Course 5 Book 5A: Complex Analysis and Conformal GeometryChapter 4

Harmonic Functions and Conformal Mapping

The Dirichlet problem, Poisson’s kernel and the Riemann mapping theorem.

27 min read · Updated Oct 2, 2026

Read with Stein and Shakarchi, Complex Analysis, chapter 8, "Conformal Mappings" (conformal equivalence and examples, the Schwarz lemma and automorphisms of the disc and upper half-plane, the Riemann mapping theorem, conformal mappings onto polygons). The Poisson kernel is in their Fourier Analysis, chapter 2, section 5, and in chapter 3, section 7 of Complex Analysis.

In this chapter · 7 sections

4.1Steady temperature in a plate
4.2Harmonic functions are real parts
4.3The Poisson kernel
4.4The Schwarz lemma
4.5The Riemann mapping theorem
4.6History
4.7Exercises

The real part of a holomorphic function is harmonic (5A.1 Holomorphic Functions Are Conformal), and in a simply connected region every harmonic function arises this way. So the whole machinery of the last two chapters applies to the Laplace equation in the plane. Two consequences make this chapter. First, harmonic functions can be transplanted by conformal maps: if you can solve the Laplace equation on a disc, you can solve it on any region that is conformally equivalent to the disc. Second, by the Riemann mapping theorem, that is every simply connected region other than the plane itself. Steady temperatures, electrostatic potentials and ideal flows in any such region reduce to a formula on the disc: the Poisson integral.

The Riemann mapping theorem is also the first uniformization theorem: it says that, up to conformal equivalence, there is only one simply connected proper region of the plane. 5A.5 Uniformization and the Two-Dimensional Ricci Flow extends it to all surfaces, and that is where curvature enters.

By the end of this chapter you will be able to:

find a harmonic conjugate, and use conformal invariance to move harmonic functions between regions;
solve the Dirichlet problem on the disc with the Poisson kernel, and explain the kernel as an approximate identity;
state and prove the Schwarz lemma, and use it to find all conformal self-maps of the disc;
state the Riemann mapping theorem and follow its proof by normal families;
recognise a Schwarz–Christoffel map, and say what conformal flattening of the brain's surface does and does not claim.

Steady temperature in a plate

In the world Model Hot and cold edges

A thin metal plate is held at fixed temperatures along its edge and left until nothing changes. The temperature $u$ then satisfies the heat equation with $u_t = 0$ , which is the Laplace equation $\Delta u = 0$ , with $u$ prescribed on the boundary: the Dirichlet problem. The same problem describes the electrostatic potential in a region bounded by conductors held at fixed voltages, and the velocity potential of an ideal flow (5A.2 Cauchy’s Theorem and Its Consequences).

Take the upper half-plane, with the negative real axis held at temperature $1$ and the positive real axis at $0$ . The function $\arg z$ , with values in $(0, \pi)$ , is the imaginary part of the holomorphic function $\log z$ , hence harmonic; it equals $\pi$ on the negative axis and $0$ on the positive one. So

u(z) = \frac1\pi\arg z

is the temperature, and the isotherms are the rays from $0$ . Now take a disc whose upper boundary semicircle is held at $1$ and lower at $0$ . The Möbius map $\phi(z) = i\frac{1 + z}{1 - z}$ sends the unit disc conformally onto the upper half-plane, the upper semicircle onto the negative real axis and the lower onto the positive one (Exercise 4.8). So the temperature in the disc is

u(z) = \frac1\pi\arg\Big(i\frac{1 + z}{1 - z}\Big),

and the isotherms are the images of rays: arcs of circles through the two points $\pm1$ where the boundary temperature jumps (Figure 4.1, right). At the centre the temperature is $\frac12$ , the average of the boundary values, as the mean value property says it must be.

This is the method in general: map the region conformally to one where the answer is known, solve there, and map back. Engineers used it for a century to compute fields around conductors, flows around obstacles and stresses around holes, and it still underlies some numerical methods. Its limits are that it works only in two dimensions, only for the Laplace equation and its relatives, and only when a conformal map can be found.

