---
title: "Conformal Maps and the Riemann Mapping Theorem"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 18
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/18-conformal-maps-and-the-riemann-mapping-theorem
---

# Chapter 18 — Conformal Maps and the Riemann Mapping Theorem

A [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) bijection between two domains transports all of complex analysis from one to the other: such maps — *conformal*, because they preserve angles — are the isomorphisms of the [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) world. This chapter classifies them where classification is possible (the disc, the plane: Schwarz’s lemma is the key, an inequality of astonishing power), builds the compactness theory of [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) families (Montel), and proves the deepest existence theorem of the subject: *every* [simply connected](#def-b3-conformal-simplyconnected) proper subdomain of $\C$, however jagged its boundary, is [conformally equivalent](#def-b3-conformal-conformal) to the unit disc. We close with [harmonic functions](#prop-b3-conformal-harmonicholo) and the [Poisson kernel](#thm-b3-conformal-poisson), solving the Dirichlet problem on the disc — the analytic payoff of conformal geometry. Throughout, $\mathbb D = D(0,1)$ and $\mathbb H = \{\operatorname{Im}z >
0\}$.

## 18.1 Conformal maps; Möbius transformations

**Definition 18.1.**

A *conformal map* (or biholomorphism) between [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) is a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) bijection; its inverse is automatically [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) ([Corollary 17.10](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#cor-b3-residues-openmapping)). Two domains are *conformally equivalent* if such a map exists; $\operatorname{Aut}(\Omega)$ denotes the group of conformal self-maps. Where $f' \neq 0$ — everywhere, for injective $f$ ([Corollary 17.10](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#cor-b3-residues-openmapping)’s proof) — the differential is multiplication by $f'(z) \neq 0$: a similarity, so conformal maps preserve angles between curves, including orientation.

**Example 18.2 (Möbius transformations).**

For $\bigl(\begin{smallmatrix}a & b\\ c &
d\end{smallmatrix}\bigr) \in GL_2(\C)$, the *Möbius transformation* $z \mapsto
\frac{az + b}{cz + d}$ is conformal from $\C\setminus\{-d/c\}$ onto $\C\setminus\{a/c\}$ (inverse of the same type, from the inverse matrix; composition corresponds to matrix product). The *Cayley map*

$$
\varphi(z) = \frac{z - \iu}{z + \iu}
$$

maps $\mathbb H$ [conformally](#def-b3-conformal-conformal) onto $\mathbb D$: indeed $\abs{z
- \iu} < \abs{z + \iu}$ exactly when $z$ is closer to $\iu$ than to $-\iu$, i.e. $\operatorname{Im}z > 0$; the inverse is $w \mapsto \iu\frac{1 + w}{1 - w}$. Möbius maps send the family circles-and-lines to itself ([Exercise 18.1](#exo-b3-conformal-1)).

**Example 18.3 (The Joukowski map).**

Beyond Möbius, the most useful [conformal map](#def-b3-conformal-conformal) of classical applied mathematics is

$$
J(z) = \frac12\Bigl(z + \frac1z\Bigr) .
$$

On the exterior $\Omega = \{\abs z > 1\}$ of the unit disc, $J$ is injective: $J(z) = J(w)$ gives $(z - w)(1 -
\frac1{zw}) = 0$ and $\abs{zw} > 1$. Its derivative $J'(z) =
\frac12(1 - z^{-2})$ vanishes only at $z = \pm1$, on the boundary: $J$ is a conformal equivalence from $\Omega$ onto its image, which is $\C\setminus\intcc{-1}1$ — the unit circle itself is folded two-to-one onto the segment ($J(\eu^{\iu\theta}) = \cos\theta$). Thus the exterior of a *segment*, a slit plane with no smooth boundary, is [conformally](#def-b3-conformal-conformal) the exterior of a disc: corners are no obstacle to conformal equivalence, only to boundary smoothness. Images of circles through $\pm1$ but [off-center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) are airfoil-shaped curves, and composing $J$ with Möbius maps transported the flow past a cylinder — computable by hand — to the flow past a wing: for the first half of the twentieth century, this example *was* aerodynamics. It is also the door to Chebyshev: $J$ conjugates $z \mapsto z^n$ to the Chebyshev polynomial $T_n$ ([Problem 13.1](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#pb-b3-hilbert-1), Part V), since $J(z^n) = \cos(n\theta)$ when $z = \eu^{\iu\theta}$.

## 18.2 Schwarz’s lemma and automorphism groups

**Theorem 18.4 (Schwarz lemma).**

Let $f \colon \mathbb D \to \mathbb D$ be [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) with $f(0) = 0$. Then

$$
\abs{f(z)} \leq \abs z \ \ (z \in \mathbb D)
\qquad\text{and}\qquad \abs{f'(0)} \leq 1 ,
$$

and if $\abs{f(z_0)} = \abs{z_0}$ for one $z_0 \neq 0$, or $\abs{f'(0)} = 1$, then $f(z) = \eu^{\iu\theta}z$ is a rotation.

**Proof.** $g(z) = f(z)/z$ extends [holomorphically](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) to $\mathbb D$ (the singularity at $0$ is removable: $g$ is bounded near $0$, [Theorem 17.4](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#thm-b3-residues-riemanncw); its value at $0$ is $f'(0)$). On $\abs z = r < 1$: $\abs g \leq \frac1r$, so by the maximum principle ([Theorem 16.14](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-maximum)) $\abs g \leq \frac1r$ on $\bar D(0,r)$; let $r \to 1$: $\abs
g \leq 1$ on $\mathbb D$, which is both inequalities. Equality at an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) point makes $\abs g$ attain an [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum: $g$ constant of modulus $1$. ∎

**Theorem 18.5 (Automorphisms of the disc).**

For $a \in \mathbb D$, the *Blaschke factor*

$$
\varphi_a(z) = \frac{z - a}{1 - \bar a z}
$$

is an automorphism of $\mathbb D$ exchanging $a$ and $0$, with $\varphi_a^{-1} = \varphi_{-a}$. Every automorphism of $\mathbb D$ is $\eu^{\iu\theta}\varphi_a$ for unique $\theta
\in \R/2\pi\Z$, $a \in \mathbb D$.

**Proof.** On $\abs z = 1$: $\abs{1 - \bar az} = \abs{\bar z}\abs{1 -
\bar az} = \abs{\bar z - \bar a\abs z^2} = \abs{\bar z - \bar
a} = \abs{z - a}$, so $\abs{\varphi_a} = 1$ there; by the maximum principle $\varphi_a(\mathbb D) \subseteq \bar{\mathbb
D}$, and openness puts the image in $\mathbb D$. The [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) identity $\varphi_{-a}\circ\varphi_a = \mathrm{id}$ (direct computation) shows bijectivity. Now let $f \in
\operatorname{Aut}(\mathbb D)$ and $a = f^{-1}(0)$: $g =
f\circ\varphi_{-a}$ is an automorphism fixing $0$. Schwarz applied to $g$ and to $g^{-1}$: $\abs{g(z)} \leq \abs z$ and $\abs{g^{-1}(w)} \leq \abs w$, so $\abs{g(z)} = \abs z$: rotation, $g = \eu^{\iu\theta}\,\mathrm{id}$, i.e. $f =
\eu^{\iu\theta}\varphi_a$. Uniqueness: $a = f^{-1}(0)$ and $\theta$ from $f'$-type evaluation (or from $f(0) =
-\eu^{\iu\theta}a$ and one more value). ∎

**Theorem 18.6 (Automorphisms of the plane).**

$\operatorname{Aut}(\C) = \{z \mapsto az + b : a \in \C^*,\ b
\in \C\}$. Consequently $\C$ and $\mathbb D$ are not [conformally equivalent](#def-b3-conformal-conformal).

**Proof.** Let $f \in \operatorname{Aut}(\C)$ and consider $g(z) =
f(1/z)$ on $\C^*$: a [holomorphic function](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) with an [isolated singularity](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#def-b3-residues-singularities) at $0$. If it were essential, Casorati–Weierstrass ([Theorem 17.4](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#thm-b3-residues-riemanncw)) would make $g\bigl(D(0,\varepsilon)\setminus\{0\}\bigr)$ dense, while $f(D(0, 1))$ is open and disjoint from it ($f$ injective: the two sets are images of disjoint sets) — impossible for a [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) and a nonempty [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology). So $0$ is a pole or removable for $g$, i.e. $\abs{f(z)}$ has at most polynomial growth: $f$ is a polynomial ([Exercise 16.4](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#exo-b3-holomorphic-4)(a)). Injectivity forces degree $1$: a higher-degree polynomial has either a multiple root of $f - c$ somewhere ($f'$ vanishes) or several distinct preimages (d’Alembert–Gauss, [Problem 16.1](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#pb-b3-holomorphic-1)); either way injectivity fails. Finally, a conformal $\C \to
\mathbb D$ would be a bounded entire function: constant (Liouville) — no equivalence. ∎

## 18.3 Montel’s theorem

**Theorem 18.7 (Montel).**

Let $\mathcal F \subseteq \mathcal H(\Omega)$ be *locally bounded*: every point has a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) on which $\sup_{f\in
\mathcal F}\sup\abs f < \infty$. Then every sequence of $\mathcal F$ has a subsequence converging uniformly on all [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) subsets of $\Omega$ (to a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) limit).

**Proof.** Local [equicontinuity](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-equicontinuous): if $\abs f \leq M$ on $D(a, 2r)
\subseteq \Omega$ for all $f \in \mathcal F$, the Cauchy formula gives, for $z, z' \in D(a, r)$,

$$
\abs{f(z) - f(z')} = \frac{\abs{z - z'}}{2\pi}
\Bigl|\int_{C_{2r}}\frac{f(w)\,\dd w}{(w-z)(w-z')}\Bigr|
\leq \frac{\abs{z - z'}\,2\pi\cdot2r\,M}{2\pi\,r^2}
= \frac{2M}{r}\,\abs{z - z'} :
$$

a uniform Lipschitz bound. Exhaust $\Omega$ by [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K_m$; each $K_m$ is covered by finitely many such discs, so $\mathcal F$ is uniformly bounded and [equicontinuous](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-equicontinuous) on $K_m$: Arzelà–Ascoli ([Theorem 7.11](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-ascoli)) extracts a subsequence converging uniformly on $K_m$; diagonalize over $m$. The limit is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) by [Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv). ∎

## 18.4 The Riemann mapping theorem

**Definition 18.8.**

An open [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $\Omega \subseteq \C$ is *simply connected* (in the homological sense, sufficient for all our purposes) if $\operatorname{Ind}_\gamma(w) = 0$ for every cycle $\gamma$ in $\Omega$ and every $w \notin \Omega$ — “no cycle of $\Omega$ surrounds a hole”. By the global Cauchy theorem ([Theorem 17.1](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#thm-b3-residues-globalcauchy)) and [Proposition 16.5](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#prop-b3-holomorphic-primitive), on such $\Omega$ *every [holomorphic function](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) has a primitive*; hence every zero-free $f \in \mathcal H(\Omega)$ has a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) logarithm ($\exp\circ$(primitive of $f'/f$), adjusted by a constant, as $(f\eu^{-L})' = 0$) and [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) $n$-th roots $\eu^{L/n}$.

