Lectures on Complex Analysis M. Pollicott

Home , Argument principle, Schwarz lemma, Wirtinger derivatives

M. Pollicott

Contents

1. Introduction 5 1.1. A little history 5 1.2. Notation 6 1.3. Some useful reference books 7 2. A few basic ideas 7 2.1. The Riemann sphere 7 2.2. Möbiusmaps 9 2.3. Two applications 12 2.4. Classification of Möbiusmaps and their behaviour 15 2.5. cross-ratios of quadruples of complex numbers 18 2.6. Problems 20 3. Analyticity 21 3.1. Ingredients for the first definition (using power series) 22 3.2. Ingredients for the second definition (using complex differentiability) 22 3.3. Ingredients for the third definition (using the Cauchy-Riemann Equations) 23 4. Definitions of analyticity 24 4.1. The main result 24 5. Integrals and Cauchy’s Theorem 28 5.1. Integration on piecewise continuous curves 28 5.2. parameterization of the curve of integration 28 5.3. Theorems of Gousart and Cauchy 32 5.4. Immediate applications of Cauchy’s Theorem 36 5.5. Converse to Cauchy’s Theorem 38 5.6. Problems 39 6. Properties of analytic functions 40 6.1. Logarithms and roots 40 6.2. Location of zeros: Argument Principle and Rouché’sTheorem 41 6.3. Liouville’s Theorem and the Fundamental Theorem of Algebra 45 6.4. Identity Theorem 46 6.5. Families of functions and Montel’s Theorem 47 6.6. Problems 49 6.7. More problems 50 7. Riemann Mapping Theorem 51 7.1. Riemann Mapping Theorem 51 7.2. Christoffel-Schwarz Theorem 53 7.3. examples 57 7.4. Generalizations of the Schwarz-Christofel theorem 60 7.5. Application: Fluid dynamics 60 7.6. Problems 60 7.7. More problems 62 8. Behaviour of analytic functions 62 8.1. The Maximum Modulus Principle 63 8.2. The Schwarz Lemma 63

3 4 CONTENTS

8.3. Pick’s lemma 64 9. Harmonic functions 65 9.1. Harmonic conjugates 66 9.2. Poisson Integral formula 67 9.3. Mean Value Property and Maximum Principle for harmonic functions 71 9.4. Schwarz reﬂection principle for harmonic functions 72 9.5. Application: Continuous extensions to the boundary of analytic bijections 73 9.6. Problems 74 10. Singularities 75 10.1. Removable singularities 75 10.2. Analogues of Cauchy’s theorem 76 10.3. Laurent’s Theorem 77 10.4. Classiﬁcation of singularities 80 10.5. Behaviour at singularities 80 11. Entire functions, their order and their zeros 82 11.1. Growth of entire functions 82 11.2. Factorization of entire functions 84 12. Prime number theorem 86 12.1. Riemann zeta function 86 12.2. A related complex function 87 12.3. A related counting function 87 12.4. Finiteness of integrals 88 12.5. Proof of the Prime Number Theorem 90 13. Further Topics 91 13.1. Weierstrass’s elliptic function 92 13.2. Picard’s Little theorem 92 13.3. Runge’s theorem 92 13.4. Conformality: Property of analytic maps 92 13.5. Functions of several complex variables 93 1. INTRODUCTION 5

1. Introduction Complex analysis is one of the classical branches in mathematics with roots in the 19th century and just prior. Complex analysis, in particular the theory of conformal mappings, has many physical applications and is also used throughout analytic number theory. In modern times, it has become very popular through a new boost from complex dynamics and the pictures of fractals produced by iterating holomorphic functions. Another important application of complex analysis is in string theory which studies conformal invariants in quantum ﬁeld theory.

1.1. A little history. The study complex numbers arose from try to find solutions to polynomial equations. Al-Khwarizmi (780-850) in his book Algebra had solutions to quadratic equations of various types. Under the caliph Al-Mamun (reigned 813-833) Al-Khwarizmi became a member of the House of Wisdom in Bagdad, The first to solve the polynomial equation x3 + px = q was Scipione del Ferro (1465-1526). On his deathbed, del Ferro confided the formula to his pupil Antonio Maria Fiore, who subsequently challenged another mathematician Nicola “Tartaglia” Fontana (1500-1557) to a mathematical contest on solving cubics. (The name Tartaglia means “stammerer” a symptom of injuries acquired aged 12 dur- ing the french attack on his home town of Bresca). The night before the contest, Tartaglia rediscovered the formula and won the contest. Tartaglia in turn told the formula (but not the proof) to an influential mathematician Gerolamo Cardano (1501-1576), provided he signed an oath to secrecy. However, from a knowledge of the formula, Cardano was able to reconstruct the proof. Later, Cardano learned that del Ferro, not Tartaglia, had originally solved the problem and then, feeling under no further obligation towards Tartaglia, proceeded to publish the result in his Ars√ Magna (1545). Cardano was also the first to introduce complex numbers a + b into algebra, but had misgivings about it. In the Ars magna he observed, for example, that the problem√ of finding two√ numbers that add to 10 and multiply to 40 was satisfied by 5 + −15 and 5 − −15 but regarded the solution as both absurd and useless. Cardan was also said to have correctly predicted the exact date of his own death (but it has also been claimed that he achieved this by committing suicide!). Rene Descartes (1596-1650), the mathematician and philospher coined the term imaginary: “For any equation one can imagine as many roots [as its degree would suggest], but in many cases no quantity exists which corresponds to what one imag- ines.” Abraham de Moivre (1667-1754), a protestant, left France to seek religious refuge in London at eighteen years of age. There he befriended Isaac Newton. In 1698 he mentions that Newton knew, as early as 1676 of an equivalent expression to what is today known as de Moivres theorem (and is probably one of the best known formulae) which states that: (cos(θ) + i sin(θ))n = cos(nθ) + i sin(nθ) where n is an integer. (De Moivre, like Cardan, is famed for predicting the day of his own death. He found that he was sleeping 15 minutes longer each night and summing the arithmetic progression, calculated that he would die on the day that he slept for 24 hours.) √ Leonhard Euler (1707-1783) introduced the notation i = −1 in his book Introductio in analysin innitorum in 1748, and visualized complex numbers as points with rectangular coordinates, but did not give a satisfactory foundation for complex numbers. In contrast, there are indications that Carl Friedrich Gauss (1777-1855). had been in possession of the geometric representation of complex numbers since 6 CONTENTS

1796, but it went unpublished until 1831, when he submitted his ideas to the Royal Society of Gottingen. It was Gauss who introduced the term complex number. Joseph-Louis Lagrange (1736-1813) showed that a function is analytic if it has a power-series expansion However, it is Augustin-Louis Cauchy (1789-1857) who really initiated the modern theory of complex functions in an 1814 memoir submitted to the French Academie des Sciences.

Although the term analytic function was not mentioned in his memoir, the con- cept is present there. The memoir was eventually published in 1825. In particular, contour integrals appear in this memoir (although Poisson had written a 1820 pa- per with a path not on the real line). Cauchy also gave proofs of the Fundamental Theorem of Algebra (1799, 1815) which, as we will see, has analytic proof . In summary, Cauchy, gave the foundation for most of the modern ideas in the field, including: (1) integration along paths and contours (1814); (2) calculus of residues (1826); (3) integration formulae (1831); (4) Power series expansions (1831); and (5) applications to evaluation of denite integrals of real functions The Cauchy- Riemann equations (actually dating back to dAlembert 1752, then Euler 1757, dAlembert 1761, Euler 1775, Lagrange 1781) are also usually attributed to Cauchy 1814-1831 and Riemann 1851. Cauchy resigned from his academic positions in France in 1830 rather than to swear an oath of allegence to the new government. However, he felt able to resume his career in France in 1848, when the oath was finally abolshed. Regarding subsequent work, Karl Weierstrauss (1815-1897) formulated analyticity in terms of existence of a complex derivative, which is the perspective taken in most textbooks, and Georg Riemann (1826-1866) made fundamental use of the notion of conformality (previously studied by Euler and Gauss). Later contribu- tions were made by Poincaréto conformal maps and Teichmüllerto quaai-conformal maps.

1.2. Notation. We denote by C the complex numbers. If z ∈ C then we can write z = x + iy where x, y ∈ R. We denote the real and imaginary parts by x = Re(z) and y = Im(z). We then write the absolute value as |z| = px2 + y2. Occasionally it is useful to write complex numbers in radial coordinates, i.e., z = reiθ where r > 0 and 0 ≤ θ < 2π. We denote by z = x − iy = re−iθ the complex conjugate. If z, w ∈ C then we write [z, w] = {αz + (1 − α)w : 0 ≤ α ≤ 1} for the line segment joining them. 2. A FEW BASIC IDEAS 7

Exercise 1.1. Show that the map φ : C → M2(R) (= 2 × 2 real matrices) x y φ : x + iy 7→ . −y x is a monomorphism. 1.3. Some useful reference books. (1) R. Churchill and J. Brown, Complex Variables and Applications (ISBN 0-07-010905-2). This is a fairly readable account including much of the material in the course. (2) I. Stewart and D. Tall, Complex Analysis (ISBN 0-52-128763-4). This is a popular and accessible book. (3) L. Alhfors, Complex Analysis: an Introduction to the Theory of Analytic Functions of One Complex Variable (ISBN 0-07-000657-1). This is a classic textbook, which contains much more material than included in the course and the treatment is fairly advanced. (4) S. Krantz and R. Greene, Function Theory of One Complex Variable (ISBN 0-82-183962-4). This is a nice textbook, which contains much more material than included in the course. (5) S. Krantz, Complex Analysis: The Geometric Viewpoint (0-88-385035-4). The ﬁrst chapter gives a nice summary of some of the ideas in the course. The rest of the book is very interesting, but too geometric for this course.

2. A few basic ideas

2.1. The Riemann sphere. It is convenient to add an extra point to C. In order to accommodate the extra point ∞, we need to extend the complex plane by adding this point in to get the Riemann sphere.

We denote by Cb = C ∪ {∞} the Riemann sphere. There is a natural “stereographic” projection between the sphere (minus the “north pole” (1, 0, 0)) and the complex plane

3 2 2 2 π : {(x1, x2, x3) ∈ R : x + x2 + x3 = 1} − {(1, 0, 0)} → C deﬁned by x1 x2 π(x1, x2, x3) = z := +i . 1 − x3 1 − x3 In particular, 0 it the image of the south pole (0, 0, −1), and the unit circle |z| = 1 is the image of the equator x3 = 0.

2 Definition 2.1. A circle on the sphere corresponds to the intersection of x1 + 2 2 x2 + x3 = 1 with a plane ax + by + cz = d. 8 CONTENTS

The following is easy to prove. Lemma 2.2. The stereographic projection of a circle on the sphere is either a circle or a line in C. Proof. The image of the intersection under the projection can be written as a(z + z) − ib(z − z) + c(|z|2 − 1) = d(|z|2 + 1) If we write z = x + iy then (d − c)(x2 + y2) − 2ax − 2by + (d + c) = 0. Case I: If c = d then this is the equation of a straight line.

Case II: If c 6= d then 2ax 2by (d + c) x2 + y2 − − + = 0 d − c d − c d − c which we can rearrange as a 2 b 2 a2 + b2 + (c2 − d2) x − + y − = . d − c d − c (d − c)2 It only remains to show that a2 + b2 + c2 − d2 > 0 to see this is the equation of a circle. However,

p 2 2 2 p 2 2 2 |α0| = |ax + by + cz| ≤ x + y + z a + b + c | {z } =1 by the usual Cauchy-Schrwartz inequality and we are done. Note that in the proof there is an equality in the last line only when (a, b, c) = λ(x1, x2, x3) for some λ, i.e., the plane is tangent to the sphere.

Remark 2.3. The inverse images of a points z ∈ C is a triple (x1, x2, x3) lying 2 2 2 2 on the sphere and satisfying |z| = (x1 + x2)/(1 − x3) = (1 + x3)/(1 − x3) (since 2 2 2 x1 + x2 + x3 = 1). We can then write z + z 2 x = since 1 − x = . 1 1 + |z2| 3 |z|2 + 1| Similarly, z − z |z|2 − 1 x = and x = . 2 1 + |z2| 3 1 + |z2| We deduce that z + z z − z |z|2 − 1 (x , x , x ) = , , . 1 2 3 1 + |z2| 1 + |z2| 1 + |z2| This completes the proof. Exercise 2.4. Prove the converse, i.e., the preimage of a circle or a straight line in C is a circle on the Riemann sphere. 2. A FEW BASIC IDEAS 9

Remark 2.5. To associate the appropriate topology to Cb we take the usual open sets in C plus the complements of compact sets union with the point ∞. This means that we can interpret zn → z in the usual sense if z 6= ∞. However, we say that zn → +∞ if for every K > 0 we have there exists N > 0 such that |z| > K. Then C is homeomorphic to the ball minus the ”north pole”. The approrpiate metric for Cb comes form the standard metric on the ball, i.e., q 0 0 0 0 2 0 2 0 2 d((x1, x2, x3), ((x1, x2, x3)) = (x1 − x1) + (x2 − x2) + (x3 − x3) q 0 0 0 = 2(1 − x1x1 + x2x2 + x3x3).

0 0 0 If z := π((x1, x2, x3)) and w := π((x1, x2, x3)) then we can deﬁne a natural metric on Cb ∪ {∞} by

0 0 0 |z − w| d(z, w) = d((x1, x2, x3), ((x , x , x )) = 1 2 3 p(1 + |z|2)(1 + |w|2)

0 0 0 where (x1, x2, x3), (x1, x2, x3) 6= (0, 0, 1) (i.e., z, w 6= ∞) and 2 d(z, ∞) := d((x1, x2, x3), (0, 0, 1)) = . p1 + |z|2 Exercise 2.6. Check all the above formulae.

2.2. Möbiusmaps. We can consider a, b, c, d ∈ C with ad 6= bc,. d az+b Definition 2.7. We define a map f : Cb − {− c } → C by f(z) = cz+d . (If ad = bc then f(Cb) = {∞}, a single point). Remark 2.8. We can also assume without loss of generality that ad − bc = 1, since we see that replacing a, b, c, d to λa, λb, λc, λd gives the same map. We will adopt this convention when convenient. Observe that if c = 0 then f(∞) = ∞, i.e., ∞ is a fixed point. On the other hand, if c 6= 0 then letting z 7→ −d/c we see that f(−d/c) = ∞. Letting z → +∞ we can write that f(∞) = a/c.

Lemma 2.9. If c 6= 0 then the Möbiusmap f is invertible, and its inverse is also a Möbiusmap (and thus f : Cb → Cb is a homeomorphism). az+b −1 dz−b Proof. If f(z) = cz+d with ad − bc = 1 then we can define f (z) = −cz+a . We can then explicitly check f ◦ f −1(z) = z and f −1 ◦ f(z) = z. For example,

dz−b a −cz+a + b f(f −1(z)) = dz−b c −cz+a + d a (dz − b) + b (−cz + a) = c (dz − b) + d (−cz + a) (ad − bc) z = = z (ad − bc)

Lemma 2.10. If f1 and f2 are M¨obius maps then so is the composition f1 ◦ f2

Proof. This follows immediately by substitution. (Exercise) In particular, the set of M¨obiusmaps forms a group. 10 CONTENTS

Example 2.11. There are four fundamental examples of M¨obiustransformations f : Cb → Cb. 1.z+b a) Translations: z 7→ z + b = 0.z+1 where b ∈ C. These are often called parabolic transformations. √ a.z+0 b) Rotations: z 7→ az = √ where |a| = 1. These are often called 0.z+ a−1 elliptic transformations. √ λ.z+0 c) Expansions (and Contractions): z 7→ λz = √ −1 with λ > 1 (or 0.z+ λ 0 < λ < 1) and λ real. These are often called hyperbolic transformations. 1 d) Inversions: z 7→ z . We can use this to deduce the following.

Lemma 2.12. Every M¨obiusmap can be written as a composition of M¨obius maps of the above type.

Proof. Every Mobius map is a combination of three types of maps f(z) = Az, f(z) = z + B and f(z) = 1/z, where A, B ∈ C, since we can write az + b a (cz + d) + b − ad a b − ad = c c = + c cz + d cz + d c cz + d which is a composition of there 1 ad 1 ad 1 a z 7→ cz 7→ cz + d 7→ 7→ b − 7→ b − + cz + d c cz + d c cz + d c The follow result is very useful. Consider circles in the complex plane of the form C := {z ∈ C : |z − z0| = r} where z0 ∈ C and r > 0. Theorem 2.13. The image of a circle or straight line under a M¨obiustransformation is a circle or straight line.

Proof. It suffices to consider three different types of Möbiustransfomations: (1) If f(z) = Az where A ∈ C with A 6= 0 then the image f(C) is a circle. (2) If f(z) = z + B where B ∈ C then the image f(C) is a circle. (3) If f(z) = 1/z then the image f(C) is a circle. We first observe if z = x+iy lies on a circle centred at z0 = x0 + iy0 of radius r > 0 then 2 2 2 (x − x0) + (y − y0) = r (1) Consider the set of z such that |z − p| = k (2) |z − q| where p, q ∈ C and k > 0. If p = u + iv and q = s + it then this becomes: (x − u)2 + (y − v)2 = k2 (x − s)2 + (y − t)2 (3) In particular, we can rewrite (1) in terms of (3) by choosing p, q, k such that (u − k2s) x = 0 1 − k2 v − k2t y = 0 1 − k2 u2 + v2 − k2(s2 + t2) u − sk2 v − tk2 r2 = − + + > 0. 1 − k2 1 − k2 1 − k2 2. A FEW BASIC IDEAS 11

Finally we can see that if z satisﬁes (2) then

1 − 1 |z − p| z p q = k ⇐⇒ = k (4) |z − q| 1 − 1 p z q 1 1 i.e., a version of (3) (with p replaced by p , q replaced by q and k replaced q by p k). In particlar, the image of the circle given by (2) is a circle given by (3).

Theorem 2.14. For distinct z1, z2, z3 ∈ Cb and distinct w1, w2, w3 ∈ Cb there exists a unique M¨obiusmap f : Cb → Cb such that f(zi) = wi, for i = 1, 2, 3. Proof. We ﬁrst prove existence and then uniqueness.

Existence. Consider the case that z1, z2, z3 6= ∞. Let (z − z )(z − z ) S(z) = 2 1 3 (z − z3)(z1 − z2) then we see that S(z1) = 1, S(z2) = 0, S(z3) = ∞. Consider (z − w )(z − w ) T (z) = 2 1 3 (z − w3)(z1 − w2) −1 then we see that T (w1) = 1, T (w2) = 0, T (w3) = ∞. If we deﬁne f(z) := S ◦T (z) −1 then we can then observe that f(zi) := T ◦ S(zi) = wi (i = 1, 2, 3). z−z2 In the case that z1 = ∞ then we let S(z) = and similarly for T . z−z3 z1−z3 In the case that z2 = ∞ then we let S(z) = and similarly for T . z−z3 z−z2 In the case that z3 = ∞ then we let S(z) = and similarly for T . z1−z2

Uniqueness. We can assume without loss of generality that w1 = 1, w2 = 0, w3 = ∞. (Otherwise we can additionally compose with a Möbiusmap g; Cb → Cb taking w1, w2, w3 to 1, 0, ∞, respectively, and then we can apply the following argument −1 −1 to show that f1 ◦ g = f2 ◦ g , which therefore gives us f1 = f2). Assume that fj : Cb → Cb are two Möbiusmaps j = 1, 2 such that fj(1) = z1, fj(0) = z2, −1 fj(∞) = z3. Since S(z) = f1 ◦ f2(z) is a Mobius transformation we can write az + b S(z) = cz + d −1 Moreover, f1 ◦ f2 : Cb → Cb fixes the three points 1, 0, ∞. In particular, we see that S(0) = b/d = 0 =⇒ b = 0 S(∞) = a/c = ∞ =⇒ c = 0, and S(1) = (a + b)/(c + d) = a/d = 1 from which we deduce that S(z) = z for all z, i.e., f1 = f2. Let us consider a could of examples of this result. Example 2.15. Find the Möbiustransformation f which maps −1, 0, 1 to the points −i, 1, i. az+b b az+b Assume that f(z) = cz+d . Since f(0) = d = 1 we have b = d and f(z) = cz+b . −ia+b ia+b Similarly, since f(−1) = −ic+b = 1 =⇒ ic − ib = −a + b and f(1) = ic+b = i =⇒ ic + ib = a + b. Adding the last two equations gives c = −ib and subtracting gives a = ib. Thus ibz + b iz + 1 i − z f(z) = = = . −ibz + b −iz + 1 i + z (Formally, we should also multiply the coefficients by constant to get ad − bc = 1) 12 CONTENTS

Example 2.16. Let D = {z ∈ C : |z| < 1} be the open unit disk unit and let H = {z ∈ C : Im(z) > 0} be the upper half-plane. Find a surjective M¨obiusmap f : H → D. Actually, the map in the previous example works (as one might guess from the three points in the boundary of H being mapped to three points in the boundary of D). To see this, let z = x + iy with y > 0 then

2 2 2 2 i − (x + iy) −x + i(1 − y)) x + (1 − y) |f(z)| = = = < 1, i + (x + iy) x + i(y + 1) x2 + (1 + y)2 i.e., f(z) ∈ D.

Example 2.17. Fix a ∈ C and then we can deﬁne a M¨obiusmap f :b C → C by

a − z f(z) = 1 − az

Notice that

|a − z|2 |a|2 − 2Re(az) + |z|2 |f(z)|2 = = |a − az|2 1 − 2Re(az) + |az|2

It is easy to check that if |z| = 1 if and only if |f(z)| = 1. Moreover, if |f(z)| < 1 is equivalent to

|a|2 + |z|2 < 1 + |az|2 < 1 ⇐⇒ (1 − |a|2)(1 − |z|2) > 0 ⇐⇒ |z| < 1 since |a| < 1. We can conclude that f : D → D.

2.3. Two applications.

Application 2.18 (Apollonion Circle Packings). This is related to taking three circles in the plane.

Theorem 2.19 (Apollonius). Given three mutually tangent circles C1, C2, C3 there are precisely two circles C, C0 tangent to all three.

Proof. We can write this in a number of steps.

Step 1: Let ξ be the point of intersection C1 ∩ C2 of two of the circles. We can 1 apply the M¨obiustransformation f(z) = z−ξ then f(ξ) = ∞ and C1 − {ξ} and C2 − {ξ} are mapped to disjoint (and thus parallel) straight lines L1 and L2. Since C1 and C3 are tangent at one point η, say, and C2 and C3 are tangent at one point ρ, say, then the image C4 = f(C3) is another circle tangent to L1 and L2. 2. A FEW BASIC IDEAS 13

Step 2: We can trivially translate C4 between the two parallel lines L1, L2 to precisely two more circles C5 and C6 of the same radius such that: C5 and C6 are both tangent to both L1 and L2 and C3.

−1 0 −1 Step 3: We then define C = f (C5) and C = f (C6). Since again Möbius transformations take circles to circles (where straight lines are understood as circles passing through infinity) we deduce that C,C0 are the only two such circles tangent to C1, C2, C3.

Theorem 2.20. Descartes Theorem: Given four mutually tangent circles C1, C2, C3 and C4 whose radii are r1, r2, r3, r4 then F (r1, r2, r3, r4) = 0 where −2 −2 −2 −2 −1 −1 −1 −1 2 F (r1, r2, r3, r4) = 2(r1 + r2 + r3 + r4 ) − (r1 + r2 + r3 + r4 ) (1)

Proof. We begin with two simple estimates. Consider a circle E of radius k. (1) We can reﬂect in it a second circle C of radius r whose centres are separated by d > r + k. In particular, the circles are disjoint and have disjoint interiors. The image circle f(E0) with have radius k2r/(d2 − r2). (2) Consider next a straight line L at a distance b > k from the centre of E. We can reﬂect it in the circle E and the image is a circle of radius k2/2b. 14 CONTENTS

Step 1. Let C1, C2, C3 and C4 be four mutually tangent circles. Choose E to be a (large) circle centred at the tangency point ξ = (x0, y0) between C1 and C2. We can invert the four circles in E by f, say, to arrive at a conﬁguration (after some rotation and translation) with: L1 = f(C1) a line given by y = 1; L2 = f(C2) 0 0 a line given by y = −1; and C3 = f(C3) and C4 = f(C4) being circles of radius 1 with centres (−1, 0) and (1, 0), respectively, i.e., the straight lines are “circles” with radius r1 = r2 = ∞ and the other two circles have radius r3 = r4 = 1. In particular, we observe that it satisﬁes (1).

