Math 116 Complex Analysis

Home , Biholomorphism, Complex logarithm, Harmonic function, Meromorphic function

Yakov Eliashberg

December 4, 2019 2 Contents

I Complex Analysis Basics 9

1 Linear algebra 11 1.1 Complex numbers ...... 11 1.2 Complex linear function from the real perspective ...... 14

2 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

3 Differential 1-forms and their integration 23 3.1 Complex-valued differential 1-forms ...... 23 3.2 Holomorphic 1-forms ...... 26 3.3 Integration of differential 1-forms along curves ...... 27 3.4 Integrals of closed and exact differential 1-forms ...... 28

4 Cauchy integral formula 31 4.1 Stokes/Green theorem ...... 31 4.2 Area computation ...... 32 4.3 Cauchy theorem and Cauchy integral formula ...... 33 4.4 Integral criterion for exactness ...... 36

3 4.4.1 Proof of Theorem 2.3 ...... 37

5 Convergent power series and holomorphic functions 41 5.1 Recollection of basic facts about series ...... 41 5.2 Power series ...... 42 5.3 Analytic vs holomorphic ...... 44

6 Properties of holomorphic functions 47 6.1 Exponential function and its relatives ...... 47 6.2 Entire functions ...... 48 6.3 Analytic continuation ...... 50 6.4 Complex logarithm ...... 52 6.5 Schwarz reﬂection principle ...... 54

7 Isolated singularities, residues and meromorphic functions 55 7.1 Holomorphic functions with isolated singularities ...... 55 7.2 Residues ...... 57 7.3 Application of the residue theorem to computation of integrals ...... 59 7.4 Complex projective line or Riemann sphere ...... 63 7.5 Residue of meromorphic diﬀerential forms ...... 66 7.6 Argument principle ...... 67 7.7 Winding number ...... 70

8 Harmonic functions 73 8.1 Harmonic and holomorphic functions ...... 73 8.2 Properties of harmonic functions ...... 75

9 Conformal mappings and their properties 79 9.1 Biholomorphisms ...... 79 9.2 Conformal mappings ...... 80

4 9.3 Examples of conformal mappings ...... 81 9.3.1 Unit disc and the upper-half plane ...... 81 9.3.2 Strips and sectors ...... 82 9.4 Schwarz lemma ...... 82 9.5 Automorphisms of the Riemann sphere, C, the unit disc and the upper-half plane . 83

9.5.1 GL(n, C), GL(n, R), PGL(n, C), PGL(n, R) and PGL+(n, R) = PS L(n, R) . . 83 9.5.2 Automorphisms of CP1 and C ...... 84 9.5.3 Automorphisms of H and D ...... 87 9.6 Summary of useful conformal maps ...... 89

10 Riemann mapping theorem 91 10.1 Functional analytic background ...... 92 10.1.1 Arzela-Ascoli´ theorem ...... 92 10.1.2 Montel’s theorem ...... 94 10.2 Proof of the Riemann mapping theorem ...... 95 10.2.1 Embedding into D ...... 95 10.2.2 Maximizing the derivative ...... 95 10.2.3 Preservation of injectivity ...... 96 10.2.4 Surjectivity ...... 96 10.2.5 Discussion: boundary regularity ...... 98 10.3 Annuli ...... 99 10.3.1 Conformal classiﬁcation of annuli ...... 99 10.3.2 Laurent series ...... 100 10.3.3 Proof of Theorem 10.15 ...... 102 10.4 Dirichlet problem ...... 103 10.4.1 Poisson integral and Schwarz formula ...... 104 10.4.2 Solution of the Dirichlet problem for the unit disc ...... 107 10.4.3 Solving the Dirichlet problem for other domains ...... 109

5 11 Riemann surfaces 111 11.1 Deﬁnitions ...... 111 11.2 Uniformization theorem (or strong Riemann mapping theorem) ...... 113 11.3 Quotient construction ...... 113 11.4 Covering maps ...... 115 11.5 Quotient construction and covering maps ...... 116 11.6 Universal cover ...... 117 11.7 Lattices ...... 119 11.8 Branched covers ...... 123 11.9 Riemann surfaces as submanifolds ...... 127 11.9.1 Aﬃne case ...... 127 11.9.2 Projective case ...... 128 11.10Linear projective transformations ...... 130

II Famous meromorphic functions 135

12 Elliptic functions 139 12.1 Elliptic integrals ...... 139 12.1.1 Motivation ...... 139 12.1.2 Elliptic curves ...... 141 12.1.3 Projectivization ...... 142

12.1.4 From a cubic curve to a torus T(ω1, ω2) ...... 144 12.1.5 Summary of the construction ...... 147 12.2 The Weierstrass ℘-function ...... 148 12.2.1 The deﬁnition ...... 148 12.2.2 Diﬀerential equation for ℘(u) ...... 149 12.2.3 Identifying Weierstrass ℘-function with the solution of the pendulum equation ...... 152

6 12.2.4 More about geometry of the Weierstrass ℘-function ...... 153

13 The Gamma function 155 13.1 Product development ...... 155 13.2 Meromorphic functions with prescribed poles and zeroes ...... 158 13.3 Some product and series developments for trigonometric functions ...... 159 13.4 The Gamma function: deﬁnition and some properties ...... 162 13.5 Integral representation of the Gamma function ...... 165

14 The Riemann ζ-function 167 14.1 Product development fo ζ(s) ...... 167 14.2 Meromorphic extension of ζ(s)...... 168 14.3 Zeroes of the ζ-function ...... 170

7 8 Part I

Complex Analysis Basics

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−1z = 1. Indeed, we can set a b z−1 = − i. a2 + b2 a2 + 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 asz ¯ = a − bi. The conjugation operation z 7→ z¯ is the reflection of C with respect to the real axis R ⊂ C. Note that 1 1 Re z = (z + z¯), Im z = (z − z¯). 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, ttrigonometric. 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| = zz¯. An important role plays the triangle inequality

|z1| − |z2| ≤ |z1 + z2| ≤ |z1| + |z2|.

Exponential function of a complex variable

Recall that the exponential function ex has a Taylor expansion X∞ xn x2 x3 ex = = 1 + x + + + .... n! 2 6 0 We then deﬁne for a complex z the exponential function by the same formula z2 zn ez := 1 + z + + ··· + + .... 2! n!

12 One can check that this power series absolutely converging for all z and satisﬁes the formula

ez1+z2 = ez1 ez2 .

In particular, we have y2 y3 y4 eiy = 1 + iy − − i + + ··· + ... (1.1.1) 2! 3! 4! X∞ y2k X∞ y2k+1 = (−1)k + i (−1)k . (1.1.2) 2k! (2k + 1)! k=0 k=0

∞ 2k ∞ 2k+1 P − k y P − k y But ( 1) 2k! = cos y and ( 1) (2k+1)! = sin y, and hence we get Euler’s formula k=0 k=0 eiy = cos y + i sin y, and furthermore, ex+iy = exeiy = ex(cos y + i sin y), i.e. |ex+iy| = ex, 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.

iφ1 iφ2 Given z1 = r1e , z2 = r2e we get iφ1 iφ2 i(φ1+φ2) z1z2 = r1e r2e = r1r2e ,

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 = rneinφ = rn(cos nφ + i sin nφ). 7→ π Note that the operation z iz is the rotation of C counterclockwise by the angle 2 . 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.

1Convergence of power series will be discussed later in Chapter 5.

13 n n Exercise 1.1. 1. Compute P cos kθ and P sin kθ. 0 1

n n n 2. Compute 1 + 4 + 8 + 12 + ....

1.2 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),

2 where z1, z2, z are any vectors from R 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   2 Thus, viewing a complex number z = x + iy as a vector   ∈ R we get y      −    a b x F(z) =     b a  y

In other words, we proved the following   a b 2 2   Lemma 1.2. A real linear map F : R → R 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   2 2 2 det   = a + b = |c| , where c = a + ib. (1.2.2) b a  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)

1 − i − Note that x = 2 (z + z¯), y = 2 (z z¯), and hence A Bi 1 1 F(x, y) = Ax + By = (z + z¯) − (z − z¯) = (A − iB)z + (A + iB)¯z = αz + βz¯, (1.2.4) 2 2 2 2 1 − 1 2 where we denoted α := 2 (A iB), β := 2 (A + iB). Note that the function l1(z) = αz is complex

linear, while the function l2(z) = βz¯ is complex anti-linear, which means that it is linear in the real

sense, but satisﬁed the condition l2(λz) = λ¯l2(z). If F is a complex linear map, then F¯ is anti-linear and vice versa. In particular, every complex anti-linear map F has the form F(z) = az¯ 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 = F + F , where F is complex linear and F is complex anti-linear. 1 2 1   2 a b   1 − 1 If F is given by a matrix   then F1(z) = αz, F2(z) = βz,¯ where α := 2 (A iB), β := 2 (A+iB), 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, z¯) form a basis of this space.

15 16 Chapter 2

Holomorphic functions

2.1 Diﬀerentiability and the diﬀerential

2 2 2 For any point z = (x, y) ∈ R we denote by Ra the space R with the origin, shifted to the point a. 2 2 Though the parallel transport allows one to identify spaces R and Ra 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, where U ⊂ R2 a domain in R2, is called diﬀerentiable at a point a ∈ U if near the point a in can be well approximated by a linear

2 2 function. More precisely, if there exists a linear map A : Ra → R such that

f (a + h) − f (a) = A(h) + o(||h||)

2 for any sufficiently small vector h = (h1, h2) ∈ R , where the notation o(t) stands for any vector- valued function such that o(t) → 0. The linear function A is called the differential of the function t t→0 2 f at the point a and is denoted by da f . In other words, f is differentiable at a ∈ U if for any h ∈ Ra there exists a limit f (a + th) − f (a) da f (h) = A(h) = lim , t→0 t

2 2 and the limit A(h) linearly depends on the vector h. By identifying Ra and with R 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) ∂u (a)  ∂x ∂y  Ja( f ) =   .  ∂v ∂v  ∂x (a) ∂y (a) 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) f (a + h) − f (a) = ch + o(|h|), for a complex number c;

(2) There exists a limit f (a + h) − f (a) lim , denoted by f 0(a) h→0 h and called the complex derivative of f at the point a;

(3) f is diﬀerentiable in the real sense and the following Cauchy–Riemann equations are satis- ﬁed at the point a: ∂u ∂v = , ∂x ∂y (2.2.1) ∂v ∂u = − . ∂x ∂y

18 (4) f is diﬀerentiable in the real sense and ! ∂ f 1 ∂ f ∂ f (a):= (a) + i (a) = 0. ∂z¯ 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) = βh¯. Rephrasing Lemma 1.3 we have ! ! 1 ∂ f ∂ f 1 ∂ f ∂ f α = (a) − i (a) h + (a) + i (a) h¯. 2 ∂x ∂y 2 ∂x ∂y If we introduce the notation ! ∂ f 1 ∂ f ∂ f (a) = (a) − i (a) , ∂z 2 ∂x ∂y ! ∂ f 1 ∂ f ∂ f (a) = (a) + i (a) , ∂z¯ 2 ∂x ∂y then we can write ∂ f ∂ f d f (h) = (a)h + (a)h¯. a ∂z ∂z¯ Hence, f is holomorphic at a point a if and only if ∂ f (a) = 0, (2.2.2) ∂z¯ and this condition is just another form of the Cauchy-Riemann equations (2.2.1). ∂ f 0 ∂ f It is important to note that f is holomorphic at a, i.e. ∂z¯ (a) = 0 then f (a) = ∂z (a). Note that if f is holomorphic at the point a then the determinant of the real Jacobi matrix is given by

∂u (a) ∂u (a) det J( f )(a) = ∂x ∂y = | f 0(a)|2, ∂v ∂v ∂x (a) ∂y (a)

19 see formula (1.2.2) The function f : U → C is called holomorphic in U if it is holomorphic at every point of U. ∂ f This is equivalent to the condition ∂z¯ = 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);

f if g(a) , 0 then 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 C1-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 C1, 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 C1 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 ∂ f directional derivative ∂v (a) 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 . ∂v t→0 t

In particular, ∂ f ∂ f (a) = d f (1), (a) = d f (i). ∂x a ∂y a ∂ f ∂ f The derivatives ∂z and ∂z cannot be interpreted as partial derivatives of the function f . However, 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. 1 − 1 Let us write z = x + iy, w = u + iv. Consider vector ﬁelds T = 2 (1, i) and S = 2 (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 F(T), (a) = d F(S ). ∂z (a,a) ∂z (a,a)

Proof. We have ! 1 1 ∂ f ∂ f ∂ f d F(T) = (d ( f )(1) − d ( f )(i)) = (a) − i (a) = (a). (a,a) 2 a a 2 ∂x ∂y ∂z ! 1 1 ∂ f ∂ f ∂ f d F(S ) = (d ( f )(1) + d ( f )(i)) = (a) + i (a) = (a). (a,a) 2 a a 2 ∂x ∂y ∂z

If f is holomorphic at a then for any vector v ∈ Ca we have

∂ f (a) = d f (v) = f 0(a)v, ∂v a

21 where we identify v with a complex number v ∈ C via a parallel transport. For instance,

∂ f (a) = d f (1) = f 0(a) · 1 = f 0(a), ∂x a ∂ f (a) = d f (i) = i f 0(a). ∂y a

∂ f 0 ∂ f ∂ f In this case ∂z (a) = f (z) = ∂x (a) and ∂z (a) = 0.

22 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

2 a ﬁeld of linear functions λz : Rz → R. Thus, a diﬀerential 1-form is a function of arguments of 2 kind: of a point z ∈ U and a vector

2 h ∈ Rz . 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,

2 where h1, h2 are Cartesian coordinates of h ∈ Rz . 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 → R2 on U its differential d f is an example of a differential form: ∂ f ∂ f d ( f )(h) = (z)h + (z)h . z ∂x 1 ∂y 2 In particular differentials dx and dy of the coordinate functions x, y are also differential 1-forms, 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

∂ f ∂ f d 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. Not every closed differential 1-form λ = Pdx + Qdy is exact. The necessary condition for

∂P ∂Q exactness is that λ is closed which by deﬁnition means ∂y = ∂x . The necessity of this condition for exactness follows from the mixed derivatives equality (assuming that the coeﬃcients P, Q are

1 ∂ f ∂ f C -smooth). Indeed, if P = ∂x and Q = ∂y . Then

∂P ∂2 f ∂Q = = . ∂y ∂x∂y ∂x

On the other hand the closedness of λ is not suﬃcient for its exactness, as it is demonstrated by an example of a 1-form dφ on R2 \ 0 (written in polar coordinates), or

1 dφ = (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 Example 3.1.

dz = dx + idy, dz¯ = dx − idy, zdz = (x + iy)(dx + idy) = xdx − ydy + i(xdy + ydx),

We have 1 i dx = (dz + dz¯), dy = − (dz − dz¯). 2 2 Hence, any complex valued 1-form λ can be written as a linear combination of forms dz and dz¯:

λ = f dz + gdz¯, which is a decomposition of λ into a sum of complex linear and complex anti-linear parts.

Lemma 3.2. The form λ = f dz + gdz¯ is closed if and only id ∂ f ∂g = . ∂z¯ ∂z Proof.

λ = f dz + gdz¯ = f (dx + idy) + g(dx − idy) = ( f + g)dx + i( f − g)dy.

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 ∂ f ∂g ∂g + i = − i , ∂x ∂y ∂x ∂y and hence ! ! ∂ f 1 ∂ f ∂ f 1 ∂g ∂g ∂g = + i = − i = . ∂z¯ 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 dz¯ = dx − idy parts.

25 Lemma 3.3. For any complex valued function f = u + iv we have ∂ f ∂ f d f = dz + dz¯. ∂z ∂z¯ In particular, when f is holomorphic we have ∂ f d f = dz = f 0(z)dz. ∂z Proof. We have ∂ f ∂ f 1 ∂ f i ∂ f d f = dx + dy = (dz + dz¯) − (dz − dz¯) ∂x ∂y 2 ∂x 2 ∂y ! ! 1 ∂ f ∂ f 1 ∂ f ∂ f ∂ f ∂ f − i dz + + i dz¯ = dz + dz¯. 2 ∂x ∂y 2 ∂x ∂y ∂z ∂z¯

3.2 Holomorphic 1-forms

A complex-valued 1-form λ is called holomorphic if λ = f dz for a holomorphic function f .

Lemma 3.4. The form f dz is closed in a domain U if and only if the function f is holomorphic in U.

∂ f Proof. According to Lemma 3.2 closedness of f dz is equivalent to ∂z¯ = 0, which, in turn, is equivalent to the holomorphicity of f .

dz Example 3.5. The holomorphic form 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 dz i.e. zn is exact. If n = 1 we have in polar coordinates dz d(reiφ) eiφdr + ireiφdφ dr = = = + idφ = d(ln r) + idφ, z reiφ reiφ r but we already discussed above that the form dφ is not exact.

26 3.3 Integration of diﬀerential 1-forms along curves

Curves as paths

A path, or parametrically given curve in a domain U ⊂ R2 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 Zb α = γ∗α.

