5.) Complex Analysis

Complex AnalysisKarl-Heinz FieselerUppsala 20101

Contents1 Complex numbers 32 Elementary functions 73 Convergence 114 Logarithm functions 145 Complex Differentiability 176 Path integrals 227 The Cauchy formula 358 Basic properties of holomorphic functions 419 Laurent series and residues 4910 Construction of holomorphic and meromorphic functions 6011 The Dirichlet Problem 6612 Conformal maps 782

1 Complex numbersLet us start with a heuristic description of the passage from the set Q of allrational numbers to the set R of all real numbers:Rational numbers can be thought of as points on an infinite (horizontal)line once one has specified the positions of 0 and 1 (with 1 lying to the righthand side of the point 0). But the set Q of all such rational points does notfill the entire line. Real numbers then are (or correspond to) the points onthat line, they form a set R ⊃ Q.Indeed it turns out that the arithmetic operations on Q, i.e. addition andmultiplication, can be extended continuously from Q to R ⊃ Q such that thefollowing rules are satisfied:1. The commutative laws: a + b = b + a, ab = ba2. The associative laws: a + (b + c) = (a + b) + c, a(bc) = (ab)c3. The distributive law: a(b + c) = ab + ac4. a + 0 = a, a · 1 = a5. The equation a + x = 0 is solvable.6. The equation ax = 1 is solvable for a ≠ 0.Remark 1.1. 1. Any set K with two distinguished elements 0, 1 ∈ Kand two binary operations + and · satisfying the above six conditionsis called a field.2. The solution of the equation a + x = 0 resp. ax = 1 in a field K isunique, we write it in the form x = −a resp. x = a −1 . Fractions thenare defined as:ab := ab−1 = b −1 a.3. In a field K there are no zero divisors:since if ab = 0 and a ≠ 0, we have:ab = 0 ⇐⇒ a = 0 ∨ b = 0,0 = a −1 · (ab) = (a −1 a)b = b.3

So it seems natural to ask whether the above extension process can becontinued:Is there a field K R the binary operations of which extend those on R?If so, K becomes an R-vector space: The scalar multiplicationR × K −→ Kis the restriction of the field multiplicationK × K −→ K.If even n = dim K < ∞ we can choose a basis e 1 := 1, e 2 , ..., e n , such thatevery element z ∈ K has the formz = a +n∑a µ e µwith unique real numbers a, a 2 , ..., a n . Then the multiplication is determinedby the valuese µ e ν ∈ K, 2 ≤ µ, ν ≤ n.For n = 2 there is only the square (e 2 ) 2 ∈ K to be considered. And by a goodchoice of the second base vector e 2 ∈ K \ R we can make it equal −1 ∈ R.Proposition 1.2. Let K R be a field. If dim K = 2 holds for the dimensionof the real vector space K, there is an element i ∈ K such thatµ=2i 2 = −1andK = R + Ri.Before we prove Prop.1.2, we remark, that the statement is true wheneverdim K < ∞, see Th.8.5.2.Proof. Since dim K = 2, we have only to hunt for an element i ∈ K withi 2 = −1. Indeed, it is sufficient to find an element j ∈ K \ R with j 2 ∈ R.In that case we have j 2 < 0: The case j 2 = c ≥ 0 is impossible, since then(j − √ c)(j + √ c) = j 2 − c = 0,4

ut the field K does not admit zero divisors. Finally we rescale j toi :=j √|c|,where c = j 2 . To find j we start with any element e = e 2 ∈ K \ R and writee 2 = a + be.We write j = x + ye and compute its square(x + ye) 2 = x 2 + 2xye + (a + be)y 2 = (x 2 + ay 2 ) + (2x + by)y · eand see that j = b − 2e will do.Of course we have still to make sure that such a field really exists: Thecomplex plane C is the real vector spaceC := R 2 ,endowed with the following multiplicationThen the map(a, b) · (c, d) = (ac − bd, ad + bc).R −→ C, a ↦→ (a, 0)preserves both addition and multiplication, and the second base vector i :=(0, 1) satisfiesi 2 = (−1, 0).Identifying a ∈ R with (a, 0) ∈ C we arrive atand thusa + bi = (a, b)C = {a + bi; a, b ∈ R}.Both field operations have geometric interpretations: The addition is theaddition of vectors in the plane, while the multiplication is best understoodin polar coordinates. For ϕ ∈ R we definee(ϕ) := cos(ϕ) + i sin(ϕ),5

any vector x + iy ≠ 0 then can be writtenx + iy = re(ϕ),where r = √ x 2 + y 2 is the length of the vector x + iy and the angle ϕ is onlydetermined up to an integer multiple of 2π, i.e.x + iy = re(ϕ) = re(ϕ + 2πk), ∀k ∈ Z.This ambiguity in the choice of the angle ϕ, as intuitive as it is, is veryimportant and should never be forgotten about: Indeed it is behind manyphenomena in complex analysis!Let us now consider the product of two vectors in polar coordinates: Weobtainre(ϕ) · se(ψ) = (rs)e(ϕ + ψ)as a consequence of the addition theorems for trigonometric functions. Thusthe length of the product of two vectors is the product of the lengths andthe angles add. With that formula in mind we easily check the above sixconditions - note in particular that(re(ϕ)) −1 = r −1 e(−ϕ).Let us add some useful remarks on the arithmetics of complex numbers: Firstof all(−i) 2 = (−1) 2 i 2 = i 2 = −1,so from the point of view of arithmetics, −i ∈ C should be as good as i ∈ C.What do we mean by this? Let us consider the mapC −→ C, z = x + iy ↦→ z := x + (−i)y = x − iy,the reflection with respect to the ”real axis” R ⊂ C, also called complexconjugation. It preserves both addition and multiplicationits fixed point set is the real line:z + w = z + w, zw = z · w,z = z ⇐⇒ z ∈ R.6

The real and imaginary part of a complex number z = x + iy are defined asRe(z) = x = z + z2 , Im(z) = y = z − z ,2iits absolute value |z| ∈ R ≥0 is nothing but its euclidean length|z| := √ x 2 + y 2 = √ zz.In particular, for z = x + iy ≠ 0 we have1z = zzz =z|z| 2 =2 Elementary functionsx|z| − i y2 |z| . 2Complex analysis deals with functions f : G −→ C defined on certain subsetsG ⊂ C. We shall discuss in this section some basic examples.Let us start with R-linear maps f : C −→ C, i.e. endomorphisms of thereal vector space C. Such a map can uniquely be writtenf(z) = f(x + iy) = ax + bywith complex coefficients a, b ∈ C. In fact a = f(1), b = f(i).If we don’t want to split up into real and imaginary part, we can as wellrewritef(z) = cz + dzwith unique complex coefficients c, d ∈ C using the identitiessuch thatx = z + z2 , y = z − z ,2ic = 1 2 (a − ib), d = 1 (a + ib).2Note that an endomorphism f : C −→ C is even C-linear, i.e.f(λz) = λf(z), ∀ λ ∈ C,7

iff d = 0. The condition is obviously sufficient; to see the necessity take λ = iand compare the coefficients of the expressions for the right and the left handside.Starting with the constant and R-linear maps and allowing all combinationsof them using sums and products we obtain the real polynomialmaps f : C −→ C; they are given as sumsf(z) = f(x + iy) =m∑µ=0 ν=0n∑a µν x µ y νwith unique complex coefficients a µν ∈ C. The function f coincides with itsTaylor expansion at every point z 0 = x 0 + iy 0 ∈ C, i.e.withf(z) = f(x + iy) =m∑µ=0 ν=0c µν = 1 ∂ µ+ν fµ!ν! ∂x µ ∂y (z 0).νn∑c µν (x − x 0 ) µ (y − y 0 ) νAs a consequence f ≡ 0 near z 0 (by that we mean that f vanishes identicallyon some open discD r (z 0 ) := {z ∈ C; |z − z 0 | < r}of positive radius r) implies that c µν = 0 for all µ, ν. So we obtain thefollowing:Theorem 2.1 (Weak Identity Theorem for R-polynomials.). Let f, g :C −→ C be real polynomial maps, and D := D r (z 0 ) the disc with center z 0and radius r > 0. Thenf| D = g| D =⇒ f = g.Proof. Apply the above reasoning to the difference f − g.Finally, a real polynomial map may be written as a (z, z)-polynomialf(z) =m∑µ=0 ν=08n∑b µν z µ z ν .

To see the uniqueness of the coefficients b µν ∈ C we may use an argumentanalogous to the above one with the ”complex partial derivatives” ∂ ∂z , ∂ ∂z tobe introduced later on.The complex polynomial maps f : C −→ C are the real polynomialswithout z-terms, i.e.m∑f(z) = a µ z µ .Though we can not draw the graphµ=0Γ f := {(z, f(z)), z ∈ C} ⊂ C 2of such a function, we want to study its ”geometry”: Here the followingconcept is quite useful:Definition 2.2. Let f : G −→ C be a complex valued function. The fiberf −1 (w) ⊂ G over a point w ∈ C is defined as the setf −1 (w) := {z ∈ G; f(z) = w},so it consists of all points z ∈ G lying ”above” w ∈ C - where one thinks ofthe map f as a sort of projection.Let us consider the power mapf(z) = z n .It is surjective: Its zero fiber consist of one pointf −1 (0) = {0},while for r > 0 the fibre{ (f −1 (re(ϕ)) = n√ ϕ r · en + k 2π n)}; k = 0, ..., n − 1contains n different points. In particular f −1 (1) ⊂ C is the set of n-th rootsof unity, which span a regular n-gon.A rational function f : G −→ C is a functionf(z) = p(z)q(z)9

with complex polynomials p, q : C −→ C and q(z) ≠ 0 for z ∈ G. Sothe maximal choice of the domain of definition for a rational function isG := C \ N(q) with the finite set N(q) := {z ∈ C; q(z) = 0} of zeros of q.The complex exponential mapexp : C −→ Cextends the real exponential function to a function on the entire plane C; itis defined byexp(z) = exp(x + iy) := e x e(y) = e x (cos(y) + i sin(y)).It satisfies the functional equationand attains all nonzero values:exp(z 1 + z 2 ) = exp(z 1 ) exp(z 2 )exp(C) = C ∗ := C \ {0}.Namely, given w = se(ψ) ∈ C ∗ the point z = ln(s) + iψ satisfies exp(z) = w.Furthermoreexp −1 (1) = 2πiZ.Now the functional equation implies that the w-fiber looks as followsexp −1 (w) = z + exp −1 (1) = z + 2πiZ,since w = exp(z 1 ) = exp(z) implies exp(z 1 − z) = 1.The complex trigonometric functionssin(z) := 1 (exp(iz) − exp(−iz))2iandcos(z) := 1 (exp(iz) + exp(−iz))2extend the real trigonometric functions from the real line R to the complexplane C. They are 2π-periodic functionsand satisfysin(z + 2π) = sin(z), cos(z + 2π) = cos(z)sin 2 (z) + cos 2 (z) = 1.But that does not imply that the complex trigonometric functions are bounded!10

3 ConvergenceHere we shall recall briefly some basic facts about convergence, continuityand open sets: Nothing of that is specific for C but rather holds in any R n .Definition 3.1. A sequence (z ν ) of complex numbers is said to converge toz 0 ∈ C, iff lim ν→∞ |z ν − z 0 | = 0.Remark 3.2. 1. lim ν→∞ z ν = z 0 iff lim ν→∞ x ν = x 0 and lim ν→∞ y ν = y 0 .That follows from the estimates|x ν − x 0 |, |y ν − y 0 | ≤ |z ν − z 0 | ≤ |x ν − x 0 | + |y ν − y 0 |.2. The sum and the product of two convergent sequences is convergent,the limit being the sum resp. product of the two limits.3. If lim ν→∞ z ν = z 0 ≠ 0, then lim ν→∞1z ν= 1 z 0.Definition 3.3. Let D := D r (z 0 ) be a disc and D ∗ := D\{z 0 } the punctureddisc. Assume f : D ∗ −→ C is a function. We writelimz→z 0f(z) = a,if for any sequence (z ν ) ⊂ D ∗ converging to z 0 we havelim f(z ν) = a.ν→∞The functions we are interested in are defined on open subsets of thecomplex plane:Definition 3.4. A set G ⊂ C is called open, if, given any point z 0 ∈ G thereis some r = r(z 0 ) > 0, such that D r (z 0 ) ⊂ G. By an open neighbourhood ofz 0 ∈ C we mean any open set U ∋ z 0 .Remark 3.5.1. Unions and finite intersections of open sets are open.2. A sequence z ν converges to z 0 iff outside any open neighbourhood U ofz 0 there are only finitely many z ν .11

Definition 3.6. Let G ⊂ C be open. A function f : G −→ C is calledcontinuous at z 0 ∈ G iflim f(z) = f(z 0 ).z→z 0It is called continuous if it is continuous at every point of G.Remark 3.7. 1. The function f : G −→ C is continuous at z 0 ∈ G if forevery open set V ∋ f(z 0 ) the inverse imagecontains some disk D r (z 0 ).f −1 (V ) := {z ∈ G; f(z) ∈ V }2. The function f : G −→ C is continuous if for every open set V theinverse image f −1 (V ) ⊂ G is open as well.In order to construct interesting functions f : G −→ C one often representsthem as limits: f(z) = lim n→∞ f n (z) with functions f n : G −→ C. Butif we want to show that certain properties of the functions f n hold as wellfor the limit f, we need that the convergence is uniform.Example 3.8. Letg(z) :={ 3|z| 2 − 2|z| 3 , if |z| ≤ 11 , if |z| ≥ 1.Though the functions f n := g n : C −→ C are continuous, indeed evendifferentiable, their pointwise limit, the function f : C −→ C with{ 0 , if |z| < 1f(z) :=1 , if |z| ≥ 1is not!Definition 3.9. A sequence of functions f n : G −→ C is said to convergeuniformly to a function f : G −→ C, if for any ε > 0 there is a naturalnumber n 0 , such that for n ≥ n 0 the estimate |f(z) − f n (z)| < ε holdssimultaneously for all z ∈ G.The uniform limit of a sequence of continuous functions is again continuous:12

Proposition 3.10. If the sequence of continuous functions f n : G −→ Cconverges uniformly to f : G −→ C, then the function f is continuous aswell.Proof. We have to show that z ν → z 0 ∈ G implies f(z ν ) → f(z 0 ). Take someε > 0. Choose n ∈ N, such that the estimate |f(z) − f n (z)| < ε/3 holdsfor all z ∈ G. Since f n is continuous at z 0 , we have f n (z ν ) → f n (z 0 ). Inparticular we can find a ν 0 ∈ N, such that |f n (z 0 ) − f n (z ν )| < ε/3 holds forν ≥ ν 0 . Now|f(z 0 ) − f(z ν )| = |(f(z 0 ) − f n (z 0 )) + (f n (z 0 ) − f n (z ν )) + (f n (z ν ) − f(z ν ))|≤ |f(z 0 ) − f n (z 0 )| + |f n (z 0 ) − f n (z ν )| + |f n (z ν ) − f(z ν )| ≤ ε/3 + ε/3 + ε/3 = ε.Example 3.11. Let g ν : G −→ C be continuous functions, assume that|g ν (z)| ≤ R ν , ∀ z ∈ Gwith reals R ν > 0, such that ∑ ∞ν=0 R ν < ∞. Then the function f : G −→ Cwith∞∑f(z) := g ν (z)ν=0is continuous, since the sequence of partial sumsf n (z) =n∑g ν (z)converges uniformly to f. Namely, given ε > 0, choose n 0 ∈ N such that∑ ∞ν=n 0R ν < ε. Then for n ≥ n 0 we have|f(z) − f n (z)| =∣∞∑ν=n+1ν=0g ν (z)∣ ≤∞∑ν=n+1|g ν (z)| ≤∞∑ν=n+1R ν < ε.Finally we need the notion of a connected open set:Definition 3.12. An open set G ⊂ C is called (path-)connected if for anytwo points a, b ∈ G there is a path (=continuous map) γ : I := [0, 1] −→ Gwith γ(0) = a, γ(1) = b. A connected open set is also called a domain.13

The following criterion is often quite useful:Proposition 3.13. An open set G ⊂ C is a domain iff it can not be writtenG = U ∪ Vas the disjoint union of two non-empty open sets U and V .Proof. The condition is necessary: We show that a point a ∈ U can not beconnected to a point b ∈ V by a path γ : I := [0, 1] −→ G with γ(0) =a, γ(1) = b. Otherwise denote t 0 ∈ I the least upper bound (supremum)of the set γ −1 (U). Necessarily γ(t 0 ) ∈ V - otherwise γ(t 0 ) ∈ U, and bycontinuity there would be some small interval ]t 0 − ε, t 0 + ε[ such that evenγ(]t 0 − ε, t 0 + ε[) ⊂ U, a contradiction to the fact that t 0 is an upper boundfor the set γ −1 (U). But if γ(t 0 ) ∈ V , then, once again by continuity, γ(]t 0 −ε, t 0 + ε[) ⊂ V for some ε > 0, so t 0 − ε is an upper bound for γ −1 (U) as welland t 0 is not the least upper bound!Now assume our criterion is satisfied. Take a point a ∈ G and defineU := {z ∈ G; there is a path in G connecting a with z}.Furthermore let V := G \ U. We show that both U and V are open andconclude G = U. Since ”being connectable by a path in G” is a transitiverelation - in fact an equivalence relation - on G, it follows that G is pathconnected.The set U is open: Take z ∈ U and choose a disc D r (z) ⊂ G. Obviouslyall points in D r (z) can be connected by a path in D r (z) ⊂ G to its center z,hence also to a ∈ G. So D r (z) ⊂ U.The complement V is open as well: Given z ∈ V choose again D r (z) ⊂ G.If some point of D r (z) could be connected by a path in G with a, then z aswell - with other words, D r (z) ⊂ V .4 Logarithm functionsThe real exponential function R −→ R >0 is bijective and has an inversefunction ln : R >0 −→ R, the natural logarithm. The complex exponentialmap is not injective due to the ambiguity of the angle in the polar coordinate14

description of a complex number. Even worse, there is no unique maximaldomain of definition for the logarithm.We start defining in general ”logarithm functions” or ”branches of thelogarithm”:Definition 4.1. Let G ⊂ C be a domain. A continuous function g : G −→ Cis called a branch of the logarithm on G if exp(g(z)) = z, ∀z ∈ G.Though the terminology: ”the logarithm” seems to indicate it there is ingeneral no preferred choice of a branch of the logarithm.Lemma 4.2. If g, h : G −→ C are branches of the logarithm on the domainG, then there is an integer l ∈ Z such that h = g + 2πil.Proof. We have exp(g − h) ≡ 1, and thus g(z) − h(z) = 2πil(z) with aninteger l(z) ∈ Z.. Since g and h are continuous, l : G −→ Z ⊂ C is aswell. But Z ⊂ C is discrete (i.e. consists of isolated points only), so bythe intermediate value theorem the continuous function t ↦→ l(γ(t)) ∈ Z isconstant along any path γ : I −→ G. Since G is connected it follows thatl : G −→ Z is constant.A necessary condition for the existence of a branch of the logarithm onG is obviously that G ⊂ C ∗ , the exponential function having no zeros. Unfortunatelythat condition is not sufficient. But let us look first at a positiveexample:Example 4.3.1. LetG := C \ R ≤0be the plane with the left real half line, i.e. the non-positive reals,removed. The principal branch of the logarithmis the functionLog : G −→ CLog(z) := ln |z| + iArg(z),where the argument function Arg : G −→ R is defined as follows⎧⎨ − arccos( x ) , if z = x + iy, y < 0|z|Arg(z) := arctan( y ) , if z = x + iy, x > 0⎩arccos( x ) , if z = x + iy, y < 0|z|15

with the inverse trigonometric functions arccos :] − 1, 1[−→]0, π[ andarctan : R −→] − π 2 , π 2 [.2. There is no branch of the logarithm g : C ∗ −→ C on the puncturedplane C ∗ : Otherwise g| G would be a branch of the logarithm on G andthus g| G = Log + 2πil. We may even assume l = 0 - replace g withg − 2πil. Butg(−r) =lim Log(z) = ln(r) + πiz→−r,Im(z)>0as well asg(−r) =lim Log(z) = ln(r) − πi.z→−r,Im(z)