Harmonic functions are real parts

Proposition 4.1 Harmonic conjugates

Let $u$ be a harmonic function on a simply connected open set $\Omega$ . Then $u = \operatorname{Re}f$ for a holomorphic $f$ on $\Omega$ , unique up to adding an imaginary constant. Its imaginary part $v$ is a harmonic conjugate of $u$ .

Proof. The function $g = u_x - iu_y$ satisfies the Cauchy–Riemann equations because $u$ is harmonic and $u_{xy} = u_{yx}$ (the computation in the last exercise of 5A.2 Cauchy’s Theorem and Its Consequences). On a simply connected set it has a primitive $f$ (5A.2 Cauchy’s Theorem and Its Consequences), and $f' = g$ forces $\operatorname{Re}f = u + \text{const}$ .

On a region with a hole this can fail: $\log|z|$ is harmonic on $\mathbb{C}\setminus\{0\}$ , but its conjugate would be $\arg z$ , which is not single-valued. Even so, every harmonic function is locally a real part, which is enough for local properties. In particular harmonic functions are smooth, even real-analytic, and satisfy the mean value property and the maximum principle (5A.2 Cauchy’s Theorem and Its Consequences). 6A.2 Harmonic Functions proves all of these again in any dimension, where there is no complex structure to lean on.

Proposition 4.2 Conformal invariance

If $\phi : \Omega \to \Omega'$ is holomorphic and $u$ is harmonic on $\Omega'$ , then $u\circ\phi$ is harmonic on $\Omega$ .

Proof. Locally $u = \operatorname{Re}f$ , so $u\circ\phi = \operatorname{Re}(f\circ\phi)$ , the real part of a holomorphic function.

In fact $\Delta(u\circ\phi) = |\phi'|^2(\Delta u)\circ\phi$ (Exercise 4.9): the Laplacian changes only by the conformal factor. This identity is the seed of the formula for the curvature of a conformal metric in 5A.5 Uniformization and the Two-Dimensional Ricci Flow.

The Poisson kernel

Theorem 4.3 The Dirichlet problem on the disc

Let $f$ be continuous on the unit circle. Then

u(re^{i\theta}) = \frac{1}{2\pi}\int_0^{2\pi}P_r(\theta - \varphi)f(e^{i\varphi})\,d\varphi, \qquad P_r(\theta) = \frac{1 - r^2}{1 - 2r\cos\theta + r^2},

is harmonic in the open disc and extends continuously to the closed disc with $u = f$ on the circle. It is the only such function.

Proof. The kernel. For $0 \leq r < 1$ , summing two geometric series (2A.7 Series),

P_r(\theta) = \sum_{n=-\infty}^\infty r^{|n|}e^{in\theta} = \operatorname{Re}\frac{1 + re^{i\theta}}{1 - re^{i\theta}}.

The second form shows $P_r(\theta)$ is the real part of $\frac{1 + z}{1 - z}$ at $z = re^{i\theta}$ , so it is harmonic in $z$ ; so is $u$ , by differentiating under the integral (3A.3 The Lebesgue Integral).

An approximate identity. $P_r$ is positive, its average over $[0, 2\pi]$ is $1$ (only the $n = 0$ term survives integration), and for every $\delta > 0$ , $P_r(\theta) \to 0$ uniformly on $\delta \leq |\theta| \leq \pi$ as $r \to 1$ , since there the denominator is at least $1 - 2r\cos\delta + r^2 \geq \sin^2\delta$ while the numerator tends to $0$ . These are the three properties of an approximate identity (3A.8 Convolution and Mollifiers), so $u(re^{i\theta}) \to f(e^{i\theta})$ uniformly as $r \to 1$ , exactly as Gaussian blur converges to the original image.

Uniqueness. The difference of two solutions is harmonic, continuous on the closed disc and zero on the circle; by the maximum principle applied to it and its negative, it is zero.

