). We get
!
X 1
1
1
0
π cot πz = + g (z) +
+ .
z
z−n n
n,0
Comparing with the expression for π cot πz from Lemma 13.9 we conclude that g0 (z) = 0, and
hence g(z) = C is a constant. Thus, from (13.3.4) we get
Y
sin πz
z nz
C
=e π
1− e .
z
n
n,0
Passing to the limit when z → 0 we get 1 = eC , and hence we obtain the required formula
sin πz = πz
Y
n,0
1−
z nz
e .
n
Combining terms with n and −n we get the second expression for the right-hand side:
!
∞
Y
z2
sin πz = πz
1− 2 .
n
n=1
(13.3.5)
(13.3.6)
175
13.4
The Gamma function: definition and some properties
Denote
G(z) :=
∞ Y
n=1
z − nz
1+ e .
n
Then (13.3.5) can be rewritten as
sin πz
= zG(z)G(−z).
π
(13.4.1)
Note that we have G(0) = 1.
The value
γ = − log G(1).
is called Euler’s constant. Thus,
−γ
e
=
∞
Y
n=1
Note that the partial product Πn =
!
1 − n1
1+ e
n
(13.4.2)
1
n Q
1 + 1k e− k is equal to
k=1
−
Πn = (n + 1)e
n
P
1
1
k
.
(13.4.3)
Hence,
!
1 1
1
γ = lim 1 + + + · · · + − log n = 0.57722 . . .
n→∞
2 3
n
(13.4.4)
We define now the Gamma function by the formula
z
∞
∞
e−γz
e−γz Y z −1 nz e−γz Y ne n
Γ(z) =
=
1+
e =
.
zG(z)
z n=1
n
z n=1 z + n
(13.4.5)
Thus, Γ(z) is a meromorphic function without zeroes on C with poles at 0, −1, −2, . . . .
Lemma 13.11 (Euler’s formula).
n!nz
(nz = ez ln n ).
n→∞ z(z + 1)(z + 2) . . . (z + n)
Γ(z) = lim
176
(13.4.6)
Proof. By definition,


z
1
n
 n!e−γz Y
j 

e
n!e−γz ez(1+···+ n )



Γ(z) = lim 
 = lim
n→∞
z
z + j  n→∞ z(z + 1)(z + 2) . . . (z + n)
j=1
But according to (13.4.4) we have
−γz
e
z −z
= lim n e
n
P
1
1
k
n→∞
,
and hence we get the required formula (13.4.6).
Theorem 13.12. Properties of the Gamma function.
1. Γ(z + 1) = zΓ(z); in particular, Γ(n + 1) = n!;
√
2. Γ(z)Γ(1 − z) = sinππz ; in particular, Γ( 12 ) = π;
0 (z) P
∞
1
= (z+n)
3. dzd ΓΓ(z)
2.
0
Proof.
1. Using Euler’s formula (13.4.6) we get
!
n!nz+1
n
n!nz
Γ(z + 1) = lim
= z lim
lim
= zΓ(z).
n→∞ (z + 1)(z + 2) . . . (z + n + 1)
n→∞ z + n + 1
n→∞ z(z + 1)(z + 2) . . . (z + n)
Part 2 follows immediately from (13.4.1)and Part 1, while Part 3 can be proven by a direct
computation.
We will need below the following estimate for Γ(z).
Lemma 13.13. Let z = x + iy, x > 0. Then
p
π|y|
xΓ(x)
|Γ(x + iy)| ≥ p
.
p
x2 + y2 sinh π|y|
In particular, for x ∈ [1, 2] we have
|Γ(x + iy)| ≥ Ce−
3π|y|
4
.
Proof. We have
|Γ(x + iy)| = p
e−γx
x2 + y2
∞
Y
1