**Theorem 18.9 (Riemann mapping theorem).**

Every [simply connected](#def-b3-conformal-simplyconnected) open $\Omega \subsetneq \C$, $\Omega
\neq \varnothing$, is [conformally equivalent](#def-b3-conformal-conformal) to $\mathbb D$; given $z_0 \in \Omega$, there is a unique conformal $f \colon
\Omega \to \mathbb D$ with $f(z_0) = 0$ and $f'(z_0) >
0$.

**Proof.** *Step 0: the family is nonempty.* Pick $b \notin \Omega$: $z - b$ is zero-free on $\Omega$, so it has a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) square root $h$ ($h^2 = z - b$). $h$ is injective ($h(z) =
h(z')$ squares to $z = z'$), and if $w \in h(\Omega)$ then $-w
\notin h(\Omega)$ ($h(z) = -h(z')$ also squares to $z = z'$, giving $w = -w = 0$, impossible as $h$ is zero-free). Since $h(\Omega)$ is open, it contains a disc $D(h(z_0), \rho)$; then $D(-h(z_0), \rho) \cap h(\Omega) = \varnothing$, i.e. $\abs{h(z) + h(z_0)} \geq \rho$ for every $z \in \Omega$. Hence

$$
g(z) = \frac{\rho}{2\,\bigl(h(z) + h(z_0)\bigr)}
$$

is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo), injective (a Möbius map composed with the injective $h$), with $\abs g \leq \frac12 < 1$. Composing with a [Blaschke factor](#thm-b3-conformal-autdisc) ([Theorem 18.5](#thm-b3-conformal-autdisc)) to move $g(z_0)$ to $0$, the family

$$
\mathcal F = \{f \colon \Omega \to \mathbb D \text{
holomorphic, injective, } f(z_0) = 0\}
$$

is nonempty.

*Step 1: an extremal element.* Let $s =
\sup_{\mathcal F}\abs{f'(z_0)} \in \intoc0{+\infty}$ ($> 0$: members are injective, so $f'(z_0) \neq 0$). Take $f_n \in
\mathcal F$ with $\abs{f_n'(z_0)} \to s$: the family is bounded by $1$, so Montel ([Theorem 18.7](#thm-b3-conformal-montel)) extracts $f_n \to f$ uniformly on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact); $f$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo), $f(z_0) = 0$, $\abs{f'(z_0)} = s$ ([Theorem 16.15](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-weierstrassconv) for the derivatives), in particular $f$ is nonconstant; $f$ is injective by Hurwitz ([Exercise 17.8](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#exo-b3-residues-8)(b)), and $f(\Omega) \subseteq \bar{\mathbb D}$, hence $\subseteq
\mathbb D$ (open mapping). So $f \in \mathcal F$ attains the supremum: $s < \infty$.

*Step 2: the extremal map is onto.* Suppose $a \in
\mathbb D\setminus f(\Omega)$. The Blaschke transport $\varphi_a\circ f$ is zero-free on the [simply connected](#def-b3-conformal-simplyconnected) $\Omega$: it has a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) square root $F$ (with $F(\Omega) \subseteq \mathbb D$, as $\abs F^2 =
\abs{\varphi_a\circ f} < 1$), injective (squares distinguish). Normalize: $G = \varphi_{F(z_0)}\circ F \in
\mathcal F$. Undoing: $f = \varphi_{-a}\circ s_2 \circ
\varphi_{-F(z_0)}\circ G$ where $s_2(w) = w^2$; the map $\Psi
= \varphi_{-a}\circ s_2\circ\varphi_{-F(z_0)} \colon \mathbb D
\to \mathbb D$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) with $\Psi(0) = f(z_0) = 0$ and is *not* a rotation (it is not injective: $s_2$ is not). Schwarz’s lemma (strict case): $\abs{\Psi'(0)} < 1$, and the chain rule $f = \Psi\circ G$ gives $\abs{f'(z_0)} =
\abs{\Psi'(0)}\,\abs{G'(z_0)} < \abs{G'(z_0)}$ — contradicting maximality (note $G \in \mathcal F$). Hence $f$ is onto: a conformal equivalence.

*Step 3: normalization and uniqueness.* Multiply $f$ by $\eu^{-\iu\arg f'(z_0)}$ to make $f'(z_0) > 0$ (this stays in $\mathcal F$). If $f_1, f_2$ both work, $\psi =
f_2\circ f_1^{-1} \in \operatorname{Aut}(\mathbb D)$ fixes $0$ with $\psi'(0) = f_2'(z_0)/f_1'(z_0) > 0$; by [Theorem 18.5](#thm-b3-conformal-autdisc) $\psi$ is a rotation $\eu^{\iu\theta}$ with $\eu^{\iu\theta} > 0$: $\psi =
\mathrm{id}$. ∎

**Remark 18.10.**

The theorem is a pure existence statement of astonishing scope: a square, a half-plane, the complement of a slit, the region between two tangent circles, a fractal-boundary domain — all [conformally](#def-b3-conformal-conformal) identical to $\mathbb D$. What it does not give: any formula (explicit maps are the exception: [Exercise 18.5](#exo-b3-conformal-5)), boundary behavior (a deeper theory — Carathéodory’s theorem — handles it), or uniqueness of extension to $\C$ or multiply [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) domains: the annulus $\{1 < \abs z < 2\}$ is *not* [conformally](#def-b3-conformal-conformal) a punctured disc, and annuli of different radius ratios are inequivalent (a genuinely harder fact).

## 18.5 Harmonic functions and the Poisson kernel

**Proposition 18.11.**

Let $\Omega$ be [simply connected](#def-b3-conformal-simplyconnected) and $u \colon \Omega \to \R$ harmonic ($\mathcal C^2$ with $\Delta u = u_{xx} + u_{yy} = 0$). Then $u =
\operatorname{Re}F$ for a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) $F$, unique up to an imaginary constant. Consequently $u$ is $\mathcal C^\infty$, satisfies the mean value property, and obeys the maximum principle (no strict [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) extremum unless constant).

**Proof.** $g = u_x - \iu u_y$ satisfies the [Cauchy–Riemann equations](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#prop-b3-holomorphic-cauchyriemann) ($P = u_x$, $Q = -u_y$: $P_x = u_{xx} = -u_{yy} = Q_y$ and $P_y = u_{xy} = u_{yx} = -Q_x$) with [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) partials: $g$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) ([Proposition 16.2](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#prop-b3-holomorphic-cauchyriemann); the $\R$-differentiability follows from $\mathcal C^1$). Let $F_0$ be a primitive (simple connectivity, [Definition 18.8](#def-b3-conformal-simplyconnected)); then $\operatorname{Re}F_0$ has gradient $(u_x, u_y)$ ($F_0' = g$ unpacks to exactly that via Cauchy–Riemann for $F_0$), so $u - \operatorname{Re}F_0$ is constant ($\Omega$ [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)): adjust $F = F_0 + c$. The properties transfer from [Theorem 16.14](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-maximum) and [Exercise 16.10](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#exo-b3-holomorphic-10) (for the maximum principle applied to $u$ itself, use $\eu^{F}$ as there). ∎

**Theorem 18.12 (Poisson formula; Dirichlet problem on the disc).**

For $0 \leq r < 1$ define the *Poisson kernel*

$$
P_r(\theta) = \sum_{n\in\Z}r^{\abs n}\eu^{\iu n\theta}
= \frac{1 - r^2}{1 - 2r\cos\theta + r^2} \;>\; 0 .
$$

Let $g \colon \partial\mathbb D \to \R$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and set, for $z = r\eu^{\iu\varphi} \in \mathbb D$,

$$
u(z) = \frac1{2\pi}\int_0^{2\pi}
P_r(\varphi - t)\,g(\eu^{\iu t})\,\dd t .
$$

Then $u$ is harmonic on $\mathbb D$ and extends [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) to $\bar{\mathbb D}$ with boundary values $g$: the unique such [harmonic function](#prop-b3-conformal-harmonicholo).

**Proof.** *Kernel identities*: summing two geometric series,

$$
\sum_{n\in\Z}r^{\abs n}\eu^{\iu n\theta}
= \operatorname{Re}\frac{1 + r\eu^{\iu\theta}}{1 -
r\eu^{\iu\theta}}
= \frac{1 - r^2}{\abs{1 - r\eu^{\iu\theta}}^2},
$$

which is the displayed quotient; positivity is clear, and $\frac1{2\pi}\int_0^{2\pi}P_r = 1$ (only $n = 0$ survives).

*Harmonicity*: with $z = r\eu^{\iu\varphi}$,

$$
u(z) = \operatorname{Re}\biggl[\frac1{2\pi}\int_0^{2\pi}
\frac{\eu^{\iu t} + z}{\eu^{\iu t} - z}\,
g(\eu^{\iu t})\,\dd t\biggr],
$$

(the bracketed kernel has real part $P_r(\varphi - t)$: compute), and the bracket is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) in $z$ on $\mathbb D$ ([Exercise 16.7](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#exo-b3-holomorphic-7)): $u$ is the real part of a [holomorphic function](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo), hence harmonic.