5 Complex DifferentiabilityThe notion of differentiability of a function f : G −→ C in a point z ∈ G isliterally the same as in real analysis of one variable. The difference is thatdifferentiability of a complex valued function is quite a strong condition and,in contrast to the real situation, has striking consequences.Definition 5.1. The function f : G −→ C is called differentiable (or C-differentiable) in z ∈ G, iff(z + h) − f(z)limh→0 hexists. In that case we denote the limit f ′ (z) and call it the (complex)derivative of f at z. The function f is called holomorphic at z, iff it isdifferentiable at every point of an open disc D r (z) ⊂ G around z. It is calledholomorphic or differentiable, iff it is differentiable at every point in G. WedenoteO(G) := {f : G −→ C; f holomorphic}the set of all holomorphic functions on G.1. A constant function f ≡ c is differentiable with deriva-Example 5.2.tive f ′ ≡ 0.2. The function f(z) = z is differentiable with derivative f ′ ≡ 1.3. The complex conjugation f(z) = z is nowhere differentiable. (Notethat functions of that type also exist in real analysis, but they are notthat easy to define!). Indeed for h = re iϕ we havez + h − zh= h h = e−2iϕ ,i.e. along different rays through z there are different limits! So there isno limit in the whole!4. The derivative preserves sums and satisfies the Leibniz rule: Assumethat f, g : G −→ C are differentiable at z ∈ G. Then so are the sumf + g and the product fg, and we have:(f + g) ′ (z) = f ′ (z) + g ′ (z), (fg) ′ (z) = f ′ (z)g(z) + f(z)g ′ (z).17

In particular we obtain by induction on n ∈ N >0 that the power functionf(z) := z n is differentiable with derivative f ′ (z) = nz n−1 . Complexpolynomials are differentiable on C.5. An R-linear map f(z) = cz + dz is differentiable at 0 ∈ C iff it isdifferentiable everywhere iff d = 0 iff f is even C-linear.6. The set O(G) is closed with respect to both sums and products offunctions.7. If f, g : G −→ C are differentiable at z ∈ G and g(z) ≠ 0, so isf: G \ N(g) −→ C (where N(g) ⊂ G denotes the set of zeros of g),gand( ) ′ f(z) = f ′ (z)g(z) − f(z)g ′ (z).gg(z) 28. The chain rule holds: Let f : G −→ G ′ ⊂ C and g : G ′ −→ C befunctions. If f is differentiable at z ∈ G and g at w = f(z), then thecomposition g ◦ f : G −→ C is differentiable at z ∈ G and(g ◦ f) ′ (z) = g ′ (w)f ′ (z) = g ′ (f(z))f ′ (z).Let us view at the above situation from the point of view of real analysis inthe two variables x, y, taking C as the real plane. The functions differentiablein the sense of real analysis we shall call here R-differentiable. We recall:Definition 5.3. A function f : G −→ C is R-differentiable at a point z 0 ∈ G,if there is an R-linear map L : C −→ C, such thatf(z 0 + h) − f(z 0 ) − L(h)limh→0|h|= 0,or equivalently, h being a complex number (and not only a vector in R n ),f(z 0 + h) − f(z 0 ) − L(h)limh→0h= 0.The linear map L is uniquely determined by the above condition: It iscalled the differential of f at z 0 and also denotedDf(z 0 ) := L.18

Writing h = s + it the differential L : C −→ C of an R-differentiable mapf : G −→ C at some point z 0 ∈ G takes the formL(s + it) = as + bt,where the coefficients a, b ∈ C are the partial derivativesOn the other hand we may writea = ∂f∂x (z 0), b = ∂f∂y (z 0).L(h) = ch + dhwith c = 1(a − ib), d = 1 (a + ib). We are thus led to the following definition2 2of ”complex” partial derivatives:Definition 5.4. For a function f : G −→ C, which is R-differentiable atz 0 ∈ G with differential L(h) = ch + dh we define∂f∂z (z 0) := c,∂f∂z (z 0) := d.Indeedand∂f∂z = 1 ( ) ∂f2 ∂x − i∂f ∂y∂f∂z = 1 ( ) ∂f2 ∂x + i∂f .∂yThe relation between real and complex differentiability is as follows:Proposition 5.5. A function f : G −→ C is C-differentiable at z 0 ∈ G, iffit is R-differentiable at z 0 and∂f∂z (z 0) = 0.19

Proof. If f is C-differentiable at z 0 , then L(h) := f ′ (z 0 )h is the R-linearmap L : C −→ C satisfying the above condition. On the other hand, given afunction, which is R-differentiable at z 0 ∈ G with differential L(h) = ch+dh,we havef(z 0 + h) − f(z 0 )= L(h)hh+ ε(h),where lim h→0 ε(h) = 0. So a function, which is R-differentiable at z 0 ∈ G, isthere C-differentiable iff its differential L : C −→ C at z 0 is C-differentiableat 0 ∈ C, or equivalently: L(h) = ch for some c ∈ C, see 5.2.5.In terms of ordinary partial derivatives we obtain:Corollary 5.6. For a function f : G −→ C and a point z 0 ∈ G the followingstatements are equivalent:1. f is C-differentiable at z 0 .2. f is R-differentiable at z 0 and there∂f∂x = −i∂f ∂yholds. With other words: The differential quotient of f at z 0 takenalong the line z 0 + R coincides with that taken along z 0 + iR.3. f = u + iv is R-differentiable at z 0 and there the Cauchy-Riemannequations∂u∂x = ∂v∂y , ∂u∂y = −∂v ∂x .hold.Remark 5.7. A function f : G −→ C is C-differentiable at z 0 ∈ G iffBut be careful: We do not have∂f∂z (z 0) = 0.∂f∂z (z f(z 0 + h) − f(z 0 )0) = lim,h→0 hsince the right hand side in general does not exist, but if it exists, then theequality holds and furthermore ∂f∂z (z 0) = 0.20

The calculus for the first order differential operators ∂ , ∂ is analogous∂z ∂zto that of partial derivatives: They satisfy∂z ∂z ∂z ∂z= 1, = 0, = 0,∂z ∂z ∂z ∂z = 1,are linear, satisfy the Leibniz rule and even the chain rule as for usual partialderivatives.Example 5.8.either is1. A real polynomial functionm∑ n∑f(z) = a µν z µ z ν ,µ=0 ν=0(a) a complex polynomial and thus holomorphic everywhere(b) or nowhere holomorphic.First of all we havea µν = 1 ∂ µ+ν fµ!ν! ∂z µ ∂z ν (0).Thus the coefficients a µν are uniquely determined by f.Now the z-derivativem∂f∂z = ∑µ=0 ν=1n∑νa µν z µ z ν−1 ,being a real polynomial as well, is either ≡ 0 - that means that all itscoefficients vanish resp. that f was a complex polynomial - or does notvanish identically on any open subset, with other words, f is nowhereholomorphic.2. The exponential function exp : C −→ C is holomorphic and exp ′ = exp.Indeed∂ exp∂x (z) = ex e(y), ∂ exp∂y (z) = ex ie(y).3. A branch of the logarithm is holomorphic - to see that it is sufficientto look at branches on C \ R ≥0 e iϑ resp. at the principal branch of thelogarithm Log. Now the chain rule applied to z = exp(g(z)) gives1 = exp ′ (g(z))g ′ (z) = exp(g(z))g ′ (z) = zg ′ (z)i.e. g ′ (z) = 1 z . 21

6 Path integralsThe values of a holomorphic function f ∈ O(G) on the interior D of a closeddisc D ⊂ G can be computed from its restriction f| ∂D . This is the essentialdifference between differentiable functions on open parts of the real line andholomorphic functions on domains of the complex plane. If something similarwould hold in the real situation (with closed intervals replacing closed discs),every differentiable function would be an affine linear function f(x) = ax+b!In order to compute f(z) for z ∈ D from f| ∂D , we need path integrals:This section is devoted to a careful discussion of that subject. Indeed we donot only treat it from the point of view of complex analysis, but explain thegeneral background from real analysis in two variables as well.First let us fix some notation:Definition 6.1. For a domain G ⊂ C we denoteC k (G)the set of all complex valued functions, admitting continuous partial derivativesup to order k. In particularC(G) := C 0 (G)is the set of all complex valued continuous functions on G.The objects to be integrated along ”paths” are not functions, but ”differentialforms”. We start with a preparatory remark:Remark 6.2. We denoteL(C) := {L : C −→ C; L R-linear}the complex vector space of all R-linear maps from C to itself (with the scalarmultiplication (λL)(h) := λL(h)). Recall that it admits two natural basesL(C) = C · Re ⊕ C · Im,where Re, Im : C −→ C satisfy Re(x + iy) = x, Im(x + iy) = y, andwhere id C is complex conjugation.L(C) = C · id C ⊕ C · id C ,22

Definition 6.3. A C k -differential form on a domain G is a functionω : G −→ L(C), z ↦→ ω z ,such that the function G −→ C, z ↦→ ω z (h), is a C k -function for all h ∈ C. Wedenote D k (G) the set of all C k -differential forms on G, with the conventionD(G) := D 0 (G).Remark 6.4. 1. Let us try a physical interpretation of differential forms:Differential forms can be regarded as rules associating to any smalldisplacement vector h ∈ C at a point z ∈ G the variation ω z (h) ∈ C ofsome (complex) ”magnitude”; and h being small we may assume thatω z (h) is R-linear in h ∈ C.2. Differential forms can be added, and multiplied with functions f ∈C k (G) as follows(fω) z := f(z)ω z .Definition 6.5. The total differential of a function F ∈ C 1 (G) is defined asthe continuous differential formdF : G −→ L(C)associating to a point z ∈ G the differential of F at z, i.e.dF z := DF (z).On the other hand, given a differential form ω ∈ D(G), any function F ∈C 1 (G) with dF = ω is called a primitive function for ω.In the following example we see that differential forms are determined bytwo complex valued coefficient functions:Example 6.6. 1. The total differential of an R-linear map F = L : C −→C is the map L itself:dF ≡ L.In particularwhiledx ≡ Re, dy ≡ Im,dz ≡ id C , dz ≡ id C .23

As a consequence every differential form ω ∈ D k (G) has a unique representationω = fdx + gdywith functions f, g ∈ C k (G) as well asω = h 1 dz + h 2 dzwithh 1 = 1 2 (f − ig), h 2 = 1 (f + ig).2To see that use the relationsdz = dx + idy, dz = dx − idyanddx = 1 2 (dz + dz), dy = 1 (dz − dz).2i2. The total differential of a function F ∈ C 1 (G) then takes the formdF = ∂F ∂F ∂F ∂Fdx + dy = dz +∂x ∂y ∂z ∂z dz.Remark 6.7. If dF = d ˜F on a domain G, then ˜F = F + c with somec ∈ C, i.e. two primitive functions differ only by a constant. To prove thatwe consider d( ˜F − F ) = 0. So it is sufficient to show that dF = 0 impliesF ≡ c ∈ C. It is clear that F ≡ F (z 0 ) on any disc D r (z 0 ) ⊂ G, since F isconstant on the line segments [z 0 , x+iy 0 ] and [x+iy 0 , z]. Now consider somevalue c ∈ F (G), TakeU := {z ∈ G; f(z) = c}, V := {z ∈ G; f(z) ≠ c}.Then G is the disjoint union G = U ∪ V , the set V is open, since F iscontinuous, but U is as well by the above argument. Since G is connectedand U ≠ ∅, we conclude G = U.Definition 6.8. A differential form ω ∈ D(G) is called1. integrable or exact, if it admits a primitive function F ∈ C 1 (G), i.e.ω = dF ,24

2. locally integrable, if for every point z 0 ∈ G there is an open disc D =D r (z 0 ), such that ω| D admits a primitive function F ∈ C 1 (D).Example 6.9. Let G ⊂ C ∗ be a domain. Then the differential formω := dzz ∈ D(G)is locally integrable, but not necessarily integrable: Indeed, it is integrableiff there is a branch of the logarithm log : G −→ C. We have already seenthat ω = d log. On the other hand, if ω = dF , thend(ze −F ) = zde −F + e −F dz = −ze −F dF + e −F dz = 0and thus ze −F (z) ≡ c with some c ∈ C ∗ . If then e d = c, we see that log(z) :=F (z) − d is a branch of the logarithm on G. The above integrability criterionapplies to discs D ⊂ C ∗ : If D = D r (z 0 ), then D ⊂ C ∗ \ R ≥0 (−z 0 ), the latterdomain admitting a branch of the logarithm. Thus ω is locally integrable onany domain G ⊂ C ∗ .Differential forms can be integrated along paths:Definition 6.10. Let G ⊂ C be a domain.1. A path in G is a continuous map γ : I = [a, b] −→ G from a closedinterval I = [a, b] to G. The point γ(a) ∈ G is called the start point,γ(b) ∈ G the end point of γ. We call γ a closed path or a loop ifγ(a) = γ(b).2. A smooth path in G ⊂ C is a C 1 -map γ : I = [a, b] −→ G, i.e. bothcomponent functions α, β of γ = α+iβ have continuous first derivatives.We denote ˙γ : I −→ C the derivative of γ, i.e. ˙γ = ˙α + i ˙β, and call˙γ(t) the tangent vector on γ at t ∈ I.3. The length L(γ) of a smooth path is defined asits trace is the setL(γ) :=∫ ba|γ| := γ(I).| ˙γ(t)|dt,25

4. A piecewise smooth path γ in G is a sequence of smooth paths γ 1 , ..., γ r ,such that the end point of γ i , i = 1, ..., r − 1, coincides with the startpoint of γ i+1 . Its length isL(γ) :=r∑L(γ i ),i=1its tracer⋃|γ| := |γ i |.i=1Some standard paths: By [z 1 , z 2 ] we mean the smooth path[0, 1] −→ C, t ↦→ z 1 + t(z 2 − z 1 ).Furthermore, given a coordinate rectangle R = [a, b] + i[c, d] we denote ∂Rthe piecewise smooth loop consisting of the smooth pieces [a + ic, b + ic],[b + ic, b + id], [b + id, a + id], [a + id, a + ic].We denote ∂D r (z 0 ) the smooth loop[0, 2π] −→ C, t ↦→ z 0 + re it ,i.e. the (counterclockwise) circle with center z 0 and radius r > 0.As a general rule if a path surrounds a domain, one calls it positivelyoriented if one has that domain always on ones left hand side when followingthe path.Given a path γ : I −→ G its inverse path isγ −1 : −I −→ G, t ↦→ γ(−t),the inverse of a piecewise smooth path γ = (γ 1 , ..., γ r ) is γ −1 := (γr−1 , ..., γ1 −1 ).Definition 6.11. 1. Let γ : [a, b] −→ G be a smooth path. For a differentialform ω we define∫ ∫ bω := ω γ(t) ( ˙γ(t))dt.γa26

2. If γ is a piecewise smooth path, say γ = (γ 1 , ..., γ r ), we set∫r∑∫ω := ω.γγ kRemark 6.12.k=11. If γ = α + iβ and ω = fdx + gdy, we have∫ ∫ bω = (f(γ(t)) ˙α(t) + g(γ(t)) ˙β(t))dt.γa2. For ω = h 1 dz + h 2 dz we have∫ ∫ b ()ω = h 1 (γ(t)) ˙γ(t) + h 2 (γ(t)) ˙γ(t) dt.γa3. If τ : [c, d] −→ [a, b], s ↦→ t = τ(s), is a C 1 -map with τ(c) = a, τ(d) = b,and ˜γ := γ ◦ τ, then ∫ ∫ω = ωas a consequence of the chain rule. Furthermore˜γγL(˜γ) = L(γ)if dτ ≥ 0. So, the value of the integral of a differential form along a pathdsdoes not change under a reparameterization! But if you are hesitatingwhile going from γ(a) to γ(b) and even go backwards a little bit beforecontinuing to γ(b) then the length of your way is bigger: L(˜γ) > L(γ).4. We have ∫ ∫ω = −γ −1for any piecewise smooth path γ.5. The fundamental theorem of calculus together with the chain rulefor two variables gives∫γωddt (F ◦ γ)(t) = dF γ(t) ( ˙γ(t))γdF = F (γ(b)) − F (γ(a)),27

so the path integral of a differential form admitting a primitive functiondepends only on the start and end point of the path. In particular∫dF = 0for any (piecewise smooth) loop λ.λExample 6.13. 1. Let D = D r (z 0 ) and D := {z ∈ C; |z − z 0 | ≤ r}. Wewant to prove that∫ {dz 2πi, , if a ∈ Dz − a = 0, , if a ∉ D .∂DIf a ∈ D, we decomposeC \ {a} = G 1 ∪ G 2into two parts, where ω = dzz−a admits a primitive function F i : G i −→C. TakeG 1 := C \ (a + R ≤0 ), G 2 := C \ (a + R ≥0 ).On G 1 the function F 1 (z) := Log(z − a) is a primitive function ofdz/(z − a), on G 2 we can take F 2 (z) := Log(a − z). Now let γ 1 : I −→G 1 be the arc of ∂D on the right hand side of the line a + iR, andγ 2 : J −→ G 2 the complementary arc (assuming ∂D to start and endat the lower point of ∂D ∩ (a + iR)). Denote a + b the start point ofγ 1 , and a + c the end point of γ 1 . Then∫ ∫ ∫dzz − a = dzz − a + dzz − a∂Dγ 1= (F 1 (a + c) − F 1 (a + b)) + (F 2 (a + b) − F 2 (a + c))= (Log(c) − Log(b)) + (Log(−b) − Log(−c))= (Log(−b) − Log(b)) + (Log(c) − Log(−c)) = πi + πi = 2πi.Finally, if a ∉ D the differential form dz has a primitive function onz−aC \ (a + R ≤0 (z 0 − a)) ⊃ D, hence the integral vanishes.28γ 2