In Fourier terms (2B.7 Fourier Series and the First Heat Equation): if $f(e^{i\theta}) = \sum c_ne^{in\theta}$ , then $u(re^{i\theta}) = \sum c_nr^{|n|}e^{in\theta}$ . Each Fourier mode is extended inward as $r^{|n|}e^{in\theta}$ , which is $z^n$ or $\bar z^{|n|}$ , and high frequencies die off fast towards the centre. Compare the heat equation on the ring in 2B.7 Fourier Series and the First Heat Equation, which multiplies the $n$ -th mode by $e^{-n^2t}$ : in both cases high frequencies are damped, which is why both equations smooth. At $r = 0$ only $c_0$ survives, the mean value property again.

Figure 4.1. Left: the Poisson kernel

P_r(\theta)

for

r = 0.5

0.75

and

0.9

(computed). It has average

1

and concentrates at

\theta = 0

r \to 1

, which is why the boundary values are recovered. Right: the steady temperature in a disc with upper edge at

1

and lower edge at

0

; the isotherms

u = 0.1, 0.2, \dots, 0.9

are arcs of circles through

\pm1

The Schwarz lemma

Lemma 4.4 The Schwarz lemma

Let $f$ be holomorphic from the unit disc $\mathbb{D}$ to itself with $f(0) = 0$ . Then $|f(z)| \leq |z|$ for all $z$ , and $|f'(0)| \leq 1$ . If $|f(z_0)| = |z_0|$ for some $z_0 \neq 0$ , or $|f'(0)| = 1$ , then $f$ is a rotation, $f(z) = e^{i\theta}z$ .

Proof. Since $f(0) = 0$ , $g(z) = f(z)/z$ has a removable singularity at $0$ , with $g(0) = f'(0)$ (5A.3 Residues and Fourier Transforms). On the circle $|z| = r < 1$ , $|g| \leq 1/r$ ; by the maximum modulus principle (5A.2 Cauchy’s Theorem and Its Consequences) $|g| \leq 1/r$ on the whole disc $|z| \leq r$ . Let $r \to 1$ : $|g| \leq 1$ , which is both inequalities. If equality holds at an interior point, $|g|$ has an interior maximum, so $g$ is a constant of modulus $1$ .

Corollary 4.5 Automorphisms of the disc

The conformal bijections of $\mathbb{D}$ onto itself are exactly the maps

z \mapsto e^{i\theta}\frac{z - a}{1 - \bar az}, \qquad |a| < 1,\ \theta \in \mathbb{R}.

Proof. These maps are bijections of the disc (an exercise in 5A.1 Holomorphic Functions Are Conformal). Conversely, let $F$ be a conformal bijection with $F(a) = 0$ , and compose with $\phi_a(z) = \frac{z - a}{1 - \bar az}$ to get $G = F\circ\phi_a^{-1}$ , a bijection fixing $0$ . The Schwarz lemma applied to $G$ and to $G^{-1}$ gives $|G'(0)| \leq 1$ and $|1/G'(0)| \leq 1$ , so $|G'(0)| = 1$ and $G$ is a rotation.

Where this goes The Schwarz lemma is a curvature comparison

The Schwarz lemma has a geometric form, due to Georg Pick: a holomorphic map of the disc to itself does not increase distances in the hyperbolic metric $\frac{4|dz|^2}{(1 - |z|^2)^2}$ , which has curvature $-1$ (5A.5 Uniformization and the Two-Dimensional Ricci Flow). Lars Ahlfors (1938) turned it into a statement about curvature alone: a conformal metric on the disc with curvature at most $-1$ is everywhere at most the hyperbolic one. His proof is a maximum principle argument applied to the logarithm of the ratio of the two metrics. Comparison by curvature, proved with a maximum principle, is exactly the method of Hamilton's Ricci flow estimates (11A.4 Maximum Principles under Ricci Flow), and Schwarz–Ahlfors-type lemmas are a standard tool in the Kähler–Ricci flow (5A.6 Where This Track Leads).