0 0 Step 2. We can apply the two estimates to the lines L1,L2 and the circles C3, C4 to obtain k2 k2 k2 k2 r(C3) = 2 2 , r(C4) = 2 2 , r(C2) = and r(C1) = x0 − 2x0 + y0 x0 + 2x0 + y0 2(y0 − 1) 2(y0 + 1) The result follows by substituting. Proceeding inductively we can construct circle packings.

az+b Exercise 2.21. If a Möbiusmap f(z) = cz+d maps real numbers to real numbers then show that a, b, c, d are all real numbers. Show that f : H → H is a bijection. Application 2.22 (Hyperbolic Half-plane). The upper half plane H has the Poincarémetric written as (dx2 + dy2)/y2, i.e., it is similar to the usual Euclidean 2. A FEW BASIC IDEAS 15 metric, except at z = x + iy the distance is scaled by 1/y. We claim that this is preserved under Möbiusmaps. We want to show that any path σ has the same length as its image fσ If we let az+b f(z) = cz+d then we can write a c(az + b) ad − bc 1 f 0(z) = − = = cz + d (cz + d)2 (cz + d)2 (cz + d)2 and az + b (az + b)(cz + d) 1 Im(g(z)) = Im = Im = Im(z) cz + d |cz + d|2 |cz + d|2 By the chaine rule Z |(g ◦ σ0(t)| Z |(g0(σ0(t))||σ0(t)| length(gγ) = dt = dt Img ◦ σ(t) Img ◦ σ(t) Z 1 1 = |σ0(t)||cσ(t) + d|2 dt |cσ(t) + d|2 Imσ(t) Z |σ0(t)| = dt = length(γ) Imσ(t)

Alternatively, if z1, z2 ∈ H then can show that

−1 z1 − z2 d(z1, z2) = 2 tanh z1 − z2 and show that this is preserved by M¨obiusmaps, i.e.,

az1 + b az2 + b |g(z1) − g(z2)| = − cz1 + d cz2 + d

(az1 + b)(cz2 + d) − (az2 + b)(cz1 + d) = (cz1 + d)(cz2 + d)

(ad − bc)(z1 − z2) = (cz1 + d)(cz2 + d) Thus

|g(z1) − g(z2)| |z1 − z2| (cz1 + d)(cz2 + d) = |g(z1) − g(z2)| |z1 − z2| (cz1 + d)(cz2 + d) | {z } =1 2.4. Classification of Möbiusmaps and their behaviour. We would like to understand better the behavious different types of Möbiusmaps We begin with a simple result. Lemma 2.23. Mobius maps other than the identity must fix at least one point and at most two.

az+b 2 Proof. Assume that f(z) = cz+d = z then this is equivalent to cz + (d − a)z − b = 0. The solution to this quadratic equation is p z = ((a − d) ± (a − d)2 + 4bc)/2c ∈ C. Unless f is the identity (a = c and b = d = 0) this has two roots (possibly a single 2 root repeated if (a − d) + 4bc = 0). We would like to classify M¨obiusmaps up to conjugacy.

Definition 2.24. We say that two Möbiusmaps f1, f2 are conjugate if f2 = −1 g f1g for some Möbiusmap g. az+b Definition 2.25. We can associate to the Möbiusmap f(z) = cz+d (satisfying ad − bc = 1) the value tr(f) := a + d. 16 CONTENTS

Lemma 2.26. The value tr(f) is preserved by conjugacy, i.e., if f1, f2 are conjugate then tr(f1) = tr(f2) Proof. Since every Möbius map g can be written as a composition of basic Möbiustransformations (translation, inversion and rotation) it suffices to show the result where g is of each type. Translations: if g(z) = z + β then we can explicitly write a(z − β) + b f (z) = g−1f g(z) = + β 2 1 c(z − β) + d a(z − β) + b + β(c(z − β) + d)) = c(z − β) + d z(a + βc) + (b + βd − βa − β2c) = cz + (d − βc) and we see that tr(f2) = (a + βc) + (d − βc) = a + d = tr(f1). Inversions: If g(z) = 1/z then a/z + b f (z) = g−1f g(z) = 2 1 c/z + d a + bz = c + dz and we see that tr(f2) = d + a = tr(f1). Rotations: If g(z) = αz then 1 aαz + b f (z) = g−1f g(z) = 2 1 α cαz + d az + b/α = cαz + d and we see that tr(f2) = a + d = tr(f1). Definition 2.27. We say that a rational map is parabolic it it has precisely one fixed point in Cb In particular, being parabolic is easily seen to be a conjugacy invariant (i.e., −1 if f1 has a single fixed point z0, say, then f2 has a single fixed point g z0. In particular, if f1 has a fixed point f1(z0) = z0 we can choose g so that g(∞) = z0 and then the conjugate map f := f2 fixes ∞. Lemma 2.28. Let f be a Möbiusmap with f(∞) = ∞. (1) Then f is of the form f(z) = αz + β; (2) f has a second distinct fixed point (i.e., it is not parabolic) if and only if α 6= 1; (3) if the second fixed point is 0 (i.e., f(0) = 0) then f(z) = αz.

Proof. (1) Since f(∞) = ∞ this immediately implies that f2(z) = αz + β, for some α, β ∈ C. (2) If f has a second point z0, say, (other than ∞) then f(z0) = αz0 + β = z0, i.e, β z0 = α−1 with α 6= 1. In particular, if α 6= 1 then f has a second ﬁxed point (in addition to ∞). Conversely, if α = 1 then f has no other ﬁxed points except ∞ since f(z) = z + β is a straightforward translation. (In the degenerate case β = 0 this is just the identity.) (3) Since f(0) = 0 we see that β = 0 and the result follows. 2. A FEW BASIC IDEAS 17

Remark 2.29. In fact Möbiustransformations f1, f2 are conjugate if and only 2 2 if tr(f1) = tr(f2) The non-identity Möbius transformations are commonly classified into three types: (1) parabolic (conjugate to z 7→ z + β with a single fixed point ∞); (2) elliptic (conjugate to z 7→ αz with |α| = 1 with points 0, ∞); and (3) loxodromic (everythingelse). The hyperbolic transformations are those conjugate to the expanding and con- tracting maps z 7→ λz (with 0 < λ < 1 or 1 < λ < +∞) being a subclass of the loxodromic ones Lemma 2.30. Let f : Cb → Cb be a non-trivial Möbiustransformation. It is (1) a parabolic Möbiustransformation if and only if tr(f) = ±2; (2) an elliptic Möbiustransformation if and only if −2 < tr(f) < 2; (3) a hyperbolic Möbiustransformation if and only if tr(f) > 2. Proof. Assume that f(∞) = ∞ (else replace by a conjugate map with this property). Case I: f is parabolic . By definition, in this case f has no other fixed points and we see from the lemma see that f(z) = z + β. This always corresponds to tr(g) = 1 + 1 = 2 (or −2 since we recall that multiplying a, b, c, d by −1 gives the same map). Case II: f is not parabolic. In this case, f has a second fixed point. Assume that f(0) = 0 (else we can replace f by a conjugate map with this property)√ and then by the lemma we can write f(z) = αz. Then we require a = d−1 = α (to satisfy ad − bc = ad = 1). 1 We can now consider the different values of the trace tr(f) = a + a . (1) If tr(f) = ±2 then a + 1/a = ±2 and so a(a2 ± 2a + 1) = (a ± 1)2 = 0. But a = ±1 which means that f(z) = z, the identity map, which is excluded by the non-triviality assumption. The proofs of parts (2) and (3) are slightly jumbled up. (a) More generally, if a = reiθ then 1 tr(f) = a + a e−iθ (1) = reiθ + r = (r + 1/r) cos θ + i(r − 1/r) sin θ If we assume that the trace in (1) is real (i.e., Im(tr(f)) = (r − 1/r) sin θ = 0) then this is equivalent to having either r = 1/r(= 1) or θ = 0, π or both. In the first case, if |a| = r = 1 then f is elliptic, and tr(f) = 2 cos θ ∈ (−2, 2). In the second case, if θ = 0 or π then f is hyperbolic and tr(f) = r + 1/r ≥ 2. (b) If we assume that the trace in (1) has a non-zero imaginary part (i.e., tr(f) 6∈ R) then this is equivalent to requiring both r 6= 1/r and θ 6= 0, π. But then since |a| = r 6= 1 this means that the map is not elliptic (as well as not being parabolic). Thus by definition it is loxodromic. Remark 2.31. A loxodromic Mobius map is a composition of a hyperbolic and elliptic map. In addition, a loxodromic Mobius map has that tr(f) ∈ C − [−2, 2]. Exercise 2.32. Show that if f is hyperbolic then f n(z) converges to one of the fixed points. 18 CONTENTS

2.5. cross-ratios of quadruples of complex numbers. We begin with the deﬁnition of a very elegant expression.

Definition 2.33. Given four distinct complex numbers z0, z1, z2, z3 ∈ Cb we deﬁne the cross ratio by

(z0 − z2)(z1 − z3) (z0, z1, z2, z3) = . (z0 − z3)(z1 − z2) Recall that we have already proved a theorem to the effect that can find a az+b (unique) Möbiusmap f(z) = cz+d that maps any three distinct points z1, z2, z3 ∈ Cb, say, to the reference points 1, 0, ∞ in the same order.

Lemma 2.34. Let z0, z1, z2, z3 ∈ Cb be four distinct points. We have that

(z0, z1, z2, z3) = f(z0) i.e., the image of z0 under the unique Möbiusmap which takes z1, z2, z3 to 1, 0, ∞. The following is trivial. Example 2.35. (z, 1, 0, ∞) = z since the associated Möbiusmap is simply f(z) = z. The next lemma shows that the cross ratio is preserved by Möbiusmaps.

Lemma 2.36. If g : Cb → Cb is (another) M¨obiusmap then

(z0, z1, z2, z3) = (gz0, gz1, gz2, gz3). Proof. Exercise. The M¨obiustransformation has an interesting geometric interpretation:

Theorem 2.37. If z1, z2, z3 are three distinct points in Cb then let C denote the unique circle(or line, if one of the point is equal to ∞) passing through them. Then (z0, z1, z2, z3) ∈ R if and only if z0 ∈ C.

Proof. Let f(z) = (z, z1, z2, z3) be the associated Mobius map, then we want to show that f −1(R ∪ {∞}) = C. Suppose w ∈ f −1(R ∪ {∞}) or, equivalently, that f(w) ∈ R ∪ {∞}. Case 1 (f(w) = ∞): In this case since z − z z − z f(z) = 2 1 3 z − z3 z1 − z2 we deduce that this is equivalent w = z3 ∈ C.

Case 2 (f(w) 6= ∞): In this case w 6= z3 then since by assumption f(w) ∈ R we see that f(w) = f(w) and thus aw + b aw + b = ⇐⇒ (ac − ac)|w|2 + (ad − bc)w + (bc − ad)w + (dd − db) = 0. cw + d cw + d This brings us to two subcases. Case 2a (ac = ac): In this case the equation is of the form Aw − Aw + ik = 0 where k is real and A = ad − bc. Rewriting this as iAw − iAw + k = 0 this is the equation of straight line (cf. the usual form Bw + Bw + C = 0). Case 2b (ac 6= ac): If ac = ac then we can write (ad − bc) (bc − ad) dd − db |w|2 + w + w = = 0. ac − ac ac − ac ac − ac 2. A FEW BASIC IDEAS 19

Completing the square we can write 2 2 (ad − bc) 2 (ad − bc) (bc − ad) (ad − bc) w − = |w| + w + w + ac − ac ac − ac ac − ac ac − ac 2 dd − db (ad − bc) 2 = + (= R ). ac − ac ac − ac In particular, this is seen to be the equation of a circle of radius R provided the right hand side is positive. We can rewrite this as dd − db (ad − bc) (da − ca) |a|2|d|2 + |b|2|c|2 − abcd − abcd + = ac − ac ac − ac ca − ca |ac − ac| Thus it suffices to observe that we can write the numerator as (ad − bc)(ad − bc) = |ad − bc|2 > 0 using that ad − bc 6= 0. −1 In either case we see that f (R ∪ {∞}) is a circle or straight line As an immediate corollary we have another presentation of the proof that Möbiustransformation’s preserve circles (although the proof is essentially the same as before). Corollary 2.38. If f is a Möbiustransformation then it maps circles (or lines) to circles (or lines).

Proof. Let C be a circle (or line) determined by three distinct points z1, z2, z3 ∈ C. Then z0 ∈ C ⇐⇒ (z0, z1, z2, z3) ∈ R ⇐⇒ (f(z0), f(z1), f(z2), f(z3)) ∈ R 0 ⇐⇒ (f(z0), f(z1), f(z2), f(z3)) ∈ f(C) =: C 0 where C is a circle (or line). In particular, the circle (or line) for (z1, z2, z3) is mapped to the circle (f(z1), f(z2), f(z3)). Finally, the following is an interesting geometric interpretation of cross ratios. Remark 2.39 (Cross ratios and hyperbolic geometry). We let 3 H = {(z, t): z = x + iy ∈ C, t > 0} denote the three dimensional upper half-space with metric dx2 + dy2 + dt2 ds2 = . t2 We can consider a geodesic γ : R → H3 with end points

z1 = lim γ(t) := γ(−∞) and z2 = lim γ(t) := γ(+∞) t→−∞ t→∞

More precisely, for any two distinct t1 < t2 the curve [t1, t2] 3 t 7→ γ(t) is the 2 3 shortest path between γ(t1) and γ(t2) in H . It appears a semi-circular arc in H . 0 0 Given z3, z4 ∈ C we can consider the geodesic γ with end points z3 = γ (−∞) 0 and z4 = γ (+∞). We denote the Hausdorﬀ distance between the curves γ and γ by 0 0 d(γ, γ ) = inf d(γ(t1), γ (t2)). t1,t2∈R The connection between the cross ratio of the four complex numbers z1, z2, z3, z4 (i.e., the end points) and the geometry is the following.

0 Lemma 2.40 (Fenchel and Ahlfors). |(z1, z2, z3, z4)| = tanh(d(γ, γ )) 20 CONTENTS

The argument of (z1, z2, z3, z4) also has a geometric interpretation. It corresponds to the change in the angle by parallel transporting a frame along the geodesic realizing this distance.

3. Analyticity

We want to recall the deﬁnition(s) of analyticity of functions f : U → C.

Definition 3.1. Assume that U ⊂ C is an open set, i.e., for every z0 ∈ U there exists > 0 such that

B(z0, ) = {z ∈ C : |z − z0| < } ⊂ U. It is usually also convenient to additionally assume that: (1) U is path connected (i.e., for any two points z, w ∈ U we can ﬁnd a continuous path γ : [0, 1] → U such that γ(0) = z and γ(1) = w).

(2) U is simply connected (i.e., any closed path γ : [0, 1] → U with γ(0) = γ(1) can be contracted to a single point or, equivalently, U is homeomorphic to the unit disk).

We will assume that U always has these properties, unless we explicitly state otherwise, and frequently call it a domain. We will present three equivalent deﬁnitions once we have introduced three new lots of notation.

3.1. Ingredients for the first definition (using power series). We want to start by recalling the more intuitive definition of analytic functions. 3. ANALYTICITY 21

∞ Definition 3.2. Let z0 ∈ C and let (an)n=0 be a sequence of complex numbers. P∞ n We say that an infinite series n=0 an(z − z0) has a radius of convergence 0 ≤ R ≤ +∞ (at z0) if 1 1/n = lim sup |an| . (1) R n→+∞ In particular, we have the folowing: Lemma 3.3. Assume that R > 0. For any 0 < r < R the series ∞ X n fz0 (z) := an(z − z0) n=0 converges (uniformly) on B(z0, r) and defines a function. 3.2. Ingredients for the second definition (using complex differentiability). We begin by what it means for a function to be differentiable as a complex function. Definition 3.4. Let U be a domain. We say that f : U → C is complex differentiable at z0 if the limit

0 f(z) − f(z0) f (z0) := lim z→z0 z − z0 0 exists, i.e., there exists f (z0) ∈ C such that for all > 0 there exists δ > 0 such that if |z − z0| < δ then 0 |f(z) − f(z0) − (z − z0)f (z0)| ≤ |z − z0|.

Remark 3.5. The key point is that z can approach z0 in “all directions” as a complex number. It is easy to see the following result.

Lemma 3.6. If f is complex diﬀerentiable at z0 then it is continuous at z0.

Proof. Given > 0 there exists δ > 0 such that whenever |z − z0| < δ then we have that

0 f(z) − f(z0) f (z0) − < . z − z0 In particular, for |z − z0| < δ we can bound 0 0 |f(z) − f(z0)| ≤ (|f (z0)| + )|z − z0| ≤ (|f (z0)| + )δ, which can be made arbitrarily small by choosing δ > 0 appropriately small. 3.3. Ingredients for the third deﬁnition (using the Cauchy-Riemann Equations). We can write a complex function f : U → C in terms of its real and imaginary parts f(x + iy) = u(x, y) + iv(x, y) where z = x + iy ∈ U and u(x, y), v(x, y) ∈ R. 22 CONTENTS

Definition 3.7. We say that f : U → C satisﬁes the Cauchy-Riemann equations if for each z0 = x0 + iy0 ∈ U the partial derivatives

∂u ∂u ∂v ∂v (x , y ), (x , y ), (x , y ) and (x , y ) ∂x 0 0 ∂y 0 0 ∂x 0 0 ∂y 0 0 all exist and satisfy

( ∂u ∂v ∂x (x0, y0) = ∂y (x0, y0) and ∂u ∂v ∂y (x0, y0) = − ∂x (x0, y0) for each z0 = x0 + iy0 ∈ U. Equivalently, we can write f(z) = u(x, y) + iv(x, y) and

∂f ∂u ∂v ∂f ∂u ∂v i = i + i and = + i . ∂x ∂x ∂x ∂y ∂y ∂y | {z } ∂v ∂u i ∂y + ∂y using the Cauchy-Riemann equations. Thus we see that

∂f ∂f i = ∂x ∂y

∂f Remark 3.8. We can write this as ∂z = 0 using the Wirtinger derivatives

∂f 1 ∂f ∂f ∂f 1 ∂f ∂f = + i and = − i . ∂z 2 ∂x ∂y ∂z 2 ∂x ∂y

4. Deﬁnitions of analyticity We are now in a position to give the equivalent deﬁnitions of analyticity.

4.1. The main result. We say that a function f : U → C is analytic if is satisﬁes any (and thus all) of the following three equivalent conditions.

Theorem 4.1. The following are equivalent: ∞ (1) For each z0 ∈ U we can choose > 0 and (an)n=0 such that the function f(z) can be represented as a convergent power series about the point z0, i.e.,

∞ X n f(z) = an(z − z0) n=0

for z ∈ B(z0, ) ⊂ U and with radius of convergence R > > 0; (2) The function f : U → C is complex diﬀerentiable at each z0 ∈ U; ∂u ∂u ∂v ∂v (3) The partial derivatives ∂x , ∂y , ∂x , ∂y all exist, are continuous and satisfy the Cauchy-Riemann equations.

We are not yet in a position to prove the theorem completely (since we will still need to recall the Cauchy theorem for integrals). However, we will proceed with the proofs of the parts that we can prove, and then we will return to the remaining parts once we have established Cauchy’s Theorem. 4. DEFINITIONS OF ANALYTICITY 23

Proof. (1) =⇒ (2). The proof of part (2) assuming part (1) is easy. More precisely, given z0 ∈ U we can write down its expansion ∞ X n f(z) = an(z − z0) n=0 for z ∈ B(z0, r) in a sufficiently small neighbourhood of z0. (i.e., r < R) P∞ n Let R > 0 be the radius of convergence of a series f(z) = n=0 an(z − z0) . We can choose 0 < r < R then for h ∈ C sufficiently small f(z + h) − f(z ) f(z + h) − a 0 0 − a = 0 0 − a h 1 h 1 ∞ ∞ (1) X n−1 X = anh =: gn(h) n=2 | {z } n=2 gn(h) where we denote ( n−1 anh if h 6= 0 gn(h) = 0 if h = 0 for n ≥ 2. In the disc {h ∈ C : |h| < r} we want to show that ∞ X lim gn(h) = 0, h→0 n=2 P∞ (i.e., n=2 gn(h) is continuous at 0) from which we deduce, together with (1), that n−1 f(z) is differentiable at z0. For n ≥ 2, we can bound |gn(h)| ≤ |an|r . Now since r < R we have that ∞ X n−1 |an|r < +∞ n=2 By uniform convergence we can interchange the summation and the limit to get ∞ ∞ X X n−1 lim gn(h) = lim anh = 0. h→0 h→0 n=2 n=2

(2) =⇒ (1). To prove (1) assuming (2) requires the Cauchy theorem, so we will return to this later (after we recall that theorem). 24 CONTENTS

(2) =⇒ (3). If f(z) = u(x, y) + iv(x, y) is a complex diﬀerentiable function of a complex number z = x + iy ∈ C then the derivative at z0 = x0 + iy0 is given by

f(z0 + h) − f(z0) 0 lim = f (z0) h→0 h

If we choose h = t ∈ R to be real then this becomes

f(z0 + t) − f(z0) ∂f 0 lim = (z0)(= f (z0)) t→0 t ∂x ∂u ∂v = (x , y ) + i (x , y ) ∂x 0 0 ∂x 0 0

If we choose h = it where t ∈ R to be purely imaginary then this becomes

f(z0 + it) − f(z0) 1 ∂f 0 lim = (z0)(= f (z0)) t→0 it i ∂x ∂u ∂v = −i (x , y ) + i (x , y ) ∂y 0 0 ∂y 0 0

Complex diﬀerentiability means that there is only one derivative and we can equate the real parts and imaginary parts of these identities and write ∂u ∂v (x , y ) = (x , y ) ∂x 0 0 ∂y 0 0 (from the real parts) and ∂v ∂u (x , y ) = − (x , y ) ∂x 0 0 ∂y 0 0 (from the imaginary parts). This corresponds to the Cauchy Schwartz indentities. We postpone the proof of continuity of the derivatives (until after we recall Cauchy’s theorem). 4. DEFINITIONS OF ANALYTICITY 25

(3) =⇒ (2). We can write f(z) − f(z ) (u(x, y) − u(x , y )) + i(v(x, y) − v(x , y )) 0 = 0 0 0 0 . z − z0 z − z0

For each |x − x0|, |y − y0| suﬃciently small we can write ∂u ∂u |u(x, y) − u(x , y ) − (x − x ) (x, y ) − (y − y ) (x , y )| 0 0 0 ∂x 0 0 ∂y 0 0

≤ |x − x0|1(x0, y0) + |y − y0|1(x, y0), and ∂v ∂v |v(x, y) − v(x , y ) = (x − x ) (x , y ) + (y − y ) (x, y )| 0 0 0 ∂x 0 0 0 ∂y 0

≤ |x − x0|2(x0, y0) + |y − y0|2(x, y0) by using the triangle inequality

|u(x, y) − u(x0, y0)| ≤ |u(x, y) − u(x, y0)| + |u(x, y0) − u(x0, y0)| and

|v(x, y) − v(x0, y0)| ≤ |v(x, y) − v(x, y0)| + |v(x, y0) − v(x0, y0)| and the mean value theorem, where 1(·), 2(·) are bounds on the partial derivatives. Moreover, continuity of the partial derivatives means that 1(·), 2(·) → 0 as z → z0. Since |x−x0|, |y −y0| ≤ |z −z0| then for |z −z0| suﬃciently small we can bound the diﬀerences by ∂u ∂u |u(x, y) − u(x , y ) − (x − x ) (x , y ) − (y − y ) (x, y )| 0 0 0 ∂x 0 0 0 ∂y 0

≤ |z − z0| (1(x0, y0) + 1(x, y0)) , and (1) ∂v ∂v |v(x, y) − v(x , y ) = (x − x ) (x , y ) + (y − y ) (x, y )| 0 0 0 ∂x 0 0 0 ∂y 0

≤ |z − z0| (2(x0, y0) + 2(x, y0)) where continuity of the derivatives means that 1(·), 2(·) → 0 as (x, y) → (x0, y0). By the Cauchy-Riemann equations we have that ∂u ∂u ∂v ∂v (x − x ) (x , y ) + (y − y ) (x, y ) + i (x − x ) (x , y ) + (y − y ) (x, y ) 0 ∂x 0 0 0 ∂y 0 0 ∂x 0 0 0 ∂y 0 ∂u ∂v ∂u ∂v = (x − x ) (x , y ) + i (x , y ) + (y − y ) (x , y ) + i (x , y ) 0 ∂x 0 0 ∂x 0 0 0 ∂y 0 0 ∂y 0 0 | {z } ∂v ∂u =:− ∂x (x0,y0)+i ∂x (x0,y0) ∂u ∂v ∂v ∂u = (x − x ) (x , y ) + i (x , y ) + (y − y ) − (x , y ) + i (x , y ) 0 ∂x 0 0 ∂x 0 0 0 ∂x 0 0 ∂x 0 0 ∂u ∂v ∂v ∂u = ((x − x ) + i(y − y )) (x , y ) + i (x , y ) + i (x , y ) + (x , y ) 0 0 ∂x 0 0 ∂x 0 0 ∂x 0 0 ∂x 0 0 ∂u ∂v = (z − z0) (x0, y0) + i (x0, y0) | {z } ∂x ∂x (x−x0)+i(y−y0) (2) using the Cauchy-Riemann equations. Comparing (1) and (2) we have that f(z) − f(z0) ∂u ∂v |z − z0| = (x0, y0) + i (x0, y0) + (1(·) + i2(·)) z − z0 ∂x ∂x z − z0 ∂u ∂v → (x , y ) + i (x , y )(= f 0(z )) ∂x 0 0 ∂x 0 0 0 as z → z0. 26 CONTENTS

Example 4.2 (Three ways to see f(z) = ez is analytic at z = 0). We let f : C → C be the exponential map f(z) = ez where z = x + iy ∈ C. P∞ n zn (1) We can write f(z) = n=0(−1) n! ; (2) We have a complex derivative f 0(z) = ez (which makes sense as a power series); and (3) We can f(z) = u + iv = ex cos y + iex sin y. In particular, ∂u ∂v ∂u ∂v = ex cos y = and = −ex sin y = − . ∂x ∂y ∂y ∂x The continuity of the partial derivatives in part (3) of the theorem is crucial. Without the assumption that partial derivatives are continuous, then satisfying the Cauchy-Riemann equations is not enough to deduce that f is complex diﬀerentiable.