γ 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, so that Z Zb Zb α = γ∗α = (P(x(t), y(t))x0(t) + Q(x(t), y(t))y0(t))dt.

γ a 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 3.6. Let a path eγ be obtained from γ :[a, b] → U by a reparameterization, i.e. R R eγ = γ ◦ φ, where φ :[c, d] → [a, b] is an orientation preserving diﬀeomorphism. Then α = α. eγ γ R Thus the integral α depends only on the curve γ as an oriented submanifold and not on a γ particular parameterization which is compatible with the orientation. For instance, given a unit R circle S 1 = {|z| = 1} oriented counter-clockwise we can compute dφ = 2π. Indeed, the circle can S 1 be parameterized by the angular coordinate φ ∈ [0, 2π], and this parameterization is compatible R R2π with the counter-clockwise orientation. Hence, dφ = dφ = 2π. S 1 0

R dz Exercise 3.7. Compute z S 1

27 iφ dz Solution. Let us parameterize the circle by polar coordinates: z = e , φ ∈ [0, 2π]. Then z = 2π R dz R R d(ln r) + idφ and z = d(ln r) + dφ = 0 + 2πi = 2πi. Here we used the fact that the integral S 1 S 1 0 of an exact form d(ln r) over the circle S 1 is 0, see Theorem 3.8 below.

3.4 Integrals of closed and exact diﬀerential 1-forms

Theorem 3.8. 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 H In particular, if γ is a loop then α = 0. γ Similarly for an oriented curve Γ ⊂ U with boundary ∂Γ = B − A we have Z α = f (B) − f (A).

R Rb Rb Proof. We have d f = γ∗d f = d( f ◦ γ) = f (γ(b)) − f (γ(a)) = f (B) − f (A). γ 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 3.9. 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.

28 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,

Z Z1 Z1 F(z) = α = P(tx, ty)d(tx) + Q(tx, ty)d(ty) = (xP(tx, ty) + yQ(tx, ty))dt (3.4.1)

−→ 0 0 I0,z

Taking partial derivatives of the integral with respect to x and y we get

1 ! 1 ∂F Z ∂P ∂Q Z (x, y) = tx (tx, ty) + ty (tx, ty) dt + P(tx, ty)dt; ∂x ∂x ∂x 0 0 1 ! 1 ∂F Z ∂P ∂Q Z (x, y) = tx (tx, ty) + ty (tx, ty) dt + Q(tx, ty)dt ∂y ∂y ∂y 0 0

∂P ∂Q By our assumption the form α is closed, and hence ∂y = ∂x . Using this we can further write

1 ! 1 ∂F Z ∂P ∂P Z (x, y) = tx (tx, ty) + ty (tx, ty) dt + P(tx, ty)dt; ∂x ∂x ∂y 0 0 1 1 1 1 Z ∂P Z 1 Z Z = t (tx, ty)dt + P(tx, ty) = tP(x, y) − P(tx, ty)dt + P(tx, ty)dt = P(x, y); ∂t 0 0 0 0 0 1 ! 1 ∂F Z ∂Q ∂Q Z (x, y) = tx (tx, ty) + ty (tx, ty) dt + Q(tx, ty)dt ∂y ∂x ∂y 0 0 1 1 1 1 Z ∂Q Z 1 Z Z = t (tx, ty)dt + Q(tx, ty) = tQ(x, y) − Q(tx, ty)dt + Q(tx, ty)dt = Q(x, y); ∂t 0 0 0 0 0

Thus

dF =Pdx + Qdy = α

29 R Given a diﬀerential 1-form α = Pdx + Qdy we will deﬁne |α| as Γ

Z Zb |α| := |P(x(t), y(t))x0(t) + Q(x(t), y(t))y0(t)| dt, Γ a R R where (x(t), y(t), t ∈ [a, b], is a parameterization of Γ. Unlike α the integral |α| is non-negative Γ Γ and does not depend on the orientation of Γ. Clearly, we have

Z Z α ≤ |α|,

Γ Γ and Z Z Z |α + β| = |α| + |β|.

Γ Γ Γ

30 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 it 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 C1-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 "

31 Corollary 4.2. Suppose a 1-form α is closed in a domain U. Then

Z α = 0.

∂U

∂Q ∂P Indeed, closedness of α = Pdx + Qdy just means that ∂x − ∂y = 0.

4.2 Area computation

∂Q ∂P If for α = Pdx + Qdy we have ∂x − ∂y = 1 in the closure U of the domain U, then Green’s formula yields Z Pdx + Qdy = dxdy = Area(U).

∂U U " − 1 − For instance this is the case for α = xdy, ydx or 2 (xdy ydx). Another example of such 1-form α is the form i i 1 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 neighborhood and f : Ω → C a holomorphic function such that f (Γ) ⊂ C bounds a domain U ⊂ C. Suppose f (Γ) is oriented as the boundary of U and Γ is oriented accordingly and f preserves these orientations. Then

i Z Area(U) = − f (z) f 0(z)dz. 2 Γ

Proof. Using Green’s formula together with the change of variable formula we get

i Z i Z i Z Area( f (U)) = − zdz = − f (z)d f (z) = − f (z) f 0(z)dz. 2 2 2 f (Γ) Γ Γ

32 4.3 Cauchy theorem and Cauchy integral formula

Corollary 4.4 (Cauchy theorem). Let f be a function, holomorphic in the domain U and continuous up to the boundary. Then Z f (z)dz = 0.

∂U

Proof. According to Lemma 3.4 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 2πi, n = 1, dz  n =  z  ∂U 0, otherwise.

dz Indeed, for n > 1 the 1-form zn 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 dz Z dz Z dz 0 = = − . z z z ∂(U\{D }) ∂U ∂D

R dz But we already computed that z = 2πi, and therefore {|z|=})

Z dz = 2πi. z ∂U

Theorem 4.6 (Cauchy integral formula). Suppose that f : U → C is a continuous function which is holomorphic in U. Then for any u ∈ U we have

1 Z f (z)dz = f (u). 2πi z − u ∂U

33 f (z) Proof. Let Dδ(u) denote the disc {|z − u| < } centered at u, where 0 < δ < | − |u|. The function 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 f (z)dz Z f (z)dz Z f (u + w)dw = = . z − u z − u w ∂U ∂Dδ(u) |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,

2π Z f (u + w)dw Z f (u)dw Z | f (u + w) − f (u)|δdφ − ≤ ≤ 2π. w w δ |w|=δ |w|=δ 0

Note that according to Example 4.5

Z f (u) Z dw = f (u) = 2πi f (u). w w |w|=δ |w|=δ

Thus

Z f (z)dz − 2πi f (u) ≤ 2π z − u ∂U for any > 0. But the left-hand side is independent of , and therefore

1 Z f (z)dz = f (u). 2πi z − u ∂U As the first application of the Cauchy integral formula we prove the infinite differentiability of a holomorphic function.

34 Corollary 4.7. Any holomorphic in a domain U function is inﬁnitely diﬀerentiable at every point. Its derivatives can bu computed by the formula

k! Z f (z)dz f (k)(u) = . 2πi (z − u)k+1 ∂U

R f (z)dz f (z)dz Proof. The variable u enters the integral z−u as a parameter. The integrand z−u is differntiable ∂D 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 ! Z 0 d  1 f (z)dz 1 f (z)dz 1 f (z)dz f (u) =   = = . du 2πi z − u  2πi z − u 2πi (z − u)2 ∂U ∂U ∂U

R f (z)dz 00 Applying the same argument to the integral (z−u)2 we compute f (u), etc. ∂U

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

Mn! | f (n)(z )| ≤ , 0 rn

where M := max | f (z)|. |z−z0|=r

Proof. By the Cauchy integral formula we have

n! Z f (z + ζ)dζ f (n)(z ) = 0 . 0 2πi ζn+1 |ζ|=r

Therefore, 2π n! Z Mrdθ Mn! | f (n)(z )| ≤ = . 0 2π rn+1 rn 0

35 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 Z f (x + t, y) − f (x, y) = α,

lt where lt is a straight interval connecting the point (x, y) with the point (x + t, y). Hence

Z Zx+t f (x + t, y) − f (x, y) = α = P(u)du.

lt x

Hence   Zx+t ∂ f f (x + t, y) − f (x, y) d   (x, y) = lim =  P(u)du = P(x, y). ∂x t→0 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

36 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|).

∂ f ∂ f And this is by deﬁnition means that f is diﬀerentiable and that d f = ∂x dx+ ∂y dy = Pdx+Qdy = α.

R Remark 4.10. The analysis of the above proof shows that it is enough to assume that α = 0 only γ 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.

4.4.1 Proof of Theorem 2.3

Green’s theorem required the form to be C1-smooth. Hence, in order to deduce from it Cauchy’s theorem we had to assume C1-smoothness of a holomorphic function. We prove below Looman- Menchoﬀ’s theorem 2.3, which shows that the C1-condition follows from holomorphicity, i.e. complex diﬀerentiability at every point of the domain. The key in the proof is the following Lemma 4.11 which is due to R. Narasimhan, see Section 1.6 in the book “Complex Analysis on One Vari- able” by R. Narasimhan and Y. Nievergelt, Birkhauser,¨ 2001.

Lemma 4.11. 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

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.

R Suppose the converse, i.e. that f (z)dz = a > 0. Let us partition the rectangular Q into 4

∂Q L rectangulars Q1, Q2, Q3 and Q4 of perimeter 2 . Then

Z X4 Z f (z)dz = f (z)dz. (4.4.1) 1 ∂Q ∂Q j

Indeed, each side S of Q j which is not contained in ∂Q is also a side of another rectangular 0 Q j0 , j , 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

Z X4 Z a = f (z)dz ≤ f (z)dz . (4.4.2)

1 ∂Q ∂Q j

Hence, there exists j0 ∈ {1, 2, 3, 4} we have

Z a f (z)dz ≥ . 4

∂Q j0

L Now we partition the rectangular Q j0 into 4 rectangulars Q j01, Q j02, Q j03, Q j04 of perimeter 4 and again conclude that there exists j1 ∈ {1, 2, 3, 4} such that

Z a f (z)dz ≥ . 16

∂Q j0 j1 Continuing this process we ﬁnd a sequence of rectangulars

Q j0 ⊃ Q j0 j1 ⊃ · · · ⊃ Q j0 j1... jk ⊃ ...,

L such the perimeter of ∂Q j0 j1... jk is equal to 2k and

Z a f (z)dz ≥ . (4.4.3) 4k

∂Q j0 j1... jk

38 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 L | f (z) − f (z ) − f 0(z )(z − z )| < |z − z | < , (4.4.4) 0 0 0 0 2k

− | ≤ L because z z0 2 < L. Note that Z 0 ( f (z0) + f (z0)(z − z0))dz = 0,

∂Q j0 j1... jk

0 1 because the inhomogeneous linear function f (z0) + f (z0)(z − z0) is holomorphic and C -smooth. Therefore, (4.4.5) implies that

Z L2 f (z)dz ≤ . (4.4.5) 4k

∂Q j0 j1... jk

a Choosing < L2 we get a contradiction with (4.4.3).

Proof. [Proof of Theorem 2.3] In view of Theorem 4.9 and Remark 4.10 Lemma 4.11 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.

39 40 Chapter 5

Convergent power series and holomorphic functions

5.1 Recollection of basic facts about series

P∞ Let us recall that a series bk, where b j are complex numbers, is called converging if there exists 0 a ﬁnite limit of partial sums XN S := lim S N = lim bk. N→∞ N→∞ k=0 P∞ In this case we write bk = S . A necessary and suﬃcient condition for convergence is given by 0 Cauchy criterion: n+m P For any > 0 there exists N such that for any n ≥ N and m > 0 we have bk < . k=n P∞ P∞ A series bk is called absolutely converging if the series |bk| is converging. Absolute con- 0 0 vergence implies convergence, as it it immediately follows from the Cauchy criterion and the inequality

Xn+m Xn+m b ≤ |b |. k k k=n k=n An important tool for establishing an absolute convergence (or divergence) is the following comparison criterion:

41 P P Lemma 5.1. Let an and 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.

5.2 Power series

∞ P n A power series is a series of the form anz , an, z ∈ C. 0 A remarkable fact about power series is existence of a radius of convergence. ∞ P n Proposition 5.2. For any power series anz there exists R (which could be = 0 or ∞) such that 0 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 |a | n . R n Proof. The proof follows from the comparison with a geometric series P rn which converges when

1 r < 1 and diverges when r ≥ 1. Indeed, for any r < R we have |an| < rn for a suﬃciently large n. n ∞ n |z| P n Hence, if |z| < r then |an||z| < r , and therefore the power series anz is absolutely converging 0 ∞ P |z| n due to the comparison with the geometric series r . But r is any number < R, and hence 0 ∞ P n anz is absolutely converging for all |z| < R. If |z| > R then there exists inﬁnitely many n such 0 ∞ n 1 n P n that |an| > |z| , and hence for these values of n we have |an||z| > 1. This implies that anz 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 zn is equal to 1. On the boundary of {|z| = 1} 0 of the disc of convergence the series diverges at every point. ∞ P zn 2. The radius of convergence of the geometric series n is also equal to 1. However, the behavior 1 on the boundary of the disc of convergence is diﬀerent: the series is convergent at every point except z = 1.

42 ∞ P zn ∞ 3. The radius of convergence of the exponential series n! is equal to , i.e. the series is absolutely 0 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 any 0 z , 0.

∞ ∞ P n P n Exercise 5.4 (Operations on converging power series). Suppose power series anz and bnz 0 0 ∞ ∞ ∞ P n P n P n are converging for |z| < R. Denote A(z):= anz ,B(z):= bnz . Then the series (an + bn)z 0 0 n=0 ∞ n ! P P n and akbn−k z are also converging for |z| < R and n=0 k=0

∞ X n A(z) + B(z) = (an + bn)z , n=0   X∞ Xn   n A(z)B(z) =  akbn−k z . n=0 k=0

∞ P n Proposition 5.5. Suppose the series f (z) = anz is converging in the disc DR := {|z| < R}. Then 0

∞ ∞ P n−1 P n−1 0 a) The series nanz is also converging in DR and nanz = f (z). Thus, the sum of a 1 1 power series is holomorphic in the disc of its convergence.

∞ P an n+1 0 b) The series n+1 z is converging in DR to a function F(z) such that F (z) = f (z). 0

∞ ∞ P n P n−1 Proof. First observe that according to the Hadamard criterion the power series anz , nanz 0 1 ∞ P an n+1 and n+1 z have the same radius of convergence. Indeed, 0

1 ! n 1 1 |an| lim sup(n|a |) n = lim sup |a | n = lim sup . n n n + 1

∞ P n−1 0 Let us prove that nanz = f (z). Recall an identity 1

un − vn = un−1v + un−2v2 + ··· + uvn−1. u − v

43 Diﬀerentiating both sides with respect to v we get −nvn−1(u − v) + un − vn = un−1 + 2un−2v + ··· + (n − 1)uvn−2. (u − v)2

Taking u = z + h, v = z ∈ DR− we get

n n (z + h) − z − − − − − nzn 1 ≤ |h| (R − )n 1 + 2(R − )n 1 + ··· + (n − 1)(R − )n 1 h n(n − 1)(R − )n−1|h| = . 2 Hence, we have

| − X∞ f (z + h) f (z) n−1 − nanz h 1

X∞ n − n ! (z + h) z n−1 ≤ an − nz h 1 X∞ n(n − 1)(R − )n−1|h| ≤ |a | ≤ C()|h|. n 2 1 ∞ f (z+h)|− f (z) P n−1 Hence, h converges to nanz when h → 0. 1 ∞ P an n+1 To prove b) we apply a) to the series F(z) = n+1 z . 0

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.

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 f 00(z ) X∞ f (n)(z ) f (z + u) = f (z ) + f 0(z )u + 0 u2 + ··· + 0 un + ..., 0 0 0 2 n! 0

44 for |u| < .

Indeed, according to Proposition 5.5 a converging power series can be diﬀerentiated term-wise ∞ P n in the disc of its convergence. Hence, if f (z0 + u) = anu then 0

f (z0) = a0, ∞ 0 X n−1 f (z0 + u) = nan−1u , and hence 1 0 f (z0) = a1. Continuing this process we get f (n)(z ) a = 0 . n 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

1 Z f (ζ)dζ f (u) = . 2πi ζ − u {|ζ|=r}

We have f (ζ) f (ζ) 1 f (ζ) X∞ un = = . ζ − u ζ 1 − u ζ ζn ζ 0 Let us prove that the power series in the right hand side can be integrated term-wise and thus we get   ∞ Z f (ζ)dζ X Z f (ζ)dζ  =   un. (5.3.1) ζ − u  ζn+1  0   |ζ|=r |ζ|=r

Then for any u ∈ D, |u| < r we have

f (ζ) Mr ≤ , ζn+1 rn+1

45 where we denoted Mr := 2π max | f (ζ)|, and hence the power series in the right-hand side absolutely |ζ|=r converges in D. Let us choose any ρ < r. Then for any u ∈ D, |u| ≤ ρ we have     Z XN Z  X∞ Z  f (ζ)dζ  f (ζ)dζ  n  f (ζ)dζ  n −   u ≤   |u| ζ − u  ζn+1   ζn+1  0   N+1   |ζ|=r |ζ|=r |ζ|=r ∞ n N+1 N+1 Mr X ρ Mrρ Mrρ ≤ = ρ = → 0. r r rN+2(1 − ) rN+1(r − ρ) N→∞ N+1 r

This proves formula (5.3.1) for any u ∈ D because ρ is an arbitrary number < r.