2. As a consequence of the above discussion the differential formdzz − a∈ D(C \ {a})is locally integrable, but does not admit a primitive function: Thereare loops with non zero integral.Proposition 6.14. For a differential form ω ∈ D(G) the following statementsare equivalent1. ω is locally integrable.2. For every coordinate rectangle R = [a, b]+i[c, d] ⊂ G we have ∫ ∂R ω = 0.3. ω| D is integrable on any disc D = D r (z 0 ) ⊂ G.Proof. ”1) =⇒ 2)”: Let R ⊂ G be a coordinate rectangle. For every n ∈ Nwe may subdivide R into n 2 rectangles R ij , 1 ≤ i, j ≤ n, similar to R, but of1/n the size of R. Then∫ ∫∂Rω = ∑ i,j∂R ijωfor all ω ∈ D(G), since any edge of an R ij inside R shows up twice, but withopposite orientation. For n sufficiently large any R ij is contained in a discD ij ⊂ G, where ω admits a primitive function, and thus ∫ ∂R ijω = 0.”2) =⇒ 3)”: We construct a primitive function on D = D r (z 0 ) ⊂ G asfollows: We have∫∫∫∫ω +ω = ω +ω =: F (z).[z 0 ,x+iy 0 ][x+iy 0 ,x+iy][z 0 ,x 0 +iy]With other words, if ω = fdx + gdy, thenF (x+iy) =∫ xx 0f(s+iy 0 )ds+∫ yy 0g(x+it)dt =∫ y[x 0 +iy,x+iy]y 0g(x 0 +it)dt+∫ xx 0f(s+iy)ds.To check ∂F = g consider the first expression for F : The first term does not∂ydepend on y and for the second apply the fundamental theorem of calculus.For ∂F = f the analogous argument applies to the second expression.∂x”2) =⇒ 3)”: Obvious.29

The rectangle criterion for local integrability is not very convenient tobe checked, but it is of interest since it may be used to show the followingextension property:Proposition 6.15. If a differential form ω ∈ D(G) is locally integrable onG \ {z 0 }, then it is locally integrable on G as well.Proof. We show that ∫ ω = 0 for every coordinate rectangle R ⊂ G. For∂RR ∌ z 0 that is clear. With the notation of the proof of Prop. 6.14 we havez 0 ∈ R ij for at most 4 such rectangles, and the ω-integrals over the boundariesof these rectangles tend to zero for n ↦→ ∞: We have∫∣ ∣ ≤ M · L(∂R ij) = M n L(∂R)∂R ijωwith M = max{|f(z)|, |g(z)|; z ∈ R}, if ω = fdx+gdy. Hence ∫ ∂R ω = 0.The following result is interesting from a theoretical point of view; thereader mainly interested in a straight forward introduction to complex analysismay skip it. Indeed we shall prove it once more soon under slightlystronger assumptions.Theorem 6.16. For f ∈ O(G) the differential form ω = f(z)dz is locallyintegrable.Proof. 1) Let R ⊂ D ε (z 0 ) ⊂ G be a coordinate rectangle contained in a smalldisc D ε (z 0 ). We are hunting for an estimate of∫∣ f(z)dz∣ .Write∂Rf(z) = f(z 0 ) + f ′ (z 0 )(z − z 0 ) + h(z)(z − z 0 )with a continuous function h : G −→ C. For z ≠ z 0 we may simply solve forh(z), and define h(z 0 ) := 0. LetWe haveM(ε) := max{|h(z)|; |z − z 0 | ≤ ε}.(f(z 0 ) + f ′ (z 0 )(z − z 0 ))dz = d(f(z 0 )z + f ′ (z 0 ) 1 )2 (z − z 0) 230

and thuswhence∣ ∣∣∣∫∂R∫∂R∫f(z)dz =∂Rh(z)(z − z 0 )dz,f(z)dz∣ ≤ L(∂R)M(ε)ε ≤ 8εM(ε)ε = 8ε2 M(ε).2) Consider a coordinate rectangle R ⊂ G. Assume A := | ∫ f(z)dz| ̸= 0.∂RWe shall find a decreasing sequence of rectangles R n shrinking to a pointz 0 ∈ R: Each R n is similar to R and has 2 −n the size of R, and we have∫∣ f(z)dz∣ ≥ 4−n A.R nTake R 0 := R. If R n is found with the given estimate, we consider thesubdivision of R into four coordinate rectangles having the barycenter of R nas one of its vertices. The integral of ω over ∂R n is the sum over the ω-integrals along the boundaries of the subdividing rectangles, hence at leastthe absolute value of one of these integrals has to be ≥ (1/4) · 4 −n A =4 −(n+1) A. Take as R n+1 the corresponding rectangle. We have then∞⋂R n = {z 0 }n=0for some point z 0 ∈ C. Then R n ⊂ D ε (z 0 ) for ε = 2 −n D, where D > 0denotes the length of the diagonal of R. Thus, comparing the upper boundof the first part with the lower bound just found, we obtainresp.8 · 2 −2n D 2 · M(2 −n D) ≥ 4 −n A,M(2 −n D) ≥ A8D 2 ,but the left hand side converges to 0, a contradiction.Local integrability of differential forms ω ∈ D 1 (G) can also be expressedin terms of derivatives:31

Definition 6.17. The differential form ω = fdx + gdy ∈ D 1 (G) is said tobe closed or to satisfy the (local) integrability condition if∂f∂y = ∂g∂x .Example 6.18. We leave it to the reader to check that ω = h 1 dz + h 2 dz isclosed iff∂h 1∂z = ∂h 2∂zholds. So the differential form fdz ∈ D 1 (G) is closed if and only if f isholomorphic.Proposition 6.19. A differential form ω ∈ D 1 (G) is locally integrable if andonly if it is closed.Proof. Let F be a primitive function on the disc D ⊂ G. Since F ∈ C 2 (D),we have( )∂ ∂F= ∂ ( ) ∂F.∂y ∂x ∂x ∂yOn the other hand, if D = D r (z 0 ) and ω = fdx + gdy we defineThenF (x + iy) :=∫ xx 0f(s + iy 0 )ds +∂F(z) = g(x + iy)∂y∫ yy 0g(x + it)dt.according to the fundamental theorem of calculus, the first term not dependingon y. On the other hand∂F∂x (z) = f(x + iy 0) +∫ yy 0∂g(x + it)dt∂x∫ y∂f= f(x+iy 0 )+y 0∂y (x+it)dt = f(x+iy 0)+(f(x+iy)−f(x+iy 0 )) = f(x+iy).Corollary 6.20. For a function f ∈ C 1 (G) the following statements areequivalent:32

1. The differential form f(z)dz is closed.2. The differential form f(z)dz is locally integrable.3. f ∈ O(G).We remark that the equivalence of the second and the third statementholds for any f ∈ C(G). The implication ”3) =⇒ 2)” then is nothing butTh. 6.16, while ”2) =⇒ 3)” is known as Morera’s theorem: It follows fromthe fact that the derivative of a holomorphic function again is holomorphic,see Th.7.4.On an open disc every closed differential form has a primitive function. Inthe remaining part of this section we are looking for more domains enjoyingthat property. We define:Definition 6.21. A domain G ⊂ C is called simply connected if every locallyintegrable differential form ω ∈ D(G) admits a primitive function F ∈ C 1 (G).Example 6.22.1. Open discs are simply connected.2. The union G = ⋃ ∞n=1 G n of an increasing sequence G 1 ⊂ G 2 ⊂ ... ofsimply connected domains G n , n ∈ N, is simply connected.3. If G 1 , G 2 ⊂ C are simply connected and the intersection G 1 ∩ G 2 isconnected, then G := G 1 ∪ G 2 is simply connected as well: A locallyintegrable form ω ∈ D(G) has on G i a primitive function F i ∈ C 1 (G i )for i = 1, 2. So d(F 1 − F 2 ) = 0 on the domain G 1 ∩ G 2 , hence F 1 − F 2 ≡c ∈ C, and we may define a primitive function F ∈ C 1 (G) by F | G1 := F 1and F | G2 := F 2 + c.4. A starshaped domain G is simply connected: A domain G ⊂ C is calledstarshaped, if there is a point z 0 ∈ G, such that for all z ∈ G the linesegment from z 0 to z is contained in G as well: For a locally integrableform ω the integral∫F (z) := ωdefines a primitive function.We mention without proof the following characterization of simply connecteddomains.33[z 0 ,z]

Theorem 6.23. For a domain G ⊂ C the following statements are equivalent:1. G is simply connected.2. The domain G ”has no holes”: The complement C\G can not be writtenC \ G = A ˙∪ Kas the disjoint union of a compact set K ≠ ∅ and a closed set A ⊂ C.3. Every continuous loop γ : I −→ G can within G be contracted to apoint, i.e. there is a continuous mapH : I × I −→ Gwith H(t, 0) = γ(t), H(t, 1) ≡ z 0 ∈ G and H(0, s) = H(1, s) for alls ∈ I.Remark 6.24. We remark, that the third characterization of simply connecteddomains is usually taken as definition. In that form it also applies toarbitrary topological spaces. - The above definition Def. 6.21 is less intuitive,but sometimes it is easier to handle, e.g. the proof of Rem. 6.22.3 becomesquite easy then.We conclude this section by looking at some nonsimply connected domains,namely annuliA ϱ,r (a) := {z ∈ C; ϱ < |z − a| < r}with inner radiur ϱ and outer radius r, where 0 ≤ ϱ < r ≤ ∞.Proposition 6.25. Let ϱ < s < r. A locally integrable differential formω ∈ D(A ϱ,r (a)) is integrable if and only if∫ω = 0.∂D s(a)Proof. ”=⇒”: Obvious.”⇐=”: We may assume a = 0. The dissected annulusA ∗ := A ϱ,r (0) \ (−r, −ϱ)34

is simply connected: To see that one can note that the restricted exponentialexp : (ln ϱ, ln r) + i(−π, π) −→ A ∗is a homeomorphism (a bijective map, continuous in both directions) froma rectangle to our dissected annulus, and simple connectedness is preservedunder such maps - use the third characterization of simply connected domainsin Th.6.23. Alternatively we may decompose A ∗ into starshaped segements:For n ∈ N letA n := (ϱ, r) exp(πi(− 1 n , 1 )n ) .If cos( π)r > ϱ, then A n n is starshaped with respect to z 0 := ϱ cos −1 ( π) and nn−1⋃( ) πiνA ∗ = exp A nnν=−n+1is the union of such segments, such that the intersection of one of them withthe union of the previous ones is connected.So on both A ∗ and −A ∗ there is a primitive function F resp. ˜F of ω.SinceA ∗ ∩ (−A ∗ ) = A + ∪ A −with the connected sets A + = {z ∈ A ϱ,r (0); Im(z) > 0}, A − := −A + , we have(F − ˜F )| A± ≡ c ± . Thus integrating separately over the right and the left halfarc of ∂D s (0) we obtain:∫( ( i0 = ω = F − F −s)i ) (+s˜F − i ) (−s˜F i= c + − c − .s)∂D s(0)So F may be extended to a function O(A ϱ,r (0)) by defining it on −A ∗ as˜F + c with c := c + = c − .7 The Cauchy formulaAs an immediate consequence of the previous section let us note:Theorem 7.1 (Cauchy’s integral theorem). Let G ⊂ C be a simplyconnected domain. Then for f ∈ O(G) we have∫f(z)dz = 0for every piecewise smooth loop λ in G.λ35

Proof. The differential form ω = fdz is locally integrable, since f ∈ O(G),and G being simply connected it has a primitive function F (which is itselfholomorphic) and thus the integral of ω = dF over any piecewise smoothloop vanishes.If we cannot guarantee that an integral of the above form vanishes, itoften will be useful to have an estimate instead:Proposition 7.2. Let f ∈ C(G) and γ be a piecewise smooth path in G.Then we have ∣∫∣∣∣ f(z)dz∣ ≤ ||f|| γ · L(γ).Hereγ||f|| γ := max{|f(z)|; z ∈ |γ|}denotes the maximum of |f| along the trace of γ.Proof. We may assume that γ is smooth, say γ : [a, b] −→ G. For a complexvalued, continuous function g : [a, b] −→ C we have∫ b∫ b∣ g(t)dt∣ ≤ |g(t)|dt,since∣ ∣∣∣∣ s∑g(t i )(t i − t i−1 )∣ ≤i=1aas∑|g(t i )| · (t i − t i−1 ),where t 0 = a < t 1 < ... < t s = b, and, with shrinking interval lengths t i −t i−1 ,the sum on the right hand side converges to the corresponding integral, andthe same is true for the left hand side.Now ∣∫∣∫ ∣∣∣ ∣∣∣ b∫f(z)dz∣ = bf(z) ˙γ(t)dt∣ ≤ |f(z) ˙γ(t)|dtγai=1∫ b≤ ||f|| γ | ˙γ(t)|dt = ||f|| γ · L(γ).aaThe next theorem is behind a lot of nice results about holomorphic functions:36

Theorem 7.3 (Cauchy’s integral formula). Let G be a domain and f ∈O(G), furthermore D an open disc with D ⊂ G. Then we have for any pointa ∈ D (a need not be the center!):f(a) = 12πi∫∂Df(z)dzz − a .So the values of a holomorphic function on D are already determined by itsvalues on the boundary circle ∂D.Proof. Since f is C-differentiable at a ∈ G, we haveω :=f(z) − f(a)dz ∈ D(G).z − a∈ O(G \ {a}), the differential form ω is locally integrablein G \ {a}, hence locally integrable in G according to Prop. 6.15. SinceD = D r (z 0 ) satisfies D ⊂ G, we have even D ϱ (z 0 ) ⊂ G for some ϱ > r andthus ω = dF on D ϱ (z 0 ) resp. ∫ ω = 0. Consequently∂DSince f(z)−f(a)z−aaccording to Example 1.∫1 f(z)dz2πi ∂D z − a = 1 ∫f(a)dz2πi ∂D z − a = f(a)With the traditional choice of letters the Cauchy formula readsf(z) = 1 ∫f(ζ)dζ2πi ζ − z ,where z ∈ D is arbitrary. The letter z suggesting a variable we obtain:Theorem 7.4 (Generalized Cauchy’s integral formula). A holomorphicfunction has complex derivatives of every order, indeed for z ∈ D ⊂ G wehavef (n) (z) = n! ∫f(ζ)dζ2πi (ζ − z) . (n+1)∂DProof. We use induction on n ∈ N. For n = 0 nothing has to be shown. Nowassume f has partial derivatives up to order n and the generalized Cauchyformula holds for n. We shall use37∂D

Remark 7.5. Let γ be a smooth path andK : D × |γ| −→ C, (z, ζ) ↦→ K(z, ζ),be a continuous function, R-differentiable with respect to z and its partialderivatives∂K∂z , ∂K : D × |γ| −→ C∂zbeing continuous as well. Then the function∫D −→ C, z ↦→ K(z, ζ)dζis R-differentiable and(∫∂∂zas well aswithTaking γ = ∂D andγγ) ∫K(z, ζ)dζ =γ(∫ ) ∫∂K(z, ζ)dζ =∂z γγ∂K∂zK(z, ζ) :=f(ζ)(ζ − z) n+1f(ζ)(z, ζ) = n(ζ − z) (n+2)we obtain with Rem. 7.5 the case n + 1.We can do even better:∂K(z, ζ)dζ.∂z∂K(z, ζ)dζ.∂zTheorem 7.6. Let f ∈ O(G) and D r (z 0 ) ⊂ G be an open disc. Then theTaylor series of f represents f on D r (z 0 ), i.e.:f(z) =∞∑a n (z − z 0 ) n , where a n = f (n) (z 0 )/n!,n=0for all z ∈ D r (z 0 ), the right hand side converging uniformly on every closeddisc D ϱ (z 0 ), ϱ < r.38

A function (of one or several real or complex) variables is called (realor complex) analytic if near any point of its domain of definition it can bewritten as a power series. So Th. 7.6 tells us that holomorphic functions arecomplex analytic functions.Proof. We may assume that D r (z 0 ) ⊂ G. We write1ζ − z = 1(ζ − z 0 ) − (z − z 0 ) == 1ζ − z 011 − z−z 0ζ−z 0= 1ζ − z 0∞∑n=0Note that, for |z − z 0 | ≤ ϱ, ζ ∈ ∂D r (z 0 ), we have∣ z − z 0 ∣∣∣∣ ≤ ϱ ζ − z 0 r < 1,n=0( z − z0ζ − z 0) n.so the geometric series converges uniformly on D ϱ × ∂D r (z 0 ). So we mayinterchange summation and integration and obtain∞∑∫1f(ζ)dζf(z) =2πi (ζ − z 0 ) · (z − z 0) n .n+1∂D r(z 0 )Corollary 7.7 (Weak Identity Theorem for holomorphic functions).Let f, g ∈ O(G) be holomorphic functions on the domain G. If f| D = g| D forsome open disc D ⊂ G, then already f = g. Or, equivalently, the restrictionmapO(G) −→ O(D), f ↦→ f| Dis injective for every open disc D ⊂ G.Proof. We may assume g ≡ 0. So let f| D ≡ 0. Then the nonempty setis obviously open, butU := {z ∈ G; ∃ r > 0 : f| Dr(z) ≡ 0}V := {z ∈ G; ∃ n ∈ N : f (n) (z) ≠ 0}is as well, and G = U ∪V , since a point, where all derivatives vanish, belongsto U. Since G is connected and U non-empty, we conclude U = G.39

Remark 7.8. Assume (a n ) n∈N is a sequence of complex numbers such thatthe sequence (|a n |ϱ n ) n∈N is bounded by M > 0. Then the power series∞∑a n (z − z 0 ) nn=0converges uniformly on every closed disc D s (z 0 ) with s < ϱ, since there wehave( n s|a n (z − z 0 ) n | ≤ R n := Mϱ)with ∑ n R n < ∞ (geometric series). In particular it defines a continuousfunction f : D ϱ (z 0 ) −→ C. As a consequence we see that a complex analyticfunction is holomorphic: The functionf(z) − f(z 0 )z − z 0=∞∑a n (z − z 0 ) n−1n=1is continuous at z 0 . But even better: The above power series defines afunction f ∈ O(D ϱ (z 0 )), as we shall see in the next section.Corollary 7.9. Assume D ϱ (z 0 ) ⊂ G. For the coefficients a n of the Taylorseries of f ∈ O(G) at z 0 ∈ G we havein particulara n = 1 ∫f(ζ)dζ2πi ∂D ϱ(z 0 ) (ζ − z 0 ) , n+1|a n | ≤ M(f; ϱ, z 0)ϱ nwhere M(f; ϱ, z 0 ) := max{|f(ζ)|; |ζ − z 0 | = ϱ}.Proof. Th. 7.4 with z = z 0 and Prop. 7.2 with L(∂D ϱ (z 0 )) = 2πϱ.40