The Riemann mapping theorem

Two open sets are conformally equivalent if there is a holomorphic bijection between them (its inverse is then automatically holomorphic). The plane $\mathbb{C}$ is not conformally equivalent to the disc: a holomorphic map $\mathbb{C} \to \mathbb{D}$ is bounded and entire, hence constant (Liouville). That is the only exception.

Theorem 4.6 The Riemann mapping theorem

Let $\Omega \neq \mathbb{C}$ be a non-empty simply connected open subset of the plane, and $z_0 \in \Omega$ . There is a unique conformal bijection $F : \Omega \to \mathbb{D}$ with $F(z_0) = 0$ and $F'(z_0) > 0$ .

The proof finds $F$ as the solution of an extremal problem: among all injective holomorphic maps $\Omega \to \mathbb{D}$ sending $z_0$ to $0$ , take one that stretches most at $z_0$ . It needs one more fact, which follows from the argument principle (Exercise 4.13).

Lemma 4.7 Hurwitz's theorem

If injective holomorphic functions $f_n$ on a connected open set converge uniformly on compact sets to $f$ , then $f$ is injective or constant.

Hurwitz's theorem is proved in Exercise 4.13. We now prove Theorem 4.6.

Proof. Uniqueness. If $F_1$ , $F_2$ both work, $F_2\circ F_1^{-1}$ is an automorphism of the disc fixing $0$ with positive derivative there, hence the identity by Corollary 4.5.

Step 1: the family is not empty. Let $\mathcal F$ be the set of injective holomorphic $f : \Omega \to \mathbb{D}$ with $f(z_0) = 0$ . Pick $\alpha \notin \Omega$ . Since $\Omega$ is simply connected and $z - \alpha \neq 0$ on it, there is a holomorphic square root $h(z) = \sqrt{z - \alpha}$ on $\Omega$ (5A.2 Cauchy’s Theorem and Its Consequences), and $h$ is injective. It cannot take both values $w$ and $-w$ , so $h(\Omega)$ misses a whole disc around $-h(z_0)$ , since it contains a disc around $h(z_0)$ by the open mapping property. A Möbius map sending the complement of that disc into $\mathbb{D}$ , followed by a disc automorphism, gives an element of $\mathcal F$ .

Step 2: an extremal map exists. Let $s = \sup_{f\in\mathcal F}|f'(z_0)|$ ; it is finite by the Cauchy estimates, since all $f$ are bounded by $1$ , and positive by step 1. Take $f_n \in \mathcal F$ with $|f_n'(z_0)| \to s$ . The family is bounded by $1$ , so by Montel's theorem (5A.2 Cauchy’s Theorem and Its Consequences) a subsequence converges uniformly on compact sets to a holomorphic $F$ , with $|F'(z_0)| = s > 0$ (derivatives converge too, by the Cauchy formula). So $F$ is not constant, hence injective by Hurwitz, and $|F| \leq 1$ with $|F| < 1$ by the maximum modulus principle. So $F \in \mathcal F$ .

Step 3: the extremal map is onto. Suppose $F$ misses a point $a \in \mathbb{D}$ . Then $\phi_a\circ F$ misses $0$ , and on the simply connected $\Omega$ it has an injective square root $g$ , mapping into $\mathbb{D}$ . Let $G = \phi_{g(z_0)}\circ g \in \mathcal F$ . Unwinding, $F = \Psi\circ G$ , where $\Psi = \phi_a^{-1}\circ\sigma\circ\phi_{g(z_0)}^{-1}$ with $\sigma(w) = w^2$ ; $\Psi$ maps $\mathbb{D}$ to itself, fixes $0$ , and is not injective (because of the squaring), so it is not a rotation. By the Schwarz lemma $|\Psi'(0)| < 1$ , and $|F'(z_0)| = |\Psi'(0)||G'(z_0)| < |G'(z_0)|$ , contradicting the maximality of $|F'(z_0)|$ . Finally, rotate $F$ so that $F'(z_0) > 0$ .