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. R Proposition 5.9. Let f (z) be a continuous function in U. Suppose that f (z)dz = 0 for any piece- γ 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 .

46 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 holomorphic 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

X∞ z2n+1 sin z = (−1)n , (2n + 1)! 0 X∞ z2n cos z = (−1)n , (2n)! 0 ez − e−z X∞ z2n+1 sinh z = = , 2 (2n + 1)! 0 ez + e−z X∞ z2n cosh z = = 2 (2n)! 0

47 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

Functions which are holomorphic on the whole C 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 a inﬁnite radius of convergence.

Theorem 6.2 (Liouville’s theorem). If an entire function is bounded it is a constant.

48 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

Mk! | f k(0)| ≤ for any R > 0. Rk

Hence f k(0) = 0 for k > 0. Therefore,

1 f (z) = f (0) + f 0(0)z + f 00(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.

n Theorem 6.4 (Fundamental theorem of algebra). Any polynomial P(z) = a0+a1z+···+anz , an , 0, of degree n > 0 has a root.

∈ 1 Proof. Suppose P(z) , 0 for all z C. Then g(z):= P(z) is an entire holomorphic function. On the other hand, ! |a | |a | |a | |P(z)| ≥ |zn| |a | − n−1 − n−2 − · · · − 0 . n |z| |z|2 |z|n

|an−2| |a0| But 2 + ··· + n → 0, and therefore |z| |z| |z|→∞

|a | |a | |a | n−2 + ··· + 0 ≤ n |z|2 |z|n 2

for |z| suﬃciently large. But then ! |a | |a | |a | |a ||z|n |P(z)| ≥ |zn| |a | − n−1 − n−2 − · · · − 0 ≥ n , n |z| |z|2 |z|n 2

49 and therefore 2 | | ≤ → g(z) n 0. |an||z| |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.

50 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:

2 f (a + u) = c1u + c2u + ....

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

k ck+1 ck+2 2 f (a + u) = cku (1 + g(u)), where g(u) = u + u + .... ck ck

The function g is holomorphic in a neighborhood of u = 0 and we have g(0) = 0. Hence, there | | 1 | | exists r > 0 such that g(u) < 2 for u < r. Therefore, 1 | f (a + u)| ≥ |c ||u|k, for |u| < r. 2 k

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 za → 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.

∞ P n 1 Example 6.6. 1) The radius of convergence of the series z is 1. However the function f (z) = 1−z 0 ∞ provides a holomorphic extension of P zn from the unit disc {|z| < 1} to C \ 1. 0

51 6.4 Complex logarithm

dz Consider a closed diﬀerential 1-form 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,

dz where R is the ray R = {z ∈ C; Rez ≤ 0, Imz = 0}. The primitive of 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 Z dz log z = , z Γz where Γz is any path connecting 1 with the point z. Then

dz 1 d log z = , or (log z)0 = . z z

dz As we already computed above, z = d ln r + idφ, and therefore Z dz Z Z log z = = d(ln r) + i dφ z Γz Γz 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). 1) elog z = z

∞ P n zn 2) log(1 + z) = (−1) n for |z| < 1. 1

log z log r+iφ iφ dz 1 Indeed, e = e = re = z. To prove 2) we observe that d(log(1 + z)) = 1+z . But 1+z = ∞ ∞ P n n P n zn (−1) z for |z| < 1. Hence, by integrating both parts of this equality we get log(1 + z) = (−1) n 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 z1z2 = log z1 + log z2 holds only up to a multiple of 2πi.

52 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 logU 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 ≤ U 0, Imz = 0}. Let log : 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).

U R dz R dz R Indeed, log z − ln z = z , where λ a loop, but z = i dφ = 2kπi. λ λ λ Corollary 6.9. elogU z = z.

Indeed, if z ∈ U \ R = U ∩ U0 then according to Lemma 6.8 we have

U elog z = eln z+2kπi = z.

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.

1 √ √ n 1 n n n In particular, if a = n then a branch of z = z deﬁned this way satisﬁes z = z.

53 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

Ub := U ∪ I ∪ U.

Let f : U → C be a holomorphic function which extends continuously to I and takes real values on I. Then f holomorphically extends to U.b

Proof. Deﬁne f (z) = f (¯z) for each z ∈ U and extend it by continuity to I. To show that f is holomorphic consider any piecewise linear loop γ ⊂ Ub. The function f is holomorphic in U and U R and extends continuously to I from both sides. Hence, the integral f (z)dz is equal to 0 over loops γ δ in U ∪ I and U ∪ I. But any loop γ in Ub is split by the interval I into several loops, each one is R either in U ∪ I or U ∪ I. Hence, f (z)dz = 0, and by Proposition 5.9 the function f is holomorphic γ in Ub.

54 Chapter 7

Isolated singularities, residues and meromorphic functions

7.1 Holomorphic functions with isolated singularities

Let U ⊂ C be an open domain. A closed subset Z ⊂ U is called discrete of 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.

55 Then g is diﬀerentiable at a and g0(a) = 0. Indeed

g(z) |g0(a)| = | lim | ≤ lim | f (z)|z − a|| ≤ |z − a|1−σ = 0, z→a 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)2h(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 g(z) f (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 P(z) are all non-essential is called meromorphic. For instance, any rational function 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 g(z) Proof. If u is a pole then near u we can write f (z) = (z−u)k , where g is a holomorphic function and 1 g(u) , 0. Hence when z → u we have |g(z)| → |g(u)| , 0 and k → ∞. Hence, lim | f (z)| = ∞. |z−u| z→u 1 Suppose now that | f (z)|z→u → ∞. Then h(z):= → 0. Thus, u is a removable singularity for f (z) z→u h and u is its zero of some order k. Then we can write h(z) = (z − u)kh˜(z), where h˜(u) , 0. Hence,

1 1 ˜ f (z) = = h(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.

56 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, Birkhauser,¨ 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 1 Z Res f = f (z)dz. (7.2.1) u 2πi |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 k = 1; 1  Res0 = zk  0, 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 singulartiies. Suppose f extends continuously to Γ. Then   Z X   f (z)dz = 2πi  Resu f  . u∈Z Γ

57 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  . u∈Z Γ ∂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.

g(z) Proposition 7.8. Suppose f has f has a pole of order n at a point u, i.e. f (z) = (z−u)n , where g is a holomorphic function such that g(u) , 0. Then 1 Res f = g(n−1)(u)), where g(z) = (z − u)n f (z). (7.2.2) u (n − 1)! In particular, if u is a pole of order 1 we have

Resu f = g(u) = lim(z − u) f (z). (7.2.3) z→u 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 not simple. Proof. Consider the Taylor expansion of g at the point u: g(n−1)(u) g(n)(u) g(z) = g(u) + g0(u)(z − u) + ··· + (z − u)n−1 + (z − u)n + .... (n − 1)! (n)! Hence, g(z) g(u) g0(u) g(n−1)(u) f (z) = = + + ··· + + 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 ! 1 Z g(u) g0(u) g(n−1)(u) Res f = + + ··· + + h(z) dz u 2πi (z − u)n (z − u)n−1 (n − 1)!(z − u) |z−u|=r 1 Z g(n−1)(u)dz 1 = = g(n−1)(u). 2πi (n − 1)!(z − u) (n − 1)! |z−u|=r

58 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+x4 . 0 1 R Consider a meromorphic function f (z) = 1+z4 and compute f (z)dz, where ΓR

ΓR = ∂{z; Imz ≥ 0, |z| ≤ R}.

Then R Z Z dx Z f (z)dz = + f (z)dz, 1 + x4 −R + ΓR S R + where S R = {z; Imz ≥ 0, |z| = R}. We have

π Z Z Rdφ πR f (z)dz ≤ = → 0. R4 − 1 R4 − 1 R→∞ + S R 0

Therefore, ∞ Z Z dx f (z)dz → = . R→∞ 1 + x4 ΓR −∞

iπ/4 3iπ/4 The function f has simple poles at the points z1 = e , z2 = e ∈ {z; Imz ≥ 0, |z| ≤ R}, provided

that R > 1. The residues of f at z j, j = 1, 2 are equal to

z − z 1 1 j . lim 4 = 4 0 = 3 z→z j | 1 + z (1 + z ) z=z j 4z j

59 Hence, ! √ Z e−3πi/4 + e−9πi/4 π π 2 f (z)dz = 2πi = π sin = . 4 4 2 ΓR Therefore, ∞ √ Z dx 1 Z π 2 = lim f (z)dz = . 1 + x4 2 R→∞ 4 0 ΓR

R∞ Remark 7.9. Similar techniques applies to integrals R(x)dx, when this integral is converging, −∞ P(x) where R(x) = Q(x) is a rational function. ∞ R cos xdx 2. Compute 1+x2 . 0 Consider a function eiz f (z) = . 1 + z2 Denote

UR,a := {z; 0 ≤ Im z ≤ a, |Re z| ≤ R}.

Then R a Z eizdz Z exdx Z eiRe−ydy = + 1 + z2 1 + x2 1 + (R + iy)2 ∂UR,a −R 0 0 R Z e−iRe−ydy Z e−aeixdx + + = I + I + I + I . 1 + (−R + iy)2 1 + (x + ia)2 1 2 3 4 a −R Then we have a ∞ Z −y Z e dy 1 −y 1 |I2| ≤ ≤ e dy ≤ → 0. R2 − 1 R2 − 1 R2 − 1 R→∞ 0 0 Similarly,

I3 → 0. R→∞ Furthermore, 2Re−a |I | ≤ . 4 a2 − 1

60 Choose a = ln R then 2 |I4| ≤ → 0. (ln R)2 − 1 R→∞ Hence,

∞ R Z eixdx Z eixdx Z eizdz = lim = lim 1 + x2 R→∞ 1 + x2 R→∞ 1 + z2 −∞ R ∂UR,ln R eizdz π = 2πiRes = . i 1 + z2 e Therefore,   Z∞ Z∞ Z∞ cos xdx 1 cos xdx 1  eixdx  π = = Re   = . 1 + x2 2 1 + x2 2  1 + x2  2e 0 −∞ −∞

∞ R sin x 3. Compute x 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   Z Z ZR Z eizdz  eixdx eixdx eizdz 0 = =  +  − + z  x x  z ∂UR,ln R, −R ∂D ∩{Im z≥0} ln R 0 R Z eiRe−ydy Z e−iRe−ydy 1 Z eixdx + + . R + iy −R + iy R x + i ln R 0 ln R −R As in Example 2 the last three terms converge to 0 when R → ∞. On the other hand, Z eizdz eiz −→ πiRes0 = πi. z →0 z ∂D ∩{Im z≥0 Finally, R R ∞ Z eixdx Z eix Z eix Z eix + dx = 2 dx −→ dx. x x x R→∞ x −R →0 0 Hence,   Z∞ Z∞ sin x 1  eix  π dx = Im  dx = . x 2  x  2 0 0

61 R∞ Similar techniques applies to integrals of the form eixR(x)dx, (and in particular to integrals −∞ ∞ ∞ R R P(x) (cos x)R(x)dx and (sin x)R(x)dx) when this integral is converging, where R(x) = Q(x) is a −∞ −∞ rational function.

4. Compute the integral π Z dθ I = , a > 1. a + cos θ 0

iθ z+z−1 z2+2az+1 { | | } Denote z = e then cos θ = 2 , and hence a + cos θ = 2z . Denote D := z; z < 1 , Consider the integral

2π Z dz 1 Z idθ J = = z2 + 2a + 1 2 a + cos θ ∂D 0 π Z dθ = i = iI, a + cos θ 0

i.e. I = −iJ.

On the other hand, we can apply the residue theorem to compute the integral J. the function √ 1 − ± 2 − | | f (z) = z2+2az+1 have simple poles at the points A± := a i a 1 and we have A+ < 1 and 1 1 |A−| > 1. We have ResA f = = √ . + A+−A− 2 a2−1 Hence, π I = −iJ = −i(2πi)ReszA+ = √ . a2 − 1

Remark 7.10. The same method applies to integrals of the form

Z2π R(cos t, sin t)dt,

where R is a rational function.

62 7.4 Complex projective line or Riemann sphere

n Consider the space C 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:

0 0 0 0 (z1,..., zn) + (z1,..., zn) = (z1 + z1,..., zn + zn),

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.

2 Any non-zero vector z = (z1, z2) ∈ C generates the 1-dimensional complex subspace, or complex line 2 lz = Span(z) = {λz; λ ∈ C} ⊂ C .

1 The line lz can be viewed as a point of CP . Any proportional vectorz ˜ = µz, µ ∈ C, generates the 1 2 same line: lz˜ = lz. Hence, we equivalently can deﬁne CP as the space of points in C \ 0 up to a complex proportionality.

2 2 Let us ﬁx an aﬃne line L1 = {z2 = 1} ⊂ C . Any line in C 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

z1 1 it intersects L1 at the point (u = , 1). So one can view CP as C with one point added “at infinity”. z2 On the other hand there is nothing special in this point at infinity. If instead we take an affine line

2 L2 = {z1 = 1} ⊂ C 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

z2 1 (v = , 1). So in CP1 \ (l(0,1) ∪ l(1,0) we have two coordinates u and v related by the formula u = . z21 v 1 1 1 Thus, CP \{l(1,0)} and CP \{l(0,1)} can be identiﬁed with C, and we can say that CP is 1 1 obtained by gluing two copies of C along C \ 0 using the gluing map z 7→ z . It follows that CP 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

63 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 = StN(p), where the bar denote the complex conjugation. Then one can check that for p ∈ Σ \ (S ∪ N) coordinates u

1 1 and v are related by u = v , exactly as we had seen above in CP . This leads to the following interpretation of CP1. We add to C one extra point ∞. Disc com-

1 plements Ur := {|z| > r} form a system of neighborhoods of ∞. The function u = z can be viewed as a coordinate at this neighborhood which is equal to 0 at inﬁnity. Given a holomorphic function

1 g : Ur → C we say that it extends to ∞ if the function g( u ) extends as a holomorphic function to 0. 1 We say that g has a pole or zero of order m at inﬁnity if so does the function g( u ). For instance, any n n−1 polynomial P(z) = z + a1z + ··· + an of degree n has a pole of order n at ∞. Indeed, replacing 1 z = u we observe that the function ! 1 1 + a u + a u2 + ... a un f (u):= P = 1 22 n u un

1 has a pole of order n at 0. Similarly, the function P(z) 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

1 point u by setting f (u) = ∞. Moreover, from the point of view of the coordinate u = 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.

P(z) 1 Theorem 7.11. Any rational function R(z) = Q(z) is meromorphic on CP , and conversely any meromorphic function on CP1, i.e. a holomorphic map CP1 → CP1 is rational, i.e. the ratio of 2 polynomials.

64 P(z) n n−1 ··· m n−1 ··· Proof. Let R(z) = Q(z) , where P(z) = a0z + a1z + + an, Q(z) = b0z + b1z + + 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

1 1 1 the function f : CP → CP has ﬁnitely many poles and zeroes. Let q1, q2,..., qk ∈ C ⊂ CP

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

r1 rk R(z):= (z − q1) ... (z − qk)

The functions R(z) and f (z) have zeroes and poles of the same multiplicities at the same points in f (z) ∞ C. Then h(z):= R(z) has no poles and zeroes in C. We argue that is also not a pole and hence h 1 is a non-zero constant C (according to Liouville’s theorem). Indeed, if ∞ is a pole then for h it is a 1 zero and then h has to be a non-zero constant which implies that h(∞) , 0. Therefore, f (z) = CR(z) is rational.

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.

P(z) Proof. According to Theorem 7.11 f (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.

65 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 Z Res f = f (z)dz. a 2πi ∂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 1 1 instance, Res0 z = 1, but if we make a change of coordinate z = 2u then Res0 2u = 2 . 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 : U0 → U where h is a biholomorphism such that h(0) = 0. Then

0 Res0( f (z)dz) = Res0( f (h(u)dh = f (h(u))h (u)du).

Proof.

1 Z 1 Z Res ( f (h(u))h0(u)du = f (h(u))h0(u)du = f (z)dz. 0 2πi 2πi ∂D ∂h(U)

But for a biholomorphism h which preserves 0 we have

1 Z 1 Z f (z)dz = f (z)dz = Res ( f (z)dz)). 2πi 2πi 0 ∂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

66 1 u = z as a coordinate and deﬁne   ! !! f 1 1 1  u  Res∞( f (z)dz):= Res0 f d = −Res0  du . u u  u2 

For instance, ! ! ! dz du du Res = −Res = −1; and Res (zdz) = −Res = 0. ∞ z 0 u ∞ 0 u3

1 It is interesting to observe that while the meromorphic function z has a 0 and infinity and z has a dz pole at infinity the meromorphic differential form z has a simple pole at infinity, while zdz has at dz 1 1 − du − du infinity a pole of order 3. Indeed, z = u d u = u and zdz = u3 . We also observe that that for 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 C1-extends to the boundary ∂U without poles and zeroes on ∂U.