8 Basic properties of holomorphic functionsIn this section we present the most important consequences of Cauchy’s integralformula resp. the fact that holomorphic functions are analytic.Definition 8.1. By an entire function one means a holomorphic functionf ∈ O(C).Theorem 8.2 (Liouville). A bounded entire function f ∈ O(C) is constant.Proof. We havef(z) =∞∑a n z nn=0for all z ∈ C, the coefficients a n satisfying the estimate|a n | ≤M(f; r, 0)r n ≤ M r n ,where M > 0 is an upper bound for f, i.e. |f(z)| ≤ M for all z ∈ C. Sincethe right hand side holds for any r > 0, we conclude a n = 0 for n > 0.Theorem 8.3 (Fundamental Theorem of Algebra). Every complex polynomialn∑f(z) = a ν z νof degree n can be factorized as a product of n linear factors:f(z) = a nν=0n∏ν=1(z − b ν ).Note that the zeros b ν need not be pairwise distinct!Proof. We do induction on n, the case n = 1 being trivial. Clearly we mayassume a n = 1. If f has no zero, the function g(z) := 1/f(z) is a boundedentire function: Since g as a continuous function is bounded on every discD r (0), it suffices to show that |f(z)| ≥ R > 0 in case |z| > r for a suitabler > 0. Indeed,|f(z)| =∣ zn (1 +n∑ν=1∣ ∣∣∣∣ z ) = |z n n∑|ν ∣ 1 +a n−ν41ν=1a n−νz ν∣ ∣∣∣∣

≥ |z n n∑|(1 −∣= |z| n (1 −n∑ν=1ν=1a n−νz ν∣ ∣∣∣∣) ≥ |z| n (1 −|a n−ν ||z| ) ≥ ν |z|n (1 −n∑ν=1n∑ν=1|a n−ν |)|z ν ||a n−ν |) ≥ |z|n|z| 2 ,if |z| ≥ r := max(1, 2 ∑ nν=1 |a n−ν|).Now Th. 8.2 says that the bounded entire function g and thus as well fis constant, a contradiction.So, there is a zero z 0 ∈ C. Then let us writef(z) = f((z−z 0 )+z 0 ) =n∑n∑c ν (z−z 0 ) ν = (z−z 0 )· c ν (z−z 0 ) ν−1 = (z−z 0 )h(z)ν=1ν=1and apply the induction hypothesis to the polynomial h(z).Corollary 8.4. Every real polynomialf(z) =n∑a ν z νν=0of degree n can be factorized as a product of monic linear or quadratic polynomials:r∏f(z) = a n f ν (z),ν=1with the quadratic polynomials having no real zeros.Proof. We do induction on n = deg f. For n = 1 everything is clear. For n >1 take z 0 ∈ C with f(z 0 ) = 0. If z 0 ∈ R, wee may write f(z) = (z − z 0 )g(z)with a polynomial g with real coefficients; otherwise z 0 is an other zero off and f(z) = q(z)g(z) with q(z) = (z − z 0 )(z − z 0 ) = z 2 − 2Re(z 0 )z + |z 0 | 2and a polynomial g(z) with real coefficients. Finally apply the inductionhypothesis to g(z).For algebraically minded readers we remark:Theorem 8.5. 1. Let K ⊃ C be a field the arithmetic operations of whichextend the addition and multiplication of complex numbers. If then K,as a complex vector space, has finite dimension, then K = C.42

2. Let K R be a field the arithmetic operations of which extend theaddition and multiplication of real numbers. If then K, as a real vectorspace, has finite dimension, then K = R + Ri with an element i ∈ K,such that i 2 = −1.Proof. 1.) Take an element c ∈ K. Since dim K < ∞, the powers c ν , ν ∈ N,are not linearly independent. So there is a nontrivial relationn∑a ν c ν = 0ν=0with complex coefficients a ν ∈ C. Of course we may assume a n = 1. But,according to the fundamental theorem of algebra:n∑n∏f(z) := a ν z ν = (z − b ν ).ν=0The above identity holds for all z ∈ C and thus may be understood as anidentity for the coefficients of the polynomials on the right and left hand side(when writing the RHS as a linear combination of powers of the variable z),hence holds even for all z ∈ K. Taking z = c we obtainn∏0 = f(c) = (c − b ν ).Since K as a field does not contain zero divisors we obtain that c = b ν ∈ Cfor some ν.2.) Take c ∈ K \ R. As in the first part we see that f(c) = 0 for some realpolynomial f(z). Now by Cor.8.4 we see, that we may assume that f is aquadratic polynomial without real zeros. We leave it to the reader to checkthat there is an element i ∈ R + Rc with i 2 = −1. Using the first part weconclude that K = R + Ri.ν=1ν=1Th. 8.3 may also be formulated by saying thatf(C) = Cholds for polynomial functions of degree > 0. For transcendent functions(=entire nonpolynomial functions) that is not true any longer, sinceBut we can proveexp(C) = C ∗ .43

Theorem 8.6 (Weak Casorati-Weierstraß). For any entire function f ∈O(C) \ C we havef(C) = C,i.e. given any point w 0 ∈ C there are points of the form f(z), z ∈ C, arbitrarilyclose to w 0 . With other wordsfor every open disc D ⊂ C.f(C) ∩ D ≠ ∅Proof. Assume D r (z 0 ) ∩ f(C) = ∅. Theng(z) :=1f(z) − w 0is a bounded entire function: |g(z)| ≤ 1/r. Hence g as well as f are constantfunctions.But much more is true, the exponential function is more or less typical.We state without proof:Theorem 8.7 (Weak theorem of Picard). For a non-constant entire functionf ∈ O(C) we havef(C) = Corwith some a ∈ C.f(C) = C \ {a}Let us now come to the local behaviour of holomorphic functions. It iscompletely determined by the following invariant:Definition 8.8. The multiplicity of a holomorphic function f ∈ O(G) \ C atz 0 ∈ G is the number n ∈ N >0 , such thatf(z) = a 0 + (z − z 0 ) n h(z), h(z 0 ) ≠ 0with a holomorphic function h ∈ O(G), or equivalently∞∑f(z) = a 0 + a ν (z − z 0 ) ν , a n ≠ 0,near z 0 ∈ G.ν=n44

Note that f ∈ O(G) \ C implies, according to Th.7.7, that f| D ∉ C for alldiscs D ⊂ C. Hence the multiplicity is well defined for every point z 0 ∈ G.Lemma 8.9. The zeros of a holomorphic function f ∈ O(G), f ≢ 0, areisolated, i.e. given any zero z 0 ∈ G of f, there is an open disc D = D r (z 0 ) ⊂G, such that f has no zeros on the punctured disc D ∗ := D \ {z 0 }.Proof. We use the notation of Def.8.8. We have a 0 = 0, and the functionh ∈ O(G) is continuous, hence h(z) ≠ 0 on some disc D r (z 0 ) and thusf(z) = (z − z 0 ) n h(z) ≠ 0 for z ∈ D r (z 0 ) ∗ .Theorem 8.10 (Identity theorem). Let A ⊂ G be a subset with a pointz 0 of accumulation within G and f, g ∈ O(G). If f| A = g| A , then f = g.Proof. The zeros of the function f − g ∈ O(G) are not isolated, since bycontinuity, f(z 0 ) − g(z 0 ) = 0. As a consequence of the previous lemma wehave f − g ≡ 0.In order to understand the local mapping properties of a holomorphicfunction we rewrite the presentation given in Def.8.8:Theorem 8.11. Let f : G −→ C be a holomorphic function of multiplicityn at z 0 ∈ G. Then there is an open disc D = D r (z 0 ) together with aholomorphic function g ∈ O(D) with g(z 0 ) = 0, g ′ (z 0 ) ≠ 0, such thatf(z) = a 0 + g(z) n .Proof. Write f(z) − a 0 = (z − z 0 ) n h(z). Choose g(z) := (z − z 0 ) n√ h(z).Here n√ z := exp(log(z)/n) with some branch of the logarithm log : V −→ Cdefined on a neighbourhood V of h(z 0 ) ≠ 0. Finally choose D r (z 0 ) ⊂ G withh(D r (z 0 )) ⊂ V .Definition 8.12. A bijective map f : G −→ G ′ between domains G, G ′ iscalled biholomorphic if both f and f −1 are holomorphic.Here is a geometric reformulation of Th.8.11:Theorem 8.13. Let f : G −→ C be a holomorphic function, z 0 ∈ G a point,where f is of multiplicity n. Then there is an open neighbourhood U ⊂ G ofz 0 ∈ G, such that there is a factorizationf| U = p n ◦ g : U g−→ D ϱ (0) pn−→ D r (w 0 ),where w 0 := f(z 0 ), g is biholomorphic, r = ϱ n and p n (ζ) = w 0 + ζ n .45

In particular, a bijective holomorphic map is biholomorphic.Proof. We use the notation of Th.8.11. We have g ′ (z 0 ) ≠ 0. Since theJacobian of g at z 0 has determinant |g ′ (z 0 )| 2 ≠ 0, the inverse function theoremtells us, that there is an open neighbourhood U 0 of z 0 , such that g(U 0 ) is openand g| U0 : U 0 −→ g(U 0 ) is bijective with an R-differentiable inverse g(U 0 ) −→U 0 . Indeed, it is even holomorphic, as is easily checked, hence biholomorphic.Now take ϱ > 0 with D ϱ (0) ⊂ g(U 0 ) and U := (g| U0 ) −1 (D ϱ (0)).Corollary 8.14 (Open mapping theorem). A non-constant holomorphicfunction f : G −→ C on a domain G is an open map, i.e. the image f(U)of an open set U ⊂ G is again open.Proof. Since we may replace f with f| U , it suffices to show the f(G) is open.And that is an immediate consequence of Th.8.13: Take w 0 = f(z 0 ) ∈ f(G).As a nonconstant holomorphic function f has finite multiplicity at z 0 andthus, with the notation of Th.8.13, D r (w 0 ) ⊂ f(G).Corollary 8.15 (Maximum principle). Let G be a domain and f ∈ O(G).If there is a point z 0 ∈ G with |f(z 0 )| ≥ |f(z)| for all z ∈ G, then f ≡ f(z 0 ).With other words: A non-constant holomorphic function on a domain doesnot attain its maximum.Proof. Since f(G) ⊂ C is open, for any point w 0 = f(z 0 ) ∈ f(G) there isan open disc D r (w 0 ) ⊂ f(G), in particular w = (1 + ε)w 0 ∈ f(G) for anysufficiently small ε > 0, and thus |f(z 0 )| < |f(z)|, if w = f(z).Another fundamental property of the class of all holomorphic functionsis that it is stable under locally uniform limits:Definition 8.16. A sequence of functions f n ∈ C(G) is said to convergelocally uniformly to f ∈ C(G), if every point z 0 ∈ G is the center of an opendisc D = D r (z 0 ) ⊂ G, such that the restricted functions f n | D converge uniformlyto f| D . Or equivalently, if for any compact set K ⊂ G the restrictionsf n | K converge uniformly to f| K .Example 8.17. Assume (a n ) n∈N is a sequence of complex numbers such thatthe sequence (|a n |ϱ n ) n∈N is bounded. Then the power series∞∑a n (z − z 0 ) nn=046

converges locally uniformly on the open disc D ϱ (z 0 ). Hence, by the nextresult, it defines a holomorphic function f ∈ O(D ϱ (z 0 )).Proposition 8.18. Assume the sequence of functions (f n ) n∈N ⊂ O(G) convergeslocally uniformly to the function f : G −→ C. Then1. f ∈ O(G) and2. for all k ∈ N the sepuence (f n(k) ) n∈N of k-th derivatives converges locallyuniformly to f (k) .Note that a locally uniform limit of real analytic functions need not be areal analytic function. Indeed, every continuous function on R is the locallyuniform limit of polynomials! Second: The sequence of real analytic functionsf n (x) = 1 sin(nx) converges even uniformly to 0, but their derivatives do notnconverge locally uniformly to 0.On the other hand, in Prop.8.18 we can not replace locally uniform convergenceby pointwise convergence: The pointwise limit of holomorphic functionsneed not even be continuous, but examples for that phenomenon arenot that easy at hand, cf. Ex.8.21.Proof. We show that f| D ∈ O(D) for every open disc with D ⊂ G. Indeedf n converges uniformly on D to f| D ∈ C(D). Hence∫1 f n (ζ)dζf(z) = lim f n (z) = lim= 1 ∫f(ζ)dζn→∞ n→∞ 2πi ζ − z 2πi ζ − z ,since (f n ) converges uniformly to f on γ := ∂D, cf. the below lemma 8.20.Finally use Rem. 7.5.In order to see that the f n(k) converge uniformly on D = D r (z 0 ) to f (k)we use the generalized Cauchy formulah (k) (z) = k! ∫h(ζ)dζ2πi (ζ − z) k+1for h = f n − f and obtain the estimate∂D∂D r+ε (z 0 )∂D||f n (k) − f (k) || D ≤ k!||f n − f|| ∂Dr+ε (z 0 )· (r + ε).ε k+147

Remark 8.19. In the above proof we have only used, that the sequence(f n ) n∈N converges uniformly on the boundary circle ∂D: Indeed, if a sequenceof functions f n ∈ O(G) converges uniformly on ∂D to a function f ∈ C(∂D),then they converge as well on D uniformly to a continuous function ˆf ∈C(D) ∩ O(D) extending f. This is due to the fact that||f n − f m || D = ||f n − f m || ∂Das a consequence of the maximum principle.Lemma 8.20. Let γ be a piecewise smooth path. If the continuous functionsf n : |γ| −→ C converge uniformly to the function f : |γ| −→ C, then∫∫f n (z)dz = f(z)dz.Proof. We have∫∫∣ f n (z)dz −γγlimn→∞γγγ∣∫∣∣∣f(z)dz∣ = (f n (z) − f(z))dz∣ ≤ ||f n − f|| γ · L(γ).We conclude this section with an example of a pointwise convergent sequenceof holomorphic functions with a non-continuous limit function:Example 8.21. We start with an increasing sequence of compact sets K nexhausting the complex plane C, while their interiors only yield the planewithout the real line, i.e. ⋃ ∞n=1 ˚K n = C \ R. LetwithandK n := K + n ∪ K − nK + n := D n (0) ∩ (R + iR ≥1/n ).K − n := D n (0) ∩ (R + iR ≤0 ).Now the function f : C −→ C with{ 0 , if Im(z) > 0f(z) =1 , if Im(z) ≤ 048

is holomorphic in a suitable neighbourhood U n ⊃ K n for every n ∈ N ≥1 ,and Runge’s theorem Th.10.1 provides a polynomial p n : C −→ C with||f − p n || Kn < 1 . Then the sequence (p n n) n≥1 converges pointwise to the noncontinuousfunction f. We remark that the behaviour of the sequence (p n ) n≥1near points on the real line becomes more and more ”chaotic”, necessarilywe have||p n || K → ∞for any compact set K having an interior point on the real line.9 Laurent series and residuesOne of the most impressive applications of complex analysis is the help itprovides in computing certain real integrals. Indeed, they are reduced tocomplex loop integrals ∫f(z)dzλfor loops λ in the domain G of holomorphy of the function f.First of all, the integral does not change, if we replace λ with a loop ˜λhomotopic to λ in G.Definition 9.1. Two continuous loops λ, ˜λ : [a, b] −→ G are called homotopicin G, if there is a continuous mapH : [a, b] × I −→ Gwith H(t, 0) = λ(t), H(t, 1) = ˜λ(t) and H(a, s) = H(b, s) for all s ∈ I :=[0, 1].Remark 9.2. For a locally integrable differential form ω ∈ D(G) we maydefine∫ωγfor any path γ : [a, b] −→ G as follows: Let t ν := a + ν b−a , ν = 0, ..., n.nThen, for n ≫ 0 the images γ([t ν−1 , t ν ]) are contained in open sets, whereω admits a primitive function F ν , ν = 1, ..., n. Then we set∫n∑ω := F ν (γ(t ν )) − F ν (γ(t ν−1 )).γν=149

We leave it to the reader to check that the given definition does not dependon n ∈ N.Proposition 9.3. Let λ, ˜λ be loops in G. If they are homotopic in G, then∫ ∫ω = ωfor any locally integrable differential form ω ∈ D(G).˜λProof. Denote H : R := [a, b] × I −→ G a homotopy between the loops λand ˜λ. Let R = ⋃ 1≤i,j≤n R ij be the decomposition of R into n 2 congruentrectangles of size 1 the size of R. For sufficiently big n ∈ N we haven∫ ∫ ∫ω − ω = ω = ∑ ∫ω = 0,λ ˜λ H(∂R) i,j H(∂R ij )since the ”vertical edges” H(b × [0, 1]) and H(a × [1, 0]) of the ”rectangle”H(∂R) are inverse one to the other and for n ≫ 0 any ”rectangle” H(∂R ij )is contained in an open set, where ω admits a primitive function.Definition 9.4. A cycle λ in a domain G is a finite sequence of piecewisesmooth loops λ 1 , ..., λ s in G. It is called nullhomologous in G if and only if∫s∑∫ω := ω = 0λλ jholds for all locally integrable differential forms ω ∈ D(G).Remark 9.5. Loops are considered to be cycles of length s = 1.j=1Example 9.6. 1. A loop λ, such that |λ| ⊂ G 0 ⊂ G with a simply connecteddomain G 0 is nullhomologous in G.2. A cycle λ = (˜λ, λ −1 ) with (in G) homotopic loops ˜λ, λ is nullhomologousin G.3. Let G = C \ {± i }, D := D 2 1(0) and λ + resp. λ − the loops consisting ofthe upper resp. lower half circle (in counterclockwise orientation) and[−1, 1] resp. [1, −1]. Then∫ω =∂D∫ω +λ +∫50λλ −ω

holds for any differential form ω ∈ D(G) (not only locally integrableones). Thus(∂D, λ −1+ , λ −1− )is nullhomologous in G.The framework in which we consider loop integrals is the following:1. A domain G ⊂ C is given as well as2. a function f ∈ O(G \ S), where S = S f ⊂ G is a finite set, and3. a piecewise smooth loop λ in G \ S nullhomologous in G.We shall see that the integral over a loop nullhomologous in G is determinedby local contributions arising from the points in S = S f . First of allwe have to take into account how many times the loop λ winds around apoint a ∈ S.Definition 9.7. Let λ be a cycle in C, a ∈ C \ |λ|. Thenind λ (a) := 1 ∫dz2πi z − a ∈ Zis called the index of a with respect to the cycle λ or the winding number ofλ with respect to a.The index ind λ (a) is an integer: If γ : I −→ C is a smooth path witha ∉ |γ|, the functionI −→ C, t ↦→ 1 12πi γ(t) − ahas a primitive function F : I −→ C satisfying e 2πiF (t) = γ(t) − a. Now ifλ = (γ 1 , ..., γ r ) with γ j : I j = [a j , b j ] −→ C, we may choose F j : I j −→ Csuch that F j+1 (a j+1 ) = F j (b j ) for j = 1, ..., r − 1. Then we haveind λ (a) = F r (b r ) − F 1 (a 1 ), e 2πiFr(br) = e 2πiF 1(a 1 ) ,whence the result follows. The generalization from loops to cycles is immediate.We shall need a global version of Cauchy’s integral formula:51λ