Example 4.3. If f : C → C is deﬁned by ( 0 if z = 0 f(z) = exp(−1/z4) otherwise Thus all the expressions in the Cauchy-Riemann equations are zero, but one can show that f is not diﬀerentiable.

5. Integrals and Cauchy’s Theorem The most useful theorem in Complex analysis is probably Cauchy’s theorem.

5.1. Integration on piecewise continuous curves. Given a continuous curve γ :[a, b] → C we can take the real and imaginary parts γ(t) = u(t) + iv(t), where u, v :[a, b] → R are real valued.

We say that f(t) is differentiable if u, v :[a, b] → R are differentiable and then we define f 0(t) = u0(t) + iv0(t).

Definition 5.1. If f :[a, b] → C is continuous then we deﬁne Z b Z b Z b f(t)dt = u(t)dt + i v(t)dt. a a a

Moreover, for a piecewise continuous function f(t) on a = a0 < a1 < ··· < an = b we can still deﬁne the integral by n−1 X Z ai+1 fdt. i=0 ai Lemma 5.2. For f, g piecewise continuous on [a, b] and α, β ∈ C we have that: R b R b R b (1) a (αf(t) + βf(t))dt = α a f(t)dt + β a f(t))dt R b R b (2) | a f(t)dt| ≤ a |f(t)|dt 5. INTEGRALS AND CAUCHY’S THEOREM 27

Proof. The first part follows easily from the definitions. R b For the second part, assume that a f(t)dt 6= 0 (otherwise the result is trivial) iθ R b R b R b −iθ and we can write re = a f(t)dt. Thus r = | a f(t)dt| = a e f(t)dt ∈ R. Therefore, Z b ! Z b Z b Z b −iθ −iθ −iθ r = Re e f(t)dt = Re e f(t) dt ≤ e f(t) dt = |f(t)| dt a a a a and thus Z b Z b | f(t)dt| ≤ |f(t)| dt a a 5.2. parameterization of the curve of integration. If φ :[c, d] → [a, b] is differentiable map with continuous positive derivative so that φ(t) is strictly increasing then Z b Z d f(s)ds = f(φ(t))φ0(t)dt a c by the substitution law and the change of variables law with s = φ(t). Definition 5.3. A contour (or curve or arc or path) is a continuous image by z :[a, b] → C of a closed interval [a, b] in C with the orientation provided by the ordering on [a, b]. Thus z(a) is the first point on the contour and z(b) is the last point on the contour. Any continuous map giving the curve and orientation is called a parameterization. The contour is called smooth if there is a parameterization z such that z is differentiable in its interval of definition [a, b] and if z0(t) is continuous in [a, b]. The contour is called piecewise smooth if there exists a parameterization z : [a, b] → C for the contour and a partition a = a0 < a1 < ··· < an = b such that z|[ai, ai+1] is smooth. A contour is simple if can be given by an injective parameterization (i.e., z(a) = z(b) =⇒ a = b) A contour is closed and simple if it can be given by an injective parameterization z :[a, b) → C (i.e., z(a) = z(b)a = b) and or z(a) = z(b).

Remark 5.4. A piecewise smooth curve γ may be given by many parameterizations. If φ :[c, d] → [a, b] is strictly increasing then w;[c, d] → C defined by w(t) = z(φ(t)) is another parameterization. In complex analysis we are usually concerned with piecewise smooth curves and piecewise smooth parameterizations. A piecewise smooth curve γ may be given by many parameterisations, i.e., if φ :[c, d] → [a, b] is strictly increasing then 28 CONTENTS w :[c, d] → C defined by w(t) = z(φ(t)) is also piecewise smooth if z is piecewise smooth and φ is differentiable with continuous and positive derivative. Definition 5.5. We say that two parameterizations z and w are equivalent if there exists such a φ. Exercise 5.6. Show that this is an equivalence relation on parameterizations. Hint. Clearly we can write w ◦ φ−1 = z and φ−1 is continuous with positive derivative. Transitivity follows simply. If z :[a, b] → γ ⊂ C is a piecewise smooth parameterization of γ then we obtain a piecewise parameterization of γ (in reverse) by z− :[−b, −a] → C with z−(t) = z(−t). Let us denote by −γ the reverse of γ.

Definition 5.7. If γ1 is piecewise smooth and γ2 is piecewise smooth and if the last point of γ1 is the same as the first point of γ2 then we define the composition γ1γ2 as follows: If z1 :[a, b] → γ1 ⊂ C is a parameterization of γ1 and z2 :[c, d] → γ2 ⊂ C is a parameterization of γ2 we define z3 :[a, b + d − c] → γ3 ⊂ C by ( z1(t) if t ∈ [a, b] z3(t) = z2(φ(t)) if t ∈ [b, b + d − c] where φ is the translation φ(t) = t + (b − c). We are interested in integration around a piecewise smooth simple contour.

Definition 5.8. Given smooth curves zi :[ai, ai+1] → γi ⊂ C for

a = a0 < a1 < ··· < an = b we define n Z X Z f(z)dz = f(z)dz. γ i=1 γi (Once we define the integral for smooth contours we can then extend it to piecewise smooth contours.) If γ is a smooth contour parameterized by z :[a, b] → γ ⊂ C then we define Z Z b f(z)dz := f(z(t))z0(t)dt γ a R R Exercise 5.9. Show that −γ f(w)dw = − γ f(z)dz

R R −a 0 Proof. We can write −γ f(z)dz = −b f(z(−t))z (−t)dt. Substituting s = −t R b 0 R this becomes: − a f(z(s))z (s)ds = − γ f(z)dz. 5. INTEGRALS AND CAUCHY’S THEOREM 29

Remark 5.10. More generally, integration with respect to arc length along a simple contour γ is rectifiable if it is given by a continuous map z :[a, b] → C which is of bounded variation, i.e., the summations n−1 X |z(ai+1) − z(ai)|, i=0 ranging over all partitions a = a0 < a1 < ··· < an = b, is bounded. The supremum of these numbers is by defintion the length of γ. In particular, if γ is smooth R b 0 represented by z :[a, b] → C then γ is rectifiable and l(γ) = a |z (t)|dt. The following upper bound is useful. Lemma 5.11 (Integration along a straight line). If γ is piecewise smooth and f is continuous on γ and |f(z)| ≤ M on γ then Z

f(z)dz ≤ Ml(γ). γ Proof. Let z :[a, b] → γ ⊂ C be a parameterization. Then

Z Z b 0 f(z)dz = f(z(t))z (t)dt γ a Z b ≤ |f(z(t))|.|z0(t)|dt a Z b ≤ M |z0(t)|dt = Ml(γ) a as required. It is useful to illustrate this by a couple of examples .

n Example 5.12 (Integration of z along a straight line). Let γ = [w1, w2] and consider the aﬃne parameterization z : [0, 1] → C given by

z(t) = w1(1 − t) + w2t, for 0 ≤ t ≤ 1, 0 then z (t) = (w2 − w1) and by deﬁntion Z Z 1 f(z)dz = f (w1 + t(w2 − w1)t)(w2 − w1)dt. [w1,w2] 0

Consider the speciﬁc example f(z) = zn, then if n 6= −1 then Z 1 n+1 n+1 n w2 w1 (w1 + t(w2 − w1)) (w2 − w1)dt = − 0 n + 1 n + 1 Example 5.13 (Integration around a simple closed curve). Let γ be the θ-arc in the unit circle with z : [0, θ] → C given by z(t) = eit for 0 ≤ t ≤ θ 30 CONTENTS then z0(t) = ieit and by deﬁnition Z Z θ f(z)dz = enitieitdt. γ 0

Consider the speciﬁc example f(z) = zn then Z Z θ i h iθ ei(n+1)θ − 1 f(z)dz = enitieitdt = ei(n+1)t = γ 0 (n + 1)i 0 n + 1 if n 6= −1. On the other hand, if n = −1 then one can easily see that Z 1 Z θ Z θ dz = e−itieitdt = idt = θi γ z 0 0 In particular, if θ = 2π then ( Z Z 2π 0 if n 6= −1 znd = enitieintdt = γ 0 2πi if n = −1. 5.3. Theorems of Gousart and Cauchy. The last example is a special case of the following important theorem.

Theorem 5.14 (Gousart’s Theorem). Let f : U → C be an analytic function. R Let γ ⊂ U be a closed simple curve then γ f(z)dz = 0. There are two basic approaches to proving this. One approach uses Stokes’ theorem, but for variety we will describe the other approach (at least in the context of γ being a rectangle).

Proof. We begin with a proof of this result in the case that γ is the boundary of a rectangle R0 = [a, b] × [c, d]. In particular

∂R0 = [a, b] × {c} ∪ {b} × [c, d] ∪ [b, a] × {d} ∪ {a} × [d, c]. We proceed as follows.

Step 1. We can subdivide R0 into four similar sub-rectangles: (1) R0 = [a, (a + b)/2] × [c, (c + d)/2] (2) R0 = [(a + b)/2, b] × [c, (c + d)/2)] (3) R0 = [a, (a + b)/2] × [(c + d)/2, d] (4) R0 = [(a + b)/2, b] × [(c + d)/2, d] and then we can write 4 Z X Z f(z)dz = f(z)dz. (i) ∂R0 i=1 ∂R0 (where the integrals along the same curves in diﬀerent directions cancel). 5. INTEGRALS AND CAUCHY’S THEOREM 31

In particular. we can bound Z Z

f(z)dz ≤ 4 max f(z)dz . 1≤i≤4 (i) ∂R0 ∂R0

Let us denote by R1 the subrectangle which maximizes the last term.

Step 2. We can repeat this iteratively to construct a sequence of subrectangles

R0 ⊃ R1 ⊃ R2 ⊃ R3 ⊃ · · · ⊃ Rn ⊃ · · · (b−a) (d−c) such that Rn, n ≥ 1, is a sub-rectangle of size 2n × 2n and we can bound Z Z n f(z)dz ≤ 4 f(z)dz . (a) ∂R0 ∂Rn

∞ We can choose z0 ∈ ∩n=0Rn (which implicitly uses compactness of R0). We now want to use that f(z) is complex diﬀerentiable (at z0). In particular, for any > 0 we an choose δ > 0 so that providing |z − z0| < δ then 0 |f(z) − f(z0) + f (z0)(z − z0)| ≤ |z − z0|. (b) Provided n is suﬃciently large that the diameter 1 diam(R ) = p(b − a)2 + (d − c)2 (c) n 2n of Rn is smaller than δ then we have from (b) and (c) that for all z ∈ Rn we can write that 0 |f(z) − f(z0) − f (z0)(z − z0)| ≤ × diam(Rn) (d) ≤ p(b − a) + (d − c) 2n 32 CONTENTS

Moreover, since 1 length(∂R ) = (2(b − a) + (d − c)) (e) n 2n we can use (a), (d) and (e) to bound Z 0 (f(z) − f(z0) − f (z0)(z − z0))dz ∂Rn 0 ≤ sup {|f(z) − f(z0) − f (z0)(z − z0)|} × length(∂Rn) z∈Rn | {z } ≤ 1 (2(b−a)(d−c)) | √ {z } 2n (f) 2 2 ≤ 2n (b−a) +(d−c) ≤ p(b − a)2 + (d − c)22((b − a) + (d − c)) . 4n | {z } =:C

Step 3. Moreover, we have the following estimates: For each rectangle Rn: R (1) We can evaluate ∂R 1dz = 0; R n (2) We can evaluate (z − z0)dz = 0. ∂Rn Proof. The first part is easy, since the contribution to the integrals from opposite sides cancel. For the second part, we can simply explicitly do the integral. Comparing (a) and (f) and the above claim we can write: Z Z n f(z)dz ≤ 4 f(z)dz ∂R0 ∂Rn Z Z n 0 0 = 4 | (f(z) − f(z0) − f (z0)(z − z0)) dz − (f(z0) − f (z0)(z − z0)) dz | ∂Rn ∂Rn | {z } = 0 (by claim) C ≤ 4n = C 4n Thus, since can be chosen arbitrarily small, we finally deduce that R f(z)dz = 0, ∂R0 as required. Remark 5.15. For more general simple curves we can approximate γ by a piecewise linear curve γ0 consisting of horizontal and vertical lines. We can then fill in the region inside the curve by rectangles and apply the result above.

Corollary 5.16. Let f : U → C be analytic (i.e., complex diﬀerentiable). Let R 1 γ ⊂ U be a simple closed curve containing z0 ∈ U. Then dz = 2πi. γ z−z0 5. INTEGRALS AND CAUCHY’S THEOREM 33

Proof. We can assume for simplicity that z0 = 0 (otherwise we simply change to the translated function for z 7→ f(z − z0)). We can also assume that γ contains the unit disk D, since otherwise we can scale by z 7→ f(zR), for large enough R. In the previous section we saw that the result holds with γ replaced by the unit circle γ0. However, this suﬃces, because we can change the integral from one along γ to one along γ0 by introducing an extra arc c joining the two. Then the integrals R 1 diﬀer by −1 = 0, since the function is analytic inside the closed curve cγ0c γ z−z0 −1 cγ0c γ.

We now deduce Cauchy’s Theorem from this.

Theorem 5.17 (Cauchy’s Theorem). Let f : U → C be an analytic function (i.e., complex diﬀerentiable). Let γ ⊂ U be a simple closed curve. Assume that z0 lies inside the curve γ, then 1 Z f(z) f(z0) = dz (1) 2πi γ z − z0

Proof. Step 1. Consider the function

( f(z)−f(z0) if z 6= z0 z−z0 g(z) = 0 f (z) if z = z0

Moreover, by continuity of f at z0,

g(z)(z − z0) = lim (f(z) − f(z0))(z − z0)) = 0 z→z0 as z → z0. √ Given > 0 we can choose δ > 0 such that for |z − z0| < 2δ we have

|g(x) − g(z0)| ≤ |z − z0|.

Step 2. Let z0 = x0 + iy0 Given a rectangle R we want to divide it into nine subrectangles, including 0 R = [x0 − δ, x0 + δ] × [y0 − δ, y0 + δ] 34 CONTENTS then cancellations between the integrals on the boundaries of the eight rectangles and applying Gousart’s theorem to each of these we get Z Z g(z)dz = g(z)dz ∂R ∂R0 Moreover, since for z ∈ ∂R0 we can write |g(z) − g(z0)| ≤ ≤ |z − z0| δ we can write Z 0 8δ g(z)dz ≤ length(∂R ) sup |g(z)| ≤ sup |f(z)| ∂R0 z∈∂R0 δ z∈∂R0 which can be made arbitrarily small be choice of .

Step 3. We now apply Gousart’s Theorem’s to deduce that Z Z Z Z f(z) − f(z0) f(z) 1 0 = g(z)dz = dz = dz − f(z0) dz γ γ z − z0 γ z − z0 γ z − z0 | {z } 2πi 5.4. Immediate applications of Cauchy’s Theorem. Cauchy’s theorem has a number of immediate corollaries.

Corollary 5.18. Let f : U → C be analytic and let U be a simply connected domain. Let z1, z2 ∈ U. Then for any piecewise smooth curve γ connecting z1 to z2 R the integral γ f(z)dz is independent of γ.

Proof. Let γ1, γ2 be two piecewise smooth curves leading from z1 to z2 contained in U then γ1 ∪ −γ2 is a closed curve and thus Z Z Z f(z)dz = 0 = f(z)dz − f(z)dz γ1∪−γ2 γ1 γ2 and the result follows. Corollary 5.19. There is a formula for the derivatives of the form Z 0 1 f(z) f (z0) = 2 dz (∗) 2πi γ (z − z0) where z0 is inside the simple closed curve γ. More generally, the kth derivative takes the form Z (k) k! f(z) f (z0) = k+1 dz (∗∗) 2πi γ (z − z0) Proof. We need to justify diﬀerentiating under the integral sign. Using Cauchy’s theorem we can write, (for suﬃciently small h ∈ C): f(z + h) − f(z ) 1 Z 1 1 1 0 0 = − f(z)dz h 2πi γ h (z − z0 − h) (z − z0) 1 Z 1 = f(z)dz 2πi γ (z − z0 − h)(z − z0) We claim that the limit exists as h → 0 and that it equals Z 0 1 f(z) f (z0) := 2 .dz 2πi γ (z − z0) 5. INTEGRALS AND CAUCHY’S THEOREM 35

To see this we can bound Z f(z0 + h) − f(z0) 0 h 1 − f (z0) ≤ 2 f(z)dz h 2πi γ (z − z0 − h)(z − z0) |h| length(γ) 2 sup |f(z)| 2π (d(z0, γ) − |h|)d(z0, γ) z∈γ where d(z0, γ) = infz∈γ |z0 − z| and observe that the Right Hand Side tends to zero as |h| → 0. The higher derivative formula comes similarly (cf. next application).

Application 5.20 (Completing the equivalence of the definitions of analyticity I: Writing analytic functions as power series). We claim that a complex differentaible function always has a power series expansion. Let f : U → C be a complex differentiable function function. Let z0 ∈ U and choose > 0 such that

B(z0, ) = {z ∈ C : |z − z0| < R} ⊂ U

Choose z ∈ B(z0, ) and then ρ := |z0 − z| < and let r be such that ρ < r < . Then by Cauchy’s theorem 1 Z f(ξ) f(z) = dξ 2πi C(z0,r) ξ − z where C(z0, r) = {z ∈ C : |z − z0| = r} is the circle of radius r about z0.

For ρ < |ξ − z0| = r < we have that 1 1 = ξ − z (ξ − z0) − (z − z0) 1 = z−z0 (ξ − z0) 1 − ξ−z0 ! 1 z − z z − z 2 = 1 + 0 + 0 + ··· (ξ − z0) ξ − z0 ξ − z0

z−z0 ρ which is uniformly convergent on the circle C(z0, r) since = < 1. So by ξ−z0 r Cauchy’s theorem integrating around C(z0, r) gives 1 Z f(z) f(z0) = dξ 2πi C(z0,r) (ξ − z0) 1 Z f(z) 1 Z f(z) = dξ + (z0 − z0) 2 dξ 2πi C(z0,r) (ξ − z0) 2πi C(a,r) (ξ − z0) Z 2 1 f(z) + (z0 − z0) 3 dξ + ··· 2πi C(a,r) (ξ − z0) In particular, we can write this as 2 f(z0) = a0 + a1(z0 − a) + a2(z − a) + ··· 36 CONTENTS where 1 Z f(z) an = n+1 dξ 2π C(a,r) (ξ − z0) 1/n−1 and we know the series has Radius of convergence R = lim supn→+∞ |an| > r. Thus f(z) is a power series about a. Application 5.21 (Completing the equivalence of the definitions of analyticity II: Continuity of the partial derivatives). Let f : U → C be complex differentiable. For z0 ∈ U we can γ ⊂ U be a simple closed curve containing z0. We see from 0 (*) that f (z0) depends continuously on the point z0. Furthermore, since complex differentability implies that for f(x + iy) = u(x, y) + iv(x, y) we can write ∂u ∂v ∂v ∂u f 0(z ) = + i = − i 0 ∂x ∂x ∂y ∂y from which we can deduce that the partial derivatives are continuous.

Corollary 5.22. Let f : U → C be analytic and let z0 be a unique zero (of multiplicity one). Let γ ⊂ U be a simple closed curve. Then 1 Z f 0(z) z0 = z dz 2πi γ f(z)

Proof. We can write f(z) = (z − z0)g(z), where g : U → C where g : U → C f 0(z) is non-zero and analytic (and, in particular, g(z0) 6= 0). The function f(z) is of the form 1 + ψ(z) where ψ : U → is analytic and given by ψ(z) = g0(z)/g(z). z−z0 C Thus, by Cauchy’s theorem we can write 1 Z f 0(z) 1 Z 1 1 Z z dz = z dz + zψ(z)dz = z0. 2πi γ f(z) 2πi γ z − z0 2πi γ | {z } | {z } =z0 =0 Application 5.23 (Perturbation Theory for eigenvalues of matrices). Assume that A is a matrix and λA is a simple eigenvalue. Then letting fA(z) = det(zI − A) then we see that z0 is a simple zero for f(z). By the previous corollary, we see that Z 0 1 fA(z) λA = z dz 2πi γ fA(z)

However, it is easy to see that the coeﬃcients of the polynomials fA(z) and 0 fA(z) depend smoothly on the entries in the matrix A. This is true on γ and thus we can deduce the same for λA. 5.5. Converse to Cauchy’s Theorem. The following is a converse to Cauchy’s Theorem which is often useful in proofs of later results.

Theorem 5.24 (Moreira’s Theorem). Let U be a domain and let f : U → C be R continuous and satsify ∂∆ f(z)dz = 0 for all simple closed curves ∆ ⊂ U. Then f is analytic.

Proof. Fix a point z0 ∈ U. We can deﬁne F : U → C by associating to any z ∈ U a piecewise smooth curve γ parameterized by z : [0, 1] → γ ⊂ C where z(0) = z0 and z(0) = z and then deﬁning Z F (z) = f(z)dz. γ 6. PROPERTIES OF ANALYTIC FUNCTIONS 37

We ﬁrst observe that since U is a (simply connected) domain this is well deﬁned, since if γ1 is a second path from z0 to z then we have that Z Z Z Z Z f(z)dz − f(z)dz = f(z)dz + f(z)dz = f(z)dz = 0 γ γ1 γ −γ1 γ(−γ1) since γ(−γ1) is a picewise closed curve, and F : U → C is analytic, and thus we can apply Cauchy’s theorem.

We can now deduce that F 0(z) = f(z), i.e., f is the complex derivative of F . But if F it is complex diﬀerentiable then it is analytic, and thus all of the derivatives exist, and us f is also complex diﬀerentiable.

6. Properties of analytic functions We want to ﬁnd some way to understand better properties of analytic functions. We will begin with properties of individual functions and move onto families of functions. However, ﬁrst we have to recall some basic facts.

6.1. Logarithms and roots. Consider the exponential function z 7→ ez for z ∈ C. This is analytic for every z ∈ C. We would like to deﬁne the inverse map, the logarithm, in the complex plane but this has a few complications that need to be addressed. Lemma 6.1. Let z = x + iy and w = u + iv. Then ez = ew when there exists n ∈ Z such that y = v + 2πn. Proof. If ezex+iy = ew = eu+iv then taking the modulus gives |ez| = eu = ex = |ew| gives that u = x. This leaves that eiy = eiv and thus y − v is an integral multiple of 2π.

In particular, exp : z 7→ ez maps any strip R × [α, α + 2π) bijectively onto C − {0}. However, the inverse to exp(·) is not, strictly speaking, a well deﬁned function since there are many points mapped to a single point. However, by the lemma these points all diﬀer by values 2πin, n ∈ Z. 38 CONTENTS

We would like to write w = log z when z = ew. If w = u + iv and exp w = z then z eueiv = |z| = |z|eiθ |z| for some θ since z = 1. Hence eu = |z| and v = θ + 2πn for any n ∈ . |z| Z Therefore u = log |z| and w = log |z| + i(θ + 2πn) for some n ∈ Z. Definition 6.2. For z 6= 0 deﬁne the function arg(z) = arg(z/|z|) = θ if z/|z| = eiθ and −π ≤ θ < π.