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 Z f 0(z)dz Xk Xl = r − s . 2πi f (z) j i 1 1 Γ

Proof. According to the Cauchy theorem

Z f 0(z)dz Xk Z f 0(z)dz Xl Z f 0(z)dz = + , f (z) f (z) f (z) 1 1 Γ |z−q j|= |z−pi|=

67 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

r j f (z) = (z − q j) g j(z),

where g j(z) , 0, and in D(pi) we have

−si f (z) = (z − p j) hi(z),

where hi(z) , 0. Then

Z 0 Z r j−1 r j 0 f (z)dz r j(z − q j) g j(z) + (z − q j) g j(z))dz = r f (z) (z − q j) j g j(z) |z−q j|= |z−q j|= Z Z 0 r jdz g j(z) = + = 2πr ji. z − q j g j(z) |z−q j|= |z−q j|=

0 g j The second integral is equal to 0 because the function is holomorphic in the whole disc D(q j). g j Similarly, we get

Z 0 Z −s j−1 −s j 0 f (z)dz (−si(z − p j) h j(z) + (z − p j) h j(z))dz = −s f (z) (z − p j) j h j(z) |z−p j|= |z−p j|= Z Z 0 sidz h j(z) = + = −2πs ji. z − p j h j(z) |z−p j|=− |z−p j|=

Hence, 1 Z f 0(z)dz Xk Xl = r − s . 2πi f (z) j j 1 1 Γ

Corollary 7.16 (Rouche’s´ theorem). Let U be a domain with a piecewise smooth boundary. Let f, g : U → C be holomorphic functions which C1-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.

68 Exercise 7.17. Prove that it is suﬃcient to assume that f, g continuously, and not necessarily C1 extend to U.

Proof. Consider a 1-parametric family of functions ft := f + tg, t ∈ [0, 1]. By assumption

| ft||∂U ≥ | f |∂U − t|g|∂U > 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 1 ft (z) 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:

2 f (z) = a1z + a2z + ....

Let ak be the ﬁrst coeﬃcient which is not 0. Then

k k+1 k f (z) = akz + ak+1z + ··· = z (ak + h(z)) ,

k | | ≤ |ak| | | ≤ |ak|r where ak , 0 and h(0) = 0. If r is small enough then h(z) 2 for z r. Choose ρ < 4 . We

claim that the disc Dρ = {|z| < ρ} is contained in f (U). Indeed, for any point v ∈ Dρ the equation k k akz = v has exactly k solutions in Dr(u). In other words, the function g(z) = akz − v has k zeros in

Dr(0). Note that for z ∈ Dr(0) and v ∈ Dρ we have |a |rk |zkh(z)| = rk|h(z)| < k , 2

69 while for |z| = r we have |a |rk |a zk − v| ≥ |a |rk − k > |zkh(z)|. k k 4 Hence, according to Rouche’s´ theorem the function

k k f (z) − v = (akz − v) + z 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 Z dz Z w(γ):= = dθ. 2πi z γ 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.

Proposition 7.20. Let U ⊂ C be a domain with a piecewise smooth boundary and f : U → C is a

1 holomorphic function which extends as a C -function to U. Let γ1, . . . , γk : [0, 1] → ∂U parame- terizing the boundary components of U according to their orientation as boundary components of U. Denote

eγ j; = f ◦ γ j, j = 1,... k.

Then Xk w(eγ j) = n, 1 where n is the number of zeroes of the function f counted with multiplicities.

70 Proof. Using Theorem 7.15 we get

1 Z f 0(z)dz Xk 1 Z d f Xk 1 Z dz Xk n = = = = w(γ ). 2πi f (z) 2πi f 2πi z ej 1 1 1 Γ γ j f ◦γ j

71 72 Chapter 8

Harmonic functions

8.1 Harmonic and holomorphic functions

∂2 ∂2 The Laplace diﬀerential operator ∆ on functions on domains in C is deﬁned as ∆ := ∂x2 + ∂y2 , i.e.

∂2 f ∂2 f ∆ f = + ∂x2 ∂y2 for any C2-function f . The Laplace operator can be rewritten in the complex notation as

∂ ∂ ∂2 f ∆ = 4 , i.e. ∆ f = 4 . ∂z ∂z¯ ∂z∂z¯

Indeed, ! ! ! ∂2 ∂ ∂ 1 ∂ ∂ ∂ ∂ 1 ∂2 ∂2 1 = = − i + i = + = ∆, ∂z∂z¯ ∂z ∂z¯ 4 ∂x ∂y ∂x ∂y 4 ∂x2 ∂y2 4

because mixed derivative terms cancel. A real- or complex-valued C2-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.

73 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 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 ∂z¯

∂ f because ∂z¯ = 0, and hence ∆ f = 0 and ∆u = ∆v = 0.

Conversely, suppose that u is harmonic in a simply connected domain U. Then g := ux − iuy

is holomorphic. To see this we verify the Cauchy-Riemann equations for g. We have ∆u = uxx +

uyy = 0, and hence (ux)x = (−uy)y. We also have (ux)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 = eu + iev : U → C such that d f = g(z)dz, or ! 1 ∂ ∂ 1 − i (u + iv) = u + v + i(v − u ) = g(z) = u − iu . 2 ∂x ∂y e e 2 ex ey ex ey x y Taking into account the Cauchy-Riemann equations we get 1 (u + v ) = u = u , 2 ex ey ex x 1 (u − v ) = u = u . 2 ey ex ey y

Hence, deu = du and therefore eu = u + C, and we can choose C = 0. Thus the function 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.

74 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 p harmonic function u = ln r = ln x2 + y2 = ln |z| on C \ 0, z = reiφ = x + iy. Then its harmonic zdy−ydx conjugate is φ = arg z, which is multivalued. At the same time the form dφ = x2+y2 .is 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, u˜ : U → R on a connected domain U which coincide on a subdomain U0 ⊂ U coincide in U

Proof. Take a point a ∈ U0 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, v˜ be the harmonic congugate of u and eu on V. Holomorphic functions f := u + iv, ef = eu + iev 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 = ef near a. But then by uniqueness of the holomorphic continuation we have f = ef on V, and in

particular f (a) = ef (a) and hence u(a) = Re f (a) = Re ef (a) = eu(a).

An important fact is that the notion of a harmonic function is invariant with respect to a holomorphic change of coordinate.

Lemma 8.5. Let h : U → C be a C2-function and f : Ue → U a holomorphic function. Then

∆(h ◦ f )(z) = (∆h)( f (z))| f 0(z)|2.

In particular, if h is harmonic then so is h ◦ f .

Proof.

75 !  !  ∂ ∂(h ◦ f ) ∂  ∂h 0 ∂h ∂ f  ∆(h ◦ f )(z) = 4 (z) = 4  ( f (z)) f (z) +  ∂z¯ ∂z ∂z¯  ∂z ∂z¯ ∂z  ! ! ! ∂ ∂h ∂2h ∂ f¯ ∂2h ∂ f = 4 ( f (z)) f 0(z) = 4 ( f (z)) f 0(z) + f 0(z) ∂z¯ ∂z ∂z¯∂z ∂z¯ ∂z∂z¯ ∂z¯ = ∆h( f (z)) f 0(z) f 0(z) = ∆h( f (z))| f 0(z)|2,

∂ f ∂ f¯ ∂ f¯ 0 because ∂z¯ = ∂z = 0 and ∂z¯ = f (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

2π 1 Z h(a) = h(a + reit)dt. 2π 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

2π 1 Z f (z)dz 1 Z f (a + reit)ireit f (a) = = dt 2πi z − a 2πi reit ∂Dr(a) 0 2π 1 Z = f (a + reit)dt. 2π 0 Taking the real part of this equality we get the required formula for the function h.

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.

76 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

2π 1 Z f (a + |z |eit) < f (a), 2π 0 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.

77 78 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 00(a) f (z) − f (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 | − | ≤ | | |c| | | |c|k − k such that if z a then g(z) < 2 . Take any b , 0, b < 2 . Note that the equation c(z a) = b

has exactly k solutions in the disk D(a) = {|z − a| < }. For any z ∈ ∂D(a) we have |c|k |c(z − a)k − b| > > k|g(z)| = |(z − a)kg(z)|. 2

Note that f (z) − ( f (a) + b) = (c(z − a)k − b) + (z − a)kg(z).

79 Hence, Rouche’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 0 proves that f (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

0 1 h ( f (z)) = 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→ eiθz. Note that any complex linear map C → C has the form z 7→ cz, i.e. z 7→ reiθz, where c = reiθ. Geo- metrically 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

80 biholomorphic or conformally equivalent.

9.3 Examples of conformal mappings

9.3.1 Unit disc and the upper-half plane

We begin with exploring the inversion operation inv : C \ 0 → C \ 0, which in polar coordinates is

1 −1 given by the formula r 7→ r . In complex notations we have inv(z) = (z) . { } {| − i | 1 } Lemma 9.3. Image inv(l) of the line l = Im z = d is the circle z 2d = 2d .

1 x+id Proof. We have inv(x + id) = x−id = d2+x2 . Therefore, i x + id i inv(x + id) − = − 2d d2 + x2 2d √ 2id2 + 2dx − id2 − ix2 d4 + x4 − 2d2 x2 + 4d2 x2 1 = = = d2 + x2 2d(d2 + x2) 2d 7→ 1 { } {| − i | 1 } Therefore, the map z z maps the half-plane Im z > d onto the open disc z 2d < 2d . Consequently,

Proposition 9.4. The map 2 iz + 1 z 7→ + 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 i − z z 7→ i + z deﬁnes another conformal equivalence H → D, and that the inverse map D → H is given by 1 − z z 7→ i . 1 + z It is called Joukovsky’s map.

1 Exercise 9.6. Show that the map z 7→ z + z is a conformal isomorphism of C \ D onto C \ {−2 < Re z < 2; Im z = 0}.

81 9.3.2 Strips and sectors

z The map z 7→ e 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.

f (z) Proof. The equality f (0) = 0 implies that g(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) , and hence f (z) = cz and |c| = f (z0) = 1, i.e. f z z0 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 f (z) = g(0), and and if | f 0(0)| = |g(0)| = 1, then again the maximum z→0 z 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) = eiθz, θ ∈ R.

| 0 | ≤ −1 | −1 0 | 1 ≥ Proof. By 9.7(iii) we have f (0) 1, and applying 9.7(iii) to f we get ( f ) (0) = | f 0(0)| 1. Hence, | f 0(0)| = 1, and applying again 9.7(iii) we conclude that f is a rotation.

9.5 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. 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 called complex projective transformations. Note that two transformations A, Ae : Cn → Cn deﬁne the same transformation of CPn−1 if and only if they are proportional: Ae = cA, c ∈ C. Projective

83 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 fac-

tor. 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 sub-

group 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 SL(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 SL(n, R) by identifying matrices A and −A. Clearly we get the same thing by identifying matrices A and −A in SL(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 az + b z 7→ . cz + d

7→ az+b 1 → 1 Lemma 9.9. The function z fA(z) = cz+d is a conformal automorphism CP CP .

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

84 az+b uniquely solve the equation cz+d = w: dw − b z = , (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.

1 2 Recall that CP is the space of complex lines in C , or equivalently the space of pairs (z1, z2) , 1 (0, 0) of complex numbers up to proportionality (z1, z2) ∼ (λz1, λz2), λ ∈ C. CP can be covered by

z1 1 z2 two coordinate charts, with a coordinate u = , where z2 , 0 and v = = , where z1 , 0.   z2 u z1 a b   2 A matrix A =   ∈ GL(n, C) acts on C by c d

    az1 + bz2 z 7→ Az =   .   cz1 + dz2

This action deﬁnes the action on CP1, which in terms of the coordinate u has the form

z az + bz au + b u = 1 7→ 1 2 = , 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

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 P(z) 7.11 are rational. Let us analyze when a rational function f (z) = Q(z) deﬁnes an bijective map

85 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 k as c(z − z1) (1 + h(z)), where k > 1, c , 0 and h(0) = 0. Then arguing as in the proof of Lemma 9.1 1 we conclude that f is not 1-1 on this neighborhood. Applying the same argument to the function 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 iso- morphic to PGL(2, C). The elements of PGL(2, C) act on CP1 by fractional linear transformations.

1 Proposition 9.13. For any 3 points z0, z1, z2 of CP there exists a unique automorphism f ∈

PGL(2, C) such that f (0) = z0, f (1) = z1, f (∞) = z2.

az+b Proof. Let f (z) = 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.

Thus, c = c−z0 , a = (c−z0)z2 . z2−z1 z2−z1

Given 4 points z1, z2, z3, z4 their cross ratio is deﬁned as

(z3 − z1)(z4 − z2) (z1, z2; z3, z4) = . (z3 − z2)(z4 − z1)

86 For instance, we have (∞ − 1)(0 − z) (1, z; ∞, 0) = = 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?). az+b ∞ ∞ a Hence, f (z) = cz+d , and f ( ) = implies that c = 0. Therefore, f (z) = Az + B, where A = d , B = b d . 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 leaves the upper-half plane H invari- cz+d     a b ant (i.e. f (H) = H) if and only if f ∈ PS L(2, R), i.e. when the matrix   is proportional to a 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 + c2y2 and ! az + b ax + b + iay ay(cx + d) − cy(ax + b) y(ad − bc) Im = = = > 0. cz + d cx + d + icy (cx + d)2 + c2y2 (cx + d)2 + c2y2

87 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 = dp − 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 d = p2 + q2 −q c = p2 + q2 Note that

0 1 q = > 0, −q p p2 + q2 p2+q2 p2+q2 and hence the fractional linear transformation belongs to PS L(2, R).

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.

88 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 = eiθg−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).

az+b Exercise 9.20. Prove that a fractional linear transformation f (z) = cz+d leaves the unit disc D invariant (i.e. f (D) = D) if and only if f has the form

α − z f (z) = eiθ , (9.5.1) 1 − α¯z

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

H → D iz + 1 i − z z 7→ or z 7→ . z + i i + z z − a General form : z 7→ eiθ , a ∈ H, θ ∈ [0, 2π). z − a This automorphism sends the point a to the center of the disc.

89 D → H 1 − z z 7→ i . 1 + z

D → D

α − z z 7→ eiθ , α ∈ D, θ ∈ [0, 2π). 1 − α¯z 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 π 1 z → zα maps the sector 0 < arg z < onto H, α > α 2

Complement of an interval onto the complement of a disc

1 z 7→ z + , z see Exercise 9.3.1.

90 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 10.8. 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 10.10.

Step 3: Finding a converging sequence of injective maps fn : U → D with fn(z0) = 0 maximizing 0 the modulus of the derivative | fn(z0)|. See Proposition 10.9. Note that Step 2 implies that f is injective.

91 Step 4: Prove that f constructed in Proposition 10.9 is surjective. See Proposition 10.11. This concludes the proof of Theorem 10.1.thm:RMT. Before proving theorem we need to develop some additional tools.

10.1 Functional analytic background

10.1.1 Arzela-Ascoli´ theorem

The Arzela-Ascoli´ theorem is a general fact, not speciﬁc to the context of holomorphic functions.

n m Let K ⊂ R be a compact set, and fn : K → R a sequence of continuous maps. m We say that fn uniformly converges to a map f : K → R 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.

92 We say that fn is equicontinuous if for any > 0 there exists δ > 0 such that | fn(x) − fn(y)| < δ for all n and any x, y ∈ K such that |x − y| < δ.

Lemma 10.3. Consider domains V ⊂ U ⊂ C such that K := V is compact and K ⊂ U. Let

fn : U → C be sequence of holomorphic functions with uniformly bounded derivatives on K. Then

fn is equicontinuous on K.

Proof. By continuity of the derivative of a holomorphic function there exists an open domain U0, K ⊂ U0 ⊂ U such that | f 0(z)| < C for all z ∈ U0. Now the claim follows from the intermediate value theorem applied to real and imaginary parts of the function f .

Theorem 10.4 (Arzela-Ascoli)´ . If the sequence fn : K → X is equicontinuous and uniformly

bounded, that there is a subsequence fnk which uniformly converges to a continuous function f : K → X.

Proof. The proof follows the standard diagonal argument. We take a squence of all points in K with rational coordinates. This is a countable set, so we can enumerate them: x1, x2,..., ∈ K. 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 on points x1,..., xm. We claim that fkk converges on the

whole compact set K. Hence the diagonal subsequence fkk converges on all rational points xm, m = 1,... . This means for any > 0 and any m there exists N = Nm such that | fnn (xm)− fkk (xm)| < for all n, k ≥ N. The equicontinuity of fn implies that there exists δ > 0 such that | fnn (x)− fnn (y)| < for any n 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.