Theorem 9.8. Let λ be a nullhomologous cycle in G. Then for a holomorphicfunction f ∈ O(G) and a ∈ G \ |λ| the following Cauchy formulaholds∫1 f(ζ)dζ2πi ζ − a = ind λ(a) · f(a).Proof. The differential formλω :=f(ζ) − f(a)dζ ∈ D(G)ζ − ais locally integrable, hence ∫ ω = 0 and thusλ∫λf(ζ)dζζ − a = f(a) ∫λdζζ − a = 2πi · ind λ(a)f(a).The local contributions from points a ∈ S are called residues:Definition 9.9. Let a ∈ C.O(D r (a) ∗ ) at a is defined asRes a (f) := 12πiThe residue of a holomorphic function f ∈∫∂D ϱ(a)f(z)dz,where 0 < ϱ < r. Note that the right hand side does not depend on ϱ > 0because of Ex.9.6.2: The loops ∂D ϱ (a) and ∂D ϱ ′(a) with 0 < ϱ, ϱ ′ < r arehomotopic in D r (a) ∗ .Remark 9.10. If f ∈ O(D r (a) ∗ ) admits a continuous extension to D r (a),then we haveRes a (f) = 0as a consequence of Prop. 6.15 applied to the differential form ω = f(z)dz.Theorem 9.11 (Residue Theorem). Let λ be a nullhomologous loop inG, S ⊂ G \ |λ| a finite set and f ∈ O(G \ S). Then∫f(z)dz = 2πi ∑ ind λ (a) · Res a (f).λa∈S52

Proof. The key point in the proof of Th.9.11 is a ”generalized partial fractiondecomposition”: We writewhere1. g ∈ O(G) andf(z) = g(z) + h(z) + ∑ a∈Sc az − a ,2. h ∈ O(C \ S) is the derivative of a function H ∈ O(C \ S),3. c a = Res a (f) for a ∈ S.If G = C and f(z) = p(z) is a rational function and S denotes the set of zerosq(z)of the polynomial q(z), then g(z) is the polynomial part of the partial fractiondecomposition of f and h(z) the sum of all singular terms of multiplicity > 1.Given the decomposition we obtain the residue formula as follows: Forsufficiently small ε > 0 integration over D ε (a) yields c a = Res a (f), whileintegration over λ gives the desired formula: We have ∫ g(z)dz = 0, since λλis nullhomologous in G and ∫ h(z)dz = 0 because of hdz = dH.λFirst we derive the above decomposition in the case where G = D r (a)and S = {a}, thereby obtaining a new interpretation of the residue Res a (f).Indeed the situation we consider is slightly more general. DenoteA ϱ,r (a) := {z ∈ C; ϱ < |z − a| < r}, 0 ≤ ϱ < r ≤ ∞.the annulus with center a, inner radius ϱ and outer radius r.Theorem 9.12. Given a function f ∈ O(A ϱ,r (a)) there are unique functionsf + ∈ O(D r (0)) and f − ∈ O(D ϱ −1(0)) (where ”0 −1 = ∞” and D ∞ (0) = C)with f − (0) = 0, such that( ) 1f(z) = f + (z − a) + f − .z − aIn particular there are complex numbers a n , n ∈ Z, such that∞∑∞∑f(z) = a n (z − a) n := a n (z − a) n +n=−∞n=0∞∑a −l (z − a) −l .l=153

A series of the above type is called a Laurent series, the first part is calledthe power series part of the Laurent series, the second one its principal part.Indeeda n = 1 ∫2πi ∂D s(a)f(z)dz, ϱ < s < r.(z − a)n+1Corollary 9.13. Let G ⊂ C be a domain, a ∈ G. If the function f ∈O(G \ {a}) has the Laurent expansion f(z) = ∑ ∞n=−∞ a n(z − a) n in D r (a) ∗ ,then its residue at a satisfiesRes a (f) = a −1 .Let us first finish the proof of Th.9.11. In the case of G = D r (a), S = {a},the decomposition of Th. 9.11 is given by considering D r (a) ∗ = A 0,r (a) andtakingg(z) = f + (z − a), h(z) =−∞∑n=−2a n (z − a) n , c a = a −1 .Note that g ∈ O(D r (a)) and h ∈ O(C \ {a}). In the general case of Th.9.11let c a := Res a (f) for a ∈ S and denote h a (z) the principal part of the Laurentseries of f in some annulus A 0,r (a) minus c a /(z − a). Then we seth(z) := ∑ a∈Sh a (z), g(z) := f(z) − h(z) − ∑ a∈Sc az − a .Obviously all the h a are derivatives and thus h as well. This finishes theproof of Th. 9.11.Proof of Th.9.12. Fix z ∈ A ϱ,r (a), choose ε > 0 with ϱ + ε < |z| < r − ε.According to Ex.9.6.2 the cycle λ = (∂D r−ε (a), ∂D ϱ+ε (a) −1 ) is nullhomolgousand ind λ (z) = 1, hencef(z) = 1 ∫f(ζ)dζ2πi λ ζ − z .Definef + (z) := 1 ∫f(ζ)dζ2πi ∂D r−ε (a) ζ − z54

andf − (z) := − 1 ∫f(ζ)dζ2πi ∂D ϱ+ε (a) ζ − z .The power series expansion for f + (z) is obtained as in the proof of Th. 7.6,while for f − we write− f(ζ)ζ − z = 1z − a · f(ζ)1 − ζ−az−a=∞∑n=0f(ζ)(ζ − a) n(z − a) n+1and integrate with respect to ζ over ∂D ϱ+ε (a). Note that the integrals donot depend on the choice of ε.The discussion in the lectures1. how to compute explicitly a residue Res a (f), and2. how to relate certain real integrals to loop integrals in the complexplane,will be quite close to that one in the text book, pp. 314 - 354; so we don’tgive any comments here.Instead we discuss briefly isolated singularities of holomorphic functions:Definition 9.14. Let G ⊂ C be a domain. A point a ∈ C \ G is called anisolated boundary point of G if D r (a) ∗ ⊂ G for sufficiently small r > 0.Note that for an isolated boundary point a of G the union G a := G ∪ {a}is a domain as well. There are three essentially different possible behavioursof functions f ∈ O(G) near an isolated boundary point:Definition 9.15. Let f ∈ O(G) be a holomorphic function on the domainG. An isolated boundary point a ∈ C of G is called1. a removable singularity of f if there is a function ̂f ∈ O(G a ) extendingf.2. a pole of f if a is not a removable singularity, but there is some n ∈N >0 , auch that the function g(z) = (z − a) n f(z) has at a a removablesingularity.3. an essential singularity of f otherwise.55

Theorem 9.16. Let f ∈ O(G) be a holomorphic function on the domain Gand a ∈ C an isolated boundary point of G with D r (a) ∗ ⊂ G, write f(z) =∑ ∞n=−∞ a n(z − a) n . Then the point a is1. a removable singularity iff f| Dϱ(a) ∗ is bounded for some ϱ > 0 iff a n = 0for n < 0,2. a pole iff lim z→a f(z) = ∞ iff a n = 0 for finitely many, but at least onen < 0 ,3. an essential singularity iff f(D ϱ (a) ∗ ) is dense in C for all ϱ < r iffa n ≠ 0 for infinitely many n < 0.Proof. Obviously a is a removable singularity of f if the principal part of itsLaurent series vanishes, and in that case f is bounded near a. On the otherhand, if f is bounded, then (z −a)f(z)dz ∈ D(D ϱ (a)) is integrable on D ϱ (a) ∗resp. D ϱ (a) and thus (z − a)f(z) = F ′ (z) with a function F ∈ O(D ϱ (a)).We must have F ′ (a) = 0 — otherwise f would not be bounded near a — andthus f(z) is holomorphic near a as well.Second, the principal part f − is a nonzero polynomial in (z − a) −1 iff(z − a) n f(z) extends to G a for some n ∈ N. In that case choosing n minimalyields a −n ≠ 0, i.e. f(z) = g(z)/(z − a) n , g(a) ≠ 0. Hence lim z→a f(z) =∞. On the other hand, if that is true, we have f(D ε (a) ∗ ) ⊂ C \ D 1 (0) forsome ε > 0, hence 1/f is a holomorphic function bounded near a, and thusholomorphic near a. So f itself has at most a pole at a.Third, assume f − is not a polynomial in (z−a) −1 and f(D ϱ (a) ∗ )∩D ε (b) =∅. Then g := 1 is bounded on D b−f ϱ(a) ∗ , hence holomorphic near a. But thenf = b − 1 has at most a pole at a.gThe third point of Th.9.16 indicates a quite irregular behaviour of a holomorphicfunction near an essential singularity. We mention without proofthe famous Picard Theorem:Theorem 9.17. Let a ∈ ∂G be an isolated boundary point of G and anessential singularity of the function f ∈ O(G). Then for all b ∈ C withpossibly one exception we havefor all ε > 0.|f −1 (b) ∩ D ε (a) ∗ | = ∞56

The notion of a rational function, being a quotient of polynomials, has anatural generalization:Definition 9.18. Let G be a domain. A meromorphic function on G is afunction f : G −→ Ĉ := C ∪ {∞} such that1. the set P := f −1 (∞) ⊂ G has no points of accumulation in G,2. f| G\P ∈ O(G \ P ), and3. the points a ∈ f −1 (∞) are poles of f.We denoteM(G) := {f : G −→ Ĉ meromorphic}the set of all meromorphic functions on G.Remark 9.19. Meromorphic functions can be added, multiplied, and, if ≢ 0,inverted: Given f, g we define f + g and fg first as holomorphic functions onG \ (f −1 (∞) ∪ g −1 (∞)), the points in f −1 (∞) ∪ g −1 (∞) being either polesor removable singularities. Accordingly we extend it to G. If f ≢ 0, its zerosform a set without accumulation points in G, they become the poles of 1 , fwhile its poles become the zeros of 1 . So the set M(G) is a field!fThe local behaviour of a meromorphic function f ∈ M(G) \ {0} near apoint a ∈ G is determined by its order:Definition 9.20. Let f ∈ M(G) \ {0} be a meromorphic function. Theorderord a (f) := l ∈ Zof f at a ∈ G is defined as the integer l such thatf(z) =∞∑a n (z − a) nn=lwith a l ≠ 0, or equivalentlyf(z) = (z − a) l g(z)with a function g holomorphic near a and g(a) ≠ 0.57

So the poles of a meromorphic function are treated as zeros of negativeorder. Furthermore for function holomorphic at a ∈ G its multiplicity thereis nothing but ord a (f − f(a)).For a function f ∈ M(G) \ {0}, the meromorphic function f ′is calledfthe logarithmic derivative of f. It has only simple poles, namely at the zerosand poles of f.Proposition 9.21.1. For f ∈ M(G) \ {0} one hasRes a (f ′ /f) = ord a (f).2. If G is simply connected, then every function g ∈ M(G) with onlysimple poles and integral residues is the logarithmic derivative of somefunction f ∈ M(G).Proof. 1) We writeand applyWe obtainf(z) = (z − a) l h(z), h(a) ≠ 0(gh) ′gh = g′g + h′h .f ′f =lz − a + h′h ,whence the result, the second term being holomorphic at a.2) Fix a point z 0 ∈ G 0 := G\P , where P is the set of poles of the meromorphicfunction g. Then for a path γ z : [a, b] −→ G 0 from z 0 to z the integral∫γ zg(ζ)dζdepends, as a consequence of the residue theorem, up to an integer multipleof 2πi only on z 0 and z. Thus we may define a function f ∈ O(G 0 ) by(∫ )f(z) := exp g(ζ)dζ .γ z58

Definition 9.22. A continuous loop λ : [a, b] −→ C is called simple or asimply closed path or a Jordan curve if λ| [a,b) is injective.Though intuitive, the following remark is nontrivial:Remark 9.23. 1. The Jordan curve theorem tells us, that the complementof the trace of a simple loop λ is the union of two disjoint connectedopen sets:C \ |λ| = int(λ) ∪ ext(λ),called the interior resp.the exterior of λ, where int(λ) is bounded andext(λ) is not.2. We haveand eitherind λ | ext(λ) ≡ 0ind λ | int(λ) ≡ 1 or ind λ | int(λ) ≡ −1.In the first case we say that λ is positively oriented.3. A simple loop λ : [a, b] −→ G is nullhomologous in G iff int(λ) ⊂ G.The condition is necessary: For a point a ∈ int(λ) \ G the functionf(z) = 1 is holomorphic on G and satisfies ∫ f(z)dz ≠ 0. On thez−a λother hand sufficiency follows from Th.10.1 applied to K := |λ|∪int(λ):Any function f ∈ O(G) is on K the uniform limit of a sequence ofpolynomials p n : C −→ C. But ∫ f(z)dz = lim ∫λ n→∞ p λ n(z)dz =lim n→∞ 0 = 0.The above remark is mainly of theoretical interest; in actual computationsthe statements are easily verified. So we do not comment on the demandingproof. —The power function f n ∈ M(C) with f n (z) = z n has at 0 a zero of order nand transforms the simple closed loop λ = ∂D r (0) into the loop λ n := f n ◦ λwinding around the origin n times:ind fn◦λ(0) = n.In the next theorem that formula is generalized to meromorphic functionsdefined on a neighbourhood of |λ| ∪ int(λ), with no zeros or poles on |λ|.59

Theorem 9.24 (Argument Principle). Let f ∈ M(G) and λ be a simplepositively oriented loop in G with int(λ) ⊂ G, such that no poles and no zerosof f lie on |λ|. Thenind f◦λ (0) = 12πi∫λf ′ (z)dzf(z)= ∑a∈int(λ)ord a (f),i.e. the winding number of the image loop f ◦ λ around the origin equals thenumber of the zeros and poles of f inside the loop λ counted with multiplicities.Proof. Apply Th.9.11 and Prop.9.9.Theorem 9.25 (Rouché’s Theorem). Let f, h ∈ O(G) and λ be a nullhomologoussimple loop in G such that no zeros of f lie on |λ|. If |h(z)| < |f(z)|for z ∈ |λ|, then f and g := f + h have the same number of zeros (countedwith multiplicities) in int(λ).Proof. The paths f ◦ λ and g ◦ λ are homotopic in C ∗ with the homotopy(t, s) ↦→ f(λ(t)) + sh(λ(t)).Hence1∫f ′ (z)dz2πi λ f(z)= ind f◦λ (0) = ind g◦λ (0) = 1 ∫g ′ (z)dz2πi λ g(z) .10 Construction of holomorphic and meromorphicfunctionsIn this section we discuss briefly two famous theorems dealing with the distributionof zeros resp. singularities of holomorphic resp. meromorphic functionson a domain G. First we need:Theorem 10.1 (Theorem of Runge). Let K ⊂ C be a compact set withconnected complement C \ K. Then every holomorphic function f : U −→ Cdefined on an open neighbourhood U ⊃ K (it need not be connected!) canon K be uniformly approximated by polynomials: Given ε > 0, there is apolynomial function p(z) = ∑ nν=0 a νz ν with ||f − p|| K < ε.60

Remark 10.2. 1. Th.10.1 is trivial if K = D r (a) is a closed disc. Thenwe may take p(z) = ∑ nν=0 c ν(z − a) ν , where f(z) = ∑ ∞ν=0 c ν(z − a) ν isthe Taylor expansion of f around a and n ≫ 0.2. A compact set K is called polynomially convex if C \ K is connected.To any compact set K we can associate its polynomially convex hullˆK: It is defined asˆK := C \ E,where E is the unbounded connected component of C \ K. So ˆK is theunion of K and the holes in K, i.e. the bounded connected componentsof C \ K.3. By Mergelyan’s theorem we know that Runge’s theorem holds evenfor functions f ∈ C(K) ∩ O( ˚K), i.e. functions continuous on K andholomorphic in the interior ˚K of K.Sketch of proof. First we need that there is a cycle λ in U \ K, such thatind λ (z) = 1 for all z ∈ K. As a consequencef(z) = 1 ∫f(ζ)dζ2πi ζ − zholds for z ∈ K and thus f| K may be uniformly approximated by functionsr∑ν=1λc νa ν − z ,where a 1 , ..., a r ∈ |λ|. Namely: The above integral is a finite sum of integralsf γ (z) := 1 ∫f(ζ)dζ2πi ζ − zover smooth paths γ : [a, b] −→ U \K, and their Riemann sums are functionson K of the above type and approximate the function f γ uniformly on K.Sof = ∑ f γγis as well the uniform limit of linear combinations of degree one Laurentmonomials.61γ

Now for b ∈ C \ K denoteL b (K)the set of all functions on K being the uniform limit of polynomials in 1z−b .We show that1a − z ∈ L b(K)holds for all a ∈ C \ K, or, equivalently,In particular we obtainL a (K) ⊂ L b (K).f| K ∈ L b (K).Now choose a closed disc D := D r (0) ⊃ K and b ∉ D. Denote q(z) apolynomial in 1 with ||f − q|| z−b K < ε/2 and take p(z) as the n-th Taylorpolynomial of q around 0. Then ||q − p|| D < ε/2 for n ≫ 0 and thus||f − p|| K < ε.1It remains to show that all degree one Laurent monomials with a ∈a−zC \ K belong to L b (K). We consider the set{V b := a ∈ C \ K;Since b ∈ V b , we have V b ≠ ∅. Furthermore1a − z ∈ L b(K)a ∈ V b =⇒ D r (a) ⊂ V bwith r = dist(a, K). As a consequence a path in C \ K starting at b neverleaves V b , i.e. V b = C \ K.Indeed let c ∈ D r (a). Then, for |z − a| > |c − a| we have}.∞1c − z = − ∑ (c − a) ν(z − a) ∈ L a(K) ⊂ L ν+1 b (K),ν=0since the right hand side converges uniformly on C \ D ϱ (0) for any ϱ >|c − a|.Theorem 10.3 (Theorem of Mittag-Leffler). Let G ⊂ C be a domain.Given a sequence (a n ) n∈N ⊂ G without accumulation point in G and polynomials∑r nc nkp n (z) =(z − a n ) kk=162