The proof is a model of the direct method (4A.6 Weak Convergence and the Direct Method): define the quantity to optimise, use compactness (here Montel) to get an optimiser, then show by a variation (here the square-root trick and Schwarz) that the optimiser has the property you want. Riemann's own argument in 1851 went through the Dirichlet principle, minimising energy, which is the same strategy with a different compactness theorem; it was only made rigorous once that compactness was understood.

The theorem says nothing about how to find $F$ . For polygons there is a formula: a conformal map from the disc onto a polygon with interior angles $\alpha_k\pi$ is

F(z) = A + C\int_0^z\prod_k\Big(1 - \frac{\zeta}{z_k}\Big)^{\alpha_k - 1}d\zeta,

the Schwarz–Christoffel formula, for suitable points $z_k$ on the unit circle (the prevertices) and constants $A$ , $C$ . Each factor turns the boundary direction by the right amount at $z_k$ and keeps it straight elsewhere. For a square, by symmetry the prevertices are $\pm1$ , $\pm i$ , and $F(z) = \int_0^z\frac{d\zeta}{\sqrt{1 - \zeta^4}}$ (Figure 4.2). For general polygons the prevertices must be found numerically; Toby Driscoll's Schwarz–Christoffel Toolbox is a widely used implementation.

Figure 4.2. The Schwarz–Christoffel map

F(z) = \int_0^z(1 - \zeta^4)^{-1/2}d\zeta

from the disc onto a square with corners

F(\pm1)

F(\pm i)

(computed from the power series). The polar grid on the left becomes the curved grid on the right, still meeting at right angles. Near the corners, at the four prevertices, the map squeezes a quarter-turn of the circle into a corner of angle

\pi/2

In the world In use Flattening the brain

The cerebral cortex is a folded sheet, and most of it is hidden in the folds. To display activity on it, or to compare brains with each other, neuroscientists flatten it: map the cortical surface, extracted from an MRI scan as a triangulated surface, onto a plane, a sphere or a disc. No flattening can preserve both angles and areas (Gauss again, 5A.1 Holomorphic Functions Are Conformal), so one has to choose, and a conformal map is a natural choice because it keeps local shapes. For a surface of the topological type of a disc or sphere, the uniformization theorem (5A.5 Uniformization and the Two-Dimensional Ricci Flow, with the Riemann mapping theorem as its planar case) guarantees that such a map exists, and fixes it up to a Möbius transformation.

Two research groups made this practical around 2004. Monica Hurdal and Ken Stephenson used circle packings, a discrete version of conformal maps in which circles packed on the surface are rearranged into a packing in the plane with the same tangencies ("Cortical cartography using the discrete conformal approach of circle packings", NeuroImage, 2004; "Discrete conformal methods for cortical brain flattening", NeuroImage, 2009). Xianfeng Gu, Yalin Wang, Tony Chan, Paul Thompson and Shing-Tung Yau computed conformal maps of the cortex onto the sphere by minimising a harmonic energy ("Genus zero surface conformal mapping and its application to brain surface mapping", IEEE Transactions on Medical Imaging, 2004). These are research tools for visualisation and for registering surfaces to each other, used in brain-mapping studies; this guide makes no claim about routine clinical use. A later method from the same school, discrete surface Ricci flow, appears in 5A.5 Uniformization and the Two-Dimensional Ricci Flow.

History

Siméon Denis Poisson wrote down his integral formula in the 1820s, in work on heat and potentials. Riemann stated the mapping theorem in his 1851 thesis, with a proof based on the Dirichlet principle; after Weierstrass showed in 1870 that the principle needed justification (4A.6 Weak Convergence and the Direct Method), William Fogg Osgood gave the first rigorous proof in 1900, by potential theory. Constantin Carathéodory's 1912 proof was the first to use only function theory, with Montel's normal families, and was soon simplified by Paul Koebe. The extremal proof given here, maximising $|f'(z_0)|$ , is due to Leopold Fejér and Frigyes Riesz and was published by Tibor Radó in 1923. The lemma is named after Hermann Amandus Schwarz, who used a form of it in his work on conformal mapping; Georg Pick's metric form dates from the 1910s and Ahlfors's curvature version from 1938. Elwin Bruno Christoffel (1867) and Schwarz (1869) found the polygon formula independently.