93 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.

n Corollary 10.5. Let U ⊂ R 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 uniformly on all compact sets converges to a continuous function f : U → X.

S Proof. One can exhaust U by a sequence of compact subsets K1 ⊂ K2 ⊂ ... U, K j = U, apply j Arzela-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 (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 (why?) that the family of derivatives { f 0; f ∈ F } is also uniformly bounded, but this implies the equicontinuity via the mean-value theorem, see Lemma 10.2.

Hence, we can apply Arzela-Ascoli´ to extract a subsequence fk j uniformly converging on compact sets. But the same arguments applies to the derivatives, so we conclude that (possibly after passing to a subsequence), the derivatives f 0 also converge uniformly on compact sets, but then k j the limit function is holomorphic.

Remark 10.7. 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.

94 10.2 Proof of the Riemann mapping theorem

10.2.1 Embedding into D

Proposition 10.8. 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 elogU (z) = z, and in particular, the map logU is injective. U U We claim that there exists > 0 such that the disc D(log (z0)+2πi) is contained in C\log (U). U U U Indeed, if this is not true then there exists a sequence log (zn) ∈ log (U) ∩ D 1 (log (z0) + 2πi), n U U U U log (zn) 2πi+log (z0) n = 1, 2,..., and hence lim log (zn) = log (z0) + 2πi. Therefore, e → e = z0. But n→∞ U log (zn) U U U e = zn, and therefore lim zn = z0. But then log (zn) → log (z0) , log (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) = logU (z) − 2πi of U into D.

10.2.2 Maximizing the derivative

In Proposition 10.8 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 0 inequality | f (z0)| is uniformly bounded by some constant C (which depends on the distance of z0 0 to ∂U) for all f ∈ F . Let Cmax := sup | f (z0)|. Thus Cmax ≤ C. f ∈F 0 Proposition 10.9. There exists f ∈ F such that | f (z0)| = Cmax.

0 Proof. Take a sequence { fn} of functions from F with | fn(z0)| → Cmax. The set { fn} is uniformly bounded on compact subsets (because of Cauchy inequality, see the argument in the proof of

95 Proposition 10.8), and hence it is normal). Therefore, there is a subsequence uniformly on compact 0 sets converging to a function f which satisﬁes | f (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 10.10, which we prove below, that f is injective, and thus belongs to F .

10.2.3 Preservation of injectivity

Lemma 10.10. 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 Z g0 (z) 1 Z g0(z) lim n dz = dz ≥ 1. n→∞ 2πi gn(z) 2πi g(z) ∂D ∂D (z2)

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 10.9 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.

96 0 Proposition 10.11. Let f : U → D be a map from F which satisﬁes | f (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 ef ∈ F 0 0 0 with | ef (z0)| > | f (z0)|, which would contradicts our assumption that | f (z0)| = Cmax. Consider a

conformal automorphism ψa : D → D given by the formula

a − z ψ (z) = . a 1 − az¯

It interchanges points 0 and a, i.e. ψa(0) = a and ψa(a) = 0. Hence, the inverse fractional linear −1 transformation ψa also interchanges the points 0 and a. 0 0 0 0 Consider the domain U = ψa( f (U)). Then 0 < U , but a ∈ U . The domain U is simply- connected because it is biholomorphic to a simply connected domain U. Consider a logarithm 0 √ branch logU : U0 → C in U0 and deﬁne the z branch

0 1 logU (z) s(z):= e 2 .

It satisﬁes (s(z))2 = z. Consider the function

ef (z) = ψs(a) ◦ s ◦ ψa ◦ f.

0 We claim that ef ∈ F and | ef (z0)| > Cmax. Indeed, we have

ef (z0) = ψs(a) (s(ψa( f (z0)))) = ψs(a) (s(ψa(0))) = ψs(a)(s(a)) = 0.

To verify the injectivity of the function ef we observe that it is a composition of injective functions

√ f , ψa, s and ψ a. 0 To estimate | ef (z0)| consider the diagram

f z }| { −1 −1 f ψ s ψs(a) ψs(a) z7→z2 ψ U → f (U) →a U0 → D → D → D → D →a D. | {z } | {z } h ef

97 −1 2 −1 Taking into account that ψs(a) ◦ ψs(a) = Id, (s(z)) = z and ψa ◦ ψa = Id we conclude that f = h ◦ ef . The function h : D → D is deﬁned by the formula

−1 −1 2 h(z) = ψa ψs(a)(z) .

This function satisﬁes

−1 −1 2 h(0) = ψa ((ψs(a)(0)) ) = ψa(a) = 0, but is not injective, because the function z 7→ z2 is not injective. Hence, the Schwarz lemma implies that |h0(0)| < 1.

0 0 On the other hand, we have f = h ◦ ef . Using the chain rule we compute f (z0) = h ( f (z0)) · 0 0 0 ef (z0) = h (0) · ef (z0) and hence

| f 0(z )| | ef 0(z )| = 0 > | f 0(z )| = C 0 |h0(0)| 0 max because |h0(0)| < 1. This contradiction concludes the proof of Proposition 10.11, 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 Ck-submanifold if for any a ∈ Γ there exists > 0

k k k and a C -diﬀeomorphism (i.e. a bijective C -map h with a C -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 ef .

98 Theorem 10.12. If the boundary ∂U of a simply connected domain U is a Ck-submanifold of C and f : D → U a conformal equivalence. Then f extends to a Ck−1-diﬀeomorphism f between the closures: f : D → U. If ∂U is a C0-submanifold, then f extends to D as a homeomorphism.

The proof of this theorem (which was ﬁrst proven in a weaker form by P. Painleve)´ goes beyond this course.

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 10.13. There exists a biholomorphism h : A(r, R) → A(r, R) which switches the boundary circles, i.e. h({|z| = r}) = {|z| = R}.

rR Proof. This is done by the map z 7→ z . 0 0 R R0 Clearly, any two annuli A(r, R) and A(r , R ) with r = r0 are conformally equivalent. Indeed, 0 0 R0 the required conformal equivalence A(r, R) → A(r , R ) is the linear map z 7→ R z. It turns that this suﬃcient condition together with the one arising from Lemma 10.13 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.

1 Note that the map z 7→ z establishes a conformal equivalence of A(0, 1) and A(1, ∞). However,

Lemma 10.14. 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

99 f to 0 and hence we get a conformal equivalence D → C, which is impossible. In the latter case

1 we ﬁrst compose f with the automorphism C \ 0 → C \ 0 given by the function z 7→ z and then repeat the previous argument.

Theorem 10.15. 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 . r r0 R The quantity ln r 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, ez X∞ zn = , 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 ∞ −n 1 X z e z = , z , 0. n! 0 A series of the form ∞ X k S (z) = akz −∞ ∞ ∞ P −k P k is called Laurent series. The series is a sum of two series S −(z) = a−kz and S +(z) = akz , 1 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 1 1 variable u = z and it converges in the disc |u| < 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

100 Proposition 10.16. Any holomorphic function S : A(r, R) → C can be presented as the sum of an ∞ P k absolutely converging in A(r, R) Laurent series, S (z) = akz . −∞

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 f (ζ)dζ Z f (ζ)dζ Z f (ζ)dζ f (z) = = − . ζ − z ζ − z ζ − z 0 0 ∂A(r ,R ) ∂DR0 ∂Dr0

1 1 Changing the variable ζ = u and z = v in the second integral we get

Z f (ζ)dζ Z v f ( 1 )du = − u . ζ − z u(u − v) 1 ∂Dr0 | | ∂{ u < r0 }

Therefore,

Z f (ζ)dζ Z f (ζ)dζ f (z) = − ζ − z ζ − z ∂DR0 ∂Dr0 Z f (ζ)dζ Z v f ( 1 )du = + u . ζ − z u(u − v) 1 ∂DR0 | | ∂{ u < r0 }

Arguing as in Theorem 5.7 we can expand the ﬁrst integral in a power series in z converging for

1 |z| < R and expand the second integral in a power series in v converging for |v| < r . Changing back 1 v 7→ z = v we get the required Laurent expansion in z variable. We will also need the following formula (due to Green)

∞ P n Lemma 10.17. Let f (z) = anz 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 ∞ X 2 2n A(ρ) = π n|an| ρ . −∞

101 Proof. Using Proposition 4.3 we have

 ∞   ∞  i Z i Z X  X  − 0 −  n  m−1 A(ρ) = f (z) f (z)dz =  anz¯   mamz  2 2 −∞ −∞ |z|=ρ |z|=ρ 2π i X∞ Z 1 X∞ Z X∞ = − ma a zm−1z¯ndz = ma a ei(m−n)φρm+ndφ = π n|a |ρ2n, 2 m n 2 m n n m,n=−∞ m,n=−∞ n=−∞ |z|=ρ 0

R2π because eikφdφ = 0 unless k = 0. 0

10.3.3 Proof of Theorem 10.15

Theorem 10.15 follows from the following stronger result.

Proposition 10.18. Suppose there exists an injective holomorphic map

f : A(r0, R0) → A(r, R).

Then m(A(r0, R0)) ≤ m(A(r, R))

0 0 such that for ρ ∈ (r , R ) the circle S ρ := {|z| = ρ} bounds a domain in C which contains the disc

Dr(0) = {|z| < r}.

Proof. Without loss of generality we can assume that r0 = r = 1. We will view the annuli A(1, R) 0 0 and A(1, R ) as a subdomain of the discs DR = {|z| < R} and DR0 = {|z| < R }. Let Vρ denote 0 the closed subdomain of DR bounded by f (S ρ) for ρ ∈ (1, R ). By assumption Vρ ⊃ Dr(0). We

can assume that Vρ ⊂ Veρ if ρ < eρ. Otherwise, we can use Lemma 10.13 to switch the boundary 0 components of the annulus A(1, R ). 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 10.17 to

compute the area A(ρ):= Area(Vρ):

∞ X 2 2n A(ρ) = π n|an| ρ . −∞

102 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

∞ ∞ X 2 X 2 0 2n 2 π ≤ π n|an| ≤ π n|an| (R ) ≤ πR . −∞ −∞

The ﬁrst inequality then implies that

∞ ∞ 2 X 2 2 X 2 2n πR ≤ π n|an| R ≤ π n|an| R . −∞ −∞

But the function A(ρ) is strictly increasing. Hence the inequality

∞ ∞ X 2 0 2n X 2 2n π n|an| (R ) ≤ π n|an| R −∞ −∞ implies R0 ≤ 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 f |U = ψ.

One can make sense of Dirichlet problem even for discontinuous but integrable functions, where one requires boundary convergence at the points of continuity.

103 Note that the maximum principle for harmonic functions guarantees that the Dirichlet problem has a unique solution. Indeed, if ∆u = ∆eu = 0 and u|∂U = eu|∂U then u − eu is harmonic and

(eu − u)|∂U = 0. Hence, Corollary 8.8 implies that eu = 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 to ∂U (comp. Theorem 10.12), 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 10.19 (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   1 ζ + a dζ  u(a) = Re  u(ζ)  . (10.4.1) 2πi ζ − a ζ  |ζ|=1

Remark 10.20. 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

1 Z ζ + z dζ f (z) = u(z) 2πi ζ − z ζ |ζ|=1 is holomorphic and according to Schwarz’s formula, u(z) = Re f (z).

Proof of Proposition 10.19. Consider a fractional linear change of coordinates

v + a ζ = R(v) = , a ∈ D. 1 + av

104 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:

2π 1 Z u(a) = u(R(0)) = u(R(eiθ))dθ. 2π 0 The integral in the right-hand side can be rewritten as the complex contour integral 1 Z dv u(a) = u(R(v)) . 2πi v |v|=1 Let us change back the variable v to ζ, ζ − a v = R−1(ζ) = . 1 − aζ We have ! ! dv 1 a ζ aζ dζ = + dζ = + . v ζ − a 1 − aζ ζ − a 1 − aζ ζ Therefore, ! 1 Z dv 1 Z ζ aζ dζ u(a) = u(R(v)) = u(ζ) + 2πi v 2πi ζ − a 1 − aζ ζ |v|=1 |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 ζ a A := + ζ − a ζ − a is real, and thus   1 1  ζ 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

105 Combining all the formulas, we get ! ! 1 Z ζ aζ dζ 1 Z ζ + a dζ u(a) = u(ζ) + = Re u(ζ) . 2πi ζ − a 1 − aζ ζ 2πi ζ − a ζ |ζ|=1 |ζ|=1

1 dζ We note that i ζ = dθ is real valued, and hence   Z !  Z  1 ζ + a dζ  1 ζ + a dζ  Re u(ζ) = Re  u(ζ)  , 2πi ζ − a ζ 2πi ζ − a ζ  |ζ|=1 |ζ|=1 and we get the required formula (10.4.1).

It is useful to rewrite formula (10.4.1) in some diﬀerent forms.

Proposition 10.21. 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 Z ζ + a dζ u(a) = Re u(ζ) 2πi ζ − a ζ |ζ|=1 2π ! 1 Z eiθ + a = Re u(eiθ)dθ 2π eiθ − a . (10.4.2) 0 2π 1 Z 1 − |a|2 = u(eiθ)dθ 2π |eiθ − a|2 0 Equivalently, for a = reiφ we have

2π 1 Z 1 − r2 u(reiφ) = u(eiθ)dθ. (10.4.3) 2π 1 − 2r cos(θ − φ) + r2 0 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 10.22. 2π Z 1 − |a|2 dθ = 2π |1 − aeiθ|2 0 for any a ∈ D.

106 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

2π 1 Z 1 − |z|2 dζ 1 Z 1 − |z|2 P (z):= ψ(ζ) = ψ eiθ dθ (10.4.4) ψ 2πi |ζ − z|2 ζ 2π |eiθ − z|2 ∂D 0 .

Theorem 10.23 (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 neigh- iθ borhood of a point e ∈ ∂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

iθ lim Pψ(z) = ψ(e ), z→eiθ,z∈D

in particular if ψ is continuous on ∂D that 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 10.23 let us list some elementary properties of the integral Pψ

Lemma 10.24. (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

Pc1ψ1+c2ψ2 = c1Pψ1 + c2Pψ2 ;

(2) if ψ ≥ 0 then Pψ ≥ 0;

(3) if ψ = c is a constant them Pψ = ψ = c,

107 (4) if c < ψ < C for constants c and C then c < Pψ < C.

1−|z|2 Proof. (1) is straightforward, (2) follows from the fact that |eiθ−z|2 > 0, (3) follows from Corollary 10.22, and (4) is a corollary of (2) and (3).

Proof of Theorem 10.23.

a) To prove that Pψ is harmonic we observe that ! 1 − |z|2 ζ + z = Re , |ζ − z|2 ζ − z

and hence Pψ(z) is the real part of the holomorphic function 1 Z ζ + z dζ f (z) = u(ζ) . 2πi ζ − z ζ ∂D

Hence, Pψ(z) is harmonic. Note that the function f (z) is holomorphic everywhere except ∂D (why?). 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. 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 10.24(3) which implies that the theorem holds for constant functions ψ. Given any > 0 choose

iθ δ > 0 so small that |ψ(e )| < 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 10.24(4).

108 iθ But |z| = 1 on ∂D, and e − 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

iθ0 iθ0 |Pψ2 (z) − Pψ2 (e | = |Pψ2 (z)| < . Thus, for |z − e | < σ and z ∈ D we have

iθ |Pψ(z) − ψ(e )| = |Pψ(z)| ≤ |Pψ1 (z)| + |Pψ2 (z)| < 2.

Hence, iθ lim Pψ(z) = ψ(e ). z→eiθ,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 10.12 it extends continuously to the smooth part of the boundary ∂U and consider the function ψe := ψ ◦ f −1 : ∂D → R. If ψe is integrable on ∂D (as we will see below it is not necessarily the case), then we can consider the solution u := Pψe of the Dirichlet problem for D with the boundary value ψe. In other words, u is harmonic in D, it extends, continuously to the points of ∂U, where ψe is continuous, and coincides there with ψe. Then bu := 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}.