For log(·) there are different branches of arg with different restricted domains. The principle branch of arg is denoted Arg(z) and is such that Arg(eiθ) is the unique number of the form θ + 2πn which lies between −π and π. This is not defined on the negative real line (−∞, 0]. Returning to the definition of log:

Definition 6.3. The principle branch of log is denoted Log(z) and is of the form log |z| + iArg(z). This is not deﬁned on the negative real line (−∞, 0].

We can then use exp and Log to deﬁne complex powers of complex numbers. In particular, we can write that zw = exp(w.Log(z)).

Example 6.4. Find the value of

(−1 − i)1+i := exp ((1 + i)Log(−1 − i)) √ 3 3 We observe that arg(−i − i) = − 4 π and Log(−i − i) = log 2 − 4 πi. Thus √ 3 (−1 − i)1+i = exp ((1 + i)Log(−1 − i)) = exp((1 + i) log 2 − πi) . 4

Example 6.5. If ρ ∈ N then we can write z1/ρ = exp (log(z)/ρ) = exp ((log |z| + iArg(z))/ρ)).

If z = reiθ then

z1/ρ = exp((1/ρ)(log r + i(θ + 2πn))) = r1/ρei(θ+2πn)/ρ = r1/ρei(θe2πin)/ρ.

These take the values r1/ρ, r1/ρei2πθ, ··· , r1/ρei(θ+2π(p−1)/p. In particular 1 1 1 z1/2 = exp log(z) = exp log |z| + arg(z) = |z|1/2eiArg(z)/2. 2 2 2

More generally, for zq/p we simply replace z by zq.

Remark 6.6. If α is irrational then we can write

zα = exp (α log z) = exp(α log r + αi(θ + 2πn)) . and the set of values with n ∈ Z is inﬁnite. To see this, let z = reiθ then exp(αi(θ + 2πn)) = exp(αi(θ + 2πm)) ⇐⇒ exp(αi2π(n − m)) = 1 ⇐⇒ α2π(n − m) = 1 ⇐⇒ n = m.

(since α is irrational). 6. PROPERTIES OF ANALYTIC FUNCTIONS 39

6.2. Location of zeros: Argument Principle and Rouch´e’sTheorem. We can apply Gousart’s and Cauchy’s theorem to prove two useful results

Lemma 6.7. Suppose that f : U → C is analytic and non-zero then there exists an analytic function h : U → C such that eh(z) = f(z) for all z ∈ U. Proof. Since f is analytic we say that f 0 is analytic and since f does not vanish then f 0/f is analytic and so f 0/f has a primitive h by integrating f, i.e., h0 = f 0/f for h analytic. Thus d f(z)e−h(z) = f 0(z)e−h(z) − f(z)h0(z)e−h(z) = 0 dz −h(z) a a+h(z) so f(z)e = Constant = e 6= 0. Thus f(z) = e We can use this lemma to show the following theorem.

Theorem 6.8 (The Argument Principle). Assume f : U → C is analytic and γ ⊂ U is a simple closed curve. The general case is similar. Assume that f has no zeros on γ. Then the number of zeros N inside γ is given by 1 Z f 0(z) N = dz. 2πi γ f(z)

Proof. Assume that z0 is a zero of order n ≥ 1 for f(z). Assume for simplicity that we have no other zeros.

N h(z) Using the lemma we can then write write f(z) = (z − z0) e and then 0 N h(z) 0 n−1 h(z) f (z) = (z − z0) e h (z) + N(z − z0) e and f 0(z) N = h0(z) + f(z) z − z0

Since the ﬁrst term is analytic simple pole at z0 we can apply Gousart’s theorem to the ﬁrst term and Cauchy’s theorem to the second term to write 1 Z f 0(z) N = dz. 2πi γ f(z) With more zeroes the argument naturally generalizes. Example 6.9. Assume f : D → C has a simple zero at 0 (i.e., N = 1) and that γ = {z : |z| = r} with 0 < r < 1. Writing ∞ ∞ X n 0 X n f(z) = anz and f (z) = nan+1z n=1 n=1

(with a0 = 0 and a1 6= 0) we see that f 0(z) 1 = + g(z) f(z) z 40 CONTENTS

R where g(z) is analytic on a neighbourhood of 0. By Gousart’s theorem γ g(z)dz = 0 and a direct calculation gives

1 Z f 0(z) 1 Z 1 1 Z dz = dz + g(z)dz 2πi γ f(z) 2πi γ z 2πi γ | {z } | {z } =1 =0

Application 6.10 (localizing zeros of Riemann Zeta function). One of the major open problems in mathematics is the Riemann Hypothesis. For s ∈ C with Re(s) > 1 we deﬁne the Riemann zeta function

∞ X 1 ζ(s) = ns n=1

P∞ −Re(s) (It converges since |ζ(s)| ≤ n=1 n < +∞). Moreover, it has an analytic extension to C − {1}, i.e., there is an analytic function f : C − {1} → C such that f(s) = ζ(s) for Re(s) > 1.

Riemann Hypothesis (Conjecture, 1859) The only zeros for f(s) occur at s = −2, −4, −6, ··· and on the line

1 L := {s = + it : t ∈ }. 2 R This is Hilbert’s 8th problem from his famous list of 23 open problems from 1900, and one of the 7 Millenium problems from 2000 for which the Clay Institute oﬀered a million dollars. Computer searches for counter-examples (so far unsuccessful!) use the Argu- ment Theorem. One considers a small closed curve γ away from the line L and 1 R f 0(z) estimates the integral 2πi γ f(z) dz. One tries to ﬁnd a γ such that the integral is non-zero.

Theorem 6.11 (Rouche’s Theorem: Nearby functions and nearby zeros). As- sume that f, g : U → C are analytic. Let γ ⊂ U be a simple closed curve and assume that neither f(z) or g(z) have zeros on γ and that for each z ∈ γ we have that |g(z)| < |f(z)|. Then f and f + g have the same number of zeros in U.

Proof. This uses the ”walking the dog” method. 6. PROPERTIES OF ANALYTIC FUNCTIONS 41

g(z) Let F (z) = f(z) then if Nf and Nf+g are the number of zeros inside γ for each of the two functions then by the Argument Principle we can write 1 Z (f 0 + g0)(z) 1 Z f 0(z) Nf+g = dz and Nf = dz. 2πi γ (f + g)(z) 2πi γ f(z) Thus 1 Z (f 0 + g0)(z) 1 Z f 0(z) Nf+g − Nf = dz − dz 2πi γ (f + g)(z) 2πi γ f(z) 1 Z f 0(z)(1 + F (z)) + f(z)F 0(z) 1 Z f 0(z) = dz − 2πi γ f(z)(1 + F (z)) 2πi γ f(z) 1 Z f 0(z) F 0(z) 1 Z f 0(z) = + − 2πi γ f(z) 1 + F (z) 2πi γ f(z) 1 Z F 0(z) = dz 2πi γ 1 + F (z) But F (z) 6= −1 for z ∈ γ, since |F (z)| < 1, and so the final expression counts the number of z inside γ with F (z) = −1. Finally, observe that if 0 ≤ t ≤ 1 we can replace g(z) by tg(z) (and F (z) by tF (z)) and providing t is small enough we have that the final expression is zero. But if we assume for a contradiction that Nf+g 6= Nf then it will contradict the obvious continuity of 1 Z tF 0(z) t 7→ dz 2πi γ 1 + tF (z) s Example 6.12. Show that all four of the zeros of z4 − 7z − 1 lie in B(0, 2) = {z ∈ C : |z| < 2}. Let f(z) = z4 then all four zeros lie inside B(0, 2). In fact, z = 0 is a zero of multiplicity four. Let g(z) = −7z − 1. Let γ = C(0, 2) = {z ∈ C : |z| = 2}. For |z| = 2 we have that |g(z)| = | − 7z − 1| ≤ 7|z| + 1 = 15 < 16 = |z4| = |f(z)|. Thus by Rouché’stheorem we have that f(z) + g(z) = z4 − 7z − 1 has the same number of zeros in B(0, 2) as f(z), i.e., four. Example 6.13. Show that all five of the zeros of z5 + 3z3 + 7 lie in B(0, 2) = {z ∈ C : |z| < 2}. Let f(z) = z5 then all five zeros lie inside B(0, 2). In fact, z = 0 is a zero of multiplicity five. Let g(z) = 3z3 + 7. Let γ = C(0, 2) = {z ∈ C : |z| = 2}. For |z| = 2 we have that |g(z)| = |3z3 + 7| ≤ 3|z3| + 7 = 31 < 32 = |z5| = |f(z)|. 42 CONTENTS

Thus by Rouché’stheorem we have that f(z) + g(z) = z5 + 3z3 + 7 has the same number of zeros in B(0, 2) as f(z), i.e., five. Example 6.14. Show that z4 − 7z − 1 has precisely one zero in the unit disk D = {z ∈ C : |z| < 1}. 1 Let f(z) = −7z − 1 then f(z) has precisely one zero in D, at z = − 7 . Let g(z) = z4. Let γ = C(0, 1) = {z ∈ C : |z| = 1}. For |z| = 1 we have that |g(z)| = |z4| = 1 < 6 = 7 − 1 = |7z| − 1 ≤ | − 7z − 1| = |f(z)|. Thus by Rouché’stheorem we have that f(z) + g(z) = z4 − 7z − 1 has the same number of zeros in D as f(z), i.e., one.

Remark 6.15. There is a more symmetric version of this theorem. If f1 : U → C and f2 : U → C are analytic then |f1(z) − f2(z)| < |f1(z)| + |f(z)| for z ∈ γ then f1(z) and f2(z) have the same number of zeros inside the curve.

Corollary 6.16 (Weierstrass-Hurwitz Theorem). Assume fn : U → C are a sequence of non-zero analytic functions. If fn converges uniformly on compact sets to a continuous function f (i.e., if K ⊂ U is compact then supz∈K |f(z)−fn(z)| → 0 as n → +∞) then: (1) f : U → C is analytic (2) f either has no zeros or is identically zero. Proof. For the ﬁrst part (due to Weierstrass) one only needs to show that for any simple closed curve γ ⊂ U we have that Z Z Z

f(z)dz ≤ fn(z)dz + (f(z) − fn(z))dz γ γ γ | {z } =0 ≤ 0 + length(γ) sup |f(z) − fn(z)| z∈K | {z } →0 where the ﬁrst term is zero by Gousart’s theorem (since fn : U → C is analytic). R Thus γ f(z)dz = 0 for all simple closed curves, and so f is analytic by Moreira’s Theorem. For the second part (due to Hurwitz), if f(z) isn’t identically zero then the set N of zeros {zi}i=1 for f(z) is ﬁnite. We need to show that this set is empty. Assume for a contradiction it isn’t empty and let γ be a closed simple curve which contains one of these zeros z1, say. By Cauchy’s theorem 1 Z f 0 (z) 1 Z f 0(z) 0 = n dz → dz = 1 2πi γ fn(z) 2πi γ f(z) as n → +∞ which gives a contradiction. Application 6.17. Recall that the Riemann zeta function ∞ X 1 ζ(s) = ns n=1 was claimed to analytic for Res(s) > 1. To check this we can observe that

N N ! X 1 X f (s) = = exp (−s log n) N ns n=1 n=1 is analytic on C. For any compact set K ⊂ U := {s ∈ C : Re(s) > 1}, say, we have that sups∈K |ζ(s) − fn(s)| → 0. Thus by the theorem ζ(s) is analytic on U. 6. PROPERTIES OF ANALYTIC FUNCTIONS 43

6.3. Liouville’s Theorem and the Fundamental Theorem of Algebra. We begin with another application of the Taylor series version of analyticity. Theorem 6.18 (Liouville’s Theorem). If a function f : C → C is analytic and bounded then f is constant. Proof. We can write f(z) as a power series representation centred on 0: ∞ X n f(z) = anz . n=0

Now we use an estimate for an for any R > 0:

1 Z f(z) M 2πR M |a | = dz ≤ = n n+1 n+1 n 2πi C(0,R) z R π R where M is an upper bound for |f(z)| and taking limits as R → ∞ to get an = 0 except for n = 0, i.e., f(z) = a0. Functions which are analytic on all of U = C are called entire functions. Corollary 6.19 (Fundamental Theorem of algebra). Every non-constant polynomial has a root (i.e., a zero, that is there exists z0 such that p(z0) = 0). Proof. Suppose that the polynomial p(z) does not vanish. Then 1/p(z) is entire and since |p(z)| → +∞ as z → +∞ we have that 1/p(z) → 0 as z → +∞ and hence |1/p(z)| < for |z| > R and 1/p(z) is bounded for |z| ≤ R so 1/p(z) is bounded throughout C so 1/p(z) is constant, contradicting the fact that p(z) is non-constant. In particular, this implies that the polynomial can be written as: n p(z) = anz + ··· + a0 = an(z − z1)(z − z2) ··· (z − zn)

Let z1 be a zero and then write n n 0 n−1 0 0 p(z) = an(z−z1 +z1) +···+a0 = an(z−z1) +an−1(z−z1) +···+a1(z−z1)+a0 0 Clearly a0 = 0, by evaluation at z1. So p(z) = (z−z1)q(z) where q(z) is a polynomial of degree n − 1 (and the leading coeﬃcient of q(z) is 1). Repeating this on q(z) we obtain by induction n p(z) = anz + ··· + a0 = an(z − z1)(z − z2) ··· (z − zn). 6.4. Identity Theorem. The following is an application of the Taylor series theorem

Lemma 6.20 (Isolation of zeros). Let r > 0 and let B(z0, r) = {z ∈ C : |z−z0| < r}. The zeros of a non-trivial f : D(z0, r) → C cannot accumulate within a disk D(z0, r − ), where > 0.

Proof. Assume for a contradiction that zk → w ∈ B(z0, r − ), for some > 0, are a convergent sequence of zeros for f(z). We can consider the restriction f : B(w, ) → C and write ∞ X n f(z) = an(z − w) n=0 satisﬁes an = 0, for all n ≥ 0, i.e., f(z) is trivial contradicting the hypothesis. More precisely, assume for a contradiction to the claim that for some N ≥ 0 we have that a0 = a1 = ··· = aN−1 = 0 and |an| > 0. Since f(zk) = 0 each k ≥ 1 we can write ∞ X n f(zk) = 0 = an(zk − w) n=N 44 CONTENTS

n If |an| ≤ C(2) , say, then we can bound ∞ ∞ N X n X n 2|zk − w|) 0 = an(zk − w) ≥ |an| − C (r|zk − w|) = |an| − C 1 − 2|zk − w| n=N n=N Providing k is suﬃciently large the right hand side is positive, giving a contradiction. Theorem 6.21 (Identity Theorem). Suppose that f, g : U → C are analytic in the domain U and suppose that {zn}, z0 ∈ U and zn → z0 with zn 6= z0 with f(zn) = g(zn).

Proof. We have already proved this for a disk with centre z0, say, by the previous lemma, we see that f = g on a disk with centre z0 contained in U. Let z1 ∈ U then we can see that f(z1) = g(z1) by parameterizing a polynomial line from z0 to z1. Otherwise let w be a point on this line such that f(z) = g(z) for all points from z0 up to w and for some points arbitrarily close to w we have f(z) 6= g(z). But w is the limit of a sequence {wn} of points with f(wn) = g(wn) and so f = g in a disk cntred at w This contradicts the deﬁnition of w if w is before z1. Hence z1 = w and f(z1) = g(z1). Corollary 6.22 (Isolation of values). Let f : U → R be analytic and non- constant. Let λ ∈ C then for each z0 ∈ U there exists a punctured disc P (z0, r) = {z ∈ C : |z0 − z| < r} − {z0} such that f(z) 6= λ. Proof. Assume for a contradiction that for every small punctured disk of radius r = 1/n, say, there exists zn ∈ P (z0, 1/n) such that f(zn) = λ. But this implies f(z) = λ on U, i.e., f(z) is constant and contradicts the hypothesis. Corollary 6.23 (Open Mapping Theorem). Let f : U → R be a non-constant analytic function. Then the image f(V ) of any open set V ⊂ U is again an open set.

Remark 6.24. This is very diﬀerent to case for R. Consider f : R → R deﬁned by f(x) = x2 is not an open map since, for example, f((−1, 1)) = [0, 1), which is not open.

Proof. Let z0 ∈ V then it is enough to show that there exists > 0 such that B(f(z0), δ) ⊂ f(V ). Let g(z) = f(z) − f(z0) then g(z0) = 0. Moreover, z0 must be a simple zero (i.e., multiplicity one) for g otherwise the function g(z) (and hence f(z)) would be a constant function. In particular, there is a simple closed curve γ about z0 and within V , containing which z0 as the unique zero. Moreover, by taking the curve smaller (if necessary) we can assume that g0(z) = f 0(z) has no zeros within γ. By the Argument theorem we see that 1 Z g0(ξ) 1 Z f 0(ξ) 1 = dξ = dξ. 2πi γ g(ξ) 2πi γ f(ξ) − f(z0)

Moreover, for γ ﬁxed, this remains true if we replace f(z0) by f(z) with |f(z) − f(z0)| < δ, for suﬃciently small δ > 0. i.e, for such f(z) ∈ B(f(z0), δ) there exists a unique w inside γ, and thus in V . 6.5. Families of functions and Montel’s Theorem. Recall that a set K ⊂ C is compact if and only if it is closed and bounded. ∞ Definition 6.25. Let {fn}n=1 be a family of continuous functions fn : U → C, ∞ for n ≥ 1. We say that this family is normal if there is a subsequence {fnk }k=1 which converges uniformly on every compact subset of U to a continuous function f : U → 6. PROPERTIES OF ANALYTIC FUNCTIONS 45

C, i.e., for every compact subset K ⊂ U we have that supz∈K |fni (z) − f(z)| → 0 as ni → +∞. We can also adopt the convention that we allow uniform convergence to ∞ (consistent with the functions being viewed as being valued in the Riemann Sphere). In the case of analytic functions, we can get that the family is normal by just assuming the sequence is uniformly bounded.

Theorem 6.26 (Montel’s Theorem). Let fn : U → C, n ≥ 1, be a family of analytic functions. Assume that there is a positive constant M > 0 such that

|fn(z)| ≤ M for all z ∈ U, n ≥ 1 ∞ then {fn}n=1 is a normal family. Moreover, the limiting function f : U → C is analytic. The proof is based on a classical result in real analysis. We recall the following deﬁnition.

∞ Definition 6.27. Let K be a compact set. Let {fn}n=1 be a family of continuous functions fn : K → C, for n ≥ 1. We say that this family is equicontinuous if for every > 0 there is a δ > 0 such that whenever z, w ∈ K with |z − w| < δ then |fn(z) − fn(w)| < for all n ≥ 1.

The corresponding result from real analysis we need is the following.

Theorem 6.28 (Arzela-Ascoli). Any equicontinuous family fn : K → C has a subsequence (fnk ) which is uniformly convergent to a continuous function f : K → C

(i.e., supz∈K |fnk (z) − f(z)| → 0 as nk → +∞)

Proof. Let z0 ∈ U and let r > 0 be so that B(z0, r) ⊂ U. Then C − U and B(z0, r) are disjoint open sets. We can choose η > 0 such that if w ∈ C − U and z ∈ B(z0, r) then |z − w| > η. 46 CONTENTS

If z ∈ B(z0, η) then we can consider restriction fn : B(z0, r) → C and by Cauchy’s Theorem

1 Z f(z) M |f 0(z)| = dz ≤ 2πi (z − z )2 ρ |z0−z|=ρ 0

Thus for any z1, z2 ∈ B(z0, η) we have that ! |fn(z1) − fn(z2)| ≤ sup |f(z)| |z1 − z2|. |z|∈B(z0,η)

In particular, we see that the restrictions fn : B(z0, η) → C, n ≥ 1, form an equicontinuous family. Given > 0 we simply choose δ = C > 0.

If K ⊂ U is compact we can cover it be ﬁnitely many such balls and we again deduce that fn : K → C, n ≥ 1, forms an equicontinuous family. We can then invoke the Arzela-Ascoli theorem to deduce that there is a convergent subsequence fnk : K → C, for k ≥ 1. The analyticity of the function f : U → C follows from the Hurwitz-Weiestrauss theorem.

Remark 6.29. There is a stronger version which says that fn : U → C, n ≥ 1, is a normal value if they all omit the same three values.

Application 6.30 (Julia sets). Let f : Cb → Cb be a polynomial (or, more generally, a rational map).

Deﬁnition. We say that z ∈ C is in the Fatou set F if there exists > 0 such that n the compositions {f : B(z0, ) → C : n ≥ 1} is a normal family. We deﬁne the Juia set J to be the complement of the Fatou set J = C − F.

2 n 2n Example. If f(z) = z then f (z) = z . Thus on U∞ = {z ∈ C : |z| > 1|} we have that f n(z) → +∞, i.e., the family of iterates converge to ∞. On the other n hand, on U0 = {z ∈ C : |z| < 1|} we have that f (z) → 0, i.e., the family of iterates converge to ∞. Thus the Fatou set contains U∞ ∪ U0. But if |z| = 1 then in any disk B(z, ) those points also inside the unit disk will converge to 0 while those points also outside the unit disk will converge to ∞. i.e., f n(z) doesn’t converge to an analytic function on B(z, ). Therefore F = U∞ ∪ U0 and J = C(0, 1). The Julia set is closed, bounded and f(J) = J.

Theorem 6.31. The Julia set consists precisely of those points z ∈ C such that for every > 0 there exists w ∈ B(z, ) such that f n(w) → +∞. Proof. Assume that for every > 0 there exists w ∈ B(z, ) such that f n(w) → +∞. Since f n(z) ∈ J is bounded, we cannot have f n : B(z, ) → C isn’t normal and thus z ∈ J . 7. RIEMANN MAPPING THEOREM 47

Conversely, given z if there exists > 0 such that for all w ∈ B(z, ) we have that f n(w) is bounded. Thus we can apply Montel’s theorem to deduce that n f : B(z, ) → C is normal, i.e., z 6∈ J . Exercise: If f is a polynomial and z ∈ C then either f n(z) → ∞ or {f n(z): n ≥ 0} is a bounded set. Theorem 6.32. The Julia set is equal to the closure of the repelling periodic points. The proof uses the other version of Montel’s theorem.

7. Riemann Mapping Theorem One of the most interesting results in complex analysis is the Riemann Mapping Theorem.

7.1. Riemann Mapping Theorem. The most elegant theorem in complex analysis is the following.

Theorem 7.1 (Riemann Mapping Thoerem). Let U ⊂ C be an open subset homeomorphic to D. Then there exists an analytic bijection f : U → D. Proof. The proof is quite elaborate, so we break it down into steps. We assume for simplicity that U is bounded, i.e., U ⊂ B(z0,R) for some R > 0.

Step 1: Fix z0 ∈ U and let F be the family of analytic functions f : U → D such that (1) f is a bijection; and (2) f(z0) = 0. To show that F 6= ∅ we can construct f : U → D by (z − z ) f(z) = 0 2R then

(z − z0) |z| + |z0| R + R |f(z)| = ≤ ≤ ≤ 1 2R 2R 2R and by construction we see that f(z0) = 0 and thus f ∈ F.

Step 2: Since the functions in F are analytic and bounded (i.e., for f ∈ F and |f(z)| ≤ 1) we can apply Montel’s theorem to deduce that the family is normal.

Step 3: We deﬁne 0 M = sup{|f (z0)| : f ∈ F}.

Claim: We want to show that this is ﬁnite (i.e., M < +∞). 48 CONTENTS

Proof of Claim: Choose r > 0 suﬃciently small that B(z0, r) ⊂ U. By Cauchy’s theorem we have that for any f ∈ F,

1 Z f(z) 1 1 |f 0(z )| = dz ≤ sup |f(z)| ≤ . 0 2πi (z − z )2 r r |z−z0|=r 0 z∈D | {z } ≤1 1 Thus the result follows with M ≤ r .

Step 4: We now want to show the supremum is realized. 0 Claim: There exists f0 ∈ F such that f0(z0) = M. 0 Proof of Claim: We can chose fn ∈ F, n ≥ 1, such that |fn(z0)| → M as n → +∞. ∞ By Montel’s Theorem applied to {fn} we can choose a subsequence {fnk }k=1 and an analytic function f : U → C such that fnk → f uniformly on any compact set 0 K ⊂ C. Therefore, we see that |f (z0)| = M. iθ 0 If we replace f(z) by e f(z) then we can choose 0 ≤ θ < 2π so that f (z0) = M.

Step 5: We want to show that f : U → D is injective. Claim: f : U → D is injective.