x
∞
∞
Y
 e−γx Y
n 
ne
x
1


= 
 p
p
q
x 1 n+x
y2
(n + x)2 + y2
x 2 + y2 1
1 + (n+x)
2
x
ne n
177
Let us observe that
x
∞
e−γx Y ne n
= Γ(x)
x 1 n+x
and
∞
Y
1
1
q
1+
≥
y2
∞
Y
1
(n+x)2
1
q
1+
.
y2
n2
On the other hand, according to Lemma 13.10 we have
!
∞
y2
sinh π|y| sin πi|y| Y
=
=
1+ 2 ,
π|y|
πi|y|
n
1
and therefore
∞
Y
1
p
1
q
1+
y2
n2
= p
π|y|
sinh π|y|
.
Thus we get
p
π|y|
xΓ(x)
.
|Γ(x + iy)| ≥ p
p
x2 + y2 sinh π|y|
For x ∈ [1, 2] we have
p
π|y|
p
2π|y|
= p
≥ C1 e−
p
2
2
2
2
π|y|
−π|y|
(x + y ) sinh π|y|
(x + y )(e − e )
π|y|
2
for some constant C1 . Indeed,
p
π|y|
1 1
lim p
= ≥
y→0
(x2 + y2 ) sinh π|y| x 2
and for large |y| we have
π|y|
p
2π|y|
3π|y|
e− 2
≥ C1 p ≥ C10 e− 4 .
p
|y|
(x2 + y2 )(eπ|y| − e−π|y| )
Thus
p
3π|y|
π|y|
xΓ(x)
≥ Ce− 4 .
p
p
x2 + y2 sinh π|y|
Remark 13.14. As it is clear from the proof for x ∈ [1, 2] we have an inequality
|Γ(x + iy)| ≥ Ce−απ|y|
for any constant α > 12 .
178
13.5
Integral representation of the Gamma function
Lemma 13.15. The integral
b
Γ(z) :=
Z∞
e−t tz−1 dt
(13.5.1)
0
absolutely converges if Re z > 0. It satisfies the equation
b
Γ(z + 1) = zb
Γ(z).
(13.5.2)
Proof. Let z = x + iy, x > 0. We have
tz−1 = e(x−1+iy) ln t = t x−1 .
The integral
R1
converges because the function t x−1 is integrable when x > 0, and integral
0
R∞
con-
1
verges because the factor e−t decays faster than any power of t. Equation (13.5.2) follows from the
integration by part formula:
b
Γ(z + 1) =
Z∞
∞
e−t tz dt = −e−t tz +z
0
0
Z∞
e−t tz−1 dt = zb
Γ(z),
0
because lim tz e−t = 0 provided that Re z > 0.
t→0
Hence, the integral
R∞
e−t tz−1 dt defines a holomorphic function on the half-plane {Re z > 0}
0
which satisfy the same relation (13.5.2) as Γ(z). Moreover, b
Γ(1) = Γ(1) = 1. It turns out that
Theorem 13.16.
b
Γ(z) = Γ(z).
In particular, b
Γ(z) admits a meromorphic extension to the whole C.
Proof. The ratio
h(z) =
b
Γ(z)
Γ(z)
is a holomorphic function on the half-plane {Re z > 0}. Moreover, it is 1-periodic: h(z + 1) = h(z),
and hence can be extended by periodicity to a 1-periodic holomorphic function on the whole plane
179
C. Let us study the behavior of h(x + iy) when |y| → ∞ in the strip U := {x ∈ [1, 2]} Note the the
map g(z) = e−2πiz = e−2πy e2πix defines a biholomorphism of the strip U onto C \ {Im z = 0, Re z > 0}.
Moreover, the function f := h ◦ g−1 : C \ {Im z = 0, Re z > 0} → C extends holomorphically to
C \ 0. Indeed, both functions, g(z) end e−2πz are 1-periodic. We note that
g−1 (z) =
2πi + ln(−z)
.
2πi
Let us analyze the behavior of f near 0 and at ∞.
In view of the equality |e−t tz−1 | = |e−t t x−1 | ≤ e−t we have b
Γ(z) ≤ 1, and thus |b
Γ(g−1 (z))| ≤ 1. Let
us now estimate Γ(g−1 (z)) . We have
g−1 (reiφ ) = 1 +
Using Lemma 13.13 and plugging z = 1 +
ln r
φ
−i
.
2π
2π
φ
2π
− i ln2πr we get



!

− 38

,
Cr

3| ln r|
ln r
φ

−i
) ≥ Ce− 8 ≥ 
Γ 1+


3
2π
2π


Cr 8 ,
Thus
r ≥ 1;
r < 1.