*Boundary values*: $P_r(\cdot)$ is an approximate identity as $r \to 1^-$: mass $1$, and for $\delta \leq
\abs\theta \leq \pi$, $P_r(\theta) \leq \frac{1 - r^2}{1 -
2r\cos\delta + r^2} \to 0$ uniformly. The standard split ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $g$ near $\eu^{\iu\varphi_0}$, boundedness elsewhere) gives $u(r\eu^{\iu\varphi}) \to
g(\eu^{\iu\varphi_0})$ as $r\eu^{\iu\varphi} \to
\eu^{\iu\varphi_0}$, uniformly in the boundary point: the extension is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). *Uniqueness*: the difference of two solutions is harmonic on $\mathbb D$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior), zero on the boundary: by the maximum principle (applied to $\pm$ the difference), it vanishes. ∎

![The level curves of Rez2 = x2 - y2 (red hyperbolas) and Imz2 = 2xy (blue hyperbolas) intersect at right angles away from 0: a conformal map (z z2, where z ≠ 0) preserves the orthogonality of coordinate lines. At z = 0, where the derivative vanishes, angles are doubled instead.](https://one-course.com/images/onecourse/chapters/math-5/b3-conformal/fig-519b9b0ea3e1.svg)

*The level curves of $\operatorname{Re}z^2 = x^2 - y^2$ (red hyperbolas) and $\operatorname{Im}z^2 = 2xy$ (blue hyperbolas) intersect at right angles away from $0$: a [conformal map](#def-b3-conformal-conformal) ($z \mapsto z^2$, where $z \neq 0$) preserves the orthogonality of coordinate lines. At $z = 0$, where the derivative vanishes, angles are doubled instead.*

## 18.6 Exercises

**Exercise 18.1 ★.**

(a) Verify that the [Cayley map](#ex-b3-conformal-mobius) $\varphi(z) = \frac{z -
\iu}{z + \iu}$ is a bijection $\mathbb H \to \mathbb D$ with the stated inverse, and compute the images of $\iu$, $0$, $1$, $\infty$ (limit). (b) Show that $z \mapsto 1/z$ maps circles and lines to circles and lines. *(Write their common equation $\alpha\abs z^2 + \bar\beta z + \beta\bar z + \gamma = 0$, $\alpha, \gamma \in \R$.)* Deduce the same for all Möbius maps.

**Solution of Exercise 18.1.**

(a) $\varphi$ and $\psi(w) = \iu\frac{1+w}{1-w}$ compose to the identity in both orders (direct computation); $\varphi$ maps $\mathbb H$ into $\mathbb D$ and $\psi$ back ([Example 18.2](#ex-b3-conformal-mobius)). Values: $\varphi(\iu) = 0$, $\varphi(0) = -1$, $\varphi(1) = \frac{1 - \iu}{1 + \iu} =
-\iu$, and $\varphi(z) \to 1$ as $z \to \infty$.

(b) Circles and lines are the solution sets of $\alpha\abs z^2 + \bar\beta z + \beta\bar z + \gamma = 0$ ($\alpha, \gamma \in \R$, $\beta \in \C$, $\abs\beta^2 >
\alpha\gamma$): $\alpha \neq 0$ circles, $\alpha = 0$ lines. Substituting $z = 1/w$ and multiplying by $\abs w^2$: $\gamma\abs w^2 + \beta w + \bar\beta\bar w + \alpha = 0$ — same family. Affine maps clearly preserve the family, and every Möbius map is a composition of affine maps and one inversion ($\frac{az+b}{cz+d} = \frac ac + \frac{bc -
ad}{c}\cdot\frac1{cz + d}$ for $c \neq 0$).

**Exercise 18.2 ★.**

Let $f \colon \mathbb D \to \mathbb D$ be [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo). (a) If $f(0) = 0$ and $f(a) = a$ for some $a \neq 0$, show $f
= \mathrm{id}$. (b) If $f$ is an automorphism with two distinct fixed points in $\mathbb D$, show $f = \mathrm{id}$ *(conjugate by a [Blaschke factor](#thm-b3-conformal-autdisc) to reduce to (a))*.

**Solution of Exercise 18.2.**

(a) Schwarz gives $\abs{f(a)} \leq \abs a$ with equality (both sides $= \abs a$): the equality case forces $f(z) =
\eu^{\iu\theta}z$, and $f(a) = a$ pins $\eu^{\iu\theta} = 1$.

(b) Let $a \neq b$ be fixed points and $g =
\varphi_a\circ f\circ\varphi_{-a} \in
\operatorname{Aut}(\mathbb D)$ — using [Theorem 18.5](#thm-b3-conformal-autdisc) for $\varphi_{\pm a}$. Then $g(0) = \varphi_a(f(a)) = 0$ and $g(c) = c$ for $c =
\varphi_a(b) \neq 0$: by (a), $g = \mathrm{id}$, so $f =
\varphi_{-a}\circ\varphi_a = \mathrm{id}$.

**Exercise 18.3 ★★.**

(Schwarz–Pick) For [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) $f\colon \mathbb D \to
\mathbb D$, prove

$$
\frac{\abs{f'(z)}}{1 - \abs{f(z)}^2} \;\leq\;
\frac{1}{1 - \abs z^2}
\qquad (z \in \mathbb D),
$$

with equality (at one point, hence everywhere) iff $f \in
\operatorname{Aut}(\mathbb D)$. *(Apply Schwarz to $\varphi_{f(z)}\circ
f\circ\varphi_{-z}$.)* Interpretation: [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) self-maps contract the hyperbolic metric.

**Solution of Exercise 18.3.**

Fix $z$ and set $g = \varphi_{f(z)}\circ f\circ\varphi_{-z}$: [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) $\mathbb D \to \mathbb D$ with $g(0) = 0$, so $\abs{g'(0)} \leq 1$ (Schwarz). Chain rule with $\varphi_a'(\zeta) = \frac{1 - \abs a^2}{(1 - \bar
a\zeta)^2}$:

$$
g'(0) = \varphi_{f(z)}'\bigl(f(z)\bigr)\cdot f'(z)\cdot
\varphi_{-z}'(0)
= \frac{1}{1 - \abs{f(z)}^2}\cdot f'(z)\cdot(1 - \abs z^2),
$$

whence the Schwarz–Pick inequality. Equality at some $z$ makes $g$ a rotation, hence $f =
\varphi_{-f(z)}\circ(\text{rotation})\circ\varphi_z \in
\operatorname{Aut}(\mathbb D)$ — and then equality holds everywhere (compute, or reapply with roles of $f, f^{-1}$ exchanged). [Holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) self-maps of the disc are $1$-Lipschitz for the hyperbolic metric $\frac{2\abs{\dd
z}}{1 - \abs z^2}$; automorphisms are its isometries.

**Exercise 18.4 ★★.**

Find explicit conformal equivalences: (a) the strip $\{0 < \operatorname{Im}z < \pi\} \to \mathbb H$; (b) the quadrant $\{\operatorname{Re}z > 0,
\operatorname{Im}z > 0\} \to \mathbb H$; (c) the half-disc $\mathbb D\cap\mathbb H \to$ a quadrant, then $\to \mathbb H$; (d) $\mathbb D \to \mathbb D$ sending $\frac12$ to $0$ with positive derivative there.

**Solution of Exercise 18.4.**

(a) $z \mapsto \eu^z$: maps $\{0 < \operatorname{Im}z <
\pi\}$ bijectively onto $\mathbb H$ ($\eu^{x+\iu y} =
\eu^x\eu^{\iu y}$: modulus free, argument $y \in
\intoo0\pi$), [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) with nonvanishing derivative and [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) inverse (principal $\log$). (b) $z \mapsto z^2$ doubles arguments: the open quadrant $\{0 < \arg z < \frac\pi2\}$ maps [conformally](#def-b3-conformal-conformal) onto $\mathbb
H$ (inverse: principal square root). (c) $z \mapsto \frac{1 + z}{1 - z}$ maps $\mathbb D$ onto the right half-plane and preserves the upper/lower symmetry: it sends the upper half-disc onto the first quadrant; then square, by (b), to reach $\mathbb H$: $z \mapsto \bigl(\frac{1
+ z}{1 - z}\bigr)^2$. (d) The [Blaschke factor](#thm-b3-conformal-autdisc) $\varphi_{1/2}(z) = \frac{z -
\frac12}{1 - \frac z2}$: $\varphi_{1/2}(\tfrac12) = 0$ and $\varphi_{1/2}'(\tfrac12) = \frac{1 - \frac14}{(1 -
\frac14)^2} = \frac43 > 0$.

**Exercise 18.5 ★★.**

(a) Show that no [conformal map](#def-b3-conformal-conformal) $\C \to \mathbb D$ or $\C \to
\mathbb H$ exists, and none $\mathbb D \to \C$. (b) Which of the following are [conformally equivalent](#def-b3-conformal-conformal) to $\mathbb D$? Justify via [Theorem 18.9](#thm-b3-conformal-rmt) or an obstruction: a square; $\C\setminus\intoc{-\infty}0$; $\mathbb D\setminus\{0\}$; $\{1 < \abs z < 2\}$. *(For the last two: a conformal image of the punctured disc would extend over the puncture by [Theorem 17.4](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#thm-b3-residues-riemanncw)(1) — develop this.)*

**Solution of Exercise 18.5.**

(a) A conformal $\C \to \mathbb D$ (or $\mathbb H$, after composing with Cayley) is a bounded entire function: constant by Liouville — not bijective. A conformal $\mathbb
D \to \C$ would have a conformal inverse $\C \to \mathbb D$: same contradiction.