Proof of Claim: Let z1 6= z2 be distinct points in U. Let 0 < ρ = |z1 − z2| and deﬁne gk : B(z2, ρ) → C by

gk(z) = fnk (z) − fnk (z1)

Since the functions fnk (∈ F) are all injective, the functions gk are all non-zero on B(z2, ρ) (i.e., 6 ∃z ∈ B(z2, ρ) with f(z) = 0). By the Hurwitz-Weierstrauss theorem the limit function g(z) = f(z) − f(z1) is either identically zero or has no zeros. But, it cannot be identically zero (since we 0 assume that |f (z0)| = M > 0). Thus f(z) 6= f(z1) for all z ∈ B(z2, ρ) and thus f(z2) 6= f(z1).

Step 6: We want to show that f : U → D is surjective. Claim: f : U → D is surjective. Proof of Claim: Assume for a contradiction that f is not surjective, and thus there exists w ∈ D which is not in the image f(U). We can choose a M¨obiusmap φw : D → D by z − w φ (z) = w 1 − zw which maps w to 0. We then consider φw ◦ f : U → D. This map now doesn’t have k(z) 0 in its image. As we saw in a previous lemma, we can write φw ◦f(z) = e where k : U → C is analytic, and thus we can deﬁne g : U → C by 1/2 k(z)/2 g(z) = (φw ◦ f(z)) (=: e ). In particular, g : U → D is one-to-one. Finally, we introduce

z − g(z0) φg(z0)(z) = 1 − zg(z0) and define ρ : U → D by ρ(z) = φg(z0) ◦ g(z0). We can then compute ρ0(z ) = φ0 (g(z )) .g0(z ) (1) 0 g(z0) 0 0 | {z } 1 = 2 1−|g(z0)| 7. RIEMANN MAPPING THEOREM 49 and 20 0 g(z0) = 2g (z0).g(z0) 0 0 = φw(f(z0)).f (z0) (2) 2 0 = (1 − |w| )f (z0). (after a little calculation).) From (1) and (2) we find that: 2 0 1 1 − |w| 0 ρ (z0) = 2 f (z0) 1 − |g(z0)| 2g(z0) 2 1 1 − |w| 0 = f (z0) 1 − |w| 2g(z0) 1 + |w| 0 = f (z0) 2g(z0) p However, since w 6= 0 then 1 + |w| > 1 and g(z0) = |w| and thus ! 0 1 + |w| 0 ρ (z0) = f (z0) 2p|w| | {z } >1 0 giving |ρ (z0)| > M, a contradiction to the definition of M. However, this contradicts h(z) having maximum derivative M at z0. Remark 7.2. If U is unbounded then we can apply a more elaborate argument to prove Step 1. Let z0 ∈ C − U. The function φ : U → C is non-vanishing and we can write φ(z) = ek(z) = k(z)2, where k(z) = ek(z)/2. Moreover, k(z) is one-to-one (since φ(z) is) and there are not distinct points z1, z2 such that h(z1) = h(z2) (since then φ(z1) = φ(z2)). This is an open mapping and so we can choose B(b, r) ⊂ h(U). But then B(−b, r) ∩ h(U) = ∅. We may therefore define the holomorphic function r f(z) = . 2(h(z) + b) Since |h(z) − h(−z)| ≥ r for z ∈ U, it follows that f : U → D. Since h is one-one then so is f. Compositing f with a Möbiusmap that preserves D we can get a function which is analytic and one-on-one and bounded by 1.

Proposition 7.3 (Caratheodory’s Theorem). Let U ⊂ C be a simply connected open set whose boundary is a simple closed curve. Let f : U → D be the analytic bijection in the Riemann mapping theorem. Then φ extends to a homeomorphism from U to D. 7.2. Christoﬀel-Schwarz Theorem. The Christoﬀel-Schwarz Theorem is a more explicit version of the Riemann mapping theorem where the open region to be mapped to the unit disk is taken to be the inside of a polygon P. 50 CONTENTS

1−z We recall that the map w = f(z) = i 1+z maps the unit disk to the upper half plane H2. Thus it suﬃces to ﬁnd a map g : H → P which is analytic and then g◦f : D → P. The general result is the following:

Theorem 7.4 (Schwarz-Christoffel transform). Let P be a polygon with n sides such that the internal angles are πα1, ··· , παn. There exist w1, ··· , wn ∈ R such that if we consider the function C F (w) = 1−α 1−α 1−α (1) (w − w1) 1 (w − w2) 2 ··· (w − wn) n R z then for z0 ∈ we can choose A, C such that f(z) := A+ F (w)dw is an analytic H z0 bijection from H to the interior of the polygon P (and which extends continuously to the boundary, taking points w1, ··· , wk to the vertices of P). Here C changes the size and orientation of the image and A translates it. Remark 7.5. Since the external angles of the polygon sum to (n − 2)π we see that n X (1 − αi) = −2. i=1 Remark 7.6. It is often convenient to consider the case in which the point at infinity of the w plane maps to one of the vertices of the polygon (e.g., x1 = ∞ and the vertex with angle α1). If this is done, the first factor in the formula is effectively a constant and may be regarded as being absorbed into the constant C

The following general result is often useful. Let H = {z ∈ C : Im(z) > 0} and H− = {z ∈ C : Im(z) < 0}. Lemma 7.7 (Schwarz-reﬂection principle). If g : H → C is analytic and extends continuously to [a, b] ⊂ R such that g([a, b]) ⊂ R then g has an extension as an analytic function to H ∪ [a, b] ∪ H−.

Proof. For z ∈ H we have that z ∈ H− and we deﬁne g(z) = g(z). To deduce analyticity we can use Moreira’s theorem. For example, If γ is a closed curve that crosses [a, b] then we can consider two loops γ1, γ2 consisting of the pieces of γ in each half-plane plus a subinterval of [a, b] between their intersection points. By approximating them by curves wholly in the half planes we see that R f(z)dz = 0 γ1 and R f(z)dz = 0, and thus R fdz = R f(z)dz + R f(z)dz = 0. Thus Moreira’s γ2 γ γ1 γ2 Theorem applies and we are done. 7. RIEMANN MAPPING THEOREM 51

Lemma 7.8 (Osgood-Caratheodory Theorem). If g : P → D is an analytic bijection then it extends to a continuous function g : P → D on the boundary.

Proof. Assume that αi ≤ 1. (1) Choose one side I1 of P and use a Möbius map f1 : P → H such that S := f1(P) and f1(I) = [−1, 1] (2) We can reflect S in the real axis to get S and consider U := S ∪(−1, 1)∪S. By the Riemann mapping theorem we can find a unique analytic bijection 0 f2 : U → D with f2(0) = 0 and f2(0) > 0. However, we see that g2(z) := f2(z) satisfies the same conclusion and so g2 = f2 and we conclude that + f2([−1, 1]) ⊂ R and then f2(S) = D = {z ∈ D : Im(z) > 0} + (3) We can choose a Mobius map f3 : D → D. Then we set f(z) = f3 ◦ f3 ◦ f1(z). We thus see that f extends continuously to I1. 0 If we fix z1 ∈ P then we can choose f3 so that f(z1) = 0 and f (z1) > 0. Thus f is unique. In particular, the conclusion that f extends applies to each side.

This brings us to the main part of the Schwarz-Christoﬀel theorem.

Theorem 7.9. Every analytic bijection f : H → P is necessarily of the form (1). The existence of such maps follows from the Riemann Mapping Theorem. The mapping f(z) extends continuously to the boundary. We can then extend it to H‘[−] via one of the intervals (xi, xi+1) by the Schwarz reﬂection principle.

0 In particular, f(H) corresponds to a polygon P obtained from P by reﬂec- tion in the corresponding side. Using the reﬂection principle (for another interval 52 CONTENTS

(xj, xj+1)) gives another brach of the function, which we denote by f1(z). However, 00 00 f (z) f1 (z) the function 0 = 0 is uniquely deﬁned. f (z) f1(z)

1 α We can observe that g(z) = (f(z) − f(xi)) i maps the interval (xi − , xi + ) onto another line segment. Thus again by the reﬂection pruinciple we see that g(z) αi is analytic. We can then write that f(z) = f(zi) + g(z) , for z near to xi and thus 00 00 f (z) (αi − 1) f1 (z) 0 = + hi(z) = 0 f (z) z − xi f1(z) where hi : H → C is analytic.

Moreover, we can write globally that

00 k f (z) X (αi − 1) = + h(z) f 0(z) z − x i=1 i where h : H → C is analytic - and h extends to an entire function. However, w can deuce that it is the constant function. Schwartz - Christoffel for D. We can reformulate this in terms of the unit disk (to look more like the Riemann Mapping Theorem, except that it goes in the wrong direction). Let D = {z ∈ C : |z| < 1} denote the unit disk. Let P be an open bounded region in the complex plane with polygonal boundary and vertices V = {v1, ··· , vn}, say. The Schwarz-Christoffel theorem gives a (fairly) explicit formula for a holomorphic bijection ψ : D → P. Corollary 7.10. More precisely, the map ψ : D → P defined by: Z z ξ α1−1 ξ αk−1 ψ(z) = A + C 1 − ··· 1 − dξ. (1) z0 w1 wk where (1) The exponents α1, ··· , αk correspond to the internal angles of P; (2) the points W = {w1, ··· , wk} ∈ ∂D on the unit circle correspond to preimages of the vertices V ; 7. RIEMANN MAPPING THEOREM 53

7.3. examples.

Example 7.11 (Maps to a semi-infinite strip). Consider a semi-infinite strip P in the z-plane. (This may be regarded as a limiting form of a triangle with vertices 0, πi, and R (with R real), as R tends to infinity). Now α1 = 0 and α2 = α3 = π/2 in the limit.

Suppose we guess that x1 = −1 and x2 = 1 and then we are looking for the mapping f : H → P with f(−1) = πi, f(1) = 0 (and f(∞) = ∞). Then f(z) is given by Z z C Z z C f(z) = A + dw = A + dw p p 2 z0 (w − 1)(w + 1) z0 (w − 1) In this special case the integral can be explicitly evaluated to get f(z) = A + C cosh−1(w) where A, C ∈ C. Requiring that f(−1) = πi and f(1) = 0 gives A = 0 and C = 1. Hence the Schwarz-Christoffel mapping is given by f(z) = cosh−1(z). Example 7.12 (Maps to another semi-infinite strip). Consider the semi-infinite strip S = {x + iy : − a < x < a and y > 0} for a > 0. Find an explicit form for the holomorphic bijection f : H → S from the upper half plane to S for which extends to f(±1) = ±a and f(∞) = ∞. 54 CONTENTS

The strip has internal angles π/2. We take the two points to be mapped to the ±a to be ±1. By the Christoffel-Schwarz Theorem this takes the form Z z Z z − 1 − 1 dw f(z) = A (w − 1) 2 (w + 1) 2 dw + B = A √ + B 2 0 0 1 − w We can evaluate the indefinite integral: f(z) = B + A sin−1(z) The image −a = f(−1) of the first vertex −1 gives that Aπ −a = B + A sin−1(−1) = B − . 2 The image a = f(1) of the second vertex 1 gives that Aπ a = B + A sin−1(1) = B + . 2 from which we deduce that B = 0 and thus A = 2a/π and thus 2a sin−1(z) f(z) = π Example 7.13 (An equilateral triangle). Let f : H → C be defined by Z z dw f(z) = . 2/3 0 (w(1 − w))

√ 1 3 We claim that the image is an equilateral triangle with vertices at 0, 1, 2 + 2 . π The internal angles are all equal to 3 . It we consider the real axis as the boundary of H then we see that the interval [0, 1] is mapped to [0,K],say, since the integrand R 1 dt is real. Here K = 0 2/3 = 5.299 ··· . (t(1−t)) √ The interval [1, ∞) is mapped to the side [K, K + i 3 K) and the interval √ 2 2 K 3 (−∞, 0] is mapped to the side ( 2 + i 2 K, 0] More generally, if the triangle with angles πα1, πα2 and π(1 − α1 − α2) then let g : H → C be deﬁned by Z z dw f(z) = A + C . 1−α1 1−α2 0 (w − 1) (w + 1)

(i.e., x1 = −1, x2 = 1 (and x3 = ∞). The image is a triangle with angles πα1, πα2 (and π(1 − α1 − α2)). 55

Example 7.14 (A right angle triangle). Consider the triangle ∆ in C with vertices at −1, 0 and i and ﬁnd the analytic bijection f : H → ∆ which extends to

f(−1) = i, f(1) = −1 and f(∞) = 0.

Show that −1 − i Z z i − 1 f(z) = (z2 − 1)−3/4dz + 2κ 0 2 where Z −1 κ = (z2 − 1)−3/4dz 0 π The vertices i and −1 have internal angles of 4 and are the images of −1 and 1, respectively, We can use the Christoﬀel Schwarz Theorem to write F (z) = (z + 1)−3/4(z − 1)−3/4 = (z2 − 1)−3/4 We have that Z z f(z) = A (z2 − 1)−3/4dz + B 0 and then Z −1 Z 1 i = A (z2 − 1)−3/4dz + B and − 1 = A (z2 − 1)−3/4dz + B. 0 0 If we denote Z 1 Z 0 κ = (z2 − 1)−3/4dz = − (z2 − 1)−3/4dz 0 −1 −1−i i−1 then −Aκ + B = i and Aκ + B = −1 and we can solve for A = 2κ and B = 2 . Hence −1 − i Z z i − 1 f(z) = (z2 − 1)−3/4dz + . 2κ 0 2 Example 7.15 (Squares and Rectangles). Let g : H → C be deﬁned by Z z dw f(z) = . 1/2 0 (w(1 − w)(ρ − w)) where ρ > 1. The image is a rectangle vertices at 0, −K, −K − iK0, −iK0. 56 3

π More precisely, the internal angles are all equal to 2 . It we consider the real axis as the boundary of H then we see that the interval [0, 1] is mapped to the side [−K, 0] where Z 1 dw K = . 1/2 0 (w(1 − w)(ρ − w)) For 1 < w < ρ there is a single imaginary root and thus the integral from [1, ρ] is purely imaginary (with negative imaginary part) and the image of [1, ρ] is [−K, −K − iK0] where Z ρ 1 K0 = dw. 1/2 1 (w(w − 1)(ρ − w)) Similarly, we see that the image of [ρ, ∞) is [−K − iK0, −iK0) and the image of (−∞, 0] is (−iK0, 0]. The upper half plane H is mapped to the square by Z dw z = f(w) = pw(w − 1)(w + 1) 7.4. Generalizations of the Schwarz-Christofel theorem. More generally one can consider an analogous problem where we introducce holes into the picture. Let z1, . . . , zd ∈ D be the centres of subdisks Di = B(zi, ri) = {z ∈ D : |z − zi| < ri} where |zi| + ri < 1. We can then associate a “multiply connected d region” defined by M = D − ∪i=1Di. When d = 1 this simply corresponds to an annulus. Similarly, we can consider sub-polygons P1, ··· , Pd ⊂ P with sets of vertices d V1, ··· ,Vd, respectively. We then let Q = P − ∪i=1Pi Let Ci = ∂Di and let Ci denote the images on reflection in the unit circle 2 rj z {z ∈ : |z| = 1} = ∂ . In particular, gi(z) = zi + satisfies gi(Ci) = Ci. Let C D 1−zj z Γ = hg1, ··· , gdi be the associated Schottky group. By a result of Crowdy et al, there is an explicit formula for a holomorphic bijection ψ : D → P given by

Z z k d |Vi| Y Y Y (i) (j) αi αi ψ(z) = C(·) w(z, wi) w(z, wj ) dz. (2) z0 i=1 i=1 j=1 where w(·, ·) is the Schottky-Klein prime function and

(1) the exponents α1, ··· , αk correspond to the internal angles of P and ex- (j) (j) (j) ponents α , ··· , αn correspond to the internal angles of ; 1 |Vi| P (2) the points W = {w1, ··· , wk} ∈ ∂D on the unit circle correspond to (i) (i) (i) preimages of the vertices V and the points W = {w , ··· , w } ∈ ∂Di 1 |Vi| on the circle correspond to preimages of the vertices Vi; (3) C(·) is an explicit correction function.

7.5. Application: Fluid dynamics. The streamlines around a circular region of diameter 2a can be described by transforming the horizontal lines past a slit of width 4a by the associated conformal map f(z) = z + a2/z. More generally, given a ﬁnite number of disjoint circular regions the stream lines described by transforming the horizontal lines past a slits. ??

8. Behaviour of analytic functions There are very useful tools in complex analysis: The Maximum Principle, and the Schwarz and Pick Lemmas 8. BEHAVIOUR OF ANALYTIC FUNCTIONS 57

8.1. The Maximum Modulus Principle. The next theorem shows that the supremum of the modulus of an analytic function isn’t achieved inside U.

Theorem 8.1 (Maximum Modulus Principle). Let f : U → C be analytic. Let U be connected. If z0 ∈ U satisﬁes |f(z0)| ≥ |f(z)| for all z ∈ U then f(z) is necessarily a constant function. Proof. By multiplying by a constant, if necessary, we can assume without loss + of generality that f(z0) ∈ R . Let

S = {z ∈ U : |f(z)| = f(z0)}. Since f is continuous we see that S ⊂ U is a closed set. We claim that S ⊂ U is also open. Given z ∈ S ⊂ U, since U is open we can choose r > 0 suﬃciently small that B(z, r) ⊂ U. Then for any 0 < r0 < r we have by Cauchy’s Theorem that

1 Z f(ξ)

f(z0) = |f(z)| = dξ 2πi C(0,r0) z − ξ Z 2π 0 iθ 1 f(z + r e ) iθ = 0 iθ (ie )dθ 2πi 0 r e Z 2π 1 0 iθ = f(z + r e )dθ 2πi 0 Z 2π 1 0 iθ ≤ f(z + r e ) dθ ≤ f(z0) = |f(z)|, 2π 0 0 it where |f(z + r e )| ≤ |f(z)| = f(z0), since z ∈ S. Thus all the inequalities must be equalities and so 0 iθ |f(z + r e )| = f(z0) for all 0 ≤ θ < 2π and all 0 < r0 < r. In particular, S contains the open ball B(z, r) ⊂ S. But since S is both open and closed and U is connected we deduce that S = U and thus f takes the constant value f(z0). 8.2. The Schwarz Lemma. The next result describes analytic maps of the disk D = {z ∈ C : |z| < 1} to itself. Theorem 8.2 (Schwarz Lemma). Let f : D → D be an analytic map. (1) Let f(0) = 0 then: (a) |f(z0)| ≤ |z0|, ∀z0 ∈ D; and (b) |f 0(0)| ≤ 1. (2) Moreover, if either |f(z)| = |z| for some z ∈ D − {0} or |f 0(0)| = 1 then f is a “rotation” of the form f(z) = eiθz, for some 0 ≤ θ < 2π.

Proof. (1) The function g : D → C defined by ( f(z) if z 6= 0 g(z) = z f 0(0) if z = 0 is analytic on D. Fix > 0. We can apply the Maximum Modulus Principle to the restriction g : B(0, 1 − ) → C to deduce that for |z| ≤ 1 − we have that 1 |g(z)| ≤ . 1 − Letting → 0 we deduce that |g(z)| ≤ 1 whenever |z| ≤ 1. From the definition of g this completes the proof of the first part. (2) For the second part, if |f(z0)| = |z0| then |g(z0)| = 1. The maximum principle principle applied to g now tells us that g has a constant value, eiθ, say. This implies that f(z) is a rotation. 58 CONTENTS

Similarly, if |f 0(0)| = |g(0)| then the maximum principle principle applied to g iθ again tells us that g has a constant value, e , say.

Application 8.3. We denote a M¨obiustransformations φa : D → D by z − a φ (z) = a 1 − az

iθ where a ∈ D. We can also denote rotations ρθ : D → D by ρθ(z) = e z, for 0 ≤ θ < 2π.

Proposition 8.4. Any analytic function f : D → D which is a bijection is of the form f = φa ◦ ρθ, for some a ∈ D and 0 ≤ θ < 2π.

Proof. We ﬁrst observe that φa(∂D) = ∂D. If |z| = 1 then

z − a z − a |φa(z)| = = = 1. 1 − az z − a

By the Maximum Modulus Principle, since φa is not a constant function if |z| < 1 −1 then |φa(z)| < |z| < 1. In particular, φa(D) ⊂ D. Observe that φa = φ−a, showing that φa : D → D is a bijection. Let us denote f(0) = b, say. Consider g = φb ◦ f, which is now an analytic function g : D → D for which g(0) = 0. By the Schwarz Lemma, |g0(0)| ≤ 1. Similarly, we can apply the Schwarz lemma to g−1 : D → D to deduce that −1 0 1 0 |(g ) (0)| = |g0(0)| ≤ 1. We therefore have |g (0)| = 1. By the uniqueness part of the Schwarz Lemma we have that g(z) = ρθ for some 0 ≤ θ < 2π. But then this means that f(z) = φ−b ◦ ρθ.

8.3. Pick’s lemma. There is a generalization of Schwarz’s lemma in which we don’t require f(0) = 0

Theorem 8.5 (Pick’s Lemma). Let f : D → D be analytic. (1) For z1, z2 ∈ D,

f(z1) − f(z2) z1 − z2 ≤ (∗) 1 − f(z1)f(z2) 1 − z1z2 and 2 0 1 − |f(z1)| ) |f (z1)| ≤ 2 (∗∗) 1 − |z1|

(2) If there is either an equality in (*) for z1 6= z2, or an equality in (**) for some z1, then f must be an analytic bijection. Proof. Let z − z f(z ) − z φ(z) = 1 and ψ(z) = 1 . 1 − z1z 1 − f(z1)z Then ψ ◦ f ◦ φ satisﬁes the hypothesis of the Schwarz Lemma. Thus for any z ∈ D, |ψ ◦ f ◦ φ(z)| ≤ |z|.

Setting z = φ−1 gives the inequality (1). The Schwarz Lemma also gives that

|(ψ ◦ f ◦ φ)0(z)| ≤ 1.

Using the chain rule gives the inequality (2). 9. HARMONIC FUNCTIONS 59

9. Harmonic functions We want to consider the real valued functions.

Definition 9.1. We say that any function u : U → R is harmonic if ∆u = 0 where ∂2 ∂2 ∆ = + ∂x2 ∂y2 Remark 9.2. Any scaler multiple of linear combinations of harmonic functions are harmonic.

In particular, if f : U → C is analytic and f(x + iy) = u(x, y) + iv(x, y) satisfy the Cauchy-Riemann equations then diﬀerentiating ∂u ∂v = ∂x ∂y with respect to x implies that ∂2u ∂2v = ∂x2 ∂y∂x and diﬀerentiating ∂u ∂v = − ∂y ∂x with respect to y implies that ∂2u ∂2v = − . ∂y2 ∂y∂x Thus we have shown: Theorem 9.3. We have that u = Re(f) is harmonic. Similarly, we see that v = Im(f) is harmonic.

Example 9.4. The function f : C → C given by f(z) = z2 is analytic and thus f(z) = (u(x, y) + iv(x, y)) = (x + iy)2 = x2 − y2 + 2ixy. where u(x, y) = x2 − y2 and v(x, y) = xy are harmonic.

Example 9.5. Let f : U → C be analytic and for z = x + iy and write 1 2 2 f(z) = u(z) + iv(z). Let w(z) := log |f(z)| = 2 log(u(z) + v(z) ) then ∂w 1 ∂u ∂v = u(z) + v(z) ∂x u(z)2 + v(z)2 ∂x ∂x ∂w 1 ∂u ∂v = u(z) + v(z) ∂y u(z)2 + v(z)2 ∂y ∂y and ! ∂2w −2 ∂u ∂v 2 1 ∂u2 ∂v 2 ∂2u ∂2v = u(z) + v(z) + + + u(z) + v(z) ∂x2 (u(z)2 + v(z)2)2 ∂x ∂x u(z)2 + v(z)2 ∂x ∂x ∂x2 ∂x2 ! ∂2w −2 ∂u ∂v 2 1 ∂u2 ∂v 2 ∂2u ∂2v = u(z) + v(z) + + + u(z) + v(z) ∂y2 (u(z)2 + v(z)2)2 ∂y ∂y u(z)2 + v(z)2 ∂y ∂y ∂y2 ∂y2 Adding these, expressions, the second derivative terms vanish since u and v are harmonic. The ﬁrst order terms vanish by the Cauchy-Riemann equations. Remark 9.6. The equation ∆u = 0 is called Laplace’s equation. 60 CONTENTS

9.1. Harmonic conjugates. Definition 9.7. If u, v are harmonic and satisfy the Cauchy-Riemann equations in u then v is called the harmonic conjugate of u. Conversely, if u is harmonic then it is the real part of an analytic function.