The exponential function exp(z) = ez maps Ω conformally onto H, and composing it with a con- → i−z formal equivalence g : H D, g(z) = i+z , we get the required biholomorphism i − ez f (z) = i + ez

109 −1 1−z of Ω onto D. We have f (z) = log i 1+z − Let u : ∂Ω → R be a continuous bounded function on ∂Ω. Then eu(z) = u( f 1(z)) is a piecewise continuous function on ∂D, and we can use Theorem 10.23 to extend it to a harmonic function    Z !!  1 ζ + z 1 − ζ  dζ Pu(z) = Re  u log i  e 2πi ζ − z 1 + ζ  ζ ∂D on D. Changing back to the strip Ω we conclude that the function Πu(z):= Peu( f (z)) is harmonic on Ω. Thus we get    Z !!  1 ζ + f (z) 1 − ζ  Πu(z) = Pu( f (z)) = Re  u log i  e 2πi ζ − f (z) 1 + ζ  ∂D   Z !!  1 ez(ζ − 1) + i(ζ + 1) iζ − 1  dζ = Re  u log  2πi ez(ζ + 1) + i(ζ − 1) i − ζ  ζ ∂D Exercise 10.25. Show that the solution of the Dirichlet problem for the strip Ω = {0 < Re z < π} with the boundary data

u(x + iπ) = f1(x), u(x) = f0(x), x ∈ R, where f0, f1 are bounded continuous functions can be written in the form

∞ ∞ sin y Z f (x − t) sin y Z f (x − t) Π (x + iy) = 0 dt + 1 dt. u y cosh t − sinh y y cosh t + sinh y −∞ −∞

110 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

S RS1. S = Uλ. λ∈Λ

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

−1 n hλµ := Φµ ◦ Φλ : Φλ(Uλ ∩ Uµ) → Φµ(Uλ ∩ Uµ) ⊂ R

111 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 −1 maps between diﬀerent coordinate charts. The inverse maps Ψλ = Φλ : Gλ → Uλ are called

(local) parameterization maps. An atlas is a collection A = {Uλ, Φλ}λ∈Λ of all coordinate charts. 0 0 0 One says that two atlases A = {Uλ, Φλ}λ∈Λ and A = {Uγ, Φγ}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 = 0 0 {(Uλ, Φλ), (Uγ, Φγ)}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.

n A subset G ⊂ S is called open if for every λ ∈ Λ the image Φλ(G ∩ Uλ) ⊂ R 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 for every point a ∈ S there exist local coordinate charts (Uλ, Φλ) in S and (Ueλ, Φeλ) in Se, such that a ∈ Uλ, f (Uλ) ⊂ Ueλ and the composition map

Ψλ f Φeλ n Gλ = Φλ(Uλ) → Uλ → Ueλ → R

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.

112 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

1 coordinates z, ζ ∈ C which are on the overlap C \ 0 are related by z = ζ .

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. Poincare´ 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 11.10).

11.3 Quotient construction

An important class of examples is given by the following construction. Let S be a Riemann surface. Let us denote by Aut(S ) the group of its biholomorphisms S → S .A discrete group of biholomorphisms of S is a ﬁnite or countable subgroup G ⊂ Aut(S ) such that set G of biholomorphisms (= conformal automorphisms) of 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;

113 • 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 Ux, such that for any g ∈ G, g , Id, we have g(x) < Ux. That means that the projection π|Ux is injective and we call π(Ux) 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, ∞).

114 3. Consider another action of Z, this time on C \ 0:

z 7→ 2kz, k ∈ Z

Then the quotient (C \ 0)/Z is a 2-torus T.

z1 2 4. Choose two complex numbers ω1, ω2 ∈ C \ 0 such that R. Consider an action of Z = Z ⊕ Z z2 < 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

−1 for any point x ∈ V there exists a neighborhood Vx 3 x such that the preimage p (Vx) can be presented as a union of ﬁnitely or countably many disjoint open sets:

−1 p (Vx) = U1 ∪ U2 ∪ ...,

such that p|U j : U j → Vx is a biholomorphism for each j = 1, 2,.... If p : U → V is a covering map than U is called a cover of V1

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.

1Be 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.

115 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 ∞ −1 [ + + exp (V+) = U j , U j := {z = x + iy; y ∈ ((2k − 1)π, 2kπ)}; k=−∞ ∞ −1 [ + + exp (V−) = U j , U j := {z = x + iy; y ∈ (2kπ, (2k + 1)π)}. k=−∞ ± The restrictions exp | ± : U → V± are biholomorphisms. The inverse maps are given by branches Uk k of the logarithm.

2. Similarly, the maps πk : C \ 0 → C \ 0 given by the formula

k πk(z) = z , 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 Ux 3 x such that g(Ux) ∩ eg(Ux) = ∅ for g , eg. The tautological projection πG : X → X/G sends Ux onto its trajectory Ue := G(Ux). Ue is by −1 deﬁnition an open set containing the point Gx ∈ S/G and the preimage πG (Uex) is the union

−1 [ πG (Uex) = g(Ux) g∈G

116 of disjoint open sets g(Ux), and the restriction of the projection πG to each of these sets, πG|g(Ux) : g(Ux) → Ue is a biholomorphism. Hence, πG : S → S/G is a covering map. Given a covering p : U → V, a biholomorphism f : V → V 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(V) of V, 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

−1 means that for any v ∈ V and any two points u1, u2 ∈ p (v) there exists a deck transformation g ∈ Aut(V) 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.

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 11.9. 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.

The following theorem explains the origin of the term universal.

Theorem 11.10. 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 :

117 U1 → U2 such that the following diagram commutes:

F U1 / U2

p1 p2 f V1 / 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 11.11. 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:

F U1 / U2 p1 p2

V ~

i.e. p1 = p2 ◦ F.

Indeed, we can apply Theorem 11.10(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 (one- sheeted) cover.

Exercise 11.12. Prove that the universal cover is always regular.

Theorem 11.13 (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 11.13 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

118 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. Then Theorem ?? yields a discrete subgroup of U/G is conformally equivalent to V. But assumption, V, and hence U, is not CP1, and hence according to Theorem 11.10 U 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

2 real subspace SpanR(Λ). Because dim R = 2, any non-trivial (i.e. not consisting of only 0) lattice could be either of rank 1 or of rank 2.

Lemma 11.14. Any rank 1 lattice has the form Λ = {kω} for ω ∈ C, ω , 0, k ∈ Z. Any rank two

ω2 lattice has the form Λ = {kω1 + `ω2} where k, ` ∈ Z, ω1, ω2 ∈ C, ω1, ω2 0 and R. , ω1 <

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

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 2 2 of ω1 we can arrange that |ω − kω1| ≤ |ω1| + d 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

Λe := {kω1 + `ω2; k, ` ∈ Z}

119 generated by ω1 and ω2. Clearly, Λe ⊂ Λ. We claim the Λe = Λ. Indeed, if there exists ω ∈ Λ \ Λe,

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| <

|ω1|, but then ω + kω2 + `ω1 = 0, i.e. ω ∈ Λe.

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 11.15. 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 b g has a ﬁxed point. Indeed in that case the equation az + b = z has a solution z0 = 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.

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.

ω2 Thus T(ω1, ω2) T(1, τ = ), ω1

2. Automorphisms of a lattice. The lattice Λ(ω1, ω2) is preserved by interchanging ω1 and ω2. 2 Hence, we can always assume that ω1 and ω2 deﬁne the standard orientation of R = C (i.e. the same orientation as deﬁned by 1 and i. This is equivalent to the requirement that τ = ω2 ∈ H. ω1

120 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.

Theorem 11.16 (Conformal classiﬁcation of tori). Tori T(ω1, ω2) and T(ωe1, ωe2) are conformally ω ω equivalent if and only if τ = 2 and eτ = e2 are related by the action of an element of the modular ω1   ωe1   n m group, i.e. there exists a matrix   with integer values and det = kn − m` = 1 such that ` k 

nτ + m τ = . e `τ + k

Proof. Denote Λ := Λ(ω1, ω2), Λe = Λ(ωe1, ωe2). Let f : C/Λ → C/Λe be a conformal equivalence. Then according to the uniqueness of universal cover theorem 11.10 there exists a biholomorphism F : C → C such that the following diagram commutes:

F C / C

πΛ πΛe f C/Λ / C/Λe

, i.e. f ◦πΛ = πΛe ◦ F, where πΛ : C → C/Λ and πΛe : C → C/Λe are tautological projections. We have

F(z + Λ) = F(z) + Λe, and hence F(z + 1) = F(z) + c1, F(z + τ) = F(z) + c2, where c1, c2 ∈ Λe. Hence, F0(z) is a bi-periodic function with periods 1 and τ, and hence a constant, but then F(z) = A + Bz, where A, B ∈ C. and the two lattices Λ and Λe differ by a rotation, scaling and translation. After we used the transformation F(z) = A + Bz to identify the lattices Λ and Λe we see that (ω1, ω2) and (ω , ω ) are two different bases of the same lattice Λ, which in addition by assumption define e1 e2      k ` the same orientation of C. Hence, there exists and integer valued matrix A =   such that m n

121 ωe1 = kω1 + `ω2, ωe2 = mω1 + nω2. The matrix is invertible (as an integer valued matrix) and it is orientation preserving. Hence, det A = kn − m` = 1.

ωe2 nτ + m eτ = = . ωe1 `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 11.17. 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 11.16 given a lattice Λ = Λ(ω1, ω2) automor- 2 phisms of this lattice are real linear transformation of C = R which map a basis ω1, ω2 of Λ to

another basis ωe1, ωe2 of Λ, i.e. ωe1 = kω1 + `ω2, ωe2 = mω1 + nω2, where k, `, m, n are integers. The 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→ 1 z, we identify the ω1

ω2 lattices Λ(ω1, ω2) and Λ(1, τ = ), where τ ∈ H (due to our orientation convention). The change ω1

of a basis (ω1, ω2) 7→ (ωe1, ωe2) amounts to changing

ωe2 nτ + m τ 7→ eτ = = . ωe1 `τ + k

In other words, the action of SL(2, Z) on H via fractional linear transformations is equivalent to

the action of SL(2, Z) as the automorphism group of the lattice Λ = Λ(ω1, ω2).

Hence, our problem can be interpreted as the problem of ﬁnding an appropriate basis (ωe1, ωe2) of the lattice Λ(ω1, ω2). First, let us interpret the conditions on eτ in terms of ωe1, ωe2. The condition

122 | | ≥ | | | | | | 1 eτ 1 means ωe2 > ωe1 . To clarify geometric meaning of the condition Reeτ < 2 let us write eτ = x + iy. Then

ωe2 = eτωe1 = (x + iy)ωe1 = xωe1 + y(iωe1)

The vectors ωe1 and iωe1 are orthogonal; and have the same length. So the condition |Re ωe2| = | | ≤ 1 x 2 means that the orthogonal projection of ωe2 onto the line L := Span(ωe1) lies in the interval − ωe1 ωe1 ⊂ [ 2 , 2 ] L.

Hence, the strategy for ﬁnding the appropriate basis (ωe1, ωe2) should be the following (comp.

the proof of Lemma 11.14). Choose as ωe1 the shortest vector of Λ. This guarantees that whatever

ωe2 vector we choose as ω2 the condition τ = ≥ 1 would be automatically satisﬁed. Note that all e e ωe1 points of Λ are located on aﬃne lines L + λ, λ ∈ Λ, parallel to L. Let L1 be the line L + λ closest to

L and chosen on the appropriate side of L, i.e. to ensure that ωe1, λ deﬁne the complex (i.e. counter-

clockwise) orientation of C. Points of Λ on L1 are spaced at a distance |ωe1|. Hence, every closed

interval of length |ωe1| contains at least one point of Λ, and therefore there exists λ ∈ L such that − ωe1 ωe1 ⊂ the orthogonal projection of λ to L lies in the interval [ 2 , 2 ] L. Set ω2 := λ. Then (ω1, ω2) form a basis of Λ with the required properties.

Remark 11.18. Theorem 11.17 can be strengthen as follows:

Every torus T(ω1, ω2) is conformally equivalent to a unique torus T(1, τ) with π π τ ∈ Int U ∪ τ = eiθ, θ ∈ , . 3 2

11.8 Branched covers

We begin with a lemma describing the behavior of a holomorphic function near its critical point.

Lemma 11.19. 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

123 h : U → Ue ⊂ C deﬁned on a neighborhood U 3 0,U ⊂ Ω such that h(0) = 0 and the following diagram commutes: u7→uk Ue / C . _ ?

h f U

Proof. We can write f (z) = zkg(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

log g(z) h(z) = ze k .

We have h(0) = 0 and h0(0) =, 0. Hence, decreasing, if necessary, the neighborhood U we can assume that h is a biholomorphism of U onto its image Ue = h(U) 3 0. But we have f (z) = zkg(z) = 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 −1 −1 f |S 1\ f (B) : S 1 \ f (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

k j f |U j : U j → U can be witten in these coordinates as w = z , where k j are positive integer numbers.

Remark 11.20. 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

124 −1 −1 f |S 1\ f (B) : S 1 \ f (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 11.21. Let f : S 2 → S 1 be a branched map of order d. Suppose that a branching point Pp b ∈ B has p pre-images and (k1,..., kp) 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 j 0 such that f |U j : U j → U can be witten in these coordinates as w = z . Hence, for a point b ∈ U \ b −1 0 the preimage f (b ) has exactly k j pre-images in U j, so the total number of pre-images, which is Pp by deﬁnition is the order d of the branched cover f , is equal to k j. j=1

The set of exponents (k1,..., kp) 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 0 P nearby point p 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 11.22. The map z → zp is a branch cover C → C of order p with the unique branching point 0. Viewed as a holomorphic map CP1 → CP1 the map z 7→ zp is a branched cover of order p with branching points 0, ∞ ∈ CP1.

It turns out that any non-constant holomorphic map between two compact Riemann surfaces is a branch cover.

Theorem 11.23. 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

125 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,..., ap its pre-images. For a suﬃciently small neighborhood −1 U 3 b in S 2 its pre-image f (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

11.19 (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 j u = f (v j) = v j , j = 1,..., p, (11.8.1)

Here k j is the order of the ﬁrst non-zero derivative at the point a j. −1 Consider now a point z ∈ S 2 \ B. Then for each point a ∈ f (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 −1 of z. Hence, f (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)

−1 [ −1 f (U) = (Vea = Va ∩ f (U)) a∈ f −1z and f | : V → U Vea ea is a biholomorphism. This proves that

−1 −1 f |S 1\ f (B) : S 1 \ f (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 11.24. For any compact Riemann surface S and any non-constant meromorphic function f : S → CP1

126 is a branched cover.

1 The branching point of f are critical values of f (i.e. points f (a) ∈ CP 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

2 2 Consider the space C with complex coordinates (z1, z2). A function F : C → 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

2 Lemma 11.25. Let F : C → 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) and ∂F (p) is not equal to 0. ∂z1 ∂z2 Then S is a Riemann surface.

Proof. The holomorphic version of the implicit function theorem (which is proved in an exactly the

∂F same way as its real version) asserts that if p = (a1, a2) (p) 0 then there exists a neighborhood ∂z1 , 2 U 3 a1 in C, a neighborhood V of the point p in C and a holomorphic function h : U1 → C such

that h(a1) = a2 and

V ∩ S = {z2 = h(z1); z1 ∈ U}.

Similarly, if ∂F (p) 0 then S near the point p can be presented as the graph of a function g given ∂z2 ,

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.

2This 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.

127 We call a Riemann surface as in Lemma 11.25 a 1-dimensional complex submanifold, or a

2 smooth (aﬃne) holomorphic curve in C . If the deﬁning function F(z1, z2) is a polynomial, the the holomorphic curve S is called algebraic.

Lemma 11.26. 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 11.27. 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

3 z = (z1, z2, z3) ∈ C \ 0 up to a complex proportionality coeﬃcient:

(z1, z2, z3) ∼ (λc1, λc2, λc3), λ ∈ C \ 0.

2 z1, z2, z3 are called homogeneous coordinates in CP and usually denoted z1 : z2 : z3. 3 2 The subsets U j := {z j , 0} ⊂ C give rise to aﬃne coordinate charts Ubj in CP . Indeed, a

2 z1 z2 point z = (z1, z2, z3) ∈ U3, viewed as a point in CP , is equivalent to ( , , 1), and hence (z13 = z3 z3

z1 z2 , z23 = ) can be viewed as coordinates in this neighborhood. Similarly, in Ub1 we can introduce z3 z3

z2 z3 z1 z3 coordinates (z21 = , z31 = ), and in Ub2 coordinates (z12 = , z32 = ). z1 z1 z2 z2 2 2 We will view C as a subset of CP by identifying it with the aﬃne chart Ub3 with the aﬃne 2 coordinates (z13, z23). Hence CP2 \ C = {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.

128 Let us recall that a function F(z1, z2, z3) is called homogeneous of degree d if

d F(λz1, λz2, λz3) = λ F(z1, z2, z3)

for any λ ∈ C. A useful fact about homogeneous function is

Lemma 11.28 (Euler identity). Let F(z1, z2, z3) is homogeneous of degree k. Then

∂F ∂F ∂F z1 + z2 + z3 = kF(z1, z2, z3). ∂z1 ∂z2 ∂z3

Proof. Diﬀerentiate the identity

k F(λz1, λz2, λz3) = λ 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:

X k m d F(z1, z2, z3) = akmnz1z2 z3, akmn ∈ C. k+m+n=d

2 3 2 For instance, z3z2 − z1 + z3z1 is a homogeneous polynomial of degree 3.

Lemma 11.29. 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) 0 for at least one of j = 1, 2, 3, ∂z j , then S is compact Riemann surface. The intersection S j := S ∩ Uej is an aﬃne holomorphic curve.

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 z1 + z2 = − , 0, ∂z1 ∂z2 ∂z3

129 and hence one of two other partial derivatives have to not vanish as well. Hence the intersection 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.