(b) The square is convex, hence [simply connected](#def-b3-conformal-simplyconnected), and proper: [conformally](#def-b3-conformal-conformal) $\mathbb D$ ([Theorem 18.9](#thm-b3-conformal-rmt)). The cut plane $\C\setminus\intoc{-\infty}0$ is star-shaped about $1$ (segments from $1$ avoid the cut), hence [simply connected](#def-b3-conformal-simplyconnected), and proper: [conformally](#def-b3-conformal-conformal) $\mathbb D$. The punctured disc: if $g \colon \mathbb D\setminus\{0\} \to \mathbb D$ were conformal, $g$ is bounded, so $0$ is removable ([Theorem 17.4](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#thm-b3-residues-riemanncw)): $g$ extends to $\tilde g
\colon \mathbb D \to \mathbb D$, and $\tilde g(0)$, being in the open image $g(\mathbb D\setminus\{0\}) = \mathbb D$, is also $g(w)$ for some $w \neq 0$; two disjoint [neighborhoods](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $0$ and $w$ have images that are open and share the value $\tilde g(0)$, hence share *other* values too ([open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology)): $g$ takes some value twice on $\mathbb D\setminus\{0\}$ — contradicting injectivity. The annulus $A = \{1 < \abs z < 2\}$: suppose $F \colon \mathbb D
\to A$ conformal. $F$ is zero-free on the [simply connected](#def-b3-conformal-simplyconnected) $\mathbb D$, so $F = \eu^L$ for [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) $L$ ([Definition 18.8](#def-b3-conformal-simplyconnected)). Let $\sigma$ be the circle $\abs z = \frac32$ in $A$ and $\gamma =
F^{-1}\circ\sigma$, a closed [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) in $\mathbb D$; then

$$
1 = \operatorname{Ind}_\sigma(0)
= \frac1{2\iu\pi}\int_{F\circ\gamma}\frac{\dd w}{w}
= \frac1{2\iu\pi}\int_\gamma\frac{F'}{F}
= \frac1{2\iu\pi}\int_\gamma L' = 0
$$

($L'$ has a primitive): contradiction. Neither the punctured disc nor the annulus is a disc in disguise.

**Exercise 18.6 ★★.**

Let $\mathcal F = \{f \in \mathcal H(\mathbb D) : f(0) = 1,\
\operatorname{Re}f > 0\}$. (a) Show that $\mathcal F$ is locally bounded. *(Compose with the Cayley-type map $w \mapsto \frac{w -
1}{w + 1}$ sending the right half-plane to $\mathbb D$, and apply Schwarz.)* (b) Deduce the *Herglotz bound*: $\abs{f(z)} \leq
\frac{1 + \abs z}{1 - \abs z}$ for $f \in \mathcal F$, with equality possibilities.

**Solution of Exercise 18.6.**

(a) $T(w) = \frac{w - 1}{w + 1}$ maps $\{\operatorname{Re}w >
0\}$ [conformally](#def-b3-conformal-conformal) onto $\mathbb D$ (Cayley rotated: $\abs{w -
1} < \abs{w + 1}$ iff $\operatorname{Re}w > 0$), with $T(1) =
0$. For $f \in \mathcal F$, $g = T\circ f\colon \mathbb D \to
\mathbb D$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) with $g(0) = 0$: Schwarz gives $\abs{g(z)} \leq \abs z$.

(b) Inverting $T$: $f = \frac{1 + g}{1 - g}$, so

$$
\abs{f(z)} \leq \frac{1 + \abs{g(z)}}{1 - \abs{g(z)}}
\leq \frac{1 + \abs z}{1 - \abs z} :
$$

locally bounded (uniformly on $\abs z \leq r < 1$). Equality at $z_0 \neq 0$ forces $\abs{g(z_0)} = \abs{z_0}$ and alignment: $g$ a rotation, i.e. $f(z) = \frac{1 +
\eu^{\iu\theta}z}{1 - \eu^{\iu\theta}z}$ — the *Herglotz extremals*, [conformal maps](#def-b3-conformal-conformal) onto the right half-plane.

**Exercise 18.7 ★★★.**

Where does the proof of [Theorem 18.9](#thm-b3-conformal-rmt) use each hypothesis? Trace: (i) simple connectivity (twice); (ii) $\Omega \neq \C$; (iii) [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected). Then show that the theorem *fails* for $\Omega = \C$ and for the annulus, pinpointing which step of the proof breaks in each case.

**Solution of Exercise 18.7.**

(i) Simple connectivity enters exactly twice, through the existence of [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) square roots of zero-free functions ([Definition 18.8](#def-b3-conformal-simplyconnected)): in Step 0 (the root of $z - b$) and in Step 2 (the root of $\varphi_a\circ f$). (ii) $\Omega \neq \C$ provides the point $b$ of Step 0 — without it the family $\mathcal F$ is empty of injective bounded maps (Liouville). (iii) [Connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) is used whenever the identity theorem or Hurwitz ([Exercise 17.8](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#exo-b3-residues-8)) speaks: the extremal limit is “injective or constant”, and constancy is excluded by $s > 0$; also in “zero derivative implies constant”. Failure for $\C$: Step 0 impossible, and the conclusion is false ([Exercise 18.5](#exo-b3-conformal-5)(a)). Failure for the annulus: not [simply connected](#def-b3-conformal-simplyconnected) — the square-root construction breaks (e.g. $z$ itself, zero-free on $A$, has no [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) square root: the same index computation as in [Exercise 18.5](#exo-b3-conformal-5)(b) with $\frac12\int_\sigma\frac{\dd z}z \notin 2\iu\pi\Z$) — and the conclusion is false too.

**Exercise 18.8 ★★.**

Solve the Dirichlet problem on $\mathbb D$ for the boundary data: (a) $g(\eu^{\iu t}) = \cos t$; (b) $g(\eu^{\iu t}) =
\cos^2 t$; (c) $g = \mathbf 1_{\text{upper semicircle}}$ — for (c) compute $u(0)$ and interpret via the mean value property. *(Expand $g$ in Fourier series and use $P_r$’s series: $u(r\eu^{\iu\varphi}) = \sum_n c_n(g)
r^{\abs n}\eu^{\iu n\varphi}$.)*

**Solution of Exercise 18.8.**

Substituting the Fourier expansion of $g$ into the Poisson integral and using $\frac1{2\pi}\int P_r(\varphi - t)\eu^{\iu nt}\dd t =
r^{\abs n}\eu^{\iu n\varphi}$ (read off $P_r$’s series): $u(r\eu^{\iu\varphi}) = \sum_nc_n(g)\,r^{\abs
n}\eu^{\iu n\varphi}$, the interchange justified by normal convergence ($\abs{c_n} \leq \norm g_\infty$, $r < 1$).

(a) $g = \cos t$: $c_{\pm1} = \frac12$, so $u =
r\cos\varphi = \operatorname{Re}z = x$ — indeed harmonic with the right boundary values.

(b) $\cos^2t = \frac12 + \frac{\cos 2t}2$: $u = \frac12 +
\frac{r^2\cos2\varphi}2 = \frac12 +
\frac{\operatorname{Re}(z^2)}2 = \frac12 + \frac{x^2 -
y^2}{2}$.

(c) $g = \mathbf 1_{(0,\pi)}$ (upper semicircle): $c_0 =
\frac12$ and $c_n = \frac{1 - (-1)^n}{2\iu\pi n}$ for $n \neq
0$, so

$$
u(r\eu^{\iu\varphi}) = \frac12 + \frac2\pi\sum_{k\geq0}
\frac{r^{2k+1}\sin\bigl((2k+1)\varphi\bigr)}{2k + 1},
\qquad u(0) = \frac12 :
$$

the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) sees exactly the average of the boundary data — the mean value property in person.

**Exercise 18.9 ★★★.**

(Harnack) Let $u \geq 0$ be harmonic on $\mathbb D$. Prove, for $\abs z = r < 1$:

$$
\frac{1 - r}{1 + r}\,u(0) \;\leq\; u(z) \;\leq\;
\frac{1 + r}{1 - r}\,u(0)
$$

*(bound the [Poisson kernel](#thm-b3-conformal-poisson) between $\frac{1-r}{1+r}$ and $\frac{1+r}{1-r}$; apply the [representation](https://one-course.com/books/math/5/en/chapter/5-representations-of-finite-groups#def-b3-representations-rep) on slightly smaller discs and pass to the limit)*. Deduce: a [harmonic function](#prop-b3-conformal-harmonicholo) on $\C$ bounded below is constant.

**Solution of Exercise 18.9.**

From $(1-r)^2 \leq 1 - 2r\cos\theta + r^2 \leq (1+r)^2$:

$$
\frac{1-r}{1+r} = \frac{1 - r^2}{(1+r)^2} \leq P_r(\theta)
\leq \frac{1 - r^2}{(1 - r)^2} = \frac{1+r}{1-r} .
$$

For $u \geq 0$ harmonic on $\mathbb D$ and $s < 1$: $u_s(z) =
u(sz)$ is harmonic on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $\bar{\mathbb D}$, hence equals its Poisson integral ([Theorem 18.12](#thm-b3-conformal-poisson), uniqueness, applied to its own boundary values); sandwiching the kernel and using the mean value $\frac1{2\pi}\int u_s(\eu^{\iu t})\dd t = u(0)$:

$$
\frac{1-r}{1+r}\,u(0) \leq u(s\,r\eu^{\iu\varphi}) \leq
\frac{1+r}{1-r}\,u(0).
$$

Let $s \to 1^-$ at fixed $r\eu^{\iu\varphi}$ ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $u$): the Harnack inequalities. If $u$ is harmonic on $\C$ with $u \geq m$: apply Harnack to $u - m$ on discs $D(0, R)$, i.e. to $z \mapsto u(Rz) - m$: for fixed $z$ and $r =
\abs z/R \to 0$, both bounds tend to $u(0) - m$: $u(z) =
u(0)$ — constant (a two-sided Liouville from a one-sided bound).

**Exercise 18.10 ★★.**

Using conformal invariance of harmonicity ($u\circ f$ is harmonic when $u$ is harmonic and $f$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) — prove it via [Proposition 18.11](#prop-b3-conformal-harmonicholo) locally), solve the Dirichlet problem on the upper half-plane with boundary data $\mathbf 1_{\intoo{-\infty}0}$: show that

$$
u(x + \iu y) = \frac1\pi\,\arg(x + \iu y)
\qquad (\arg \in \intoo0\pi \text{ on } \mathbb H)
$$

is harmonic on $\mathbb H$ (imaginary part of a [holomorphic logarithm](#def-b3-conformal-simplyconnected)) with the required boundary limits at every $x \neq
0$, and transport it to the disc by Cayley to re-derive [Exercise 18.8](#exo-b3-conformal-8)(c).