Theorem 9.8. If u : U → R is a C2 function there there exists a harmonic conjugate, i.e., there exists a C2 function v : U → R such that the function f : U → C defined by f = u + iv is analytic. Proof. Consider the case U = {z = x + iy : x ∈ (a − δ, a + δ) and y ∈ (b − , b + )}, and let f, g : U → R be C1 functions defined by ∂u(z) ∂u(z) f(z) = − and g(z) = . ∂y ∂x Then since ∆u = 0 we have that ∂f ∂g ∂2u ∂2v = = − = . (1) ∂y ∂x ∂y2 ∂x2 If we define Z x Z y v(x, y) = f(t, b)dt + g(x, s)ds (2) a b 2 ∂h then v : U → R is a C function such that ∂y = g. The derivative with respect to x is a little more complicated. We can write ∂ Z x f(t, b)dt = f(x, b). (3) ∂x a For the second term in (2) we have that ∂ Z y Z y ∂g Z y ∂f g(x, s)ds = (x, s) ds = (x, s)ds = f(x, y) − f(x, b). ∂x b b ∂x b ∂y (4) | {z } ∂f = ∂y by (1) Comparing (2), (3) and (4) gives that ∂v = f = f(x, b) + (f(x, y) − f(x, b)) = f(x, y). ∂x ∂u ∂v ∂u ∂v Finally, ∂x = g = ∂y and ∂y = −f = ∂x . Exercise 9.9. Given u(x, y) = xy find the harmonic conjugate of u. In particular, ∂u ∂v 1 = y = =⇒ v(x, y) = y2 + f(x) ∂x ∂y 2

∂v ∂u 1 = f 0(x) = − = x =⇒ f(x) = − x2 + C ∂x ∂y 2 1 2 2 Therefore v(x, y) = 2 (−x + y ) + C and f(z) = u + iv (−x2 + y2) = xy + i + C 2 i = − (x + iy)2 + C 2 i = − z2 + C. 2 9. HARMONIC FUNCTIONS 61 which is analytic. One can always ﬁnd a harmonic conjugate on a disk, but not on any general region. Example 9.10 (Non-example). Consider the function u(x, y) = log(x2 + y2)1/2 on the domain U = C − {0}. Then ∂u x = ∂x x2 + y2 ∂u2 (x2 + y2) − 2x2 y2 − x2 = = ∂x2 (x2 + y2)2 (x2 + y2)2 and by symmetry ∂u2 x2 − y2 = ∂y2 (x2 + y2)2 Thus ∆u = 0, i.e., u is harmonic. Assume for a contradiction we can ﬁnd v such that f(z) = u + iv is analytic. Then ∂u ∂u ∂v ∂u x iy 1 1 f 0(z) = − i = + i = − = = ∂x ∂y ∂y ∂x x2 + y2 x2 + y2 x + iy z Thus f(z) = log z + C. But being analytic the function must be continuous contradicting the jump between branches. 9.2. Poisson Integral formula. A harmonic function on the unit disk can be written be written in terms of its values on the unit circle.

Theorem 9.11 (Poisson integral formula). Let z ∈ D and assume that u : U → C be harmonic and U ⊃ D. We can write Z 2π iθ 1 it u(re ) = Pr(θ − t)u(e )dt 2π 0 where 1 − r2 P (θ) = . r 1 − 2r cos θ + r2 Proof. Let f(z) = u(z)+iv(z) be the associated analytic function, where v(z) is the harmonic conjugate of u(z). z Given z ∈ D, let z1 = |z|2 ∈ C − D be the reﬂection in the unit circle. By Cauchy’s theorem (applied twice) 1 Z f(ξ) 1 Z f(ξ) f(z) = dξ − dξ 2πi C(0,1) z − ξ 2πi C(0,1) z1 − ξ | {z } | {z } =f(z) =0 62 CONTENTS

it iθ 1 −iθ Let us denote ξ = e and z = re and z1 = r e and then with this parameterization we can write 1 Z 2π 1 1 f(z) = f(eit) − ieitdt iθ it 1 iθ it 2πi 0 re − e r e − e 1 Z 2π 1 1 = f(eit) − dt i(θ−t) 1 i(θ−t) 2π 0 re − 1 r e − 1 Z 2π −i(θ−t) 1 −i(θ) ! 1 it 1 − re 1 − r e = f(e ) 2 − 1 2 dt 2π 0 1 + r − 2r cos(θ − t) 1 + r2 − r cos(θ − t) Z 2π −i(θ−t) 2 −i(θ−t) 1 it 1 − re r − re = f(e ) 2 − 2 dt 2π 0 1 + r − 2r cos(θ − t) r + 1 − 2r cos(θ − t) Z 2π 2 1 it 1 − r = f(e ) 2 dt 2π 0 1 + r − 2r cos(θ − t) | {z } ∈R where in the middle we multiply both expressions by the complex conjugates of their respective denimonators. Taking real and imaginary components, the result follows. Remark 9.12. The Poisson kernel can be written in different ways. For example, we can write for z = reiθ 1 − |z|2 P (θ − t) = . r |1 − ze−it|2 Remark 9.13. When z = 0 we have the mean value property: 1 Z 2π u(0) = u(eit)dt. 2π 0 Application 9.14. The Laplace equation appears in the heat equation, one physical interpretation of this problem is as follows: fix the temperature on the boundary of the domain according to the given specification of the boundary condi- tion. Allow heat to flow until a stationary state is reached in which the temperature at each point on the domain doesn’t change anymore. The temperature distribu- tion in the interior will then be given by the solution to the corresponding Laplace problem. Harmonic functions have various restrictions on their values.

Corollary 9.15 (Harnack). Assume that u : U → R is harmonic and U ⊃ D. Then for any z ∈ D we have that 1 − |z| 1 + |z| u(0) ≤ u(z) ≤ u(0). 1 + |z| 1 − |z| 9. HARMONIC FUNCTIONS 63

Proof. We observe that for any 0 ≤ t < 2π and z ∈ D: 1 − |z|2 1 − |z|2 1 + |z| ≤ = . |eit − z|2 (1 − |z|)2 1 − |z| Thus by the Poisson Integral Theorem 1 + |z| 1 Z 2π 1 + |z| u(z) ≤ u(eit)dt = u(0). 1 − |z| 2π 0 1 − |z| The other inequality is proved similarly. The following is the analogue of Montel’s theorem for analytic functions.

Corollary 9.16 (Harnack’s Principle). Assume that uj : U → R, j ≥ 1, are harmonic functions with

u1 ≤ u2 ≤ u3 ≤ · · ·

Then either uj → ∞ uniformly on compacts sets or there is a harmonic function u : U → R such that uj → u uniformly on compact sets. This is an exercise.

Corollary 9.17 (Jensen). Let f : U → R be analytic and let U ⊃ D. Let a1, ··· , aN ∈ D be the zeros of f(z) with |aj| < 1, for j = 1, ··· ,N, and assume there are no zeros on C(0, 1). Then

N X 1 Z 2π log |f(0)| = log |a | + log |f(eit)|dt j 2π j=1 0 Proof. If f(z) has no zeros then u(z) := log |f(z)| is harmonic and by the Poisson Integral Formula (at z = 0) we have that 1 Z 2π log |f(0)| = log |f(eit)|dt. (1) 2π 0

More generally, if f(z) has zeros a1, ··· , aN ∈ D then we can deﬁne

 N  Y 1 − zaj F (z) := f(z)  z − a  j=1 j which is now a zero free analytic function. We can apply (1) to F (z) to deduce that N X 1 log |F (0)| = log |f(0)| + log a i=1 i 1 Z 2π = log |F (eit)|dt 2π 0   Z 2π N it 1  it X 1 − aie  = log |f(e )| + log  dt 2π  eit − a  0  i=1 j  | {z } =1 1 Z 2π = log |f(eit)|dt 2π 0 since the maps z 7→ 1−zai preserve the unit circles. z−ai Application 9.18 (The Dirichlet problem). We can construct harmonic functions on D from a continuous function on the unit circle. 64 CONTENTS

Theorem 9.19. Let U : C(0, 1) → R be continuous on the boundary of the unit disk D. Then the function u : D → R defined by ( U(z) if z ∈ C(0, 1) u(z) = 1 R 2π it iθ 2π 0 Pr(θ − t)f(e )dt if z = re ∈ D is harmonic (i.e., ∆u = 0) Proof. We can rewrite 1 − |z|2 eit e−it P (θ − t) = = + − 1 r |z − eit|2 eit − z e−it − z where z = reiθ. Then Z 2π 1 it u(z) = Pr(θ − t)U(e )dt 2π 0 Z 2π it −it 1 e e it = it + −it − 1 U(e )dt. 2π 0 e − z e − z For the first integral we have observe that 2 Z 2π it Z 2π it ∂ e it −2e it 2 it U(e )dt = it 3 U(e )dt ∂x 0 e − (x + iy) 0 (e − (x + iy)) and 2 Z 2π it Z 2π it ∂ e it 2e it 2 it U(e )dt = it 3 U(e )dt ∂y 0 e − (x + iy) 0 (e − (x + iy)) and thus adding these two expressions gives Z eit ∆ U(eit)dt = 0. eit − (x + iy) The second integral is similarly. The final term is a constant. Thus three terms are harmonic, and thus so is the sum. To see that u(z) is continuous (on the boundary) we can write Z 2π Z 2π 2 1 it 1 1 − |z| it Pr(θ − t)U(e )dt = it U(e )dt 2π 0 2π 0 |z − e | 1 Z 2π eit + z = U it dt 2π 0 1 + ze 2 w+z 0 1−|z| using the change of variables φ−z(w) = 1+zw (with |φ−z(w)| = |1+zw|2 ) for each z ∈ D. Moreover, since for each ξ ∈ C(0, 1) we have eit + z lim = ξ z→ξ 1 + zeit

if ξ 6= −eit. The result follows from bounded convergence of integrals.

9. HARMONIC FUNCTIONS 65

More generally, we see that given a simple closed curve γ and a continuous function U :→ C we can ﬁnd a harmonic function u inside γ with agrees with U on γ. It suﬃces to solve the problem for the unit circle, as above, and then use the Riemann mapping theorem. However, for the circle we can use the Poisson integral formula. 9.3. Mean Value Property and Maximum Principle for harmonic functions. The following is a characteristic property of harmonic functions.

Theorem 9.20 (Mean Value Property). Let u : U → R be harmonic. For any B(z, r) ⊂ U we have that 1 Z 2π u(z) = u(z + reiθ)dθ. 2π 0 for any suﬃciently small r > 0.

Proof. Let r1 > r > 0 such that B(z, r1) ⊂ U. Let v : B(z, r1) → R be the harmonic conjugate for u : B(z, r1) → R and denote the analytic function f : B(z, r1) → C deﬁned by f(z) = u(z) + iv(z).

By the Cauchy theorem 1 Z f(ξ) f(z) = u(z) + iv(z) = dξ 2πi C(z,r) ξ − z Z 2π iθ 1 f(z + re ) iθ = iθ ire dθ 2πi 0 re 1 Z 2π 1 Z 2π = u(z + reiθ)dξ + i v(z + reiθ)dξ 2π 0 2π 0 Taking real parts gives the result. This is also the analogue of the maximum principle for analytic functions.

Theorem 9.21. Let u : U → R be harmonic. If there exists z0 ∈ Usuch that u(z0) = supz∈U |u(z)| then the function u is constant. Proof. We can give two proofs:

First proof (using harmonic conjugates): Let M = {w ∈ U : u(w) = sup |u(z)|}. z∈U

Clearly, z0 ∈ M= 6 ∅. It suffices to show that M ⊂ U is open, since U being connected would imply M = U. Let z ∈ M. For small r > 0 we can consider B(z, r) ⊂ U. Let v : B(z, r) → R be the harmonic conjugate for u : B(z, r) → R and denote the analytic function f : B(z, r) → C defined by f(z) = u(z) + iv(z). We define g(z) = ef(z) then 66 CONTENTS

|g(z)| = supξ∈B(z,r) |g(ξ)|. We can apply the maximum modulus principle for analytic functions to deduce that g(z) is constant on B(z, r), and thus u(z) = log |g(z)| is constant on B(z, r). Thus M is open.

Second proof (using mean value property): Let L := supz∈U |u(z)| and let M = {w ∈ U : u(w) = L}.

By hypothesis z0 ∈ M 6= ∅. Since g is continuous we see that M is closed. It suffices to show it is also open. Let z ∈ M and choose r > 0 such that B(z, r) ⊂ U and then 1 Z 2π L = g(z) = g(z + reiθ)dθ ≤ L 2π 0 and thus g(z + reiθ) = L for 0 ≤ θ < 2π, for all sufficiently small r > 0. We deduce that M is open. We can use this to get the converse to the first result.

Theorem 9.22. If h : U → R satisfies the mean value theorem then it is harmonic. Proof. Assume that B(z, r) ⊂ U. By using the Poisson Integral Theorem we have a harmonic function u : B(z, r) → R which coincides with h on the boundary C(z0, r). It suffices to show that u and h therefore coincide on all of B(z, r). Let us write g(z) = u(z) − h(z), which vanishes on C(z0, r). Moreover, it satisfies the mean value theorem since this holds for both u (since it is harmonic) and h (by hypothesis). By the previous lemma we see that g ≤ 0 (since otherwise the supremum would occur in B(z, r) and this would make the function constant, giving a contradiction). Replacing w by −w we deduce that w = 0, i.e., u(z) = h(z) on B(z, r). Since the disk is arbitrary we deduce that h(z) is analytic on U. 9.4. Schwarz reflection principle for harmonic functions. We can now formulate an analogue of the Schwarz reflection principle for functions.

Theorem 9.23. Given a domain U and suppose that U ∩ R = (a, b). Let U + = {z ∈ U : Im(z) > 0}. Assume v : U + → R is harmonic and for each − z0 ∈ (a, b) we have that limz→z0 v(z) = 0. Let U = {z : z ∈ U} and define  v(z) if z ∈ U +  w(z) = 0 if z ∈ (a, b) −v(z) if z ∈ U − Then w : U + ∪ (a, b) ∪ U − → R is harmonic. Proof. We claim that the Mean Value Theorem holds on a sufficiently small disk about any point. If z ∈ U + then it follows form u being harmonic. Similarly, it holds for z ∈ U −. If z ∈ (a, b) then for sufficiently small r > 0 we have that 1 Z 2π w(z + reiθ)dθ 2π 0 1 Z π 1 Z π = w(z + reiθ)dθ + w(z + reiθ+iπ)dθ 2π 0 2π 0 1 Z π 1 Z π = w(z + reiθ)dθ + w(z − reiθ)dθ 2π 0 2π 0 1 Z π 1 Z π = w(z + reiθ)dθ − w(z + reiθ)dθ = 0. 2π 0 2π 0 9. HARMONIC FUNCTIONS 67 since w(x + iy) = −w(x − iy). Since w(z0) = 0 the Mean Value Property holds.

9.5. Application: Continuous extensions to the boundary of analytic bijections. We can use the Schwarz reﬂection principle for analytic functions to give another (and more generalizable) proof that if f : U → D is an analytic bijection where U ⊂ C is a domain whose boundary consists of straight line segments, or, more generally, circular arcs, then it must extend continuously to the boundary (i.e., the Osgood Caratheodory theorem). We can consider the case that U = P is a polygon, the other cases being similar. Theorem 9.24. Let f : P → D be an analytic bijection then it extends continuously to the boundary.

Proof. It is easy to see that if zn → z0 ∈ ∂P then |f(zn)| → 1 as n → +∞.

(Otherwise we could ﬁnd a subsequence with f(znk ) accumulating in the interior of

D and then this would pull back to a closed set containing znk , necessarily bounded 1 away from ∂P and giving a contradiction). Let B(0, 2 ) be the closed ball about 0. −1 1 Thus K = f B(0, 2 ) ⊂ P and we can assume that zn 6∈ K for suﬃciently large n. Let us assume that z0 is not a vertex of P then we can consider a small neighbourhood D(z0, ) := B(z0, ) ∩ P which is a half-disk, provided > 0 is suﬃciently small. Moreover, provided > 0 is small enough it is disjoint K and we have that

f : D(z0, ) → C is non zero. In particular, then write u(z) + iv(z) = i log |f(z)|.

However, since |f(zn)| → 1 we have that v(z) = Im(log |f(zn)|) → 0. We can now apply the Schwarz reﬂection principle for harmonic functions to v(z) Application 9.25 (Annuli and eccentricity). Two annuli are conformally equivalent iﬀ the ratio of the outer radius and the inner radius is the same for the two. Thus each is conformally equivalent to a unique ”standard” annulus r < |z| < 1 with 0 < r < 1. Theorem 9.26 (Riemann Mapping Theorem for Annuli). There is an analytic bijection between the annuli A1 = {z ∈ C : 1 < |z| < r1} and A2 = {z ∈ C : 1 < |z| < r2} if and only if r1 = r2.

Proof. Assume we have an analytic map f : A1 → A2 and thus that f extends continuously to the closed annuli by the previous result. By using the Schwarz reﬂection principle we can show that f extends to an (1) 1 1 1 analytic map between A = {z ∈ : < |z| < r1} and A = {z ∈ : < 1 C r1 2 C r2 |z| < r2}, i.e., the annuli obtained by reﬂecting A1 and A2 in the unit circle. 68 CONTENTS

(2) We can reﬂect again to show that f extends to an analytic map between A1 = {z ∈ : 1 < |z| < r2} and A(2) = {z ∈ : 1 < |z| < r2}. Continuing iteratively, C r1 1 2 C r2 2 ∞ (n) we have an analytic map f : C − {0} → C − {0} (since C − {0} = ∪n=1A1 = ∞ (n) ∪n=1A2 ).

Moreover, we can assume that limz→0 f(z) = 0 and limz→∞ f(z) = ∞ (the other cases being similar). In particular, we can deduce that f(z) is a M¨obiusmap and thus a rotation. 10. Singularities

We have previously considered analytic functions f : U → C and their zeros. What about the possibilities of singularities (where the function isn’t analytic)? 10.1. Removable singularities. The ﬁrst thing to consider is whether we really have a singularity at all, or if we could just extend the analytic function to the missing point.

Definition 10.1. We say that f : U → C has an isolated singularity at z0 if f is deﬁned and analytic in a punctured neighbourhood P (z0, r) = B(z0, r) − {z0}, but not deﬁned at z0.

Example 10.2. Let f(z) = 1/z then z0 = 0 is a singularity. The ﬁrst result show that at singularities the function blows up, in some sense.

Theorem 10.3 (Removable Singularities Theorem). Assume f : U → C is analytic expect possibly at z0 ∈ U. However, if U is bounded on some punctured ball P (z0, ) = {z ∈ C : 0 < |z − z0| < } then f(z) is analytic at z = z0 as well.

Proof. Since f is analytic and bounded on P (z0, ) we see ! (z − z )2 Z f(ξ) 2 0 0 (z − z0) f (z) = 2 dξ ≤ 4 sup |f(ξ)| (1) 2πi |z0| (z − ξ) |ξ−z0|= 2 ξ∈P (z0,) 10. SINGULARITIES 69

2 0 Thus (z − z0) f (z) is bounded on P (z0, )

We can then deﬁne ( (z − z )2f(z) if z 6= z g(z) = 0 0 0 if z = z0

then by hypothesis we have limz→z0 g(z) = 0 and by (1) we have that 0 2 0 lim g (z) = lim (z − z0) f (z) + 2(z − z0)f(z) = 0 z→z0 z→z0 In particular, g is C1, and by Moreira’s theorem we can deduce that g(z) is analytic. P∞ n We can consider the power series expansion g(z) = n=0 an(z − z0) . Since g(0) = 0 and g0(0) = 0 we deduce that   ∞  ∞  X n 2 X n g(z) = an(z − z0) = (z − z0)  an(z − z0)    n=2 n=2  | {z } f(z)

2 and thus f(z) = g(z)/(z − z0) is analytic, as required.

Example 10.4 (Non-example). Let f(z) = sin(z)/z then this has a singularity at z0 = 0, but this is a removable singularity. 10.2. Analogues of Cauchy’s theorem. We next look for analogues of Cauchy’s theorem in a neighbourhood of the singularity (and in a neighbourhood of ∞). Assume that f is deﬁned on the annulus

A(a, r, R) = {z ∈ C : r < |z − a| < R}

where r may be zero and R may be +∞. 70 CONTENTS

Lemma 10.5 (Analogue of Cauchy’s integral formula). Let f : A(a, r, R) → C be analytic. Let r < r2 < |z0 − a| < r1 < R then 1 Z f(z) 1 Z f(z) f(z0) = dz − dz 2πi C(a,r1) z − z0 2πi C(a,r2) z − z0

Proof. Deﬁne

( f(z)−f(z0) if z ∈ A − {z0} z−z0 F (z) = 0 f (z0) if z = z0 then since f is analytic (i.e., complex diﬀerentiable) at z0 we deduce that F is continuous on A, and analytic on A−{0}. In particular, by the removable singularities theorem we see that F is analytic. Hence Z Z F (z)dz = F (z)dz (1) C(a,r1) C(a,r2)

(by joining C(a, r1) and C(a, r2) by a line, along which the integrals in diﬀerent directions cancel, and applying Cauchy’s Theorem). Observe that that Z 1 Z 1 dz = 2πi and dz = 0. (2) C(a,r1) z − z0 C(a,r2) z − z0

Then comparing (1) and (2) gives: Z Z Z f(z) − f(z0) f(z) 1 dz = dz − f(z0) dz C(a,r1) z − z0 C(a,r1) z − z0 C(a,r1) z − z0 Z f(z) = dz − 2πif(z0) C(a,r1) z − z0 Z f(z) − f(z ) Z f(z) − f(z ) = 0 dz = 0 dz C(a,r2) z − z0 C(a,r2) z − z0 Z f(z) = dz. C(a,r2) z − z0 10. SINGULARITIES 71

Rearranging these equalities gives that 1 Z f(z) 1 Z f(z) f(z0) = dz − dz 2πi C(a,r1) z − z0 2πi C(a,r2) z − z0 as required. 10.3. Laurent’s Theorem. As a consequence of the last theorem we next have the analogue of power series expansions for analytic functions.

Theorem 10.6 (Laurent’s Theorem). Suppose that f : A(a, r, R) → C is analytic. Then there exists (an)n∈Z such that ∞ X n f(z0) = an(z0 − a) n=−∞ for all z0 ∈ A(a, r, R). Moreover, for any r < r1 < R we can write Z f(z) an = n+1 dz, for n ∈ Z. C(a,r1) (z − a) Moreover, this expansion is unique. Proof. We present the proof in two parts: First the existence of the expansion and then the uniqueness.

Existence of expansion. Let r < r2 < |z0 −a| < r1 < R then (by joining C(a, r1) and C(a, r2) by a line, along which the integrals in diﬀerent directions cancel, and applying Cauchy’s Theorem) we have 1 Z f(z) 1 Z f(z) f(z0) = dz − dz 2πi C(a,r1) (z − z0) 2πi C(a,r2) (z − z0) | {z } | {z } =:I1 =:I2 where ∞ X 1 Z f(z) I = a (z − a)n where a = dz for n ≥ 0 1 n n 2πi (z − a)n+1 n=0 C(a,r1) by analogy with the proof of the power series expansion for analytic functions.

The evaluation of I2 is similar, in as much as we can write 1 1 = − z − z0 z−a (z0 − a) 1 − z0−a ! 1 z − a z − a 2 = − 1 + + + ··· (z0 − a) z0 − a z0 − a 72 CONTENTS for |z − a| = r2 < |z0 − a|. Thus ∞ 1 Z f(z) X I = dz = a (z − a)−n 2 2πi (z − z ) −n 0 C(a,r2) 0 n=1 where Z Z 1 n−1 1 n−1 a−n = f(z)(z − a) dz = f(z)(z − a) dz 2πi C(a,r2) 2πi C(a,r1) since (z−a)n−1f(z) is analytic, which is justiﬁed by by the the uniform convergence of the geometric series and integration. Hence ∞ X n f(z0) = I1 − I2 = an(z0 − a) n=−∞ 1 where an is as given.

Uniqueness of expansion. To see the uniqueness of the coeﬃcients, suppose P∞ n k+1 f(z) = n=−∞ bn(z − a) then dividing by (z − a) and integrating around a circle C with radius between r and R, i.e., ∞ 1 Z f(z) X 1 Z dz = b (z − a)n−k−1dz = b 2πi (z − a)k+1 n 2πi k C n=−∞ C showing uniqueness. Example 10.7. We can consider the function 1 1 1 f(z) = = − (z − 1)(z − 2) (z − 2) (z − 1) on various domains. Observe that we can expand ( 1 − P∞ zn if |z| < 1 = n=0 P∞ −n z − 1 n=1 z if |z| > 1 and ( 1 − P∞ 2−n−1zn if |z| < 2 = n=0 P∞ n−1 −n z − 2 n=1 2 z if |z| > 2.