1inz−a nwithout constant term, there is a meromorphic function having thepoints a n as its poles with p n (z) as the principal part of its Laurent seriesaround the point a n ∈ G.Proof. The general idea is to writef(z) =∞∑(p n (z) + g n (z)),n=0where the functions g n ∈ O(G) are chosen in such a way, that the abovesequence converges uniformly on every compact set K: That makes sense,since there is some n K ∈ N, such that all p n with n ≥ n K are holomorphicaround K. The above sum is the canonical form of f(z) in most of allcases; nevertheless in the proof we construct a, for technical reasons, slightlydifferent series.We sketch the proof for a simply connected domain G. Given a compactset K ⊂ G we have as well ˆK ⊂ G for its polynomially convex hull ˆK, cf.Rem.10.2. To see that we have to show F ⊂ G for all bounded connectedcomponents F of C \ K, cf. Th.6.23. Given such F , we writeC \ G = A ˙∪ Las the disjoint union of the compact setand the closed setL := (C \ G) ∩ FA = (C \ G) \ L.Since G is simply connected, it follows L = ∅ and thus F ⊂ G. Using thatremark one easily constructs an exhaustion of G by polynomially convexcompacts sets K ν ⊂ G, where we assume that even K ν ⊂ ˚K ν+1 . Let∑q ν (z) := p n (z).a n∈K ν+1 \K νThe function q ν being holomorphic in a neighbourhood of K ν , Runge’s theoremTh.10.1 provides a polynomial h ν : C −→ C satisfying||q ν − h ν || Kν < 2 −ν .63

Finally setf(z) =∞∑(q ν (z) − h ν (z)).ν=0Example 10.4. 1. Take G = C, a n = −n (where n ∈ N) and p n := 1Then we can choose g 0 = 0 and g n (z) = − 1 for n ≥ 1, the resultingnfunction isf(z) = 1 ∞z + ∑( 1z + n − 1 ).nn=1z+n .2. Take the above example, but replace the indexing set N with Z. Weobtain the functionπ cot(πz) = 1 z + ∑ ( 1z + n − 1 ),nn∈Z ∗where Z ∗ := Z \ {0}.3. Consider a latticeΛ = Zω 1 + Zω 2 ,where the complex numbers ω 1 , ω 2 ∈ C are linearly independent overR. We use Λ itself as indexing set, i.e. a ω = ω, takep ω (z) =1(z − ω) 2and g ω (z) = 1ω 2 . The resulting function℘(z) = 1 z 2 + ∑ ω∈Λ ∗ 1(z − ω) 2 − 1 ω 2 ,where Λ ∗ := Λ \ {0}, is called Weierstraß’ ℘-function. It is Λ-periodic,i.e. satisfies℘(z + ω) = ℘(z)for all lattice points ω ∈ Λ.64

Theorem 10.5 (Weierstraß factorization theorem). Let G ⊂ C be adomain. Given a sequence (a n ) n∈N ⊂ G without accumulation point in Gand positive integers l n ∈ N >0 , there is a holomorphic function, whose zerosare exactly the points a n ∈ G with corresponding zero order ord an (f) = l n .Proof. For a simply connected domain G we can argue as follows: Firstwe apply Th.10.3 with p n (z) := ln and obtain a function g ∈ M(G) withzonly simple poles and integer residues. Then Prop.9.21.2 provides a functionf ∈ O(G) with g as its logarithmic derivative.Example 10.6. We are looking for a function, which has simple zeros exactlyat the points in Z ≤0 . We remember the proof of Prop.9.21.2 and apply thesame strategy with z 0 = 0: The summandp n (z) + g n (z) = 1z + n − 1 nof Ex.10.4.1 is the logarithmic derivative of(1 + z )e −z/n .nThus the functionf(z) = z∞∏n=1(1 + z )e −z/nnis a solution of our problem. It satisfies the functional equationwith ”Euler’s constant”C := limm→∞Hence the Gamma functionzf(z + 1) = e −C f(z)( m∑n=1)1n − ln(m + 1) .Γ(z) := e−Czf(z)satisfiesas well as Γ(0) = 1.Γ(z + 1) = zΓ(z)65

Corollary 10.7. Every meromorphic function f ∈ M(G) can be written f =g/h as the quotient of holomorphic functions g, h ∈ O(G). With other words:The field M(G) of meromorphic functions on G is the field of fractions ofthe integral domain O(G), i.e.M(G) = Q(O(G)).Proof. Let (a n ) n∈N be the set of poles of f, l n := −ord an (f). Denote h ∈O(G) a holomorphic function, which has a zero of order l n at a n for everyn ∈ N (and, w.l.o.g., no further zeros). Take g := fh.11 The Dirichlet ProblemLet us start with the following question: Given a function u : G −→ R, whencan we find another function v : G −→ R, such that f = u + iv : G −→ C isholomorphic?Theorem 11.1. Let G ⊂ C be a simply connected domain. Then a realvalued function u ∈ C 2 (G) is the real part of some holomorphic functionf = u + iv ∈ O(G) if and only if u is harmonic, i.e.∆u := ∂2 u∂x 2 + ∂2 u∂y 2 = 0.Before we give the easy proof, we introduce for systematic reasons thefollowing conjugation operation on differential forms:Remark 11.2. For a differential form ω ∈ D(G) we define its conjugateform ω ∗ ∈ D(G) by(ω ∗ ) z (h) := ω z (−ih).Indeed forwe findω = fdx + gdy = h 1 dz + h 2 dzω ∗ = −gdx + fdy = i(−h 2 dz + h 1 dz).Proof of Th.11.1. The equation ∆u = 0 means that the differential form(du) ∗ = − ∂u ∂udx +∂y ∂x dy ∈ D1 (G)66

is closed. Since G is simply connected that is equivalent to the existence ofa primitive function v ∈ C 2 (G), i.e. such that(du) ∗ = dv,the latter equality being nothing but the Cauchy-Riemann equations for f =u + iv.Remark 11.3. We note that∆u = 4 ·∂∂z (∂u ∂z ) = 4 · ∂∂z (∂u ∂z ).In particular a real polynomial is harmonic if and only if it does not containmixed terms z µ z ν (with both µ, ν > 0). With other words, harmonicreal polynomials are those which can be written f(z) + g(z) with complexpolynomials f, g : C −→ C. Furthermore,du + i(du) ∗ = 2 ∂u∂z dzfor a real valued function u : G −→ R.Here is an example showing what may happen on non simply connecteddomains:Example 11.4. We consider the annulus A ϱ,r (a).1. The harmonic function ln |z − a| is not the real part of some functionf ∈ O(A ϱ,r (a)). But up to that failure everything is fine:2. In the annulus A ϱ,r (a) any real valued harmonic function u can bewrittenu(z) = c ln |z − a| + Re(f(z))with a holomorphic function f ∈ O(A ϱ,r (a)) and a unique c ∈ R. Forthe proof we may assume a = 0. Takec := 1 ∫(du) ∗ ∈ R.2π∂D s(0)Now we apply Prop.6.25 to the differential formω = du + i(du) ∗ − cdz (z = 2 ∂u∂z − c )dzz67

- it is closed since (du) ∗ is - and find a function f ∈ O(A ϱ,r (0)) withTaking real parts finally yieldsdf = ω.d Re(f) = du − cRe( ) dz,zand thus, w.l.o.g.,Re(f(z)) = u − c ln |z|.Harmonic functions play an important rôle in physics: They can be regardedas the potential of a stationary electric field or the temperature in asteady state-heat flow. In such a situation the following problem occurs in anatural way:Dirichlet Problem: Let G ⊂ C be a domain and u 0 : ∂G −→ R a continuousfunction. Find a continuous function u : G −→ R, harmonic in G, suchthatu| ∂G = u 0 .In this section we solve the Dirichlet problem on an open disc and thengive some comments on the general case.We start with an idea originating from the above relationship betweenharmonic and holomorphic functions and the Cauchy formula: For a holomorphicfunction f defined on a domain containing the closed disc D theCauchy formulaf(z) = 1 ∫f(ζ)dζ2πi ∂D ζ − zdescribes the values f(z), z ∈ D, in terms of the boundary values f| ∂D . Now,if we could rewrite the right hand side of the Cauchy formula in terms ofu = Re(f) only, we would obtain a good candidate for a solution u : D −→ Rof the Dirichlet problem by applying the revised Cauchy formula to the givenfunction u 0 : ∂D −→ R and then taking u = Re(f). Of course, then itremains to be shown that for z → w ∈ ∂D one has u(z) → u 0 (w).68

We study the above situation for D = D r (0). In some slightly bigger discD r+ε (0) we have∞∑f(z) = a n z n .n=0Now we express the coefficients a n , n > 0, in terms of u = Re(f) only, usingthe Fourier expansion of the function ϑ ↦→ u(re iϑ ) which is obtained fromthe above power series as follows: In the equation(u(z) = Re(a 0 ) + 1 ∞)∑a n z n + a n z n2we substitute z = re iϑ and obtain the Fourier series(u(re iϑ ) = Re(a 0 ) + 1 ∞)∑∞∑a n r n e inϑ + a n r n e −inϑ .2Its coefficients satisfyandn=1n=1n=1Re(a 0 ) = 1 ∫ 2πu(re iϑ )dϑ2π 0a n r n = 1 π∫ 2π0u(re iϑ )e −inϑ dϑfor n > 0. Substituting this in the Taylor series expansion of f and interchangingsummation and integration we obtain the desired integral formula:Theorem 11.5 (Schwarz’ integral formula). Let f = u + iv ∈ O(G) andG ⊃ D r (0). Then for z ∈ D := D r (0) we havef(z) = iv(0) + 1 ∫u(ζ) ζ + z dζ2πi ζ − z ζ∂D= iv(0) + 1 ∫ 2πu(re iϑ ) reiϑ + z2π 0 re iϑ − z dϑ.69

Proof. For fixed z ∈ D r (0) the seriesu(re iϑ ) + 2∞∑n=1( z) nu(re iϑ )re iϑconverges uniformly on the interval [0, 2π], thus integration and summationmay be interchanged and we obtainf(z) =∞∑a n z n = iv(0) + Re(a 0 ) +n=0= iv(0) + 1 ∫ 2πu(re iϑ )dϑ +2π 0= iv(0) + 1 ∫ 2πu(re iϑ )2π 0∞∑n=1(= iv(0) + 1 ∫ 2πu(re iϑ )2π 0= iv(0) + 12π∫ 2π0∞∑a n z nn=1∫1 2π( z) nu(re iϑ )e −inϑ dϑ ·π 0r)∞∑ ( z) n1 + 2dϑre iϑn=1( 21 − zre iϑ − 1u(re iϑ ) reiϑ + zre iϑ − z dϑ.)dϑFinally, taking real parts of both sides we obtain:Theorem 11.6 (Poisson integral formula). Let u : G −→ R be a harmonicfunction, G ⊃ D with D := D r (0). Thenu(z) = 1 ∫ 2πu(re iϑ ) r2 − |z| 22π 0 |re iϑ − z| dϑ 2holds for all z ∈ D r (0).= 12πi∫∂Du(ζ) r2 − |z| 2|ζ − z| 2 dζζ .And here is the solution of the Dirichlet problem on a disc:70

Theorem 11.7. Let D := D r (0) and u 0 : ∂D −→ R be a bounded function,continuous with, may be, finitely many exceptions. Then there is a uniquebounded harmonic function u : D −→ R, such thatlim u(z) = u 0(w)D∋z→wfor all points w ∈ ∂D, where u 0 is continuous. Indeedu(z) = 1 ∫ 2πu 0 (re iϑ ) r2 − |z| 22π 0 |re iϑ − z| dϑ 2holds for z ∈ D.Remark 11.8. If u 0 is continuous, it is bounded, ∂D being compact. Furthermorethe function û : D −→ R with û| D = u, û| ∂D = u 0 is continuousand thus bounded as well. So the adjective ”bounded” can be omitted inthat case.But we need it in case we admit discontinuities: Take u 0 (re iϑ ) = ϑ for0 ≤ ϑ < 2π. Then, given a bounded solution u : D −→ R another unboundedsolution is u(z) + g(z) with( )g(z) := |z|2 − 1 z + 1|z − 1| = Re .2 z − 1Proof. Uniqueness: Let r n = r−r/n. For fixed z ∈ D the uniformly boundedfunctions[0, 2π] ∈ ϑ ↦→ u(r n e iϑ ) r2 n − |z| 2|r n e iϑ − z| 2converge almost everywhere to[0, 2π] ∈ ϑ ↦→ u 0 (re iϑ ) r2 − |z| 2|re iϑ − z| 2 .So, by the Lebesgue dominated convergence theorem we knowu(z) = 1 ∫ 2πu(r n e iϑ ) r2 n − |z| 22π 0 |r n e iϑ − z| dϑ → 1 ∫ 2πu 2 0 (re iϑ ) r2 − |z| 22π 0 |re iϑ − z| dϑ. 2Existence: The function given by the Poisson formula is harmonic as the realpart of the holomorphic function f given by the Schwarz formula. We have71

to show that D ∋ z n → w ∈ ∂D implies u(z n ) → u 0 (w), if u 0 is continuousat w. We consider the sequence of functionsϕ n : ∂D −→ R, ζ ↦→ 1 r 2 − |z n | 22π |ζ − z n | . 2They form a ”Dirac sequence” with respect to w ∈ ∂D, i.e.1. ϕ n ≥ 0,2. ∫ 2π0ϕ n (re iϑ )dϑ = 1 ∀n ∈ N,3. ϕ n → 0 uniformly on ∂D \ D δ (w) for every δ > 0.The first condition is satisfied because of |z n | < r, and the second one followsfrom Poissons formula with u ≡ 1. Finallyϕ n (ζ) ≤ 1 r 2 − |z n | 22π (δ − |w − z n |) 2for ζ ∈ ∂D \ D δ (w), with the right hand side tending to 0 for n → ∞. Wehave used the following estimate for the denominator:|ζ − z n | = |(ζ − w) − (z n − w)| ≥ |ζ − w| − |z n − w| ≥ δ − |z n − w|.So we have∫u(z n ) =∂Du 0 (ζ)ϕ n (ζ) dζiζ .Now take some ε > 0. Since u 0 is continuous at w, there is a δ > 0 such that|u 0 (ζ) − u 0 (w)| < ε 2for ζ ∈ D δ (w) ∩ ∂D. Choose n 0 ∈ N such that|ϕ n (ζ)| ≤ ε4||u 0 || ,∀ζ ∈ ∂D \ D δ(w)and n ≥ n 0 . The second condition for the ϕ n impliesu(z n ) − u 0 (w) =∫ 2π0(u 0 (re iϑ ) − u 0 (w))ϕ n (re iϑ )dϑ72

(∫ )= + (u 0 (ζ) − u 0 (w))ϕ n (ζ)γ 1∫γ dζ2iζwith γ 1 = ∂D ∩ D δ and γ 2 the arc complementary to γ 1 . Now∫∣ (u 0 (ζ) − u 0 (w))ϕ n (ζ) dζγ iiζ ∣ ≤ ε 2for both i = 1 and i = 2. For i = 2 we use|u 0 (ζ) − u 0 (w)| ≤ 2||u 0 ||and the second condition for the functions ϕ n , while for i = 1 the aboveestimate |u 0 (ζ) − u 0 (w)| < ε 2 and ∫ γ 1ϕ n (ζ) dζiζ ≤ 1 applies.After having discussed the Dirichlet problem for harmonic functions ona disc one might wonder what can be said about the corresponding problemfor holomorphic functions:Theorem 11.9. Let D := D r (0). For a continuous function f 0 : ∂D −→ Cthe following statements are equivalent1. There is a continuous function f : D −→ C holomorphic in D extendingf 0 , i.e. f| ∂D = f 0 .2. We have ∫ ∂D f 0(z)z n dz = 0 for all n ∈ N.Proof. ”1) =⇒ 2)”: We have∫f 0 (z)z n dz = lim f(z)z∂Dϱ→r∫∂D n dz = lim 0 = 0.ϱ→rϱ(0)”1) =⇒ 2)”: Let f 0 = u 0 + iv 0 . We apply Th.11.7 to u 0 and v 0 and setf = u + iv with the solutions u, v of the Dirichlet problem. We have to showf| D ∈ O(D), or equivalently thatf(z) = 1 ∫f 0 (ζ)dζ2πi ∂D ζ − zholds forf(z) = 1 ∫f 0 (ζ) r2 − |z| 2 dζ2πi ∂D |ζ − z| 2 ζ .73

For ζ ∈ ∂D we findr 2 − |z| 2|ζ − z| 2 ζ − 1ζ − z =zr 2 − zζ = z r 2∞∑n=0( zζr 2 ) n,the last series converging uniformly on ∂D because of |z| < r. Thus thedifference between the above integrals is(z∑ ∞ ( ) ) n zζf(ζ)dζ =r∫∂Dz ∑ ∞z nf(ζ)ζ 2 r 2 r 2 r∫∂Dn dζ = 0.2nn=0n=0We conclude this section by giving some comments on the general case.First of all we noteRemark 11.10. 1. A harmonic function, being locally the real part of acomplex analytic function, is a real analytic function.2. Harmonic functions on a domain G, being real analytic functions, satisfythe weak identity theorem Th. 7.7. In particular, such a functionis either constant, or nonconstant on every open disc D ⊂ G.3. A nonconstant harmonic function u : G −→ R is an open function:Such a function is locally the composition u = Re◦f of two open maps,a nonconstant holomorphic function f and the projection Re : C −→ Ron the real line.4. Let G be a bounded domain. A continuous function u : G −→ R,harmonic on G, attains its maximum and minimum on ∂G (u| G beingan open map or constant).5. For u : G −→ R as in the previous point we have: u| ∂G ≡ 0 =⇒ u ≡ 0.In order to formulate a very general sufficient criterion for the solvabilityof the Dirichlet problem we extend the notion of a connected set to closedsets:Definition 11.11. A closed subset A ⊂ C is called connected if it can notbe written as the disjoint union A = A 1 ˙∪A 2 of two nonenpty closed subsetsA i ⊂ A.74