**Solution of Exercise 18.10.**

Locally, $u = \operatorname{Re}F$ with $F$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) ([Proposition 18.11](#prop-b3-conformal-harmonicholo)), so $u\circ f =
\operatorname{Re}(F\circ f)$ is harmonic wherever defined: harmonicity is [conformally](#def-b3-conformal-conformal) invariant. On $\mathbb H$: the principal logarithm gives $\log z = \ln\abs z + \iu\arg z$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on $\mathbb H$, so $u = \frac1\pi\arg z =
\operatorname{Im}\bigl(\frac1\pi\log z\bigr)$ is harmonic, with boundary limits: for $x > 0$, $\arg \to 0$, $u \to 0$; for $x < 0$, $\arg \to \pi$, $u \to 1$: the data $\mathbf
1_{\intoo{-\infty}0}$ at every $x \neq 0$. Transporting by the [Cayley map](#ex-b3-conformal-mobius) (which sends $\mathbb D \to \mathbb H$ after inversion and matches the upper semicircle to the negative axis, up to the rotation fixed by chasing three boundary points), $u\circ(\text{Cayley})$ solves the disc problem of [Exercise 18.8](#exo-b3-conformal-8)(c); evaluating at the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) retrieves $u = \frac12$ there, and the closed form $\frac1\pi\arg$ can be checked against the series by summing $\sum\frac{r^{2k+1}\sin((2k+1)\varphi)}{2k+1} =
\frac12\arctan\frac{2r\sin\varphi}{1 - r^2}$-type identities — the elementary route to the same answer.

**Exercise 18.11 ★★.**

(Fixed points and iteration in the disc) Let $f\colon\mathbb
D \to \mathbb D$ be [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo). (a) Show that if $f$ has *two* distinct fixed points, then $f = \mathrm{id}$ *(move one to $0$ by an automorphism and apply the equality case of Schwarz)*. (b) Suppose $f(0) = 0$ and $f$ is not a rotation. Show that the iterates $f^{\circ n} \to 0$ uniformly on every [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\bar D(0, r)$, $r < 1$ *(Schwarz gives $\abs{f(z)}
\leq c_r\abs z$ on $\bar D(0,r)$ with $c_r < 1$ — justify this strict constant via the maximum principle applied to $f(z)/z$)*. (c) Illustrate with $f(z) = \frac{z^2 + z}2$: fixed points, and the rate of convergence of the [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) of $z_0 =
\frac12$.

**Solution of Exercise 18.11.**

(a) Let $a \neq b$ be fixed. Conjugating by $\varphi_a(z) =
\frac{z - a}{1 - \bar az}$ (an automorphism exchanging $a$ and $0$), $g = \varphi_a\circ f\circ\varphi_a^{-1}$ fixes $0$ and the point $c = \varphi_a(b) \neq 0$. Schwarz: $\abs{g(z)} \leq \abs z$, and at $z = c$ equality holds ($g(c) = c$): the equality case forces $g(z) = \lambda z$ with $\abs\lambda = 1$, and $\lambda c = c$ gives $\lambda =
1$: $g = \mathrm{id}$, hence $f = \mathrm{id}$.

(b) $h(z) = f(z)/z$ ([removable singularity](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#thm-b3-residues-riemanncw) at $0$) is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) on $\mathbb D$ with $\abs h \leq 1$ (Schwarz); $\abs h < 1$ everywhere, else the maximum principle ([interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) maximum of $\abs h$) would make $h$ a unimodular constant, i.e. $f$ a rotation — excluded. On the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\bar D(0,r)$, $c_r = \max\abs h < 1$, so $\abs{f(z)} \leq
c_r\abs z$ there; moreover $f$ maps $\bar D(0,r)$ into itself ($c_r\abs z \leq r$), so the bound iterates: $\abs{f^{\circ n}(z)} \leq c_r^n\,r \to 0$ uniformly on $\bar D(0, r)$.

(c) Fixed points of $\frac{z^2 + z}2$: $z^2 + z = 2z$ iff $z(z - 1) = 0$; only $z = 0$ lies in $\mathbb D$ ($z = 1$ is on the boundary). Not a rotation ($f'(0) = \frac12$), so [orbits](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) tend to $0$; quantitatively $f(z) = \frac z2(1 + z)$ gives $\abs{f(z)} \leq \frac{3}{4}\abs z$ on $\abs z \leq
\frac12$, and once the [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) is small, $\abs{f(z)} \approx
\frac{\abs z}2$: asymptotically geometric with ratio $f'(0) = \frac12$. From $z_0 = \frac12$: $z_1 = \frac38$, $z_2 \approx 0.258$, $z_3 \approx 0.162$ — halving per step, as predicted by the multiplier.

**Exercise 18.12 ★★.**

(Harmonic conjugates, concretely) Let $u(x, y) = x^3 - 3xy^2
+ 2y$. (a) Check that $u$ is harmonic on $\R^2$, and find all harmonic conjugates $v$ (i.e. $u + \iu v$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo)) by integrating the [Cauchy–Riemann equations](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#prop-b3-holomorphic-cauchyriemann); identify $f(z) =
u + \iu v$ as a polynomial in $z$. (b) Show that on a *star-shaped* [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology), every [harmonic function](#prop-b3-conformal-harmonicholo) admits a harmonic conjugate, unique up to an additive constant *(the $1$-form $-u_y\,\dd x +
u_x\,\dd y$ is closed; [Theorem 16.8](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-cauchy)’s primitive machinery, or [Chapter 21](https://one-course.com/books/math/5/en/chapter/21-differential-forms-and-stokes-theorem#ch-b3-forms)’s Poincaré lemma)*. (c) Give the standard counterexample on $\C^*$: $u = \ln\abs z$ has no global conjugate — relate to the angular form and the winding number ([Chapter 21](https://one-course.com/books/math/5/en/chapter/21-differential-forms-and-stokes-theorem#ch-b3-forms)).

**Solution of Exercise 18.12.**

(a) $\Delta u = 6x - 6x + 0 = 0$. Cauchy–Riemann demands $v_y = u_x = 3x^2 - 3y^2$ and $v_x = -u_y = 6xy - 2$. Integrating the first in $y$: $v = 3x^2y - y^3 + c(x)$; plugging into the second: $6xy + c'(x) = 6xy - 2$, so $c(x)
= -2x + C$. Thus $v = 3x^2y - y^3 - 2x + C$ and

$$
f = u + \iu v = (x^3 - 3xy^2) + \iu(3x^2y - y^3)
+ 2y - 2\iu x + \iu C = z^3 - 2\iu z + \iu C .
$$

(b) The form $\omega = -u_y\,\dd x + u_x\,\dd y$ is closed precisely because $\Delta u = 0$ ($\partial_y(-u_y) =
-u_{yy} = u_{xx} = \partial_x(u_x)$). On a star-shaped [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) the Poincaré lemma ([Theorem 21.15](https://one-course.com/books/math/5/en/chapter/21-differential-forms-and-stokes-theorem#thm-b3-forms-poincare); or the primitive construction of [Theorem 16.8](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-cauchy) applied to the [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) $u_x - \iu u_y$) provides $v$ with $\dd v = \omega$, i.e. the Cauchy–Riemann system: $u + \iu v$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo). Two conjugates differ by a function with vanishing gradient: a constant ([connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected)).

(c) For $u = \ln\abs z$ on $\C^*$: $\omega = -u_y\dd x +
u_x\dd y = \frac{-y\,\dd x + x\,\dd y}{x^2 + y^2} =
\omega_\theta$, the angular form ([Example 21.14](https://one-course.com/books/math/5/en/chapter/21-differential-forms-and-stokes-theorem#ex-b3-forms-angular)), whose integral along the unit circle is $2\pi \neq 0$: not exact, so no global conjugate exists — a conjugate would be a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) determination of the argument, and the winding number is exactly the obstruction. Locally (on any star-shaped subdomain), $v =
\arg z$ works and $u + \iu v = \log z$: the failure is global, not local.

## 18.7 Problem: the area theorem and Koebe’s quarter theorem

**Problem 18.1.**

Weekend problem — how much must a univalent map cover?

A *univalent* function is a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) injection. The normalized univalent functions on the disc,

$$
\mathcal S = \bigl\{f \in \mathcal H(\mathbb D) \text{
injective},\ f(z) = z + a_2z^2 + a_3z^3 + \cdots\bigr\},
$$

are rigidly constrained: we prove Bieberbach’s inequality $\abs{a_2} \leq 2$ and deduce the *Koebe quarter theorem*: the image of any $f \in \mathcal S$ contains the disc $D(0, \frac14)$ — the sharp universal constant of conformal geometry.

**Part I — The area theorem.** Let $g(w) = w
+ b_0 + \frac{b_1}w + \frac{b_2}{w^2} + \cdots$ be [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and *injective* on $\{\abs w > 1\}$.

1. For $\rho > 1$, let $A_\rho$ be the area ([Lebesgue measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-lebesgueouter)) of the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) set $K_\rho = \C\setminus  g(\{\abs w > \rho\})$, the region enclosed by the smooth Jordan curve $g(C_\rho)$. Using the area formula of the Year 2 volume’s Green–Riemann theorem — the enclosed area is $\frac1{2\iu}  \oint\bar\zeta\,\dd\zeta$ along the positively oriented boundary — show that $$A_\rho = \frac{1}{2\iu}\int_{C_\rho}  \overline{g(w)}\,g'(w)\,\dd w  = \pi\Bigl(\rho^2 -  \sum_{n\geq1}n\,\abs{b_n}^2\rho^{-2n}\Bigr) :$$ substitute the [Laurent series](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#thm-b3-residues-laurent) of $\bar g$ and $g'$ on $C_\rho$ and integrate term by term (normal convergence; only the frequency-zero products survive).
2. Let $\rho \to 1^+$ and conclude the *area theorem*: $$\sum_{n\geq1}n\,\abs{b_n}^2 \;\leq\; 1 .$$ In particular $\abs{b_1} \leq 1$. When is $\abs{b_1}  = 1$?

**Part II — Bieberbach’s $\abs{a_2} \leq
2$.** Let $f = z + a_2z^2 + \cdots \in \mathcal S$.

3. Show that $f(z^2)/z^2$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and zero-free on $\mathbb D$ , and admits a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) square root $\varphi$ with $\varphi(0) = 1$ ; set $h(z) =  z\varphi(z^2)$ , so that $h(z)^2 = f(z^2)$ . Show that $h$ is an *odd* univalent function on $\mathbb  D$ with expansion $h(z) = z + \frac{a_2}2z^3 +  \cdots$ . *(Injectivity: $h(z)^2 = h(z')^2$ forces $z^2 = z'^2$; use oddness to finish.)*
4. Apply the area theorem to $g(w) = 1/h(1/w) = w -  \frac{a_2}{2w} + \cdots$ on $\{\abs w > 1\}$ (verify univalence and the expansion), and conclude $\abs{a_2} \leq 2$ .
5. Show that the *Koebe function* $$k(z) = \frac{z}{(1 - z)^2} = \sum_{n\geq1}n\,z^n$$ belongs to $\mathcal S$, has $a_2 = 2$, and maps $\mathbb D$ onto $\C\setminus\intoc{-\infty}{-\frac14}$ *(write $k = \frac14\bigl[\bigl(\frac{1+z}{1 -  z}\bigr)^2 - 1\bigr]$ and track the images)*: all inequalities to come are sharp.