Recall Where we stand

On simply connected regions harmonic functions are real parts of holomorphic ones, and conformal maps carry harmonic functions to harmonic functions. On the disc the Dirichlet problem is solved by the Poisson kernel, an approximate identity that damps each Fourier mode by $r^{|n|}$ . The Schwarz lemma says self-maps of the disc fixing $0$ don't stretch at $0$ , and identifies the disc's automorphisms as Möbius maps. The Riemann mapping theorem, proved by maximising $|f'(z_0)|$ over a normal family, says every simply connected proper subregion of the plane is conformally a disc. 5A.5 Uniformization and the Two-Dimensional Ricci Flow extends this to every surface, and turns it into a statement about curvature and the Ricci flow.

Exercises

Exercise 4.8 The disc and the half-plane

Show that $\phi(z) = i\frac{1 + z}{1 - z}$ maps the unit disc onto the upper half-plane, that $\phi(e^{i\theta}) = -\cot(\theta/2)$ , and hence that the upper semicircle goes to the negative real axis and the lower to the positive real axis. Check that the temperature $u(z) = \frac1\pi\arg\phi(z)$ equals $\frac12$ at $0$ .

Solution

$\phi(e^{i\theta}) = i\frac{e^{-i\theta/2} + e^{i\theta/2}}{e^{-i\theta/2} - e^{i\theta/2}} = i\frac{2\cos(\theta/2)}{-2i\sin(\theta/2)} = -\cot(\theta/2)$ , real; for $0 < \theta < \pi$ it is negative, for $\pi < \theta < 2\pi$ positive. $\phi(0) = i$ , which is in the upper half-plane, so by connectedness the disc goes into the upper half-plane, and the inverse $w \mapsto \frac{w - i}{w + i}$ (the Cayley map) shows it is onto. $\arg i = \frac\pi2$ , so $u(0) = \frac12$ .

Exercise 4.9 The Laplacian under a conformal map

Let $\phi$ be holomorphic and $u$ a $C^2$ function. Show that $\Delta(u\circ\phi) = |\phi'|^2\,(\Delta u)\circ\phi$ . (Write $\Delta = 4\partial_z\partial_{\bar z}$ , where $\partial_z = \frac12(\partial_x - i\partial_y)$ and $\partial_{\bar z} = \frac12(\partial_x + i\partial_y)$ , and use $\partial_{\bar z}\phi = 0$ .)

Exercise 4.10 The Poisson kernel as a series

Sum $\sum_{n=-\infty}^\infty r^{|n|}e^{in\theta}$ for $0 \leq r < 1$ and show it equals $\frac{1 - r^2}{1 - 2r\cos\theta + r^2}$ . Check that its average over $[0, 2\pi]$ is $1$ , and find its maximum and minimum values in $\theta$ .

Solution

$\sum_{n\geq0}(re^{i\theta})^n + \sum_{n\geq1}(re^{-i\theta})^n = \frac{1}{1 - re^{i\theta}} + \frac{re^{-i\theta}}{1 - re^{-i\theta}}$ ; over the common denominator $|1 - re^{i\theta}|^2 = 1 - 2r\cos\theta + r^2$ the numerator is $1 - re^{-i\theta} + re^{-i\theta} - r^2 = 1 - r^2$ . Maximum $\frac{1 + r}{1 - r}$ at $\theta = 0$ , minimum $\frac{1 - r}{1 + r}$ at $\theta = \pi$ .

Exercise 4.11 Harnack's inequality on the disc

Let $u \geq 0$ be harmonic on a neighbourhood of the closed unit disc. Using the bounds on $P_r$ from Exercise 4.10, show that

\frac{1 - r}{1 + r}u(0) \leq u(re^{i\theta}) \leq \frac{1 + r}{1 - r}u(0).

This is the Harnack inequality: a positive harmonic function cannot be much larger at one point than at another nearby. Deduce again that a positive harmonic function on $\mathbb{C}$ is constant (5A.2 Cauchy’s Theorem and Its Consequences).