Remark 11.12.1. Any interval [a, b] ⊂ R is connected.2. The trace |γ| of a path γ is connected.3. A path connected closed set is connected, but in contrast to the caseof open sets, a connected closed set need not be path connected: Theset{ ( }1A := [−i, i] ∪ x + i sin ; x ∈ R >0x)is connected, but not path connected.Theorem 11.13. Let G ⊂ C be a bounded domain, such that for everyboundary point a ∈ ∂G there is a compact connected set K ⊂ C\G containinga as well as points different from a. Then the Dirichlet problem is solvableon G, i.e. for every continuous function u 0 : ∂G −→ R there is a (unique)continuous function u : G −→ R, harmonic in G, such thatu| ∂G = u 0 .Before we give a sketch of the proof let us note:1. The uniqueness of the solution of the Dirichlet problem follows fromRem 11.10.5.2. For a bounded simply connected domain G the complement C \ G isconnected, hence, if G ⊂ D := D r (0), we could choose K := D \ G forevery boundary point a ∈ G.3. The above sufficient condition on G is not satisfied, if G has isolatedboundary points. And indeed, in that case there are always boundaryvalues for which the Dirichlet problem is not solvable: If a is an isolatedboundary point of G, we may write ∂G = {a} ˙∪A with the closed setA := ∂G \ {a}, and there is no u : G −→ R, harmonic in G, withu(a) = 1, u| A ≡ 0. That is a consequence of the below proposition andthe maximum principle applied to G 0 := G ∪ {a} (with ∂G 0 = A).Proposition 11.14. Let D := D r (a). A continuous function u : D −→ R,harmonic on D ∗ := D \ {a}, is harmonic on D.75

Proof. According to Rem.11.4 we can writeu(z) = c ln |z − a| + Re(f(z))with a function f ∈ O(D ∗ ) and c ∈ R. We have to show that a is a removablesingularity of f, whence even c = 0, and we are done. Assume that this is notthe case. Then the holomorphic function g(z) = e f(z) on D ∗ has an essentialsingularity at a, since g(D ε (0) ∗ ) is dense in C for every ε > 0. If f itself hasan essential singularity at a, that is clear, since exp(C) = C ∗ is dense; if f hasa pole at a, then f(D ε (0) ∗ ) ⊃ A r,∞ (0) for some r > 0 - the function 1/f isholomorphic at a and in particular open. Thus g(D ε (0) ∗ ) ⊃ exp(A r,∞ (0)) =C ∗ .On the other hand|g(z)| = |e f(z) | = e u(z) · |z − a| −c .Hence, for n > c we see that h(z) = (z −a) n g(z) satisfies lim z→a h(z) = 0, i.e.a is a removable singularity of h and at most a pole of g. Contradiction.Sketch of the proof of Th.11.13. First of all one introduces a new class ofcontinuous functions containing the harmonic functions as ”upper extremalfunctions”. We start with the observation that harmonic functions satisfythe mean value property: As a consequence of the Poisson integral formula,given a harmonic function u : G −→ R and a closed disc D r (a) ⊂ G thefunctional value at its center equals the mean value over its boundary circle:u(a) = 1 ∫ 2πu(a + re iϑ )dϑ.2π 0On the other hand one can show, that any continuous function satisfying themean value property is already harmonic.Now a continuous function u : G −→ R is called subharmonic if for anyclosed disc D r (a) ⊂ G we haveu(a) ≤ 1 ∫ 2πu(a + re iϑ )dϑ.2π 0We denote S(G) the set of all subharmonic functions on G. Here are someof their most important properties:76

1. We haveS(G) ∩ C 2 (G) = {u ∈ C 2 (G); ∆u ≥ 0}.Indeed, if ∆u is taken as a distribution, thenS(G) = {u ∈ C(G); ∆u ≥ 0}.Here for a distribution v : C ∞ 0 (G, R) −→ R we use the conventionv ≥ 0 :⇐⇒ v(C ∞ 0 (G, R ≥0 )) ⊂ R ≥0 .In particular subharmonicity is a local property.2. g, h ∈ S(G) =⇒ g + h, max(g, h), λg ∈ S(G), where λ ∈ R ≥0 .3. Subharmonic functions satisfy the maximum principle: If g ∈ S(G)attains its maximum at a point z 0 ∈ G, then g ≡ g(z 0 ).Now given a continuous function u 0 : ∂G −→ R one considers the followingfamily of subharmonic functionsS ≤u0 := { g ∈ S(G); ∀a ∈ ∂G : lim z→a g(z) ≤ u 0 (a) } .Now the candidate for a solution is the function u : G −→ R defined byu(z) := sup{h(z); h ∈ S ≤u0 } ≤ ||u 0 || ∂G ,the inequality being a consequence of the maximum principle. In order tosee that u is a harmonic function one needs the following observation: Givenany open disc D = D r (z 0 ) with D ⊂ G the “Poisson modification”P D (h) : G −→ Rof a subharmonic function h : G −→ R is defined as the unique continuousfunction satisfyingP D (h)| G\D = h, ∆(P D (h)| D ) = 0,i.e. outside D the function P D (h) coincides with h, while on D it is theunique solution of the Dirichlet problem with the boundary values h| ∂D . Itis again a subharmonic function: P D (h) ∈ S(G), lying above h, i.e.P D (h) ≥ h.77

Since the family S ≤u0 is stable under Poisson modifications, we see that uis harmonic, using the characterization of harmonic functions by the meanvalue property.On the other hand, in order to provelim u(z) = u 0(a)z→afor all boundary points a ∈ ∂G, we need that there are sufficiently manyfunctions in the family S ≤u0 . That is the case if the domain G admits at anyboundary point a ∈ ∂G arbitrarily steep barrier functions, i.e. given a discD = D ε (a) there should exist a function β ∈ S(G) satisfying1. β ≤ 0,2. lim z→a β(z) = 0,3. z ∈ G \ D =⇒ β(z) ≤ −1.In the remark 12.17 of the next section we shall explain how such barrierfunctions are constructed using the condition of Th.11.13.12 Conformal mapsIn the previous section we have seen that Dirichlet’s problem is solvable underquite general assumptions, but we have no recipe how to find the solutionexplicitly except on the unit disc. And even there the evaluation of thePoisson integral is not always that easy either. Sometimes for domains ofanother shape a good guess can lead at once to the unique bounded solution.Because of the below remark 12.1, a biholomorphic transformation of thedomain of interest could, may be, be helpful in such a situation.Remark 12.1. Let ϕ : G −→ G ′ be a biholomorphic map. Then a functionu : G ′ −→ R is harmonic iff u ◦ ϕ is. This is a consequence of the fact thatharmonic functions are locally the real part of holomorphic functions andthat compositions of holomorphic functions are again holomorphic.Example 12.2. 1. Let H := {z ∈ C; Im(z) > 0} be the upper half plane,and u 0 : ∂H = R −→ R the function{ 0 , if x < 0u 0 (x) :=1 , if x > 0 ,78

where it does not matter how we define u 0 (0), since the values of u 0 atpoints of discontinuity do not matter in the Dirichlet problem. Thenis a bounded solution, whileis an unbounded solution.u(z) := 1 − 1 π Arg(z)u(z) := 1 − 1 π Arg(z) + y2. Now consider the unit circle D = D 1 (0) and ũ 0 : ∂D −→ R with{ 1 , if Im(z) < 0ũ 0 (z) :=0 , if Im(z) > 0 .Let us look at the following biholomorphic transformationf : D −→ H, z ↦→ −i z − 1z + 1 .It extends to a homeomorphism D \{−1} −→ H, thereby transformingthe Dirichlet problem of the previous point to the given one. Consequently,the function ũ := u ◦ f is a solution of our Dirichlet problemon the unit disc.Now, in order to find the good transformation for a given Dirichlet problemon G ⊂ C, one should understand how biholomorphic maps (defined inan open neighbourhood of G, say) act on curves as possible pieces of ∂G.First of all they preserve the angle between two intersecting curves.Definition 12.3. The (oriented) angle ϑ = ∠(z, w) ∈ [0, 2π) between twononzero complex numbers is defined byw|w| = zeiϑ ·|z| .The angle between two regular paths γ, δ (i.e. smooth paths with nonvanishingtangent vector everywhere) at a common point γ(s) = δ(t) is the angle∠( ˙γ(s), ˙δ(t)) between their tangent vectors ˙γ(s), ˙δ(t).79

Lemma 12.4. A nonsingular R-linear map A : C −→ C is angle preserving,i.e.∠(A(ζ), A(η)) = ∠(ζ, η)holds for all ζ, η ∈ C ∗ , if and only ifwith some a ∈ C ∗ .A(z) = azProof. The multiplication with some nonzero complex number preserves obviouslyangles. On the other, if A is angle preserving, the map B :=a −1 A, a := A(1), is angle preserving as well and satisfies B(1) = 1. Thennecessarily B(i) = µi with some µ ∈ R >0 . Now∠(1, 1 + µi) = ∠(B(1), B(1 + i)) = ∠(1, 1 + i) = π 4 .implies µ = 1. So B = id C and A(z) = az.Definition 12.5. A C 1 -map f : G −→ C is called conformal (or anglepreserving) at z ∈ G, if its differentialDf(z) : C −→ Cis invertible, and given any regular paths γ, δ with z = γ(s) = δ(t), the anglebetween f ◦ γ and f ◦ δ at f(z) equals the angle between γ and δ at z.Lemma 12.6. A C 1 -map f : G −→ C is conformal at z ∈ G iff f is C-differentiable there with nonzero derivative: f ′ (z) ≠ 0.Proof. The tangent vector of f ◦ γ at t is f ′ (γ(t)) · ˙γ(t).As next we discuss in detail an example where a biholomorphic transformationleads to the solution of the Dirichlet problem:Example 12.7. Th.11.13 tells us that the Dirichlet problem is solvable onthe domainG := D \ [0, 1),the unit disc with a prick removed. Note that its boundary ∂G = ∂D ∪ [0, 1]is not a Jordan curve. We shall show how we can obtain a solution by findinga biholomorphic mapf : G −→ D80

etween G and the entire unit disc D. Its inverse extends to a continuousmap f −1 : D −→ G = D inducing a surjection of the boundariesf −1 : ∂D −→ ∂G,such that every point of the open prick (0, 1) has two inverse images. Nowgiven a continuous functionu 0 : ∂G −→ R,the Poisson formula provides a functionwith ũ| ∂D = u 0 ◦ f −1 . Finally defineũ : D −→ Ru : G −→ R by u| G = ũ ◦ f, u| ∂G = u 0 .The map f : G −→ D is found as the composition of several maps, whichstepwise simplify the domain G. First of all we unfold the prick: A branchof the square root√ : G −→ D + , − 1 4 ↦→ i 2maps G biholomorphically to the upper half disc D + , the intersection of theunit disc with the upper half plane. As next we want to get rid of one of thetwo corners of D + : We shift D + one length unit to the right hand side, applyinversion and shift half a length unit to the left hand side, i.e. consider themapg : D + −→ R >0 + iR

In the third step we map the fourth quadrant to the lower half planeusing the square functionFinally we arrive withR >0 + iR

In order to have more automorphisms at our disposal we enlarge thecomplex plane C, adding a point at infinity, which finally turns out to be asgood as any other point!Definition 12.10. The extended planeĈ := C ∪ {∞}is the complex plane with a point at infinity, denoted ∞, added. A subsetU ⊂ Ĉ is called open, if either U ⊂ C is open or otherwise K := Ĉ \ U iscompact (closed and bounded).Remark 12.11. There are two ways to investigate the properties of Ĉ. Thefirst one appeals to intuition, i.e. three dimensional geometry, the secondone is very effective in avoiding cumbersome and lengthy calculations: Ituses linear algebra for the complex vector space C 2 in order to understandthe ”symmetries” of the extended plane, where the point ∞ ceases to play adistinguished rôle.1. The Riemann sphere is nothing but the two dimensional unit sphereS 2 := { (w, t) ∈ C × R; |w| 2 + t 2 = 1 } .The mapS 2 \ {(0, 1)} −→ C, (w, t) ↦→w1 − tis called stereographic projection. Indeed, it associates to (w, t) ∈S 2 \ {(0, 1)} the point z ∈ C, such that (z, 0) is the intersection pointof the line through the north pole (0, 1) ∈ S 2 and (w, t) with the planeC × 0. Its inverse isz ↦→ 1|z| 2 + 1 · (2z, |z|2 − 1).So we can view Ĉ as the Riemann sphere with ∞ ∈ Ĉ correspondingto the north pole (0, 1) ∈ S 2 .2. The complex projective line P 1 (C): The points in Ĉ are in one-toonecorrespondence to the one dimensional (complex) vector subspacesof C 2 : Except C × 0 they are all of the form C(z, 1) with some z ∈ C.Hencez ↦→ L z := C(z, 1), ∞ ↦→ L ∞ := C × 083

defines a bijectionĈ −→ P 1 (C)between the extended plane Ĉ and the ”complex projective line”P 1 (C) := {L ⊂ C 2 ; L a one dimensional complex vector subspace}.Now let us define holomorphic maps of the extended plane to itself:Definition 12.12. Denote ι : Ĉ −→ Ĉ, z ↦→ 1 , 0 ↦→ ∞, ∞ ↦→ 0, the extendedzinversion.1. A map f : G −→ Ĉ defined on a domain G ⊂ C is called holomorphicif either f ∈ M(G) or f ≡ ∞.2. A map f : Ĉ −→ Ĉ is called holomorphic if both f| C and (f ◦ ι)| C areholomorphic.Proposition 12.13. The holomorphic maps f : Ĉ −→ Ĉ, f ≢ ∞, correspondto rational functions.Proof. Obviously rational functions define holomorphic maps from the extendedplane to itself. On the other hand, if f : Ĉ −→ Ĉ, f ≢ ∞, 0,is holomorphic, the meromorphic function f| C ∈ M(C) has only finitelymany zeros and poles, due to the fact that 0 is an isolated singularity off ◦ ι. Denote them a 1 , ..., a r and n 1 , ..., n r ∈ Z their multiplicities, Thenf(z) = ∏ rν=1 (z − a ν) nν · g(z) with an entire function g ∈ O(C) without zeros.Furthermore g ◦ ι ∈ M(C), s.th. g defines a holomorphic map Ĉ −→ Ĉ.If g(∞) ≠ ∞, the function g| C is bounded, hence constant according toLiouville, if g(∞) = ∞ consider 1 instead.gAnd here are some automorphisms of the extended plane:( ) a bDefinition 12.14. Let A =be an invertible matrix: det A =c dad − bc ≠ 0. Then the holomorphic mapµ A : Ĉ −→ Ĉwithµ A (z) = az + bcz + dis called a Möbius transformation.84

Remark 12.15. 1. We have µ B = µ A iff B = λA for some λ ∈ C ∗ .2. We have µ AB = µ A ◦ µ B . In particular, with B = A −1 , one obtains:Möbius transformations are automorphisms of the extended plane Ĉ.3. In terms of the projective line the Möbius transformation µ A has asimple description:µ A : P 1 (C) −→ P 1 (C), L ↦→ A(L).Indeed, ifwe have( ) a bA =c d( ) az + bA(C(z, 1)) = C(az + b, cz + d) = Ccz + d , 1 .4. Given two triples z 1 , z 2 , z 3 and w 1 , w 2 , w 3 of pairwise distinct points inĈ there is a unique Möbius transformation µ : Ĉ −→ Ĉ with µ(z i) = w ifor i = 1, 2, 3. In order to see that, we consider P 1 (C) instead of Ĉ.Existence: Given pairwise different one dimensional subspaces L i =Cu i and ˜L i = Cv i we have to find a matrix A with A(L i ) = ˜L i fori = 1, 2, 3. Since u 1 , u 2 resp. v 1 , v 2 are bases of C 2 and we may replaceu 1 , u 2 resp. v 1 , v 2 with nonzero scalar multiples, we may assume u 3 =u 1 + u 2 as well es v 3 = v 1 + v 2 . Now choose A with Au i = v i , i = 1, 2.Uniqueness: If both A, B satisfy A(L i ) = ˜L i = B(L i ), the matrixC = B −1 A has u 1 , u 2 as eigenvectors. The corresponding eigenvaluesagree – otherwise any eigenvector of C is either a multiple of u 1 or u 2– and thus C = λE.Indeed every automorphism of Ĉ is a Möbius transformation:Theorem 12.16. The automorphisms of the extended plane are the Möbiustransformations:Aut(Ĉ) = {µ A; A ∈ C 2,2 , det A ≠ 0}.85

Proof. Let f : Ĉ −→ Ĉ be an automorphism. If f(∞) = ∞, the restrictionf| C is an automorphism of the complex plane: We have f(z) = az + b andthus f = µ A withIf f(∞) = c ∈ C andsuch thatA :=B :=( a b0 1( 0 11 −c).),µ B (z) = 1z − c ,we may apply the above argument to µ B ◦ f and obtain f = µ B −1 A.Sometimes problems in C become easier to handle within Ĉ ⊃ C, sincethere we may always assume that some point of interest is the point ∞, theaction of the Möbius transformations being transitive. Here is one such case:Remark 12.17. The construction of barrier functions: Let us use thegeometry of Ĉ in order to construct barrier functions at boundary pointsa ∈ ∂G satisfying the condition of Th.11.13. We may assume a = ∞, sincewe may replace G with µ(G) with the Möbius transformation µ(z) = 1 . z−aHence, given r > 0, we have to look for a function β ∈ S(G) satisfying1. β ≤ 0,2. lim z→∞ β(z) = 0,3. z ∈ G, |z| ≤ r =⇒ β(z) ≤ −1.Now let K ⊂ Ĉ\G be a connected compact set containing ∞ and more points.Its complement is the disjoint union of simply connected domains, called theconnected components of Ĉ\K, and G, being connected, is contained in one ofthem. So there is in particular a branch of the logarithm log ∈ O(G). Beinginjective, it defines a biholomorphic map log : G −→ log(G). Replacing Gwith log(G), our problem takes the following form: Given s ∈ R, we shouldhunt for a function β ∈ S(G) satisfying1. β ≤ 0,86

2. lim Re(z)→∞ β(z) = 0,3. z ∈ G, Re(z) ≤ s =⇒ β(z) ≤ −1.Since exp | G is injective, the line s+iR through s ∈ R parallel to the imaginaryaxis intersects G in at most countably many intervals(s + iR) ∩ G =∞⋃(s + ia ν , s + ib ν ),ν∈Iwhose lengths add up to at most 2π, i.e.∑(b ν − a ν ) ≤ 2π,ν∈Iassuming a ν < b ν . For z ∈ G, Re(z) ≤ s, we setwhile for z ∈ G, Re(z) > s, we defineβ(z) ≡ −1,β(z) := ∑ ν∈Iβ ν (z),whereβ ν (z) := 1 π ∠(s + ia ν − z, s + ib ν − z) = Im(LogThen we have−1 < β(z) < 0( )s + ibν − z).s + ia ν − zand thuslim β ν(z) = −1, ∀ w ∈ (s + ia ν , s + ib ν ),z→wimplies thatlim β(z) = −1, ∀ w ∈ G, Re(w) = s,z→was well. On the other hand, some elementary geometry gives( ) πβ(x + iy) ≥ −2 arctan , ∀z = x + iy ∈ G, x > s,x − sso condition 2) is also satisfied.87