**Part III — The quarter theorem.**

6. Let $f \in \mathcal S$ and $c \notin f(\mathbb D)$. Show that $$F(z) = \frac{c\,f(z)}{c - f(z)}$$ belongs to $\mathcal S$, and compute its second coefficient: $A_2 = a_2 + \frac1c$.
7. Apply Bieberbach to both $f$ and $F$ : conclude $\abs{\frac1c} \leq \abs{A_2} + \abs{a_2} \leq 4$ , i.e. $\abs c \geq \frac14$ . *Every omitted value has modulus $\geq \frac14$* : $f(\mathbb D) \supseteq  D(0, \frac14)$ — Koebe’s quarter theorem. Check sharpness on the Koebe function.
8. Deduce a quantitative Riemann-map estimate: if $\varphi \colon \Omega \to \mathbb D$ is the Riemann map of [Theorem 18.9](#thm-b3-conformal-rmt) at $z_0$, then $$\frac{d\bigl(z_0, \partial\Omega\bigr)}{4}  \;\leq\; \frac1{\varphi'(z_0)} \;\leq\;  4\,d\bigl(z_0, \partial\Omega\bigr)$$ — prove at least the left inequality by applying Koebe to $\varphi^{-1}$ suitably normalized, and the right by Schwarz applied to $\varphi$ on the disc $D(z_0, d) \subseteq \Omega$.

**Part IV — Perspective.**

9. Bieberbach conjectured (1916) $\abs{a_n} \leq n$ for all $n$, with equality only for rotations of the Koebe function; de Branges proved it in 1985. Verify the conjecture by hand for the Koebe function and its rotations $\eu^{-\iu\theta}k(\eu^{\iu\theta}z)$. Then push question 4’s expansion one term further: writing $h(z) = z + \frac{a_2}2z^3 + c_5z^5 + \cdots$, show $c_5 = \frac{a_3}2 - \frac{a_2^2}8$ and $$g(w) = w - \frac{a_2}{2}\,w^{-1} +  \Bigl(\frac{3a_2^2}8 - \frac{a_3}2\Bigr)w^{-3} +  \cdots ,$$ so the area theorem yields the refined inequality $\bigl|\frac{a_2}2\bigr|^2 + 3\bigl|\frac{3a_2^2}8 -  \frac{a_3}2\bigr|^2 \leq 1$. Check it on the Koebe function ($a_2 = 2$, $a_3 = 3$).

**Part V — The distortion theorem.** Bieberbach’s inequality, transported around the disc by automorphisms, controls $f'$ everywhere. Fix $f \in \mathcal
S$.

10. (Koebe transform) For $z_0 \in \mathbb D$ let $\varphi(z) = \frac{z + z_0}{1 + \bar z_0z}$, a disc automorphism ([Theorem 18.5](#thm-b3-conformal-autdisc)) with $\varphi(0) = z_0$. Show that $$F(z) = \frac{f(\varphi(z)) - f(z_0)}  {f'(z_0)\,\bigl(1 - \abs{z_0}^2\bigr)}$$ belongs to $\mathcal S$ *(univalence is inherited; compute $\varphi'(0) = 1 - \abs{z_0}^2$ and check the normalization; recall $f' \neq 0$ for injective $f$, [Definition 18.1](#def-b3-conformal-conformal))*.
11. Compute the second coefficient $A_2 = \frac12F''(0)$ of $F$: $$A_2 = \frac12\Bigl[\bigl(1 - \abs{z_0}^2\bigr)  \frac{f''(z_0)}{f'(z_0)} - 2\bar z_0\Bigr] ,$$ and deduce from Bieberbach (question 4), for $z =  r\eu^{\iu\theta}$, the *fundamental inequality*: $$\Bigl|\,z\,\frac{f''(z)}{f'(z)} -  \frac{2r^2}{1 - r^2}\Bigr|  \;\leq\; \frac{4r}{1 - r^2} .$$
12. Extract the real part: $$\frac{2r^2 - 4r}{1 - r^2} \;\leq\;  \operatorname{Re}\Bigl(z\,\frac{f''(z)}{f'(z)}\Bigr)  \;\leq\; \frac{2r^2 + 4r}{1 - r^2} .$$
13. Show that $\frac{\dd}{\dd t}\log\bigl|  f'(t\eu^{\iu\theta})\bigr| = \frac1t  \operatorname{Re}\bigl(z\frac{f''(z)}{f'(z)}\bigr)$ at $z = t\eu^{\iu\theta}$ *(for a nonvanishing $\mathcal C^1$ function $g$ of a real variable, $\frac{\dd}{\dd t}\log\abs g =  \operatorname{Re}(g'/g)$)*, and integrate question 12’s bounds along the ray to obtain the *distortion theorem*: $$\frac{1 - r}{(1 + r)^3} \;\leq\; \abs{f'(z)}  \;\leq\; \frac{1 + r}{(1 - r)^3},  \qquad \abs z = r .$$
14. Deduce the *growth theorem*: $$\frac{r}{(1 + r)^2} \;\leq\; \abs{f(z)} \;\leq\;  \frac{r}{(1 - r)^2},  \qquad \abs z = r$$ *(upper bound: integrate $f'$ on the segment $[0, z]$; lower bound: if $\abs{f(z)} < \frac14$, the segment $[0, f(z)]$ lies in $f(\mathbb D)$ by question 7; pull it back by $f^{-1}$ — [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) by [Corollary 17.10](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#cor-b3-residues-openmapping) — and bound $\abs{f(z)} = \int_\gamma\abs{f'(\zeta)}  \,\abs{\dd\zeta} \geq \int_0^r\frac{1 - t}{(1 +  t)^3}\,\dd t$, using $\abs{\dd\zeta} \geq  \dd\abs\zeta$)*.
15. Verify that the Koebe function achieves equality in all four bounds, at $z = r$ for the upper ones and $z = -r$ for the lower ones: $k$ is simultaneously the most expanding and, at the antipode, the most contracting member of $\mathcal S$ .

**Part VI — Extremal rigidity.** In every inequality so far, equality identifies the Koebe function up to rotation. We prove it, then harvest.

16. Suppose $f \in \mathcal S$ has $\abs{a_2} = 2$. Chase equality through questions 2–4: the area theorem forces $g(w) = w + b_0 + \eu^{\iu\alpha}/w$; oddness of $h$ makes $g$ odd, so $b_0 = 0$; invert to find $h$, then $f$, and conclude that $$f(z) = \eu^{-\iu\theta}k\bigl(\eu^{\iu\theta}z\bigr)  \quad\text{with } \eu^{\iu\theta} =  -\eu^{\iu\alpha} :$$ the rotations of the Koebe function are the only members of $\mathcal S$ with $\abs{a_2} = 2$.
17. Show that if $f \in \mathcal S$ omits a value $c$ with $\abs c = \frac14$ exactly, then $f$ is a rotation of the Koebe function, and identify $c =  -\eu^{-\iu\theta}/4$ *(trace equality through question 7’s chain $4 = \abs{1/c} = \abs{A_2 - a_2}  \leq \abs{A_2} + \abs{a_2} \leq 4$)* : the quarter theorem’s constant is attained only by the extremal family.
18. (Coefficients on the cheap) Combine the growth theorem with the [Cauchy estimates](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-analytic) ([Theorem 16.10](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-analytic)) on the circle $\abs z = 1 - \frac1n$ to prove $$\abs{a_n} \;\leq\; \eu\,n^2 \qquad (n \geq 2) .$$ (De Branges, 1985: $\abs{a_n} \leq n$; the factor $\eu n$ is the price of elementary tools.)
19. (Covering of subdiscs) Show that for every $0 < r <  1$, $$f\bigl(D(0, r)\bigr) \supseteq  D\Bigl(0, \frac{r}{(1 + r)^2}\Bigr),$$ sharp for the Koebe function, and recover the quarter theorem as $r \to 1^-$. *(Boundary points of the open image $f(D(0,r))$ lie on $f(\partial D(0,r))$, hence have modulus $\geq  \frac{r}{(1+r)^2}$ by question 14; a segment from $0$ to a missed point of smaller modulus would have to cross that boundary.)*
20. (Koebe at every point) Let $f$ be univalent on $\mathbb D$, not necessarily normalized, and $z_0  \in \mathbb D$. Prove $$\tfrac14\bigl(1 - \abs{z_0}^2\bigr)\abs{f'(z_0)}  \;\leq\;  d\bigl(f(z_0), \partial f(\mathbb D)\bigr)  \;\leq\;  \bigl(1 - \abs{z_0}^2\bigr)\abs{f'(z_0)}$$ *(left: quarter theorem applied to the Koebe transform of question 10; right: Schwarz ([Theorem 18.4](#thm-b3-conformal-schwarz)) applied to $\psi^{-1}\circ\hat g$, where $\hat g(w) =  f^{-1}\bigl(f(z_0) + dw\bigr)$, $d$ the distance, and $\psi$ a disc automorphism sending $0$ to $z_0$)*. Why is $\partial f(\mathbb D)$ nonempty?
21. Check question 20 on $f = k$ at $z_0 = r \in  \intoo01$ : compute $d\bigl(k(r), \partial k(\mathbb  D)\bigr) = \frac{(1+r)^2}{4(1-r)^2}$ and verify that the left inequality is an *equality* : the Koebe function saturates its own theorem at every point of $\intoo01$ .
22. (The moral) In ten lines: what single principle underlies the area theorem, and how do Bieberbach, the quarter theorem, distortion, growth, and covering all flow from it? Compare with the Schwarz–Pick world of [Theorem 18.4](#thm-b3-conformal-schwarz) : in both, one [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) inequality rigidifies the whole geometry, and the extremals are unique up to rotation.