3



|b
Γ(z)| 
O(|z| 8 ), |z| → ∞,
=
| f (z)| =

3
|Γ(z)| 


O(|z|− 8 ), |z| → 0.
According to the removable singularity Theorem 7.1 this implies that f has removable singularities at 0 and ∞, and hence f (z) = const. But then h(z) = const, and taking into account that
b
Γ(1) = Γ(1) = 1 we get h(z) = 1, and hence b
Γ(z) = Γ(z).
Exercise 13.17. Prove the following identities:
1. Legendre’s duplication formula
!
1
nz− 2
1
Γ(2z) = √ Γ(z)Γ z + .
2
π
2. Gauss’ formula.
(2π)
n−1
2
Γ(z) = n
z− 21
!
!
z+1
z+n−1
Γ
Γ
...Γ
.
n
n
n
z
180
Chapter 14
The Riemann ζ-function
The series
∞
X
1
ζ(s) =
ns
1
absolutely converges if Re s > 1. Moreover it converges uniformly on {Re s > 1 + } for any > 0.
In particular, its sum is a holomorphic function on {Re s > 1}. As we will show below, it extends
to C as a meromorphic function with the unique simple pole at the point s = 1. This function was
first introduced by L. Euler in 1737 for the real values of the argument s. More than 100 years later
B. Riemann realized the importance of extension of ζ(s) to C, and it is now called the Riemann
ζ-function. This is one of the most famous functions due to its huge importance in number theory
and Mathematics in general.1
14.1
Product development fo ζ(s)
Theorem 14.1.
Y
1
=
1 − p−s , σ = Re s > 1.
ζ(s) p prime
We assume that primes in the product are ordered in the increasing order.
1
It is customary to denote the argument of the ζ-function by the letter s rather than z.
181
Proof. First we note the the product in the right-hand side absolutely converges because the series
P −s
p is converging, see Lemma 13.3. Then
p prime
ζ(s)(1 − 2 ) =
−s
∞
X
−s
n −
∞
X
1
1
X
(2n)−s =
m−s .
m is odd
Similarly,
ζ(s)(1 − 2−s )(1 − 3−s ) =
X
m−s ,
where the sum is taken over all integer m not divisible by 2 and 3. Continuing we get
ζ(s)(1 − 2−s )(1 − 3−s ) . . . (1 − p−s
n ) =
X
m−s ,
where the sum is taken over all integer not divisible by first n prime numbers: 2, 3, . . . , pn . Note
P
that m−s = 1 + T n , where T n = Nn−s + . . . , where Nn > pn . Hence, as lim T n = 0 we get
n→∞
ζ(s)(1 − 2−s )(1 − 3−s ) . . . (1 − p−s
n ) → 1,
n→∞
, i.e.
Y
ζ(s) =
1 − p−s .
p prime
14.2
Meromorphic extension of ζ(s)
Theorem 14.2. ζ(s) =
∞
P
1
n−s extends meromorphically from {Re s > 1} to C with a unique simple
pole at s = 1 with Res1 ζ = 1.
Lemma 14.3.
ζ(s)Γ(s) =
Z∞
x s−1
dx, Re s > 1.
ex − 1
0
Proof. First, we note x s−1 = e(s−1) ln x and that the integral in the right-hand-side converges at 0 and
at infinity.
182
We have
∞
X
e−x
1
e−nx .
=
=
e x − 1 1 − e−x
1
Therefore,
Z∞
0
∞
X
x s−1
dx
=
ex − 1
1
Z∞
x s−1 e−nx dx =
nx=u
0
∞
X
Z∞
u s−1 e−u du
n−s
1
0
= Γ(s)ζ(s).
We note that all the involved series converge absolutely when Re s > 1, and hence all the series
manipulations are justified.
Denote by U an infinite domain in C given by
U = {|z| < } ∪ {Re z > 0, |Im z| < }, 0 < < 2π.
Lemma 14.4.
Γ(1 − s)
ζ(s) = −
2πi
Z
∂U
(−z) s−1
dz.
ez − 1
Here (−z) s−1 = e(s−1) log(−z) , where log z is the principal branch of log defined on the complement
of the negative real semi-axis. We also note that by Cauchy theorem the integral is independent of
as far as < 2π.
Proof. The contour ∂U consists of an arc A of a circle {|z| = } and two rays B− and B+ . It is
R
s−1
straightforward to see that (−z)
dz tends to 0 when → 0 (and Re s > 1), so that the integral
ez −1
over ∂U converges to
Z
∂U
A
(−z) s−1
dz = −
ez − 1
Z∞
x s−1 e−i(s−1)π
dx +
ex − 1
0
= 2i sin(s − 1)π
Z∞
Z∞
0
x s−1
dx.
ex − 1
0
But in view of Lemma 14.3 we have
Z∞
x s−1
dx = ζ(s)Γ(s),
ex − 1
0
183
x s−1 ei(s−1)π
dx
ex − 1
and hence
Z
∂U
(−z) s−1
dz = 2i(sin π(s − 1))ζ(s)Γ(s) = −2i(sin πs)ζ(s)Γ(s).
ez − 1
According to Theorem 13.12.2 we have
(sin πs)Γ(s) =
π
.
Γ(1 − s)
Hence,
−2i(sin πs)ζ(s)Γ(s) = −
Therefore,
Γ(1 − s)
ζ(s) = −
2πi
Z
∂U
2πiζ(s)
.
Γ(1 − s)
(−z) s−1
dz.
ez − 1
Proof of Theorem 14.2. The contour integral
Z
∂U
(−z) s−1
dz
ez − 1
in Lemma 14.4 converges for all s ∈ C. It converges over the circular part A of the contour because
the circle is compact and the function is continuous, and over each ray B± the exponential term in
the denominator ez − 1 grows faster than any power of z. On the other hand Γ(1 − s) is meromorphic
with poles at 1, 2, . . . . But ζ(s) is holomorphic in {Re s > 1}. Hence, these poles are cancelled by
R dz
the zeroes of the integral. On the other hand, the pole at 1 has residue 1, while
= 1 in view
ez −1
of the residue theorem. Hence, ζ(s) has a single pole at 1 with residue 1.
14.3
∂U
Zeroes of the ζ-function
Recall that Bernoulli numbers Bn can be defined as coefficients in the Laurent expansion of the
function
1
:
ez −1
∞
1
1 1 X
Bn 2n−1
(−1)n−1
=
− +
z .
z
e − 1 z 2 n=1
2n!
184
(14.3.1)
Proposition 14.5.