Case I: For z ∈ D(0, 1) we have an expansion ∞ X f(z) = − (1 − 2−n−1)zn. n=0 This is a power series which is analytic (and with no negative powers involved),

Case II: For z ∈ A(0, 1, 2) we have a Laurent series expansion ∞ ∞ X X f(z) = z−n − 2−n−1zn. n=0 n=0

Case III: For z ∈ A(0, 2, ∞) (i.e., |z| > 2) we have a Laurent series expansion ∞ ∞ X X f(z) = z−n − (2n−1 − 1)zn n=0 n=0 This brings us to the following estimate.

1 The integrals around C(a, r1) and C(a, r2) can now be replaced by integrals around any circle of radius between r and R (by connecting the circles C(a, r1) and C(a, r2) by line segments and integrating in both directions (so that these integrals will cancel). 10. SINGULARITIES 73

Corollary 10.8. Let f : A(0, r, R) → C be analytic and have a Laurent series ∞ X n f(z) = an(z − a) n=−∞

M(r1) then for any r < r1 < R we can bound |an| ≤ n where M(r) = sup|z−a|=r |f(z)|. r1 1 Proof. Since 1 Z f(z) an = n+1 dz 2πi C(a,r1) (z − a) we have that 2πr M(r) M(r1) |an| ≤ n+1 = n . 2π r1 r1 10.4. Classiﬁcation of singularities. We can now classify the diﬀerent types of singularities. Consider f : P (a, R) → C be analytic in a punctured disk P (a, R) centred at a singularity a and let ∞ X n f(z) = an(z − a) n=−∞ be its Laurent series on a punctured disk P (a, R) = A(a, 0,R) (where R > 0).

Definition 10.9. If a−n 6= 0 for inﬁnitely many positive n then a is called an essential singularity. If a−n = 0 for only many positive n then a and the least integer −n such that a−n 6= 0 is negative then a is called a pole of order n. In the particular case that a−1 6= 0 and ak = 0 for k ≤ −2 we call a a simple pole.

Remark 10.10. If the least integer n such that an 6= 0 is non-negative is non- negative then a is called a removable singularity for in this case

n n+1 f(z) = an(z − a) + an+1(z − a) + ··· for 0 < |z − a| < R and n ≥ 0. There fore we can deﬁne (or extend) f to a by ( 0 if n > 0 f(a) = lim = z→a a0 if n = 0

And with this value we can deﬁne f : D(a, R) → C and it is analytic. We shall always adopt the convention of removing singularities, in the future.

Example 10.11. Let f(z) = sin(z)/z = 1 − z2/3! + z4/5! − · · · then we can remove the singularity at z = 0 by setting f(0) = 1.

Definition 10.12. If the least n such that that an is strictly positive then f(z) is said to have a zero at a of order n. The order of a function at a is deﬁned by −∞ if a is an essential singularity  −n > 0 if a is a pole of order n Ord(f, a) = 0 if a is a removable singularity and f(a) 6= 0  n if a is a zero of order n

Definition 10.13. A function is said to be meromorphic if it is analytic in a domain U except at a set of points which are poles (with no essential singularity). 74 CONTENTS

10.5. Behaviour at singularities. We can consider the behaviour of a function near a zero or pole Theorem 10.14. Ord(f, a) = n where n is ﬁnite if and only if ∃α, β, δ > 0 such that α|z − a|n ≤ |f(z)| ≤ β|z − a|n whenever 0 < |z − a| < δ Proof. Assume that Ord(f, a) = n, then we can write n n+1 n f(z) = an(z − a) + an+1(z − a) + ··· = (an + g(z))(z − a) where an 6= 0. Moreover, g(z) → 0 as z 7→ a and thus given 0 < < 1 we can write |g(z)| < |an| provided |z − a| < δ, provided δ is small enough. In particular, we can write −n f(z)(z − a) = an + g(z). and thus −n |an|(1 − ) ≤ |f(z)(z − a) | ≤ |an|(1 + ) if δ is small enough, giving the required bound. Conversely, suppose α|z − a|n ≤ |f(z)| ≤ β|z − a|n when |z − a| is small enough then f(z)(z − a)−n is bounded in a punctured neighbourhood of a and hence a is a removable singularity. In particular, −n 2 f(z)(z − a) = an + an+1(z − a) + an+2(z − a) + ··· in a punctured neighbourhood P (a, δ) of a (since f(z)(z − a)−n is analytic at a by the removable singularity). We want to show that an 6= 0. But from the left hand inequality |f(z)(z − a)−n| is bounded below since α > 0, which would not be the case where an is equal to zero. Hence an 6= 0 and Ord(f, a) = n. The previous result gives a rough idea of the behaviour of f near a pole (when n is negative). The next result describes the behaviour near an essential singularity (i.e., f(P (a, δ)) ⊂ C is dense). Theorem 10.15 (Casorati-Weierstrass). Let f have an essential singularity at a. Then for any b ∈ C and every , δ > 0 there exists z such that 0 < |z − a| < δ such that |f(z) − b| < . Proof. Assume for a contradiction that there exist b, , δ such that for all z such that 0 < |z − a| < δ and |f(z) − b| ≥ . Consider the function g : P (a, δ) → C deﬁned by 1 g(z) = , (1) f(z) − b which, by assumption, is bounded by 1/ in P (a, δ). Then g has a removable singularity at a (and thus is bounded). In particular, we can write n n+1 g(z) = an(z − a) + an+1(z − a) + ··· where an 6= 0 and thus g(z) = (z − a)nh(z) (2) for an analytic function h : B(a, δ) → C with h(a) 6= 0. Thus comparing (1) and (2) gives 1 f(z) − b = . (3) (z − a)nh(z) Moreover, 1/h(z) is analytic in a neighbourhood of a (since h does not vanish in a neighbourhood of a by continuity and the fact h(a) 6= 0). But (3) shows that f has a pole at a of order n, contradicting the hypothesis of an essential singularity. 10. SINGULARITIES 75

Remark 10.16 (Picard’s Big Theorem). The following is am improvement on the WeierstrassCasorati theorem. Theorem 10.17 (Picard’s Big Theorem). If an analytic function f(z) has an essential singularity at a, then on any punctured disk B(a, δ the function f(z) takes on all possible complex values, with at most a single exception, inﬁnitely often. .

1. Show that the following series deﬁne a meromorphic function on C and determine the set of poles and their orders P∞ (−1)n (1) n=0 n!(n+z) P∞ 1 (2) n=0 z2+n2 P∞ 1 (3) n=0 (n+z)2 (4) 1 + P 1 + 1 z n∈Z−{0} z−n n P∞ (−1)n (5) n=0 n!(n+z)

2. Show that ∞ X z2 f(z) = n2z2 + 8 n=1 is deﬁned and continuous for the real values of z. Determine the region of he complex plane in which this function is analytic. Determine its poles.

3. Let f : C → C be a non-constant entire function. Show that the image of f is dense in C. 4. Show that the order of a pole is invariant under analytic bijections (i.e., if f : U → C has a pole at z0 of order n and φ : V → U is an analytic bijection with φ(w) = z then w is a pole for f ◦ φ : V → C of order n.

5. Let f : C → C be a meromorphic function with lim|z|→+∞ |f(z)| = ∞. Prove that f(z) is a rational function (i.e., the ratio of two polynomials).

6. Show that the Laurent series of a function on an annulus can be diﬀerentiated term by term to give the derivative of the function on the annulus.

7. Find the Laurent series for the following functions: (1) z/(z + 2) for |z| > 2, (2) sin(1/z) for z 6= 0, (3) cos(1/z) for z 6= 0, (4) 1/(z − 3) for |z| > 3.

8. Prove the following expansions: z P∞ 1 n (1) e = e + e n=1 n! (z − 1) , P∞ n n (2) 1/z = n=0(−1) (z − 1) for |z − 1| < 1 2 P∞ n (3) 1/z = 1 + n=1(n + 1)(z + 1) for |z + 1| < 1

9. Let f(z) be analytic on the annulus 0 < r ≤ |z| ≤ R. Prove that there is a function f1(z) analytic on |z| ≤ R and f2(z) analytic on |z| ≥ r and

f = f1 + f2 on the annulus. 76 CONTENTS

11. Entire functions, their order and their zeros

11.1. Growth of entire functions. Let f : C → C be an entire function. Definition 11.1. We say that f(z) is of ﬁnite order if there exists a > 0 and r > 0 such that a |f(z)| ≤ e|z| for |z| > r The least such a (i.e the largest lower bound) is called the order of the entire function. Example 11.2. (1) Any polynomial is of order 0. (2) The function f(z) = sin(z) is of order 1. (3) The function f(z) = sin(z2) is of order 2. We can denote Mf (r) = max |f(z)|. |z|=r This function conveys a lot of information about the zeros of the entire function.

Lemma 11.3. Let f : C → C be an entire function. If N ∈ Z+ and M (r) lim inf f < +∞ r→+∞ rN then f(z) is a polynomial of degree at most N + 1. Proof. Assume that f(z) has a power series expansion (about z = 0) ∞ X n f(z) = anz n=0 PN n and consider the associated polynomial P (z) = n=0 anz . The function g : C → C deﬁned by −N−1 2 g(z) := z (f(z) − P (z)) = aN+1 + aN+2z + aN+3z + ··· . −N is analytic and not identically zero. Assume that rn Mf (rn) is bounded for some sequence rn → +∞ then

Mf (rn) N+1 → 0 and thus Mg(rn) → 0 rn as n tends to inﬁnity. In particular, we can choose R > 0 such that

sup{Mg(r) : 0 ≤ r ≤ R} > R. By the maximum modulus principle applied to g(z) the disk B(0,R) we deduce N+1 that g(z) is constant (i.e., g(z) = aN+1) and that f(z) = P (z) + aN+1z is a polynomial. There is also a close connection between Mf (r) and the of zeros of f(z). Definition 11.4. We denote by n(r)the number of zeros of f(z) in the ball B(0, r) (i.e., n(z) = Card{z ∈ B(0, r): f(z) = 0}).

Lemma 11.5. Let f : C → C be an entire function with f(0) = 1. Then M (2r) n(r) ≤ f log 2 for all r > 0. Proof. We can apply Jensen’s formula to the function F (z) := f(z/2r), say, on the unit disk D to write n(2r) X 2r 1 Z 2π 0 = log |f(0)| = − log + log |f 2reiθ |dθ (1) |ai| 2π k=1 0 11. ENTIRE FUNCTIONS, THEIR ORDER AND THEIR ZEROS 77 where a1, ··· , an(2r) are the zeros of f(z) with |ai| < 2r (and a1, ··· , an(r) are the zeros of f(z) with |ai| < r). In particular,   n(r)  r  X   n(r) log 2 ≤ log 2   |ai|  k=1 |{z} ≤1   n(2r)  r  X   ≤ log 2   |ai| k=1 | {z } ≤1 1 Z 2π = log |f 2reiθ |dθ 2π 0 | {z } by (1)

≤ log Mf (2r) As a corollary, we have the following relationship between all the zeros and the order.

Lemma 11.6. Let f : C → C be an entire function of order a with f(0) = 1. If ∞ (ai)i=1 are the zeros for f(z) ordered by modulus, i.e.,

|a1| ≤ |a2| ≤ |a3| ≤ · · · ≤ |an| ≤ · · · then ∞ X 1 < +∞. |a |a n=1 n Proof. Let > 0, then for r > 0 sufficiently large we have that M(r) ≤ exp(ra+). Thus by the previous theorem r−(a−) r−(a−) n(r)r−(r−2) ≤ M(2r) ≤ 2ra+ → 0 as r → +∞. log 2 log 2 a+2 In particular, there exists r0 such that providing r > r0 have n(r) ≤ Cr . Thus a+2 since n ≤ n(|a|n) ≤ |an| we have that a+1 1 1 a+2 a+1 ≤ . |an| n 1 Providing 0 < < 2 we have that ∞ ∞ a+1 X 1 X 1 a+2 ≤ < +∞. |a |a+1 n n=1 n n=1 11.2. Factorization of entire functions. One of the most important applications of entire functions of finite order is their representation as infinite products. This can be thought of as a generalization of the simple representation of a polyno- n mial P (z) = anz + ··· + a1z + a0 in terms of a finite product n Y z P (z) = (a a ) 1 − n 0 α i=1 i 78 CONTENTS where α1, ··· , αn are zeros of P (z). Definition 11.7. We can define the elementary entire functions z2 zp E (z) = (1 − z) exp z + + ··· + p 2 p We have the following bound.

p+1 Lemma 11.8. If z ∈ D then |Ep(z) − 1| ≤ |z| .

Proof. We leave the proof as an exercise. It requires showing that Ep(z) has a power series expansion ∞ X n Ep(z) = 1 + bnz n=p+1 P∞ P∞ where bn ≤ 0. Then since 0 = Ep(1) = 1+ n=p+1 bn we have that 1 = n=p+1 |bn| and then ∞ ∞ X n p+1 X |Ep(z) − 1| = | bnz | ≤ |z| |bn| . n=p+1 n=p+1 | {z } =1 This leads to the following.

Lemma 11.9 (Weierstrass). If an occur with multiplicity pn then ∞ Y z f (z) := E ( ) 0 pn |a | n=1 n is an entire function with zeros an having multiplicity pn. Proof. Let r > 0 then there exists N such that for n ≥ N we have that |an| > r. We can then use the previous lemma to bound z 1 pn+1 |1 − Epn ( )| ≤ | | |an| an and thus ∞ X z Epn ( − 1) < +∞. |an| n=N In particular, ∞ ∞ Y z Y z E ( ) = 1 + E ( ) − 1 pn |a | pn |a | n=1 n n=1 n which is enough to guarantee convergence. This brings us to the statement of the main result.

Theorem 11.10 (Hadamard Factorization). If f(z) is an entire function of order a and with zeros (an) then we can write ∞ Y z f(z) = eQ(z) E ( ) pn |a | n=1 n where Q(z) is a polynomial of degree at most a. 12. PRIME NUMBER THEOREM 79

Proof. The detailed proof is omitted. However, we will sketch the outline. The idea is that given f(z) we can consider its zeros and thus construct the function g(z) f0(z). Moreover e := f(z)/f0(z) will be an entire function. Let r > 0 and let a1, ··· , an be the zeros for f(z) with |ai| < r. We can apply the Poisson formula to log |f(z)| to write

n 2 Z 2π iθ X r − aiz 1 re + z iθ log |f(z)| = − log + Re log |f(re )dθ (1) r(z − a ) 2π reiθ − z i=1 i 0

∂ ∂ We can apply ∂x − i ∂y to (1) to get that

0 Z 2π iθ f (z) X 1 X ai 1 2re = − + + log |f(reiθ)dθ f(z) z − a rr − a z 2π (reiθ − z)2 i i i i 0 We can then diﬀerentiate this a times with respect to z to get

a+1 Z 2π iθ X a! X a!ai (h + 1)! 2re − + + log |f(reiθ)dθ (z − a )a+1 r2 − a z 2π (reiθ − z)a+2 i i i i 0 We want to let r → +∞ and it is possible to show that this becomes X a! − . (2) (z − a )a+1 i i In particular, we have that f 0 f 0 ga+1(z) = Da (z) − Da 0 (z) f f0 The second term on the Right Hand Side of (2) is precisely (2). We deduce that ga+1 = 0 and thus g(z) is a polynomial of degree a.

Example 11.11. We can consider sin(z) which is entire, of order 1, and zeros at πn, where n ∈ Z. Thus Y z sin(z) = C 1 − . πn n∈Z (The function g(z) = az + b must be constant by the periodicity under z 7→ z + 2π.

12. Prime number theorem Perhaps one of the most interesting applications of complex analysis to another discipline is the proof of the prime number theorem in number theory.

Theorem 12.1. Let π(x) denote the number of prime numbers less than x then x π(x) π(x) ∼ log x as x → +∞, i.e., limx→+∞ x/ log x = 1. Example 12.2. We can compute: π(10) = 4, π(100) = 25, π(1, 000) = 168, π(1, 000, 000) = 78, 498 and π(1, 000, 000, 000) = 50, 847, 534.

x In 1808 Legrendre conjecture that π(x) ∼ log(x)−1.08366 . The correct form was proved independently by Hadamard and de la Valle-Poussin in 1896, using complex analysis and the zeta function introduced by Riemann in 1859. There are proofs which do not use complex analysis, including the ”elementary” proof of Selberg and Erd?s from 1949, but these are even more diﬃcult. 80 CONTENTS

12.1. Riemann zeta function. Recall that the Riemann zeta function is P∞ −s deﬁned by ζ(s) = n=1 n , provided Re(s) > 1. 1 Lemma 12.3. ζ(s) − s−1 is analytic for Re(s) > 0. Proof. For Re(s) > 1 we can write

∞ ∞ 1 X 1 Z ∞ 1 X Z n+1 1 1 ζ(s) − = − dx = − dx s − 1 ns xs ns xs n=1 1 n=1 n and bound 1 1 Z x 1 |s| − = s du ≤ . s s s+1 Re(s)+1 n x n u n This leads to the result.

1 Remark 12.4. The zeta function ζ(s) − s−1 actually extends to an entire function. The extension to Re(s) > 0 is given at the end.

Lemma 12.5. We can write Y −1 ζ(s) = 1 − p−s for Re(s) > 1 p where the product is over all primes p.

Proof. This follows from the prime factorization of natural numbers.

12.2. A related complex function. We deﬁne X log p Φ(s) = for Re(s) > 1 ps p where the sum is over all primes p.

1 Lemma 12.6. Φ(s) − s−1 is meromorphic for Re(s) > 1/2. Moreover, the poles occur at the zeros of ζ(s) in this region.

Proof. For Re(s) > 1 we have that ζ0(s) ∂ − = log ζ(s) ζ(s) ∂s | {z } meromorphic for Re(s) > 0 ! ∂ X = log(1 − p−s) ∂s p X log p = (ps − 1) p X log p X log p = + ps ps(ps − 1) p p | {z } | {z } =:Φ(s) converges for Re(s) > 1/2

Lemma 12.7. ζ(s) has no zeros on Re(s) = 1. (In particular, Φ(s) is analytic on Re(s) = 1). 12. PRIME NUMBER THEOREM 81

Proof. We can write for s = σ + it ζ(σ)3|ζ(σ + it)|4|ζ(σ + 2it)|    ∞ X X 1 = exp  3 + 4 cos(mt log p) + cos(2mt log p) ≥ 1(1)  mpmσ   p m=1 | {z } ≥0 (since 3 + 4 cos θ + cos(2θ) = 2(1 + cos θ)2 ≥ 0). Assume for a contradiction that 1 + it0 is a zero for ζ(s). Since s = 1 is a simple pole for ζ(s) we see that 3 4 lim ζ(σ) |ζ(σ + it0)| |ζ(σ + 2it0)| = 0 σ→1 giving the contradiction to (1). 12.3. A related counting function. We denote X Θ(x) = log p for x > 0 p≤x Lemma 12.8. We can bound Θ(x) ≤ (8 log 2)x Proof. Using the binomal expansion we can bound   2n Y X 22n = (1 + 1)2n ≥ ≥ p = exp log p = eΘ(2n)−Θ(n). n   n 0, then we can now bound k X Θ(x) ≤ Θ(2k+1) ≤ Θ(2m+1) − Θ(2m) m=0 | {z } ≤2m(2 log 2) k ! X ≤ 2m 2 log 2 m=0 = 2(2k+1 − 1)2 log 2 ≤ (8 log 2)x 12.4. Finiteness of integrals. The core of the proof is a ﬁnite integral, using Cauchy’s theorem. Lemma 12.9. The following integral os ﬁnite: Z y Θ(x) − x Z ∞ Θ(x) − x lim 2 dx = 2 dx < +∞ y→+∞ 1 x 1 x Proof. We break the poof up into simpler steps.

Step 1 (Relating the integral to Φ(s)): For Re(s) > 1 we have X log p Z ∞ Θ(x) Φ(s) = = s dx. (1) ps xs+1 p 1 Moreover, by a change of variables (x = et) we can write Z ∞ Z ∞ Z ∞ Z ∞ Θ(x) −st t 1 1 −st s+1 dx = e Θ(e )dt and = s+1 dx = e dt. (2) 1 x 0 s 1 x 0 Let us denote f(t) := Θ(et)e−t − 1 then by combining (1) and (2) we have that Z ∞ Φ(s + 1) 1 g(s) := f(t)e−stdt = − 0 (s + 1) s 82 CONTENTS for Re(s) > 0. Moreover, this is analytic on a open set U ⊃ {z : Re(z) ≥ 0}.

Step 2 (g(z) as a limit): For T > 0 we can consider the entire functions Z T Z log T ! −zt Θ(x) − x gT (z) = f(t)e dt = 2 dx 0 1 x We need to show that Z T Z ∞ lim f(t)dt = lim gT (0) = g(0) = f(t)dt. T →∞ 0 T →∞ 0 Let C = C(R, δ) be the boundary of the D-shaped region deﬁned my

D = {z ∈ C : |z| ≤ R, Re(z) ≥ −δ}.

Given R, provided δ > 0 is suﬃciently small, we have that g(z) − gT (z) is analytic on a neighbourhood of D (by Step 1) and thus by Cauchy’s theorem we can write 1 Z ezT z2 g(0) − gT (0) = (g(z) − gT (z)) 1 + 2 dz. (3) 2πi C z R We can write C = C+ ∪ C− where C+ = {z ∈ C : Re(z) ≥ 0} and C− = {z : Re(z) ≤ 0}

Step 3 (Integral over C+): For z ∈ C+ we can bound the integrand of (3) using Z ∞ −zt |g(z) − gT (z)| = f(t)e dt T Z ∞ ≤ max |f(t)| f(t)e−Re(z)tdt t≥0 T (4) | {z } =:B e−Re(z)T ≤ B Re(z) and 2 zT z 1 Re(z)T 2Re(z) e 1 + = e . (5) R2 z R2 (since if z = Reiθ then |1 + z2/R2| = |1 + ei2θ| = 2 cos(θ) = 2Re(z)). Together (4) and (5) give that the integral over C+ has a bound

1 Z ezT z2 B (g(z) − g (z)) 1 + dz ≤ . (6) 2πi T z R2 R C+ 12. PRIME NUMBER THEOREM 83

Step 4 (Integral of gT (z) over C−): By analyticity of gT (z) we can write

Z ezT z2 Z ezT z2 gT (z) 1 + 2 dz = gT (z) 1 + 2 dz (7) z R 0 z R C− C−

0 where we replace the integral over C− by that over C− = {z ∈ C : |z| = R, Re(z) <

0}.

Furthermore, we can again bound

Z T Z T −Re(z)T −zt −zt Be 0 |gT (z)| = | f(t)e dt| ≤ B |e |dt = (4 ) 0 0 |Re(z)| for Re(z) < 0 (compare with (4)). The analogue of (5) is

2 zT z 1 Re(z)T 2|Re(z)| e 1 + = e . (50) R2 z R2 Together (4’) and (5’) give us a bound on the contribution of (7) to (3) of

1 Z ezT z2 B

gT (z) 1 + 2 dz ≤ (8) 2πi 0 z R R C− (compare with (6)).

Step 5 (Integral of g(z) over C−): Finally, we can bound the remaining contribution to (3) coming from the integralover C− by

Z 2 g(z) z zT 1 + 2 e dz → 0 as T → 0 (9) C− z R | {z } Independent of T since Re(z) < 0 on C−.