11.10 Linear projective transformations

Any linear map A : Cn+1 → Cn+1 induces a map Ab : CPn → CPn. Indeed, A(λz) = λz, λ ∈ C, z ∈ Cn+1. In other words, proportional vectors are mapped to proportional vectors. If A is non- degenerate then Abis called a linear projective transformation of CPn, or just a projective transformation. 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, Mobius¨ 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) = a1z1 + a2z2 + a3z3 = 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 ` ` u := 1 , u := 2 1 ` 2 ` are affine complex coordinates in the affine chart U. Every projective line in CP2, different from L intersects U in an affine line

A1u1 + A2u2 + A3 = 0.

Projective pencils. A pencil in CP2 is by deﬁnition the set of all lines passing through a given point

2 a ∈ CP . We will denote a pencil of lines through a by Πa. Note that if L is a projective line and

130 2 a ∈ L then any two lines in L1, L2 from Πa, diﬀerent from L, intersect the aﬃne chart U = CP \ 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 11.30. 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.

3 Proof. In C this is equivalent to mapping one triple of transverse planes P1, P2, P3 to another one 0 0 0 0 0 0 P1, P2, P3 by a linear transformation. Let us choose the bases v1, v2, v3 and v1, v2v3 such that

0 0 0 0 0 0 0 0 0 v1 ∈ P1 ∩ P2, v2 ∈ P2 ∩ P3, v3 ∈ P3 ∩ P1, v1 ∈ P1 ∩ P2, v2 ∈ P2 ∩ P3, v3 ∈ P3 ∩ P1.

0 0 0 0 Then the linear transformation which sends the basis v1, v2, v3 to the basis v1, v2v3 sends P1 to P1, 0 0 P2 to P3 and P3 to P3.

Quadrics. A quadric Q in CP2 is a complex curve of degree two, i.e. given by a degree two equation

2 2 2 Q = {F(z) = A11z1 + A22z2 + A33z3 + 2A12z1z2 + 2A13z1z3 + 2A23z2z3 = 0}

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

131 for each z ∈ Q. We can easily compute that

 ∂F     (z) z1  ∂z1         ∂F     (z) = A z2 ,  ∂z2         ∂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 11.31. All smooth quadrics are projectively equivalent.

Proof. Choose a point a ∈ Q. Consider the pencil Πa of lines through the point a. 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 11.30 there exists a projective linear 2 2 transformation T : CP → CP such that T(L) = {z3 = 0}, T(L1) = {z1 = 0} and T(L2) = {z2 = 0}. 2 Consider the aﬃne chart U3 = {z3 , 0}(= T(CP \ L)). To simplify the notation denote the aﬃne coordinates there by z1 z2 u1 := z13 = , u2 := z23 = . z3 z3

The equation of the aﬃne part Qe := T(Q) ∩ U3 = T(Q \ L) in U3 can be written in the form

3 2 αu2 + (bu1 + c)u2 + (ku1 + lu1 + m) = 0. (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.

c Second, we argue that b = 0. Indeed, suppose b , 0 and set u1 = − b . Then (11.10.1) implies 2 c that ku1 + lu1 + m = 0, i.e. − b is a rood of this quadratic polynomial, which means that

2 ku1 + lu1 + m = (bu1 + c)(pu1 + q),

132 and therefore, (11.10.1) takes the form

(bu1 + c)(u2 + pu1 + qu2) = 0. (11.10.2)

But this equation deﬁnes two lines c {u = − } ∪ {u = −pu − q}. 1 b 2 1

c pc−qb But the two lines intersect at the point (u1 = − b , u2 = b ), which contradicts to our assumption that the curve Q is smooth. Therefore, (11.10.1) further reduces to

2 cu2 + ku1 + 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

2 u2 = αu1 + 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

• 7→ β u1 u1 + 2 we ﬁrst kill the coeﬃcient with the linear term u1, thus getting an equation of 2 the form, u2 = αu1 + γ˜;

2 • u2 7→ u2 − eγ getting u2 = αu1, and ﬁnally √ 2 • u1 7→ α obtaining an equation u2 = u1.

Thus, any smooth quadric is projectively equivalent to the projectivization of the complex parabola

2 u2 = u1.

Exercise 11.32. Following the scheme of the proof of Theorem 11.31 show that any smooth cubic (elliptic) curve C ⊂ CP2, i.e. a curve given by a homogeneous equation of degree 3, is projectively

2 3 equivalent to the projectivization of an aﬃne curve given by the equation u2 = u1 + pu1 + q.

133 134 Part II

Famous meromorphic functions

135

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 Riemman 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.

137 138 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 x x Z dt Z p √ , 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 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) = + = 1 , a2 b2 or parametrically given by

x = a cos t,

y = b sin t.

139 Integrating the length of the vector v = (x ˙, y˙) = (−a sin t, b cos t) we compute the arc length s(τ) ∈ π between the points (x(0), y(0)) = (a, 0) and (a cos τ, b sin τ), τ (0, 2 ), by the formula τ Z p s(τ) = a2 sin2 t + b2 cos2 tdt.

0 Changing the variable sin t = u we get sin τ p sin τ r Z (a2 − b2)u2 + b2 Z 1 + k2u2 s(τ) = √ du = b du, 1 − u2 1 − u2 0 0 a2 − 2 2 2 2 − 2 − where k := 1 + b2 . Denote v := 1 u . Then we get 1 + k u = (k + 1) k v and dv = 2udu, or du = − √dv . Therefore, 2 1−v

1−sin2 τ √ 1 Z k2 + 1 − k2v s(τ) = − √ dv 2 v(1 − v) 1 1 p 1 Z v(1 − v)(k2 + 1 − k2v) = dv. 2 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

θ¨ = − sin θ and the law of conservation of energy gives 1 (θ˙)2 − cos θ = E. 2 Hence, dθ p = 2(E + cos θ), (12.1.1) dt and separating variables we get θ Z dτ t = √ , 2(E + cos τ) θ0

140 where θ0 := θ(0). Changing the variable u = cos τ we get

arccos θ 1 Z du t = √ p . 2 (E + u)(1 − u2) arccos θ0

12.1.2 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

2 3 S = {z2 = P(z1)}, where P(z1) = z1 − 1.

To verify that S is a smooth algebraic curve, and hence a Riemann surface we just need to check, according to Lemma 11.25 that both partial derivatives never vanish at the same point

∂F ∂F of S . Indeed, if = = 0 then z2 = P(z1) = 0. But this implies that z1 = ζk and hence ∂z1 ∂z2 ∂F 0 2 = −P (ζk) = −3ζ 0 for ζk = 0, 1, 2. ∂z1 k ,

dz1 2 dz1 Consider a holomorphic diﬀerential 1-form on C \{z2 = 0}. The restriction α of to the z2 z2

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.

2 Proof. To extend α to points (ζk, 0) ∈ S let us observe that the equation z2 = P(z1) which holds on 0 S implies that 2z2dz2 = P (z1)dz1, i.e. for z2 , 0 we have

dz1 dz2 = 2 0 . (12.1.2) z2 P (z1)

0 dz1 But P (ζk) 0, and hence the form can be deﬁned as a holomorphic 1-form on a neighborhood , z2

dz2 of (ζk, 0) as equal to 2 0 0. P (z1) , R Given a path γ connecting in S two points (a0, b0), (a1, b1) ∈ S the integral α depends only on γ the homotopy class of the pass γ, i.e. it remains unchanged if we continuously deform γ keeping

141 its end-points ﬁxed. This integral gives the exact meaning to the integral

a1 Z du √ u3 − 1 a0 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

2 3 S = {z2 − z1 − 1 = 0}

k we complete each monomial to degree3 by multiplying by an appropriate power z3, i.e. consider the projective curve 2 3 3 S = {z3z2 − z1 + z3 = 0}. (12.1.3)

Let us rewrite this equation in aﬃne coordinates in the charts Ub1, Ub2, Ub3. These equations are 3 3 3 obtained by dividing (12.1.3) by z1, z2 and z3, respectively.

2 3 (i) S ∩ Ub1 = {z31z21 − 1 + z31 = 0},

3 3 (ii) S ∩ Ub2 = {z32 − z12 + z32 = 0},

2 3 (iii) S ∩ Ub3 = {z23 − z13 + 1 = 0}.

In equation (iii) we recognize the equation for the Riemann surface S ⊂ C2, written in coordi-

nates (z13, z23) instead of (z1, z2).

2 Lemma 12.3. S ⊂ CP is a compact Riemann surface. It diﬀers from S ⊂ Ub2 by a unique point with projective coordinates p = (0, 1, 0).

142 Proof. Note that S \Ub3 ⊂ Ub2. Indeed, if z ∈ S \(Ub3 ∪Ub2) then z2 = z3 = 0 which in view of equation

(12.1.3) then implies that z1 = 0, which is impossible. On the other hand, S \ Ub3 = {p = (0, 1, 0)}.

In the coordinate chart Ub2 about p the equation takes the form (ii):

3 3 G(z32, z12) = z32 − z12 + z32 = 0.

We note ∂G (p) = 1 0, and hence the implicit function theorem implies that S is a compact ∂z32 , Riemann surface, which is called the projectivization of the Riemann surface S , see Lemma 11.29.

Let α be the diﬀerential 1-form which we deﬁned on S ⊂ Ub3 in Section 12.1.2 above.

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),

dz13 2dz23 dz23 α = = 0 = 2 2 . z23 P (z13) 3z13 3 0 2 Both expression holds where P(z13) = z13 − 1 , 0 and P (z13) = 3z13 , 0. Both inequalities hold when |z13| > 1. Rewriting the latter expression for α in coordinates z12 and z32 (taking into account

z12 1 that z13 = , z23 = ) we get z32 z23 2 dz − 32 α = 2 . (12.1.4) 3 z12 Recall that on S we have 3 3 z32 + z32 = z12.

Diﬀerentiating we get 2 2 (1 + 3z32)dz32 = 3z12dz12.

Plugging this expression into (12.1.4) we get dz − 12 α = 2 2 . (12.1.5) 1 + 3z32

This expression extends α to the point p with coordinates z12 = z32 = 0 as a non-vanishing holomorphic form.

143 Lemma 12.5. The aﬃne coordinate functions z | : S → C and z | : S → C extend as 13 S =S ∩Ue3 23 S 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 dz α := . 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 holomor-

phic 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 : bS → S be the universal cover of S . One can pull-back the form α to bS . The holomorphic diﬀerential 1-form β := f ∗α is deﬁned by the formula

βz(T) = αp(z) (dz p(T)) ,

where z ∈ U, T is a tangent vector to U at the point z and dpa 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 bS , but then it is exact because U is simply connected. Therefore there exists a holomorphic function f : bS → C such that β = d f . To complete the proof we need the following

144 Lemma 12.7. f : bS → 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 11.15 there exists a lattice Λ ⊂ C

−1 and a biholomorphism h : S → C/Λ such that p ◦ f = 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, ∗ −1 ∗ ∗ −1 ∗ du = ( f ) (p α) = p ◦ f 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 ∈ bS 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 ∈ bS , 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

−1 neighborhood U 3 a such that its pre-image p (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,....

145 In particular, there exists j such that U j 3 a. We denote Ue := U j. Let ζ be a local holomorphic coordinate in U (it exists if U is chosen small enough). Then eζ := ζ ◦ p can be chosen as a local coordinate on each Ue. The non-vanishing holomorphic form ∗ α|U(a) can be written as α = g(ζ)dζ, where g(ζ) , 0. Hence the form β = p α is equal g(eζ)deζ on each Ue. The form β = p∗α is exact, β = d f = f 0(eζ)deζ. Hence, on f 0(eζ) = g(eζ) , 0. Hence the map

f |Ue : Ue → C is injective if the neighborhood U is chosen small enough. The image V := f (u) is open. Hence there exists r > 0 such that Dr( f (ea)) ⊂ V. Denote

−1 −1 ∆ = f (D r ( f (a))) ∩ Ue, e∆ := f (D ( f (a))) ∩ Ue. a 2 a r

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 ∈ bS 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 (∆ ) = D r ( f (c)) ∩ f (∆ ) = D r ( f (a)) . c 2 a 2

Therefore, D r ( f (c)) ⊂ D ( f (a)). But this implies that 2 r

−1 ∆c ⊂ e∆a := f (Dr( f (a))) ∩ Ue.

Hence, b ∈ e∆ , but this contradicts to the fact that f | is injective. a e∆a

Lemma 12.9 (Step 2). The map f : bS → 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

−1 each z ∈ C the pre-image f (D r )(z) is a disjoint union of neighborhoods which are biholomor- 2

phically mapped by the map f onto D r )(z), which is the deﬁnition of a covering map. By the open 2 image theorem we know that f (U) is open. If f (U) , C then there exists a point z ∈ C \ f (U) ∈ | − | r which is a boundary point of f (U). Hence, there exists a bS such that f (a) z < 2 . But then

z ∈ D r ( f (a)) = f (∆ ) ⊂ f (U), which is a contradiction. 2 a

Lemma 12.10 (Step 3). The map f : bS → C is a biholomorphism.

146 Proof. This follows from simply connectedness of C and Theorem 11.10. Because C is simply connected, the identity map Id : C → C is also a universal cover, but then according to Theorem 11.10.2 the universal cover is unique, i.e there exists a biholomorphism F : C → bS 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.

3 2 1. For a degree 3 polynomial P(u) = u + a1u + a2u + 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 2 dz the equation w = P(z), and the diﬀerential form α = 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

ez(u) = z( f (πΛ(u))), we(u) = w( f (πΛ(u)))

then dez(u) dez(u) = du, or, = we(u). we(u) du 147 Hence, !2 dz(u) e = w(u)2 = P(u), du e i.e. the function ez(u) is a solution of the diﬀerential equation

!2 dz(u) e = P(u). (12.1.6) du

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

1 T(ω1, ω2) → CP .

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 ! X 1 1 1 ℘(u) = − + . (12.2.1) (u − λ)2 λ2 u2 λ∈Λ\0

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 1 T(ω1, ω2) → CP with a unique pole of order 2.

148 Proof. Denote = {inf |u − λ|; λ ∈ Λ}. then we have 1 1 2|λ||u| + |u|2 C − = ≤ , (u − λ)2 λ2 |λ|2|u − λ|2 |λ|3 where C depends on and R := |u|. Hence, X 1 1 X 1 − ≤ C . (u − λ)2 λ2 |λ|3 λ∈Λ\0 λ∈Λ\0 But the series in the right hand side converges, as it is clear by comparison with the converging integral ∞ 2π Z dxdy Z Z drdφ = = 2π. |u|3 r2 C\{|u|≤1} 1 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 1 ℘0(u) = −2 . (12.2.2) (u − λ)3 λ∈Λ 0 Clearly ℘ (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 ) − ℘(u) = ℘0(u + ω ) − ℘0(u) = 0, j = 1, 1, 2 du j j

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 ω j c = ℘ − ℘ − = 0, j 2 2 1 and therefore ℘(u) is doubly periodic and thus deﬁnes a meromorphic function T(ω1, ω2) → CP 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 1 ℘(u) = + au2 + bu4 + .... u2

149 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

1 X 1 1 ℘(u) − = − u2 (u − λ)2 λ2 λ∈Λ vanishes at 0. Diﬀerentiating twice we get 6 ℘00(u) = + 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 d (℘0(u))2 = 2℘0(u)℘00(u) = 12℘0(u)℘2(u) − 20a℘0(u) = 4℘3(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 2 3 (℘ (u)) = 4℘ (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.

3 Let us denote by e1, e2, e3 zeroes of the cubic polynomial 4w − g2w − g3, so that we have

3 4w − g2w − g3 = 4(w − e1)(w − e2)(w − e3).

150 By Vieta’s theorem we conclude that

e1 + e2 + e3 = 0, g e e + e e + e e = − 2 , 1 2 2 3 3 1 4 g e e e = 3 . 1 2 3 4

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 0 0 0 0 0 ℘ (ω1 − u) = −℘ (u), ℘ (ω2 − u) = −℘ (u), and ℘ (ω1 + ω2 − u) = −℘ (u).

Hence, ω ω ω + ω ℘0 1 = 0, ℘0 2 = 0, ℘0 1 2 = 0. 2 2 2 Comparing this equation with (12.2.3) we conclude that there exists the value of the ℘(u) at zeroes 0 of the derivative ℘ (u) must coincide with e1, e2, e3, i.e.

ω ω ω + ω ℘ 1 = e , ℘ 2 = e , ℘ 1 2 = e . 2 1 2 2 2 3

The following theorem gives a necessary and suﬃcient condition for a polynomial w3 + aw + b

3 to be proportional to the polynomial 4w −g2w−g3 which appears in the right hand side of equation (12.2.3).