n = 0,
− 12 ,







ζ(−n) = 
0,
n = 2k,








(−1)k Bk , n = 2k − 1.
2k
Proof. According to Lemma 14.4 we have
Γ(1 + n)
ζ(−n) = −
2πi
Z
∂U
(−z)−(n+1)
dz.
ez − 1
Hence, we get


Z
∞

(−1)n n!
1  1 1 X
m−1 Bm 2m−1 
 dz
ζ(−n) =
−
+
(−1)
z


2πi
zn+1 z 2 m=1
2m!
∂U

Z 
∞
n
X

(−1) n!
1
 1
m−1 Bm 2m−n−2 
 dz.
z
=
(−1)
 n+2 − n+1 +
2πi
z
2z
2m!
m=1
∂U
Note that though the domain U is not compact the integral
1
2πi
R
is equal to the sum of residues
∂U
in the domain U . (why?) Hence from the sum under the integral only the term with
1
z
contributes,
and hence we get the required expression from ζ(−n) through the coefficients of the Bernoulli
expansion (14.3.1) multiplied by (−1)n n!.
The values −2k, k > 0, are called trivial zeroes of ζ(s). It is clear from Theorem 14.1 that ζ(s)
has no zeroes in the half-plane {Re s > 1}. One can also show that there are no non-trivial zeroes
in the half-plane {Re s < 0} and on the lines {Re s = 0, 1}. Riemann conjectured that all non-trivial
zeroes lie on the line {Re s = 21 }. This conjecture, known as the Riemann hypothesis, is still open
and became one of the most famous mathematical problems. It is known that many fundamental
results in number theory could be deduced from the Riemann hypothesis.
185