Step 6 (Putting the bounds together): Finally, from the identity (3) and the estimates (6), (8) and (9) we have that 2B lim sup |g(0) − gT (0)| ≤ . T →+∞ R

However, since the R can be arbitrarily large we deduce that limT →∞ gT (0) = g(0). 84 CONTENTS

12.5. Proof of the Prime Number Theorem. It is actually easier to ﬁrst show the following. Lemma 12.10. Θ(x) ∼ x as x → +∞

Proof. Assume that λ > 1 we can find xn → +∞ such that Θ(xn) ≥ λxn. Since Θ(x) is monotone increasing: ∞ ! Z ∞ Θ(x) − x X Z λxk 1 Z λxk 1 dx ≥ λx dx − dx x2 k x2 x 1 k=1 xk xk ∞ ! X Z λ 1 Z λ 1 = dx − dx x2 x k=1 1 1 | {z } (By change of variables) ∞ X 1 = 1 − − log λ = +∞. λ k=1 | {z } >0 This contradicts the integral converging. On the other hand, assume that λ < 1 we can find xn → +∞ such that θ(xn) ≤ λxn we again show that the integral doesn’t converge. Either way, we see that Θ(x) ∼ x as x → +∞. This leads to the final result. Theorem 12.11 (Prime Number Theorem). x π(x) ∼ as x → +∞. log x Proof. We can bound X Θ(x) = log p p≤x X ≤ log x = π(x) log x p≤x For any > 0, X Θ(x) ≥ log p x1−≤p≤x X ≥ (1 − ) log x 1 x1−≤p≤x    1−  ≥ (1 − ) log x π(x) − π(x ) | {z } ≤x1− x Since > 0 is arbitrary, together these show that π(x) ∼ log x . Application 12.12 (More on the Riemann zeta function). Recall that we define the Riemann zeta function by ∞ X 1 ζ(s) = ns n=1 It is possible to show that this has an extension to C. We can write Z ∞ Γ(s) = xs−1e−xdx 0 13. FURTHER TOPICS 85 for σ = Re(s) and then Z ∞ n−sΓ(s) = xs−1e−nxdx. 0 Thus summing over n we have ∞ ! X 1 Z ∞ xs−1 ζ(s)Γ(s) = Γ(s) = dx. ns ex − 1 n=1 0 Lemma 12.13. For σ = Re(s) > 1, Γ(1 − s) Z (−z)s−1 ζ(s) = − z dz 2πi C e − 1 where C runs from ∞ to 0, an infinitesimal circle about 0, followed by 0 to ∞. Proof. We can write the integral as Γ(1 − s) Z ∞ xs−1e−(s−1)πi Γ(1 − s) Z ∞ xs−1e(s−1)πi − dx + dx e−(s−1)πi e−(s−1)πi 2πi 0 e − 1 2πi 0 e − 1 which is equal to 2i sin((s−1)πζ(s)Γ(s)) (using that Γ(s)Γ(1−s) = π/ sin(πs)). In particular, the Right Hand Side of the formula is meromorphic for all s. The integral is entire in s. Moreover, Γ(1 − s) is meromorphic with poles at s = 1, 2, ··· . (The poles at n ≥ 2 must actually cancel with the zeros of the integral since we already know that ζ(s) is analytic there. However, the integral at s = 1 has value 1. Corollary 12.14. ζ(s)(s − 1) is an entire function (i.e., ζ(s) has a simple zero at s = 1).

13. Further Topics There are a number of topics which would have ﬁtted nicely into the course with more time.

13.1. Weierstrass’s elliptic function. If ω1, ω2 ∈ C such that Im(ω1/ω2) then we can consider the set of points

L = {2n1ω1 + 2n2ω2 : n1, n2 ∈ Z} ⊂ C Definition 13.1. We can deﬁne the Weierstrass elliptic functions are deﬁned by 1 X 1 1 P (z) = + − z2 z − w2 w2 ω∈L

In particular, we have that P (z + 2ωi) = P (z) for i = 1, 2.. Theorem 13.2. This has a Laurent series expansion 1 1 1 P (z) = + g z2 + g z4 + O(z6). z2 20 2 28 3 where X 1 X 1 g (ω , ω ) = 60 and g (ω , ω ) = 60 2 1 2 w4 3 1 2 w6 ω∈L−{0} ω∈L−{0} −4 −6 The coeﬃcients satisfy g2(λω1, λω2) = λ g2(ω1, ω2) and g3(λω1, λω2) = λ g3(ω1, ω2), for λ > 0.

We can let τ = ω1/ω2 with Im(τ) > 0 be the period ratio and consider g2(τ) = g2(1, ω2/ω1) and g3(τ) = g3(1, ω2/ω1). 86 CONTENTS

Definition 13.3. The modular discriminant ∆ is defined as 3 2 ∆(τ) = g2 − 27g3 This is studied in its own right, as a cusp form, in modular form theory (that is, as a function of the period lattice). Theorem 13.4. The discriminant is a modular form of weight 12. That is, under the action of the modular group, it transforms as zτ + b ∆ = (cτ + d)12∆(τ) cτ + d with a, b, c, d being integers, with ad − bc = 1. 13.2. Picard’s Little theorem. These describe the values of entire functions. Theorem 13.5 (Picard’s Little Theorem). A non-constant entire function takes every value , except at most one. 13.3. Runge’s theorem. This deals with approximations on compact sets. Theorem 13.6 (Runge). Let U be a (simply-connected) domain. Then any function analytic in U can be approximated uniformly on compact sets K ⊂ U by polynomials, i.e., for each > 0 there is a polynomial p(z) with complex coefficients such that |f(z) − p(z)| < for all z ∈ K. 13.4. Conformality: Property of analytic maps. Any complex number z = reiθ where r = |z| and θ = arg(z) is defined up to a multiple of 2π. Let f : U → C be analytic and let γ be a smooth contour in U represented bt 0 a parameterization z :[a, b] → C and let a < t0 < b be such that z (t0) 6= 0. Definition 13.7. Let U be an open neighbourhood. We say that f : U → C is conformal on U if for each z0 ∈ U the function f is complex differentiable at z0 0 and f (z0) 6= 0 exists. This has a more natural interpretation. Lemma 13.8. f is conformal if and only if it preserves angles, i.e., given two smooth paths γ1, γ2 : [0, 1] → C with γ1(0) = γ2(0) we have that 0 0 0 0 arg (γ1(0) − γ2(0)) = arg ((f ◦ γ1) (0) − (f ◦ γ2) (0)) .

Proof. We can write γj(t) = xj(t) + iyj(t) and denote z0 = γ1(0) = γ2(0). 0 0 Let f(z) = f(x+iy) = u(x, y)+iv(x, y). Suppose that γ1(0) = γ2(0). We can write 0 0 (f ◦ γj) (0) = f (γj(0).γj(0) for j = 1, 2. Thus 0 0 arg((f ◦ γj) (0)) = arg(f (γj(0)) + arg(γj(0)) up to a multiple of 2π. We can then write 0 0 0 0 arg((f ◦ γ1) (0)) − arg((f ◦ γ2) (0)) = arg(γ1(0)) − arg(γ2(0)) up to a multiple of 2π.

Theorem 13.9. Analytic maps are conformal at all points at all points z0 such 0 that f (z0) 6= 0 Method I: This can be illustrated by considering two two orthogonal families families of contours such as the vertical lines and horizontal lines (or concentric circles and radial lines). Let f(z) = u(x, y) + iv(x, y) with vertical lines x = α and horizontal lines y = β. Consider the the image of 2 perpendicular curves passing through z0, i.e., z0 = α + iβ. The image of the line x = α is now z = u(α, y) + iv(α, y) for y ∈ R. The image of the line y = β is now z = u(x, β) + iv(x, β). for x ∈ R. 14. PROBLEMS 87

Method II: Consider the so-called level curves of f(z) = u(z) + iv(z), i.e., u(x, y) = α and v(x, y) = β. In particular, these curves are the ones mapped to the straight lines u = α and β = v. 0 Provided f (z0) 6= 0 we have that the angle between these preimages is θ = π/2. Example 13.10 (Method I). Let f(z) = ez then f 0(z) = ez 6= 0 everywhere. Writing z = x + iy gives f(z) = ex(cos θ + i sin θ). If x = α then u = eα cos y and v = eα cos y with y ∈ R a parameter, i.e., u2 + v2 = e2α which describes a circle. x u If y = α then u + iv = e (cos β + i sin β). Thus v = tan(β), i.e., a straight line through the origin at angle β. Example 13.11 (Method II). Let f(z) = z2 then f 0(z) = 2z =6= 0 except at z = 0. Writing f(z) = u + iv, the preimage of u = α are those x + iy such that u(x, y) = α and the preimage of v = β are those x + iy such that v(x, y) = β. If x = α then u = eα cos y and v = eα cos y with y ∈ R a parameter, i.e., u2 + v2 = e2α which describes a circle. x u If y = α then u + iv = e (cos β + i sin β). Thus v = tan(β), i.e., a straight line through the origin at angle β. Thus z2 = (x+iy)2 = (x2 −y2)+i2xy. The level sets correspond to x2 −y2 = α or 2xy = β. Remark 13.12. In practise, it can be difficult to find the appropriate constants for these mappings. This involves numerical evaluation of elliptic integrals. 13.5. Functions of several complex variables. We can consider analytic functions f : U → C, where U ⊂ C2 is an open set. We can define analyticity in terms of the convergence of power series in two variables (on poly disks 2 B(z0, w0, 1, 2) = {(z, w) ∈ C : |z − z0| < 1, |w − w0| < 2}) upon which we can apply the Cauchy theorem twice to write 1 Z Z f(z, w) f(z, w) = − 2 dzdw 4π C(z0,1 C(z0,1 (z − z0)(w − w0) There are two particularly interesting theorems: The Bochner Tube Theorem and Hartog’s Theorem

Theorem 13.13 (Bochner Tube theorem). Let Ω ⊂ R2 be a connected open set and let co(Ω) = {λx + (1 − λ)y : x, y ∈ Ω, 0 ≤ λ ≤ 1} be the convex hull. If f : Ω + iR2 → C is analytic then f : co(Ω) + iR2 → C

Theorem 13.14 (Hartog Theorem). Assume that f : B(z0, w0, 1/2, 2/2) → C is analytic in two variables and f(·, w): B(z0, r1) → C is analytic in one variable for each w ∈ B(w0, r2). Then f : B(z0, w0, 1/2, 2/2) → C is also analytic.

14. Problems 14.1. Problems 1. .

1. Show that the preimage of a circle or a straight line in C is a circle on the Riemann sphere under stereographic projection.

2. Show that the metric on Cb coming from the stereographic projection of the usual 0 0 0 Euclidean metric is given as follows: If z := π((x1, x2, x3)) and w := π((x1, x2, x3)) then 0 0 0 |z − w| d(z, w) = d ((x1, x2, x3), (x , x , x )) = 1 2 3 p(1 + |z|2)(1 + |w|2) 88 CONTENTS

0 0 0 whenever (x1, x2, x3), (x1, x2, x3) 6= (0, 0, 1) (i.e., z, w 6= ∞) and 2 d(z, ∞) := d((x1, x2, x3), (0, 0, 1)) = p1 + |z|2 whenever (x1, x2, x3) 6= (0, 0, 1) (i.e., z 6= ∞)

a b 3. Given a complex matrix M = we can associate a M¨obiusmap T : c d M az+b C → C of the form TM (z) = cz+d . Given two M¨obiusmaps, relate MT1◦T2 to MT1 and MT2 .

4. Find the M¨obiustransformation that takes: (1) The points −i, −1, i to the points −i, 0, i, respectively; (2) The points −1 + i, 0, 1 − i to the points −1, −i, i, respectively; (3) The points 0, 1 + i, 1 − i to the points 1, −i, i, respectively.

5. What is the image of the unit disk D = {z ∈ C : |z| < 1} under the following M¨obiusmaps: z+i (1) T (z) = z−i ; iz−1 (2) T (z) = i−z ; (1+i)z+(i−1) (3) T (z) = iz+1 ; (4) T (z) = 2z−1 ; √z−2 (5) T (z) = √2z−1+i ; 2z+1−i (1+i)z−1−i (6) T (z) = z+1 .

6. The cross ratio of four points z0, z1, z2, z3 ∈ Cb is given by

(z0 − z2)(z1 − z3) (z0, z1, z2, z3) = . (z0 − z3)(z1 − z2)

If λ = (z0, z1, z2, z3) show that diﬀerent permutations of the four numbers gives cross-ratios with the values 1 1 λ λ, 1 − λ, 1 − , , . λ 1 − λ 1 − λ Show that these are all possible values.

Apollation circle packing

1. Show that if E is a circle of radius k then: (1) if we reﬂect in E a circle C of radius r whose centres are separated by d > r + k then the image f(C) is a circle with radius k2r/(d2 − r2). (2) if we reﬂect in E a straight line at a distance b > k from the centre of E then the image f(C) is a circle of radius k2/2b.

az+b 2. If a M¨obiusmap f(z) = cz+d takes real numbers to real numbers (i.e., t ∈ R =⇒ f(t) ∈ R) then show that a, b, c, d are all real numbers.

Classiﬁcation of M¨obiusmaps

3. Show that any M¨obius map other than the identity must ﬁx at least one point and at most two. 14. PROBLEMS 89

4. Since we are assuming ad − bc = 1 show that we can write (a − d)2 + 4bc = (a − d)2 + 4ad − 4 = 0

5. Let f : C → C be a M¨obiusmap. Show that (1) If f(∞) = ∞ then f(z) = αz + β, for some α, β ∈ C. (2) If f(∞) = ∞ and f(0) = 0 then f(z) = αz, for some α ∈ C.

6. We say that two linear fractional transformations f1, f2 are conjugate if f2 = −1 g f1g for some linear fractional map g. We can associate to the M¨obiusmap az+b f(z) = cz+d the value tr(f) := a + d. Show that the value tr(f) is preserved by conjugacy, i.e., if f1, f2 are conjugate then tr(f1) = tr(f2).

7. Let f : Cb → Cb be a Möbius transformation. The non-identity Möbiustransfor- mations are commonly classified into three types: (1) parabolic (conjugate to z 7→ z + β with a single fixed point ∞); (2) elliptic (conjugate to z 7→ αz with |α| = 1 with two points 0, ∞); and (3) loxodromic (everything else). Show that: (1) Elliptic Möbiustransformations correspond to tr(f)2 ∈ [0, 4); (2) Parabolic Möbiustransformations correspond to tr(f)2 = 4; (3) Hyperbolic Möbiustransformations (i.e., conjugate to z 7→ λz with 0 < λ < 1 or λ ∈ (1, ∞)) correspond to tr(f)2 ∈ (4, ∞). 14.2. Problems 2. .

Here

0. Let f(z): C → C be a continuous function. Is it true that (1)

1. Show that if f : U → C is analytic and there exist a ∈ C and a convergent sequence zn → z with zn, z ∈ U such that f(zn) = a then f |U = a. What happens if z 6∈ U?

2. Show that the equivalence of parameterizations of curves is an equivalence relation.

R R 3. Show that −γ f(w)dw = − γ f(z)dz for any closed curve γ.

4. Show that if R = [a, b] × [c, d] is a rectangle then (1) We can evaluate R dz = 0; R∂R (2) We can evaluate ∂R(z − z0)dz = 0.

5. Write the Cauchy-Riemann equations in polar coordinares.

6. Let f : D → C be a function well-deﬁned in the unit disc, such that partial derivatives exist everywhere and Cauchy-Riemann equations hold true. Is it true that f is complex diﬀerentiable everywhere in the unit disc? Hint: consider f(z) = exp(−z−4). 90 CONTENTS

8. Find all holomorphic functions that map the unit disc to the real line.

9. Find domains of complex diﬀerentiablity of the following functions (1) f1(z) = |z|; (2) f2(z) = zz¯; (3) f3(z) =z ¯. 14.3. Problems 3. .

z P∞ Bn n 1. Deﬁne the Bernoulli numbers by the series ez −1 = n=0 n! z . Prove that ( B B B 1 if n = 1 0 + 1 + ··· + n−1 = n!0! (n − 1)!1! 1!(n − 1)! 0 if n > 1.

R z2 2. Compute γ ze dz where: (1) γ = [i, −i + 2]; and (2) γ = {x + ix2 : 0 ≤ x ≤ 1}. R 3. Compute γ sin(z)dz where γ = {x + ix : 0 ≤ x ≤ 1}.

4. Let f : D → D be an analytic map which is injective (i.e., f(z1) = f(z2) implies P∞ n z1 = z2) of the form f(z) = n=1 anz . Show that the image has area: ∞ X 2 2 Area(f(D)) = π n |an| . n=1

5. Let f : D × D → C be a continuous function such that (1) for w ∈ D the map D 3 z 7→ f(z, w) ∈ C is analytic; and (2) for z ∈ D the map D 3 zw 7→ f(z, w) ∈ C is analytic. P∞ P∞ n m Show that we can write f(z, w) = n=0 m=0 anmz w , which converges abso- lutely for z, w ∈ D. 6. Use Cauchy’s theorem to evaluate Z z − a2 dz C(0,1) z − b where b < 1 and C(0, 1) = {z ∈ C : |z| = 1}.

0 7. Let f : U → C be an analytic function. Let z0 ∈ U with f (z0) 6= 0. Let C be a suﬃciently small circle centred at z0 then show 2πi Z dz 0 = f (z0) f(z) − f(z0)

8. If f is analytic in U and z0 ∈ U deﬁne

( f(z)−f(z0) if z 6= z0 z−z0 g(z) = 0 f (z0) if z = z0. Show that this is analytic.

9. Use Rouche’s theorem to show that : (1) z7 − 4z3 + z − 1 = 0 has three zeros inside the unit circle; (2) z7 + 5z3 − z − 2 = 0 has three zeros inside the unit circle. 14. PROBLEMS 91

10. Use Rouche’s theorem to show that z87 + 36z57 + 71z4 + z3 − z + 1 (1) has 4 zeros inside the unit circle. (2) has 87 zeros inside the circle {z ∈ C : |z| = 2}. 14.4. Problems 4. .

1. If |a| > e then show that ez = azn has n roots in the unit disk D = {z ∈ C : |z| < 1}.

2. Consider the following polynomials: (1) How many zeros of z7 − 2z5 + 6z3 − z + 1 are in the unit disk D; (2) How many zeros of z4 − 6z + 3 are in the annulus {z ∈ C : 1 < |z| < 2}; (3) How many zeros of z4−4z3+6z2−4z+3 are in the set {z ∈ C : |z−1| < 1}.

3. Let h be analytic in the unit disk D and assume that |h(z) − h(w)| < |z − w| for all z, w ∈ D. Show that the function f : D → C deﬁned by f(z) = h(z) + z is one-to-one.

4. Let a > 0. Show that if f : C → C is analytic and satisﬁes |f(z)| < |z|a then f is a polynomial of degree at most [a].

5. Show that if f : C → C is analytic and satisﬁes |f(z)| > 1 whenever |z| > 1 then f is a polynomial.

6. Show that if f : C → C is analytic, non-constant and has no zeros then there must exist a sequence of points |zn| → ∞ such that f(zn) → 0.

7. Let f : D → C be analytic and continuous on the closed unit disk D = {z ∈ C : |z| ≤ 1}. Moreover, assume that: (1) |f(z)| ≤ M whenever |z| = 1 and Im(z) ≥ 0; and (2) |f(z)| ≤ N whenever |z| = 1 and Im(z) < 0. √ Then show that |f(0)| ≤ NM.

8. Show that if f : C → C is analytic and Im(f(z)) 6= 0 whenever |z| 6= 1 then f is constant.

9. Show that if f is a polynomial and z ∈ C then either f n(z) → ∞ as n → +∞ or {f k(z): k ≥ 0} is a bounded set.

14.5. Problems 5. 1. Consider the family of functions fn : D → C deﬁned n by fn(z) = z , n ≥ 1. Show that for all z ∈ D = {z ∈ C : |z| < 1} we have that fn(z) → 0 as n → ∞ and that the convergence is uniform on compact sets, but that it is not uniform on D.

2. Give an example of a family of analytic functions fn : D → C, n ≥ 1, for which fn(z) has precisely one zero, but fn converges uniformly on D to the function f = 0 (i.e., the function which is identically zero).

3. Give an example of a family of analytic functions fn : C → C, n ≥ 1 such that fn → ∞ uniformly on C (i.e., for all compact sets K ⊂ C and M > 0 we have that infz∈K |fn(z)| ≥ M for suﬃciently large n) but for which the derivatives 0 fn : C → C, n ≥ 1, neither tend to ∞ nor form a normal family. 92 CONTENTS

4. Given two domains U1, U2 and two points z1 ∈ U1 and z2 ∈ U2 show that there exists an analytic bijection f : U1 → U2 such that f(z1) = z2. [Hint: Use the Riemann Mapping Theorem].

5. Find the image of the unit disk with respect to the map z 7→ z + 1 . Find the D √ z image of the upper half plane H with respect to the map z 7→ z2 − 1 6. Find an analytic bijection between the following domains: (1) R × (0, π) and H = {z ∈ C : Im(z) > 0}; (2) H − [0, ir] and H; 1 (3) D and C − (−∞, − 4 ]; (4) (R × (0, 2)) − ((i − ∞, i − ] ∪ [i + , i + ∞)) and H; (5) H and the equilateral triangle.

7. Let U ⊂ C be deﬁned by ∞ [ 1 1 1 {z ∈ : 0 < Re(z) < 1, 0 < Im(z) < 1} − , + i . C 2n 2n 2 n=1 (1) Show that U is (path) connected; (2) Show that U is simply connected; (3) Show that 0 ∈ ∂U in the boundary cannot be reached by a continuous path from any point in U; (4) Show that any analytic bijection f : U → D cannot extend to a continuous map from the boundary ∂U (of U) to the unit circle ∂D.

8. Let U ⊂ C be deﬁned by [ 1 {z ∈ : Im(z) > 0 and |z| < 1} − 0, eiπp/q . C q p/q∈Q Show that every point in the boundary ∂U can be reached by a continuous path from any point in U. However, show that any analytic bijection f : U → D cannot extend to a continuous map from the boundary ∂U of U to the unit circle ∂D.

9. Consider an analytic bijection f : U → D that satisﬁes that for z0 ∈ U ∩ R we 0 have that f(z0) = 0 and f (z0) > 0. Show that if U is symmetric about the real axis then f(z) = f(z).

10. Show that there is no analytic bijection between D and C. 14.6. Problems 6. .

1. Let f : U → D be an analytic bijection. Fix z0 ∈ U. Show that if f(z0) = 0 and 0 f (z0) > 0 then f is uniquely determined (i.e., uniqueness in the Riemann Mapping Theorem).

2. Let x1 < x2 < ··· < xn be on the real line. Let 0 < α1, ··· , αn < 2. Let C F (z) = 1−α 1−α . (z − x1) n ··· (z − xn) n

Show that if xi−1 < t < xi then

Arg(F (t)) = Arg(C) − (αi + αi+1 + ··· + αn). Deduce that the image of with respect to the map f(z) = R z F (w)dw is a R z0 piecewise linear curve. Is this a closed curve? 14. PROBLEMS 93

3. Let f : P → D be continuous. Assume that f extends to a bijection f : ∂P → ∂D on the boundary. Show that f : P → D is surjective.

4. Let g : U → C be analytic. Let D ⊂ U be a closed region whose boundary curve γ = ∂D is simple. Assume that g : γ → g(γ) is one-to-one. Show that g : D → g(D) is one-to-one. (This is Darboux’s theorem). [Hint: Use the Argument Principle]

5. Do questions 2 – 4 give another proof that the Schwartz-Christoffel formula gives an analytic bijection f : H → P for a polygon P? 6. Consider the semi-infinite strip S = {x + iy : − a < x < a and y > 0} for a > 0. Use the Schwarz-Christoffel theorem to find the analytic bijection f : H → S which extends to f(±1) = ±a and f(∞) = ∞.

7. Consider the triangle ∆ in C with vertices at −1, 0 and i. Use the Schwarz- Christoﬀel theorem to ﬁnd the analytic bijection f : H → ∆ which extends to f(−1) = i, f(1) = −1 and f(∞) = 0.

8. Derive the version of the Schwarz-Christoﬀel theorem for an analytic bijection f : D → P from the unit disk to a polygon.

R z n −2/n 9. Describe the image of the unit disk D under the map f(z) = 0 (1 − ξ ) dξ, for n ≥ 3.

R z 4 −1 10. Describe the image of the unit disk D under the map f(z) = 0 (1 − ξ ) (1 + ξ4)−1/2dξ. 14.7. Problems 8. .

1. Let f : U → C be analytic. Assume that f never vanishes. If there is a point z0 ∈ U such that |f(z0)| ≤ |f(z)|, for all z ∈ U then f is a constant.

2. Suppose that both f(z) and g(z) are analytic on D with |f(z)| = |g(z)| for |z| = 1. Show that if neither f(z) nor g(z) vanishes in D then there exists λ ∈ C with |λ| = 1 such that f = λg.

3. Suppose that f(z) is analytic on D and for 0 ≤ r < 1 deﬁne A(r) = max|z|=r Ref(z). Show that unless f is a constant, A(r) is a strictly increasing function of r.

1 3 0 1 3 4. Does there exist an analytic function f : D → D with f( 2 ) = 4 and f ( 2 ) = 4 ? 5. Let f(z) = (z + 1)2 on U = {z = x + iy : 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1}. Where does |f(z)| have its maximum and minimum values?

6. Show that the function w(x, y) = ex sin(y) is harmonic.

7. Let f : U → C be analytic and non-zero. Show that log |f(z)| is harmonic.

8. Show Harnack’s principle: Let u1 ≤ u2 ≤ · · · be harmonic functions on an open set U. Then either uj → ∞ uniformly on compacts sets or there is a harmonic function u on U such that uj → u uniformly on compact sets. [Hint: Use Harnack’s Inequality].

9. Find the harmonic conjugate of u(x, y) = y3 − 3x2y. 94 CONTENTS

10. Describe suitable domains for the following harmonic functions, and ﬁnd their harmonic conjugates of the following functions: (i) 2x(1 − y); (ii) 2x − x3 + 3xy2; (iii) sinh(x) sin(y); (iv) y/(x2 + y2).

11. Show that if v1, v2 are both harmonic conjugates of u then v1 −v2 is a constant.