3 Theorem 12.12. For any polynomial P(w) = w + 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.

151 12.2.3 Identifying Weierstrass ℘-function with the solution of the pendulum equation

As we had seen in Section 12.1 the

u 1 Z du t = √ p , 2 (E + u)(1 − u2) u0 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. − E First of all we note that by the change of variable v = u 3 we kill the coeﬃcient with the square term in the cubic polynomial ! E2 E3 2E E2 E (E + u)(1 − u2) = −(u3 + Eu2 − u − E) = − v3 − Ev2 + v − + Ev2 − v + − v + − E 3 27 3 9 3 !! E2 − 2E E3 E2 2E 1 = − v3 + v + − + − := − (4v3 − g v − g ), 3 27 9 3 4 2 3

and we get v i Z dv τ := − √ t = p 3 2 4v − g2v − g3 v0 , or dτ 1 = , p 3 dv 4v − g2v − g3 and hence for the inverse function function v(τ) we get a diﬀerential equation

dv p = 4v3 − g v − g (12.2.4) dτ 2 3 or !2 dv = 4v3 − g v − g , dτ 2 3

According to Theorem 12.12 there exists a lattice Λ(ω1, ω2) such that the corresponding Weier-

strass function ℘ = wpΛ is a solution of this diﬀerential equation:

0 2 (℘ (τ)) = 4℘(τ) − g2℘(τ) − g3.

152 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 τ is chosen to satisfy the equality ℘(τ ) = v . Recalling that v = u + E = cos θ + E and τ = − √i t 0 0 0 3 3 2 we conclude that the function ! ! it E θ(t) = arccos ℘ − √ − 2 3

is the solution of the pendulum equation with the initial condition θ(0) = θ0 and where we set E v0 = cos θ0 + 3 .

12.2.4 More about geometry of the Weierstrass ℘-function

1 1 The meromorphic function ℘ : C/Λ(ω1, ω2) = T(ω1, ω2) → CP is a branch cover of CP . Recall

0 ω1 ω2 ω1+ω2 that the derivative ℘ vanishes at the points 2 , 2 and 2 . The corresponding critical values 3 e1, e2, e3 which are the roots of the cubic polynomial P(w) = 4w − g2w − g3 are branching points

153 1 of the branch cover ℘ : T(ω1, ω2) → CP . 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

1 holomorphic map ℘ : T(ω1, ω2) → CP 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).

154 Chapter 13

The Gamma function

13.1 Product development

Q∞ The inﬁnite product of complex numbers p j is called convergent if there exists lim Πn, where j=1 n→∞ Qn Πn = 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 → ∞.

Πn Indeed, pn = → 1. Πn−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.

Q∞ P∞ Theorem 13.2. The product p j converges if and only if the series log p j converges. j=1 j=1

P∞ Q∞ Proof. It is clear that if log p j converges then p j converges. Indeed, j=1 j=1

Pn n log p j Y j=1 e = Πn = p j, j=1

155 Pn so the convergence of partial sums S n := 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 Πn → that log(ab) = log a + log b. Suppose there exists a non-zero limit lim Πn = Π. Then Π 1 and

iφn pn → 1. Hence, for large n we can write pn = rne , where |φn| < π. n iΘn iΘ P Let Πn = Rne , Π = Re , 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.

| | π | | Choose n large enough, so that θn < 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 if n is large enough. Pn Thus, Θn = φ j → Θ + 2Kπ. Thus j=1 Xn Xn Xn log p j = ln r j + i φ j → log R + iΘ + 2πKi. n→∞ j=1 j=1 j=1 ∞ ∞ Q P 1 The product p j is called absolutely convergent if the series log p j absolutely converges. j=1 j=1

∞ 1 Q Warning: the absolutely convergence implies but not equivalent to the convergence of the product |p j| (why?) j=1

156 Q∞ Lemma 13.3. Denote pn = 1 + an. The product pn absolutely converges if and only if the series n=1 P∞ |an| converges. n=1

| log (1+an)| Proof. We have → 1 if |an| → 0. But both (if and only if) assumptions imply that an → 0. |an| Hence, for large n we have |a | n < | log(1 + a )| < 2|a |, 2 n n and hence both series are simultaneously converge or not.

Exercise 13.4. 1. Prove that if |z| < 1 then

1 (1 + z)(1 + z2)(1 + z4)(1 + z8) ··· = . 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

!−1 ! Y 1 Y 1 1 1 − = + + ... p p p2 p prime p prime diverges. ∞ P 1 Hint: Any integer can be uniquely factored as a product of primes and the series n diverges. n=1 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−.

157 Remark 13.5. The actual rate of growth of the function π(n) is given by the formula n π(n) π(n) ∼ , i.e. lim = 1. n→∞ 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 Vallee´ 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).

f 0(z) Proof. The function 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 f 0(z)dz dg = g0(z)dz = α = . f (z)

0 f 0(z) Thus g (z) = f (z) and ! d f 0(z) f (z)e−g(z) = ( f 0(z) − f (z)g0(z))e−g(z) = f (z) − g0(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 f (z) = zmeg(z) 1 − . a j=1 j

158 Proof. Apply Lemma 13.6 to the entire function

f (z) n m Q z z 1 − a j=1 j

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. π2 X∞ 1 = , z < Z. 2 − 2 sin πz −∞ (z n) 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

1 is the left-hand side. The coeﬃcients with (z−n)2 in the Laurent expansion about n in both sides are the same, so the diﬀerence π2 X∞ 1 h(z) = − 2 − 2 sin πz −∞ (z n) is an entire holomorphic function. Note that

1 | sin(x + iy)| = |eixe−y − e−ixey| ≤ e|y| → ∞, 2 |y|→∞

and hence

π2 → = 0. (13.3.1) sin2 π(x + iy) |y|→∞ We also note

X∞ 1 X∞ 1 = | − |2 − 2 2 −∞ x + iy n −∞ (x n) + y

159 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. (13.3.2)

1 Moreover, for x = 2 we have 2 2 1 !2 1 − n 1 1 − n 2 = − n ≤ 2 . 2 y2 2 1 2 2 1 + 4y − n + y 1 + 1 2 2 ( 2 −n) Hence,

∞ ∞ X 1 1 X 1 ≤ → 0. (13.3.3) 1 2 y2 1 2 |y|→∞ −∞ − 1 + 4 −∞ − 2 + iy n 2 n Combining (13.3.1) and (13.3.3) we conclude that ! 1 h + iy → 0, 2 |y|→∞ and therefore, in view of (13.3.2) we get h(z) = 0.

Lemma 13.9. ! 1 X 1 1 π cot πz = + + . z z − n n n,0 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  ! ∞ d 1 X 1 1  X 1  + +  = − . dz  z z − n n  (z − n)2 n,0 −∞ On the other hand, we have π2 X∞ 1 (π cot πz)0 = − = − . 2 − 2 (sin πz) −∞ (z n) Thus, ! 1 X 1 1 π cot πz = C + + − . z z − n n n,0 1 P 1 1 But both functions, π cot πz and z + z−n − n are odd (i.e. changing signs when z 7→ −z), and n,0 therefore, C = 0.

160 Lemma 13.10. ∞ 2 ! Y z z Y z sin πz = πz 1 − e n = πz 1 − . n n2 n,0 n=1 ∞ Q z z P 1 − n Proof. First note that the product πz 1 n e absolutely converges because the series n2 n,0 1 absolutely converges, see Lemma 13.3. Moreover, it converges uniformly on every compact set in

z Q z n C. The function πz 1 − n e has simple zeroes at all real integer points 0, ±1, ±2,... . But so n,0 does sin πz. Hence, the entire function

sin πz h(z) = z Q z n πz 1 − n e n,0 has no zeroes, and hence by applying Lemma 13.6 we get: g(z) Y z z sin πz = e πz 1 − e n . (13.3.4) n n,0

f 0 Let us prove that g(z) is a constant. To do this we compute the logarithmic derivative (i.e. f ) of both parts of the equation (13.3.4). We get ! 1 X 1 1 π cot πz = + g0(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 sin πz C Y z z = e π 1 − e n . z n n,0 Passing to the limit when z → 0 we get 1 = eC, and hence we obtain the required formula

Y z z sin πz = πz 1 − e n . (13.3.5) n n,0 Combining terms with n and −n we get the second expression for the right-hand side:

∞ ! Y z2 sin πz = πz 1 − . (13.3.6) n2 n=1

161 13.4 The Gamma function: deﬁnition and some properties

Denote ∞ Y z − z G(z):= 1 + e n . n n=1 Then (13.3.5) can be rewritten as

sin πz = zG(z)G(−z). (13.4.1) π

Note that we have G(0) = 1. The value γ = − log G(1). is called Euler’s constant. Thus, ∞ ! −γ Y 1 − 1 e = 1 + e n (13.4.2) n n=1

n 1 Q 1 − k Note that the partial product Πn = 1 + k e is equal to k=1

n P 1 − k Πn = (n + 1)e 1 . (13.4.3)

Hence, ! 1 1 1 γ = lim 1 + + + ··· + − log n = 0.57722 ... (13.4.4) n→∞ 2 3 n

We deﬁne now the Gamma function by the formula

−γz −γz ∞ −1 −γz ∞ z e e Y z z e Y ne n Γ(z) = = 1 + e n = . (13.4.5) zG(z) z n z z + n n=1 n=1 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 Γ(z) = lim (nz = ez ln n). (13.4.6) n→∞ z(z + 1)(z + 2) ... (z + n)

162 Proof. By deﬁnition,

 z  −γz n −γz z 1+···+ 1 n!e Y e j  n!e e ( 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 n −z P 1 −γz z k e = lim n e 1 , 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!; √ − π 1 2. Γ(z)Γ(1 z) = sin πz ; in particular, Γ( 2 ) = π; ∞ d Γ0(z) P 1 3. dz Γ(z) = (z+n)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).

− 3π|y| |Γ(x + iy)| ≥ Ce 4 .

Proof. We have

−γx ∞ x  −γx ∞ x  ∞ e Y ne n e Y ne n  x Y 1 |Γ(x + iy)| = p p =   p q 2 2 2 2  x n + x 2 2 y2 x + y 1 (n + x) + y 1 x + y 1 1 + (n+x)2

163 Let us observe that −γx ∞ x e Y ne n = Γ(x) x n + x 1 and Y∞ 1 Y∞ 1 q ≥ q . y2 2 1 1 y 1 + (n+x)2 1 + n2 On the other hand, according to Lemma 13.10 we have ∞ ! sinh π|y| sin πi|y| Y y2 = = 1 + , π|y| πi|y| n2 1 and therefore p Y∞ 1 π|y| = . q 2 p 1 y sinh π|y| 1 + n2 Thus we get p xΓ(x) π|y| |Γ(x + iy)| ≥ p p . x2 + y2 sinh π|y| For x ∈ [1, 2] we have p p π|y| 2π|y| − π|y| p = p ≥ C1e 2 (x2 + y2) sinh π|y| (x2 + y2)(eπ|y| − e−π|y|) 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

p − π|y| 2π|y| e 2 3π|y| 0 − 4 p ≥ C1 p ≥ C1e . (x2 + y2)(eπ|y| − e−π|y|) |y| Thus p xΓ(x) π|y| − 3π|y| p p ≥ Ce 4 . 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|

1 for any constant α > 2 .

164 13.5 Integral representation of the Gamma function

Lemma 13.15. The integral Z∞ bΓ(z):= e−ttz−1dt (13.5.1)

0 absolutely converges if Re z > 0. It satisﬁes the equation

bΓ(z + 1) = zbΓ(z). (13.5.2)

Proof. Let z = x + iy, x > 0. We have

− − − tz 1 = e(x 1+iy) ln t = tx 1.

R1 R∞ The integral converges because the function tx−1 is integrable when x > 0, and integral con- 0 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:

∞ ∞ Z ∞ Z − − − − bΓ(z + 1) = e ttzdt = −e ttz +z e ttz 1dt = zbΓ(z), 0 0 0 because lim tze−t = 0 provided that Re z > 0. t→0 R∞ Hence, the integral e−ttz−1dt deﬁnes a holomorphic function on the half-plane {Re z > 0} 0 which satisfy the same relation (13.5.2) as Γ(z). Moreover, bΓ(1) = Γ(1) = 1. It turns out that

Theorem 13.16. bΓ(z) = Γ(z).

In particular, bΓ(z) admits a meromorphic extension to the whole C.

Proof. The ratio bΓ(z) h(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

165 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−2πye2πix 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 2πi + ln(−z) g−1(z) = . 2πi Let us analyze the behavior of f near 0 and at ∞. In view of the equality |e−ttz−1| = |e−ttx−1| ≤ e−t we have bΓ(z) ≤ 1, and thus |bΓ(g−1(z))| ≤ 1. Let us now estimate Γ(g−1(z)) . We have φ ln r g−1(reiφ) = 1 + − i . 2π 2π φ − ln r Using Lemma 13.13 and plugging z = 1 + 2π i 2π we get   3 !  − 8 3| ln r| Cr , r ≥ 1; φ ln r −  Γ 1 + − i ) ≥ Ce 8 ≥  2π 2π  3 Cr 8 , r < 1.

Thus   3 O(|z| 8 ), |z| → ∞, |bΓ(z)|  | f (z)| = =  |Γ(z)|  − 3 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 bΓ(1) = Γ(1) = 1 we get h(z) = 1, and hence bΓ(z) = Γ(z).

Exercise 13.17. Prove the following identities:

1. Legendre’s duplication formula

z− 1 ! n 2 1 Γ(2z) = √ Γ(z)Γ z + . π 2

2. Gauss’ formula. ! ! n−1 z− 1 z z + 1 z + n − 1 (2π) 2 Γ(z) = n 2 Γ Γ ... Γ . n n n

166 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 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. 1 Y = 1 − p−s , σ = Re s > 1. ζ(s) p prime

We assume that primes in the product are ordered in the increasing order.

1It is customary to denote the argument of the ζ-function by the letter s rather than z.

167 Proof. First we note the the product in the right-hand side absolutely converges because the series P p−s is converging, see Lemma 13.3. Then p prime

X∞ X∞ X ζ(s)(1 − 2−s) = n−s − (2n)−s = m−s. 1 1 m is odd Similarly, X ζ(s)(1 − 2−s)(1 − 3−s) = m−s,

where the sum is taken over all integer m not divisible by 2 and 3. Continuing we get

−s −s −s X −s ζ(s)(1 − 2 )(1 − 3 ) ... (1 − pn ) = m ,

where the sum is taken over all integer not divisible by ﬁrst n prime numbers: 2, 3,..., pn. Note P −s −s that m = 1 + Tn, where Tn = Nn + ... , where Nn > pn. Hence, as lim Tn = 0 and thus n→∞

−s −s −s ζ(s)(1 − 2 )(1 − 3 ) ... (1 − pn ) → 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 n−s extends meromorphically from {Re s > 1} to C with a unique simple 1 pole at s = 1 with Res1ζ = 1.

Lemma 14.3. ∞ Z xs−1 ζ(s)Γ(s) = dx, Re s > 1. ex − 1 0 Proof. First, we note xs−1 = e(s−1) ln x and that the integral in the right-hand-side converges at 0 and at inﬁnity.

168 We have 1 e−x X∞ = = e−nx. ex − 1 1 − e−x 1 Therefore, ∞ ∞ ∞ Z xs−1 X∞ Z X∞ Z dx = xs−1e−nxdx = n−s us−1e−udu ex − 1 nx=u 1 1 0 0 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) Z (−z)s−1 ζ(s) = − dz. 2πi ez − 1 ∂U 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 (−z) → straightforward to see that ez−1 dz tends to 0 when 0, so that the integral over ∂U converges A to ∞ ∞ Z (−z)s−1 Z xs−1e−i(s−1)π Z xs−1ei(s−1)π dz = − dx + dx ez − 1 ex − 1 ex − 1 ∂U 0 0 ∞ Z xs−1 = 2i sin(s − 1)π dx. ex − 1 0 But in view of Lemma 14.3 we have ∞ Z xs−1 dx = ζ(s)Γ(s), ex − 1 0

169 and hence Z (−z)s−1 dz = 2i(sin π(s − 1))ζ(s)Γ(s) = −2i(sin πs)ζ(s)Γ(s). ez − 1 ∂U According to Theorem 13.12.2 we have π (sin πs)Γ(s) = . Γ(1 − s) Hence, 2πiζ(s) −2i(sin πs)ζ(s)Γ(s) = − . Γ(1 − s) Therefore, Γ(1 − s) Z (−z)s−1 ζ(s) = − dz. 2πi ez − 1 ∂U

Proof of Theorem 14.2. The contour integral Z (−z)s−1 dz ez − 1 ∂U 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 ez−1 = 1 in view ∂U of the residue theorem. Hence, ζ(s) has a single pole at 1 with residue 1.

14.3 Zeroes of the ζ-function

Recall that Bernoulli numbers Bn can be deﬁned as coeﬃcients in the Laurent expansion of the 1 function ez−1 :

1 1 1 X∞ B = − + (−1)n−1 n z2n−1. (14.3.1) ez − 1 z 2 2n! n=1

170 Proposition 14.5.  − 1  2 , n = 0,   ζ(−n) = 0, n = 2k,    − k Bk − ( 1) 2k , n = 2k 1. Proof. According to Lemma 14.4 we have

Γ(1 + n) Z (−z)−(n+1) ζ(−n) = − dz. 2πi ez − 1 ∂U Hence, by the residue theorem the values ζ(−n) are the coeﬃcients of the expansion given by formula (14.3.1) multiplied by (−1)nn!.

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 { 1 } zeroes lie on the line Re s = 2 . 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.

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