Exercise 4.12 Using the Schwarz lemma

(a) Let $f : \mathbb{D} \to \mathbb{D}$ be holomorphic with $f(0) = 0$ and $f(\frac12) = \frac12$ . Show $f(z) = z$ . (b) Show that if $f : \mathbb{D} \to \mathbb{D}$ is holomorphic with $f(0) = \frac12$ , then $|f'(0)| \leq \frac34$ . (Compose with a disc automorphism.)

Solution

(a) Equality in $|f(z)| \leq |z|$ at $z = \frac12$ , so $f$ is a rotation fixing $\frac12$ : the identity. (b) $g = \phi_{1/2}\circ f$ fixes $0$ , so $|g'(0)| \leq 1$ ; $g'(0) = \phi_{1/2}'(\frac12)f'(0) = \frac{1}{1 - 1/4}f'(0) = \frac43f'(0)$ , so $|f'(0)| \leq \frac34$ .

Exercise 4.13 Hurwitz's theorem

(a) Let $f_n \to f$ uniformly on compact sets, $f$ not identically zero, and let $C$ be a small circle around $z_0$ on which $f \neq 0$ . Show that $\frac{1}{2\pi i}\oint_C\frac{f_n'}{f_n} \to \frac{1}{2\pi i}\oint_C\frac{f'}{f}$ , and conclude that for large $n$ , $f_n$ and $f$ have the same number of zeros inside $C$ . (b) Deduce Lemma 4.7: if $f$ took a value $w$ twice, apply (a) to $f_n - w$ near both points.

Exercise 4.14 Why simple connectivity is needed

Show that the annulus $\{1 < |z| < 2\}$ is not conformally equivalent to the disc. (A conformal bijection is a homeomorphism; use the fundamental group, 7A.4 The Fundamental Group, or show directly that the disc's property "every loop can be shrunk to a point" is preserved by homeomorphisms.) In fact two annuli $\{1 < |z| < R\}$ and $\{1 < |z| < R'\}$ are conformally equivalent only if $R = R'$ , which you can try to prove with the Schwarz reflection principle: conformal geometry is much more rigid than topology.

Exercise 4.15 Rehearsal: harmonic functions are averages, in any dimension

On the disc, the mean value property followed from Cauchy's formula. In 6A.2 Harmonic Functions it is proved without complex numbers: for harmonic $u$ on $\mathbb{R}^n$ , the average $\phi(r)$ of $u$ over the sphere of radius $r$ about a point has $\phi'(r) = \frac{r}{n}\cdot(\text{average of }\Delta u\text{ over the ball}) = 0$ . Prove the two-dimensional case this way: with $\phi(r) = \frac{1}{2\pi}\int_0^{2\pi}u(re^{i\theta})\,d\theta$ , show $\phi'(r) = \frac{1}{2\pi r}\oint_{|z|=r}\partial_\nu u\,ds$ , apply the divergence theorem (1A.10 Divergence, Curl and the Integral Theorems), and conclude $\phi(r) = u(0)$ .

Solution

$\phi'(r) = \frac{1}{2\pi}\int_0^{2\pi}\partial_ru(re^{i\theta})\,d\theta = \frac{1}{2\pi r}\oint_{|z|=r}\partial_\nu u\,ds$ , since $ds = r\,d\theta$ and $\partial_r$ is the outward normal derivative. By the divergence theorem the boundary integral is $\iint_{|z|<r}\Delta u\,dA = 0$ . So $\phi$ is constant, and $\phi(r) \to u(0)$ as $r \to 0$ by continuity.

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4.1Steady temperature in a plate

4.2Harmonic functions are real parts

4.3The Poisson kernel

4.4The Schwarz lemma

4.5The Riemann mapping theorem

4.6History

4.7Exercises

Steady temperature in a plate

Harmonic functions are real parts

The Poisson kernel

The Schwarz lemma

The Riemann mapping theorem

History

Exercises