The geometry of Möbius transformations is based on the fact that theypreserve the following class of subsets of the extended plane Ĉ:Definition 12.18. A generalized circle in Ĉ is either1. a usual circle ∂D r (z 0 ) in C ⊂ Ĉ, or2. a set L ∪ {∞} with a line L ⊂ C.Remark 12.19. We comment on generalized circles both from the point ofview of Ĉ as the complex projective line and as the Riemann sphere.1. Generalized circles consist of the points in P 1 (C) corresponding to theone dimensional subspaces L ⊂ C 2 , which are contained in the zeroconeNC(σ) := {u ∈ C 2 ; σ(u, u) = 0}of a nondegenerate indefinite hermitian form σ : C 2 ×C 2 −→ C. Indeed,such a form looks as followswith a matrix H ∈ C 2,2 satisfyingσ(u, v) = u T HvH = H T , det H < 0.That means( a cH =c b), a, b ∈ R, c ∈ C, ab − |c| 2 < 0.Thus the equation σ(u, u) = 0 takes for u = (z, 1) the formazz + cz + cz + b = 0.For a = 0 we have c ≠ 0, and z ↦→ cz + cz with c ≠ 0 is the generalexpression for a surjective linear map C −→ R. Hence the equationcz +cz = −b with b ∈ R describes a line in C, and every line is obtainedin that way.If a ≠ 0, we may even assume a = 1 and rewrite our equationr 2 := |c| 2 − b = zz + cz + cz + cc = |z − (−c)| 2 ,88

describing a circle of radius r > 0 with center −c. Obviously any circlecan be obtained in that way.The above explicit discussion shows in particular that the hermitianform σ = σ (a,b,c) is determined by its zero cone NC(σ) ⊂ C 2 up to anonzero real constant.2. Generalized circles correspond via the stereographic projection to ordinarycircles in the Riemann sphere. Such a circle is the intersectionof S 2 with a plane having a distance < 1 to the origin. It consists ofthe points (w, t) ∈ S 2 satisfying an equation:cw + cw + γt = d, where c ∈ C, γ, d ∈ R, d 2 < 4|c| 2 + γ 2 .The left hand side of the equation is the general expression for anR-linear map C × R −→ R, while the inequality for the coefficientsguarantees, that the plane has distance < 1 to the origin. Now we substitutethe expression of Rem. 12.11 for the inverse of the stereographicprojection and obtain:where |2c| 2 > (d + γ)(d − γ).As a consequence we obtain:(d + γ)|z| 2 − 2cz − 2cz + (d − γ) = 0,Proposition 12.20. Möbius transformations map generalized circles to generalizedcircles, and given two such circles there is a Möbius transformationmapping the first one to the second one.Proof. If a generalized circle arises from the hermitian form σ, its image withrespect to µ A belongs to the form ˜σ(u, v) = σ(A −1 u, A −1 v).On the other hand given two forms σ, ˜σ choose a σ-ON-basis u 1 , u 2 and a˜σ-ON-basis ũ 1 , ũ 2 (the first vectors having square length 1, the second ones−1). Then A ∈ C 2 with Au i = ũ i satisfies ˜σ(u, v) = σ(A −1 u, A −1 v) and thusA(NC(σ)) = NC(˜σ).Remark 12.21. Reflections at generalized circles: Given a generalizedcircle with corresponding hermitian form σ we can define a reflection (not aMöbius transformation!) at that circle: Replacing Ĉ with P 1(C) it is definedas followsP 1 (C) ∋ L ↦→ L ⊥ .89

Here L ⊥ ⊂ C 2 denotes the σ-orthogonal complement of L. In particularL ⊥ = L for L ∈ NC(σ). Indeed for the corresponding reflection ϱ : Ĉ −→ Ĉon Ĉ, we haveϱ(z) = µ A (z)with a suitable Möbius transformation µ A . (We remark, that C 2 −→ C 2 , u ↦→Au, is not a reflection of C 2 on some real hyperplane, instead it has twoeigenvalues 1 and two eigenvalues -1.)In order to see that, let us first consider the generalized circle R ∪ {∞}.We may takeThenand thusH =( 0 −ii 0).(C(z 1 , z 2 )) ⊥ = C(z 1 , z 2 )ϱ(z) = z.In the general case take a Möbius transformation µ B mapping R ∪ {∞} tothe given generalized circle. Thenϱ(z) = µ BB−1(z).We leave it to the reader to show the following: If Z ⊂circle, then the reflection ϱ Z : Ĉ −→ ĈĈ is a generalized1. is the ordinary reflection at the line Z \ {∞}, if ∞ ∈ Z.2. is the mapif Z = D r (z 0 ).z ↦→ r 2 z − z 0|z − z 0 | 2 + z 0, z 0 ↦→ ∞, ∞ ↦→ z 0 ,As a byproduct of the above discussion we obtain that the zero cones NC(σ) ⊂C 2 are the setsS 1 · P = {λu; λ ∈ S 1 , u ∈ P },where S 1 denotes the unit circle and P ⊂ C 2 is a totally real plane, i.e. atwo dimensional real subspace, which is not a complex subspace, indeed theeigenspace of BB −1 for the eigenvalue 1.90

Dynamics of Möbius transformations: A Möbius transformation µ A ≠idĈ has either one or two fixed points, corresponding to diagonalizable resp.nondiagonalizable A ∈ C 2,2 . In the first case µ A : Ĉ −→ Ĉ is similar to amapµ(z) = azwith a ∈ C \ {0, 1}, while in the second case it is similar to a translationµ(z) = z + awith a ∈ C ∗ . In particular Ĉ is the union of the fixed points of µ A and afamily of mutually disjoint µ A -invariant curves: These curves are1. generalized circles if A is diagonalizable and both eigenvalues have thesame absolute value. For µ(z) = az our family is∂D r (0), r > 0.2. generalized circles with one point, the fixed point of µ A , removed, if Ais nondiagonalizable. For µ(z) = z + a that family is just the familyRa + tb, t ∈ Rof lines parallel to Ra. Here b ∈ C \ Ra denotes any complex numbernot on Ra.3. open segments of a generalized circle with the two fixed points of µ Aas boundary points, if A is diagonalizable and the eigenvalues of A arelinearly dependent over R. For µ(z) = az with a ∈ R we obtain thefamilyR >0 e iϑ , 0 ≤ ϑ < 2π.4. ”spirals” from one fixed point to the other one, if A is diagonalizableand the eigenvalues of A are linearly independent over R and havedifferent absolute value. In the case µ(z) = az with a = se iϕ they lookas follows (depending on the choise ϑ)S r := { z = rs t exp(itϕ); t ∈ R } , r ∈ R >0 .Here are the automorphisms of the unit circle:91

Theorem 12.22. For D := D 1 (0) we havewhereAut(D) = {µ A | D ; A T HA = H},H =( 1 00 −1More classically: The restrictions of the Möbius transformations).µ(z) = e iϑ z + z 01 + z 0 z , |z 0| < 1,are the automorphisms of the unit disc.We remark that the equation A T HA = H is equivalent to the fact that thecolumn vectors of A = (u 1 , u 2 ) constitute a σ-ON-basis for σ(u, v) := u T Hv,such that σ(u 1 , u 1 ) = 1, σ(u 2 , u 2 ) = −1, are their respective ”square lengths”.Proof. A Möbius transformation µ A maps D into itself iff A : C 2 −→ C 2satisfies σ(Au, Av) = λσ(u, v) with some λ ∈ R >0 for σ(u, v) := u T Hv.Since A is only determined up to a scalar multiple, we may assume thatλ = 1, i.e. A is a σ-isometry, equivalently A T HA = H. On the other handif we don’t want A to be a σ-isometry, we know nevertheless that the secondcolumn vector should have negative square length, so its second componentdoes not vanish, and we may even assume that it equals 1. Denoting its firstcomponent a, its square length is |a| 2 − 1 < 0, i.e. |a| < 1. The first columnvector should be σ-orthogonal to the second one and have up to sign thesame square length. So we arrive at( ) eiϑaA =e iϑ ;a 1finally take z 0 := e −iϑ a.It remains to show that every automorphism f : D −→ D is the restrictionof a Möbius transformation. Withwe obtain an automorphism(B :=1 −f(0)−f(0) 1g := µ B ◦ f92)

fixing the origin: g(0) = 0. Schwarz’ lemma, i.e. lemma 12.23, may beapplied to both g and g −1 and gives that g(z) = e iϑ z is a rotation, hencef = µ A with A = B −1 ( eiϑ00 1).Lemma 12.23 (Schwarz’ lemma). Let D = D 1 (0) be the open unit disc,and g ∈ O(D) a function with g(0) = 0, |g(z)| ≤ 1 for all z ∈ D. Then1. either g is a rotation: g(z) = e iϑ z2. or |g(z)| < |z| for all z ∈ D \ {0}.Proof. We consider the function{ g(z)/z , if z ≠ 0h(z) :=g ′ (0) , if z = 0 .It is holomorphic on D and for r < 1 the maximum principle gives||h|| Dr(0) = ||h|| ∂Dr(0) ≤ 1/r,since |g(z)| ≤ 1 for all z ∈ D. Taking the limit r → 1 we find |h(z)| ≤ 1for all z ∈ D. Now the maximum principle gives that either h(z) ≡ e iϑ - sog(z) = e iϑ is a rotation - or |h(z)| < 1 for all z ∈ D resp. |g(z)| < |z| forz ∈ D ∗Finally we want to discuss the following question: Given two domainsG, G ′ , does there exist a biholomorphic map ϕ : G −→ G ′ ?Here is a necessary criterion:Proposition 12.24. Assume f : G −→ G ′ is a diffeomorphism, i.e. bijectiveand C 1 -differentiable in both directions. Then G is simply connected iff G ′is.Proof. The statement follows immediately from Th.6.23, but we want to givehere a proof based on our definition of simple connectedness: We show thatf induces a bijection f ∗ : D(G ′ ) −→ D(G) preserving both total differentials93

and locally integrable forms: First of all we define for any C 1 -map f : G −→G ′ a pull back map on the level of functions:and on the level of differential forms:C 1 (G ′ ) −→ C 1 (G), g ↦→ g ◦ ff ∗ : D(G ′ ) −→ D(G), ω = gdx + hdy ↦→ f ∗ (ω) := (g ◦ f)du + (h ◦ f)dv,where f = u + iv. Then the chain rule impliesd(F ◦ f) = f ∗ (dF ).In particular, for a diffeomorphism f we may apply that to both f and f −1 :So differential forms with primitive function on G ′ correspond under f ∗ todifferential forms with primitive function on G. This holds as well for locallyintegrable differential forms: Indeed, a differential form on a domain G islocally integrable iff every point z ∈ G admits an open neighbourhood U,such that ω| U has a primitive function – this finally justifies the adjective”locally integrable”. The implication ”=⇒” is clear. For ”⇐=” we have toshow, according to Prop.6.14, that ∫ ω = 0 for any coordinate rectangle∂RR ⊂ G, if ω admits locally a primitive function: Take a sufficiently finesubdivision of R into n 2 congruent rectangles R ij , of size 1 the size of R. Ifnn is sufficiently big, every R ij is contained in a neighbourhood U ij , where ωhas a primitive function. Consequently∫ω = ∑ ∫ω = ∑ 0 = 0.∂R i,j ∂R ij i,jOn the other hand:Remark 12.25. There is no biholomorphic map f : C −→ D := D 1 (0)between the complex plane C and the open unit disc D. Namely, f ∈ O(C)would be a bounded holomorphic function, but such functions are constant,according to Liouville. We leave it to the reader to determine a diffeomorphismf : C −→ DBut this is the only disappointment:94

Theorem 12.26 (Riemann mapping theorem). Let G C be a simplyconnected domain. Then there is a biholomorphic map f : G −→ D = D 1 (0)from G onto the open unit disc D.Proof. Our proof is based on the following:Lemma 12.27. Given a holomorphic function f ∈ O(G) on a simply connecteddomain G ⊂ C without zeros there is a square root √ f ∈ O(G).Proof. The differential form df is closed, and hence, G being simply connected,has a primitive function F ∈ O(G). Then the function g(z) :=ff(z)e −F (z) satisfies g ′ = f ′ e −F − fF ′ e −F = 0, thus g ≡ c ∈ C ∗ . Since F isonly determined up to an additive constant, we may assume that c = 1. Sof = e F and( )√ F f := exp2is a square root of f.Construction of a biholomorphic map h : G −→ h(G) with 0 ∈ h(G) ⊂D: Take a point a ∉ G and apply in a first step the translationThen take a square rootG −→ G 1 := G − a, z ↦→ z − a.f : G 1 −→ G 2 := f(G 1 )of id G1 , i.e. f(z) 2 = z for all z ∈ G 1 . Since z ↦→ f(z) 2 = z is injective, for anopen disc D r (b) ⊂ G 2 we have D r (−b) ∩ G 2 = ∅. Theng : G 2 −→ G 3 := g(G 2 ), z ↦→rz + bmaps G 2 biholomorphically onto G 3 ⊂ D. Finally pick z 0 ∈ G 3 and composewithµ : G 3 −→ G 4 := µ(G 3 ), z ↦→ z − z 01 − z 0 z .Now given a domain G D we try to enlarge G by applying dilatations:95

Definition 12.28. Let G ⊂ C be a domain, 0 ∈ G ⊂ D. A dilatation of Gis a biholomorphic map κ : G −→ κ(G) ⊂ D, such thatWe shall prove:κ(0) = 0, |κ(z)| > |z|, ∀z ∈ G ∗ := G \ {0}.Proposition 12.29. For a simply connected domain G ⊂ C with 0 ∈ G ⊂D := D 1 (0) the following statements are equivalent:1. G = D.2. There is no dilatation κ : G −→ κ(G) ⊂ D.Proof. ”=⇒”: According to Schwarz’ lemma, i.e. Lemma 12.23, a holomorphicmap f : D −→ D with f(0) = 0 is either a rotation or satisfies|f(z)| < |z| for z ∈ D ∗ .”⇐=”: We subdivide our argument into four steps:1. For G D we construct a dilatation κ : G −→ D as a partial rightinverse of a contraction g : D −→ D, i.e. a holomorphic map satisfyingg(0) = 0 and |g(z)| < |z| for z ∈ D ∗ . More precisely, we wantAs a consequenceg ◦ κ = id G and κ(0) = 0.|z| = |g(κ(z))| < |κ(z)|for z ∈ G ∗ , i.e. κ : G −→ D is a dilatation.2. According to Schwarz’ lemma any holomorphic map g : D −→ D, g(0) =0, which is not a rotation, is a contraction.3. For a ∈ D we consider the following automorphism of the unit disc:µ a (z) := z − aaz − 1 .It satisfies µ a (a) = 0, µ a (0) = a, indeed it is involutive: µ a ◦ µ a = id D .With j(z) := z 2 the mapg a := µ a 2 ◦ j ◦ µ a : D −→ Dsatisfies g a (0) = 0, but it is not a rotation, since g a is not injective.Hence it is a contraction.96

4. Now, if a 2 ∈ D \ G, there is a function h ∈ O(G) withh 2 = µ a 2, h(0) = a.Then κ := µ a ◦ h is what we are looking for (with g = g a ).Finally let us give the proof of Riemanns mapping theorem: We mayassume 0 ∈ G ⊂ D and consider the following family of functions:F := {f ∈ O(G); f(0) = 0, f : G −→ f(G) ⊂ D is biholomorphic}.Fix a point a ∈ G ∗ := G \ {0} and setϱ := sup{|f(a)|; f ∈ F}.Obviously ϱ ≤ 1. Take a series of functions (f n ) n∈N ⊂ F with lim n→∞ |f n (a)| =ϱ. According to Th.12.30 we may assume that it converges uniformly tosome function f ∈ O(G). Then f(0) = 0 and f(a) ≠ 0 implies thatf : G −→ f(G) is biholomorphic. Now assume f(G) ≠ D. Then thereis a dilatation κ : f(G) −→ D, and since then κ ◦ f ∈ F as well, we finda contradiction.ϱ ≥ |κ(f(a))| > |f(a)| = ϱ,We have used the following two facts about locally uniform convergenceof holomorphic functions:Theorem 12.30 (Theorem of Montel). Let G ⊂ C be a domain. For afamily (i.e. a subset) F ⊂ O(G) the following statements are equivalent1. Every sequence (f n ) n∈N ⊂ F admits a subsequence (g ν ) ν∈N , i.e. g ν =f nν , converging locally uniformly to some function g ∈ O(G).2. The family F ⊂ O(G) is ”uniformly bounded”, i.e. for every compactset K ⊂ G there is a constant M K ∈ R >0 , such thatHere ||f|| K := sup{|f(z)|; z ∈ K}.||f|| K ≤ M K , ∀f ∈ F.97

For the next result recall that a holomorphic function f ∈ O(G) is injectiveif and only if f : G −→ f(G) is biholomorphic.Theorem 12.31. Let G ⊂ C be a domain and (f n ) n∈N ⊂ O(G) be a sequenceconverging locally uniformly to some function f ∈ O(G). If all functions f nare injective, so is f or f ≡ c ∈ C.We conclude this section with a discussion of the relationship betweenthe Dirichlet problem and Riemanns mapping theorem. First of all:Theorem 12.32. Let D = D 1 (0) be the unit disk, and g : D −→ G be abiholomorphic map from the unit disk to a bounded domain G ⊂ C. Thenthe following statements are equivalent1. The map g : D −→ G extends to a continuous map ĝ : D −→ G.2. ∂G = |γ|, i.e. the boundary of G is the trace of a continuous loopγ : I −→ C.Furthermore the extension ĝ is a homeomorphism if we can choose γ as aJordan curve, i.e. with I = [a, b] and γ| [a,b) injective.Example 12.33. The domain G := D \ [0, 1) satisfies the criterion, but( [1 ) ∞⋃[ ) ( ) )1 2πiG = D \2 , 1 ∪2 , 1 expndoes not.n=1Remark 12.34. 1. From Riemanns mapping theorem, Th.12.26, and Th.12.32it follows that the Dirichlet problem is solvable for domains G whoseboundary is the trace of a path: Assume u 0 : ∂G −→ R is continuous.Now there is a solution h : D −→ R of the Dirichlet problem withh| ∂D = u 0 ◦ ĝ. Define then u : G −→ R by u| G = h ◦ g −1 and u| ∂G = u 0 .2. Assume the Dirichlet problem is solvable for the simply connected domainG. Then there is a natural candidate for a biholomorphic mapf : G −→ D. We fix a point a ∈ G and denote u : G −→ R the solutionof the Dirichlet problem withu(z) = ln |z − a|, z ∈ ∂G.98

(Note that the logarithm itself, though harmonic, is not the solutionof the Dirichlet problem, since it has a singularity at a ∈ G). SinceG is simply connected there is a harmonic function v : G −→ R, suchthat h := u + iv is holomorphic. Now set f(z) := (z − a)e −h(z) . If vcan be (continuously) extended to G, then we see that |f(z)| = 1 forz ∈ ∂G. On the other hand: Since there is exactly one zero on G,Rouchés theorem implies, that every value w ∈ D is attained exactlyonce.99