**Part VII — Compactness, inverses, and a reality check.**

23. Show that the class $\mathcal S$ is *[compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)* for locally uniform convergence: it is locally bounded by the growth theorem, hence normal ( [Theorem 18.7](#thm-b3-conformal-montel) ); and a locally uniform limit of members of $\mathcal S$ is again in $\mathcal S$ *(the normalizations pass to the limit by Weierstrass convergence of derivatives; injectivity survives by Hurwitz, [Exercise 17.8](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#exo-b3-residues-8), the limit being nonconstant)* . Why does this matter for extremal problems like Bieberbach’s?
24. For $f \in \mathcal S$ , let $g = f^{-1}$ , defined near $0$ . Show $g(w) = w - a_2w^2 + O(w^3)$ , so the inverse’s second coefficient obeys the same sharp bound $\abs{A_2} = \abs{a_2} \leq 2$ , with equality exactly for the rotated Koebe functions.
25. Determine for which $a \in \C$ the polynomial $f(z)  = z + az^2$ belongs to $\mathcal S$ : show $f$ is injective on $\mathbb D$ iff $\abs a \leq \frac12$ *(factor $f(z_1) - f(z_2)$)* . Conclude: for degree-two polynomials the true coefficient bound is $\frac12$ , four times smaller than Bieberbach’s $2$ — the extremals of $\mathcal S$ are genuinely transcendental objects, and no polynomial comes close.

**Solution of Problem 18.1.**

**1.** $g$ is injective and [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo); on $C_\rho$ ($\rho > 1$) it is smooth, and the enclosed area is $\frac1{2\iu}\oint_{g(C_\rho)}\bar\zeta\,\dd\zeta$ (the Year 2 Green–Riemann area formula, applied with positive orientation). Substituting $\zeta = g(w)$, $w =
\rho\eu^{\iu\theta}$:

$$
A_\rho = \frac1{2\iu}\int_{C_\rho}\overline{g(w)}\,g'(w)\dd
w = \frac\rho2\int_0^{2\pi}\overline{g(\rho\eu^{\iu\theta})}
\,g'(\rho\eu^{\iu\theta})\,\eu^{\iu\theta}\,\dd\theta .
$$

Insert $\bar g = \rho\eu^{-\iu\theta} + \bar b_0 +
\sum_m\bar b_m\rho^{-m}\eu^{\iu m\theta}$ and $g' = 1 -
\sum_nnb_n\rho^{-n-1}\eu^{-\iu(n+1)\theta}$: after multiplying by $\eu^{\iu\theta}$, only frequency-zero products survive the $\theta$-integration (normal convergence justifies term-by-term work): the pair $(\rho\eu^{-\iu\theta})\cdot1$ contributes $2\pi\rho$, and each pair $\bar b_n\rho^{-n}\eu^{\iu
n\theta}\cdot(-nb_n\rho^{-n-1}\eu^{-\iu(n+1)\theta})$ contributes $-2\pi n\abs{b_n}^2\rho^{-2n-1}$:

$$
A_\rho = \pi\rho^2 - \pi\sum_{n\geq1}n\abs{b_n}^2\rho^{-2n}.
$$

**2.** Areas are nonnegative: $\sum_{n\leq
N}n\abs{b_n}^2\rho^{-2n} \leq \rho^2$ for every $N$; let $\rho \to 1^+$ then $N \to \infty$: $\sum_{n\geq1}n\abs{b_n}^2 \leq 1$. Equality in $\abs{b_1}
\leq 1$ forces all other $b_n = 0$: $g(w) = w + b_0 +
\eu^{\iu\alpha}/w$, which maps onto the complement of a segment of length $4$ (a Joukowski-type map): the extremals.

**3.** $f(z)/z = 1 + a_2z + \cdots$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) and zero-free on $\mathbb D$ ($f$ vanishes only at $0$, simply: injectivity), hence so is $f(z^2)/z^2$, which has a [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) square root $\varphi$ with $\varphi(0) = 1$ ([Definition 18.8](#def-b3-conformal-simplyconnected); $\mathbb D$ is convex). Then $h(z) = z\varphi(z^2)$ satisfies $h^2 =
f(z^2)$, $h(z) = z\bigl(1 + \frac{a_2}2z^2 + \cdots\bigr)$ (binomial series for the root), and $h(z) = z\varphi(z^2)$ is odd by construction ($\varphi(z^2)$ is even). Injectivity: $h(z) = h(z')$ gives $f(z^2) =
f(z'^2)$, so $z^2 = z'^2$, i.e. $z' = \pm z$; if $z' = -z$ then oddness gives $h(z) = -h(z)$, so $h(z) = 0$, forcing $z
= 0$ ($\varphi$ zero-free): $z = z' = 0$.

**4.** $g(w) = 1/h(1/w)$: for $\abs w > 1$, $1/w \in
\mathbb D\setminus\{0\}$ and $h \neq 0$ there: well defined, injective (composition of injections), with expansion

$$
g(w) = \frac{1}{\frac1w\bigl(1 + \frac{a_2}{2w^2} +
\cdots\bigr)} = w\Bigl(1 - \frac{a_2}{2w^2} + \cdots\Bigr)
= w - \frac{a_2}{2}\,w^{-1} + \cdots :
$$

of the Part I form with $b_1 = -\frac{a_2}2$. The area theorem gives $\abs{\frac{a_2}2} \leq 1$: $\abs{a_2} \leq 2$.

**5.** $k(z) = \frac z{(1-z)^2} = z\sum_{m\geq0}(m +
1)z^m = \sum_{n\geq1}nz^n$: coefficients $a_n = n$, so $a_2 =
2$. Univalence and image: $k = \frac14\bigl[w^2 - 1\bigr]$ with $w = \frac{1 + z}{1 - z}$, a [conformal map](#def-b3-conformal-conformal) of $\mathbb D$ onto the right half-plane; $w^2$ maps that half-plane [conformally](#def-b3-conformal-conformal) onto $\C\setminus\intoc{-\infty}0$; then $\frac{\cdot - 1}4$ gives $\C\setminus\intoc{-\infty}{-\frac14}$: injective at each stage, image as claimed.

**6.** $F = \frac{cf}{c - f}$: since $c \notin f(\mathbb
D)$, the denominator never vanishes: $F$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo), and injective ($w \mapsto \frac{cw}{c - w}$ is Möbius, injective off $w = c$). Expansion: with $f = z + a_2z^2 + \cdots$,

$$
F = f\cdot\frac1{1 - f/c} = \bigl(z + a_2z^2\bigr)\Bigl(1 +
\frac zc\Bigr) + O(z^3)
= z + \Bigl(a_2 + \frac1c\Bigr)z^2 + O(z^3):
$$

$F \in \mathcal S$ with $A_2 = a_2 + \frac1c$.

**7.** Bieberbach twice: $\abs{a_2} \leq 2$ and $\abs{a_2 + \frac1c} \leq 2$, so $\abs{\frac1c} \leq 4$: $\abs c \geq \frac14$. Every omitted value lies outside $D(0,\frac14)$, i.e. $f(\mathbb D) \supseteq D(0, \frac14)$. Sharp: the Koebe function omits $-\frac14$ (question 5).

**8.** Let $d = d(z_0, \partial\Omega)$ and $\psi =
\varphi^{-1} \colon \mathbb D \to \Omega$, $\psi(0) = z_0$, $\psi'(0) = 1/\varphi'(z_0) > 0$. The normalization $\tilde
f(z) = \frac{\psi(z) - z_0}{\psi'(0)}$ lies in $\mathcal S$, so its image contains $D(0, \frac14)$; scaling back, $\Omega = \psi(\mathbb D) \supseteq D\bigl(z_0,
\tfrac{\psi'(0)}4\bigr)$, so $d \geq \frac{\psi'(0)}4 =
\frac1{4\varphi'(z_0)}$: the left inequality. For the right: $\varphi\circ(z_0 + d\,\cdot)$ maps $\mathbb D$ into $\mathbb D$ with $0 \mapsto 0$; Schwarz bounds its derivative at $0$: $d\,\varphi'(z_0) \leq 1$, i.e. $\frac1{\varphi'(z_0)} \geq d$ — together

$$
d \leq \frac1{\varphi'(z_0)} \leq 4d .
$$

(The left inequality as displayed in the statement is the same chain rearranged.)

**9.** For the Koebe function $a_n = n$: equality throughout the conjecture; its rotations $\eu^{-\iu\theta}k(\eu^{\iu\theta}z) = \sum
n\eu^{\iu(n-1)\theta}z^n$ have $\abs{a_n} = n$ too. Pushing question 4: $h = z + \frac{a_2}2z^3 + c_5z^5 + \cdots$ with $h^2 = f(z^2)$; comparing $z^6$-coefficients: $2c_5 +
\frac{a_2^2}4 = a_3$, so $c_5 = \frac{a_3}2 -
\frac{a_2^2}8$. Then

$$
g(w) = \frac1{h(1/w)} = w\Bigl(1 - \frac{a_2}{2w^2} +
\Bigl(\frac{a_2^2}4 - c_5\Bigr)\frac1{w^4} + \cdots\Bigr)
= w - \frac{a_2}2w^{-1} + \Bigl(\frac{3a_2^2}8 -
\frac{a_3}2\Bigr)w^{-3} + \cdots,
$$

and the area theorem ($\sum n\abs{b_n}^2 \leq 1$) yields

$$
\Bigl|\frac{a_2}2\Bigr|^2 + 3\,\Bigl|\frac{3a_2^2}8 -
\frac{a_3}2\Bigr|^2 \leq 1 .
$$

Koebe check ($a_2 = 2$, $a_3 = 3$): $\abs{\frac{a_2}2}^2 = 1$ and $\frac{3\cdot4}8 - \frac32 = 0$: total exactly $1$ — extremal, as it must be.

**10.** $\varphi$ is a disc automorphism with $\varphi(0) = z_0$, so $f\circ\varphi$ is univalent on $\mathbb D$ (composition of injections), and $f'(z_0) \neq
0$ ([Definition 18.1](#def-b3-conformal-conformal)): $F$ is well defined and univalent. $F(0) = 0$. Quotient rule:

$$
\varphi'(z) = \frac{(1 + \bar z_0z) - (z + z_0)\bar z_0}
{(1 + \bar z_0z)^2}
= \frac{1 - \abs{z_0}^2}{(1 + \bar z_0z)^2},
\qquad \varphi'(0) = 1 - \abs{z_0}^2 ,
$$

so $(f\circ\varphi)'(0) = f'(z_0)(1 - \abs{z_0}^2)$ and $F'(0) = 1$: $F \in \mathcal S$.

**11.** With $G = f\circ\varphi$: $G'' =
f''(\varphi)\,\varphi'^2 + f'(\varphi)\,\varphi''$, and $\varphi''(z) = -2\bar z_0(1 - \abs{z_0}^2)(1 + \bar
z_0z)^{-3}$ gives $\varphi''(0) = -2\bar z_0(1 -
\abs{z_0}^2)$. Hence

$$
A_2 = \frac{G''(0)}{2f'(z_0)(1 - \abs{z_0}^2)}
= \frac12\Bigl[\bigl(1 - \abs{z_0}^2\bigr)
\frac{f''(z_0)}{f'(z_0)} - 2\bar z_0\Bigr] .
$$

Bieberbach for $F$ (question 4): $\abs{A_2} \leq 2$, i.e. $\bigl|(1 - \abs{z_0}^2)\frac{f''(z_0)}{f'(z_0)} - 2\bar
z_0\bigr| \leq 4$. Multiply through by $z_0/(1 - \abs{z_0}^2)$, whose modulus is $r/(1 - r^2)$ for $z_0 = r\eu^{\iu\theta}$, and use $z_0\bar
z_0 = r^2$:

$$
\Bigl|\,z_0\frac{f''(z_0)}{f'(z_0)} -
\frac{2r^2}{1 - r^2}\Bigr| \leq \frac{4r}{1 - r^2} .
$$

**12.** A complex number within distance $\rho$ of the real point $c$ has real part in $\intcc{c - \rho}{c +
\rho}$: apply this to $c = \frac{2r^2}{1-r^2}$, $\rho =
\frac{4r}{1-r^2}$.

**13.** For nonvanishing $\mathcal C^1$ $g$: $\log\abs g = \frac12\log(g\bar g)$, so $\frac{\dd}{\dd
t}\log\abs g = \frac{g'\bar g + g\bar g'}{2\abs g^2} =
\operatorname{Re}\frac{g'}g$. With $g(t) =
f'(t\eu^{\iu\theta})$ (zero-free: univalence), $g'(t) =
\eu^{\iu\theta}f''(t\eu^{\iu\theta})$, so at $z =
t\eu^{\iu\theta}$:

$$
\frac{\dd}{\dd t}\log\bigl|f'(t\eu^{\iu\theta})\bigr|
= \operatorname{Re}\Bigl(\eu^{\iu\theta}
\frac{f''}{f'}\Bigr)
= \frac1t\operatorname{Re}\Bigl(z\frac{f''}{f'}\Bigr)
\in \Bigl[\frac{2t - 4}{1 - t^2},\
\frac{2t + 4}{1 - t^2}\Bigr]
$$

by question 12 at radius $t$. Since $\frac{\dd}{\dd
t}\log\frac{1+t}{(1-t)^3} = \frac1{1+t} + \frac3{1-t} =
\frac{2t+4}{1-t^2}$ and $\frac{\dd}{\dd
t}\log\frac{1-t}{(1+t)^3} = \frac{-1}{1-t} - \frac3{1+t} =
\frac{2t-4}{1-t^2}$, integrating from $0$ to $r$ (all three functions vanish at $t = 0$, $f'(0) = 1$) gives

$$
\log\frac{1-r}{(1+r)^3} \leq \log\abs{f'(z)} \leq
\log\frac{1+r}{(1-r)^3} :
$$

the distortion theorem, after exponentiating.

**14.** Upper: along the segment $[0, z]$,

$$
\abs{f(z)} = \Bigl|\int_0^rf'(t\eu^{\iu\theta})
\eu^{\iu\theta}\dd t\Bigr|
\leq \int_0^r\frac{1+t}{(1-t)^3}\dd t
= \frac{r}{(1-r)^2}
$$

($\frac{\dd}{\dd t}\frac t{(1-t)^2} = \frac{1+t}{(1-t)^3}$). Lower: note $\frac r{(1+r)^2} < \frac14$ for $r < 1$ (it says $(1-r)^2 > 0$), so if $\abs{f(z)} \geq \frac14$ there is nothing to prove. Otherwise the segment $[0, f(z)]$ lies in $D(0, \frac14) \subseteq f(\mathbb D)$ (question 7), and $\gamma = f^{-1}\circ[0, f(z)]$ is a $\mathcal C^1$ [path](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) from $0$ to $z$ in $\mathbb D$ ($f^{-1}$ [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo), [Corollary 17.10](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#cor-b3-residues-openmapping)). Substituting $w =
f(\zeta)$,

$$
\abs{f(z)} = \int_{[0,f(z)]}\abs{\dd w}
= \int_\gamma\abs{f'(\zeta)}\,\abs{\dd\zeta}
\geq \int_0^1\frac{1 - \rho(s)}{(1 + \rho(s))^3}
\,\abs{\gamma'(s)}\,\dd s
$$

with $\rho = \abs\gamma$, using the lower distortion bound at radius $\rho(s)$. Since $\abs{\gamma'} \geq \rho'$ (wherever defined; $\rho$ is Lipschitz) and the integrand factor is positive,

$$
\abs{f(z)} \geq \int_0^1
\frac{1 - \rho(s)}{(1 + \rho(s))^3}\,\rho'(s)\,\dd s
= \int_0^{r}\frac{1 - u}{(1 + u)^3}\,\dd u
= \frac{r}{(1 + r)^2}
$$

($\rho(0) = 0$, $\rho(1) = r$; the substitution uses only an antiderivative, $\frac{\dd}{\dd u}\frac u{(1+u)^2} =
\frac{1-u}{(1+u)^3}$, not monotonicity).

**15.** $k'(z) = \frac{\dd}{\dd z}\,z(1 - z)^{-2} =
\frac{1 + z}{(1 - z)^3}$. At $z = r$: $k'(r) =
\frac{1+r}{(1-r)^3}$ and $k(r) = \frac r{(1-r)^2}$ — both upper bounds attained. At $z = -r$: $k'(-r) =
\frac{1-r}{(1+r)^3}$ and $\abs{k(-r)} = \frac r{(1+r)^2}$ — both lower bounds attained. The Koebe function stretches its positive axis maximally toward the far boundary and compresses the antipodal ray maximally toward the tip of its slit.

**16.** $\abs{a_2} = 2$ means $\abs{b_1} = 1$ for $g(w)
= 1/h(1/w) = w - \frac{a_2}2w^{-1} + \cdots$ (question 4); the area theorem $\sum n\abs{b_n}^2 \leq 1$ then kills every other coefficient: $g(w) = w + b_0 + \eu^{\iu\alpha}/w$ with $\eu^{\iu\alpha} = -\frac{a_2}2$. Since $h$ is odd, $g(-w) =
1/h(-1/w) = -1/h(1/w) = -g(w)$: $g$ is odd, so $b_0 = 0$. Inverting, $h(z) = 1/g(1/z) = \frac{z}{1 +
\eu^{\iu\alpha}z^2}$, and $f(z^2) = h(z)^2$ gives

$$
f(w) = \frac{w}{(1 + \eu^{\iu\alpha}w)^2}
= \frac{w}{(1 - \eu^{\iu\theta}w)^2}
= \eu^{-\iu\theta}k\bigl(\eu^{\iu\theta}w\bigr),
\qquad \eu^{\iu\theta} = -\eu^{\iu\alpha} .
$$

Conversely each rotation has $a_2 = 2\eu^{\iu\theta}$ of modulus $2$: the extremals of Bieberbach are exactly the rotated Koebe functions.

**17.** If $c$ is omitted with $\abs c = \frac14$: question 6 gives $F \in \mathcal S$ with $A_2 = a_2 +
\frac1c$, so $\frac1c = A_2 - a_2$ and

$$
4 = \Bigl|\frac1c\Bigr| = \abs{A_2 - a_2}
\leq \abs{A_2} + \abs{a_2} \leq 2 + 2 = 4 :
$$

equality throughout, in particular $\abs{a_2} = 2$. By question 16, $f = \eu^{-\iu\theta}k(\eu^{\iu\theta}\cdot)$, whose omitted set is $\eu^{-\iu\theta}
\intoc{-\infty}{-\frac14}$: the unique omitted value of modulus $\frac14$ is $c = -\eu^{-\iu\theta}/4$. (Check: $a_2 = 2\eu^{\iu\theta}$ and $\frac1c = -4\eu^{\iu\theta}$, so $A_2 = -2\eu^{\iu\theta} = -a_2$: the triangle inequality is saturated by anti-alignment, as it must be.)

**18.** The [Cauchy estimates](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-analytic) on the circle $\abs z =
r$ ([Theorem 16.10](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#thm-b3-holomorphic-analytic)), combined with the growth theorem, give

$$
\abs{a_n} \leq r^{-n}\sup_{\abs z = r}\abs f
\leq r^{1-n}\,(1 - r)^{-2} .
$$

Choose $r = 1 - \frac1n$ ($n \geq
2$): $r^{1-n} = \bigl(1 + \frac1{n-1}\bigr)^{n-1} < \eu$ (increasing sequence with limit $\eu$) and $(1 - r)^{-2} =
n^2$: $\abs{a_n} < \eu\,n^2$.

**19.** $U = f(D(0, r))$ is open ([Corollary 17.10](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#cor-b3-residues-openmapping)) and contains $0$. Boundary: if $p \in \partial U$, write $p = \lim f(z_k)$ with $z_k \in D(0, r)$; a subsequence gives $z_k \to z_\infty
\in \bar D(0, r)$, and $f(z_\infty) = p \notin U$ forces $z_\infty \in \partial D(0, r)$: $\partial U \subseteq
f(\partial D(0, r))$, so every boundary point of $U$ has modulus $\geq \frac r{(1+r)^2}$ (question 14). Now let $\abs w < \frac r{(1+r)^2}$ and suppose $w \notin U$. The segment $[0, w]$ is [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected), meets $U$ (at $0$) and its complement (at $w$), so it meets $\partial U$; but all its points have modulus $\leq \abs w < \frac r{(1+r)^2}$: contradiction. Hence $D\bigl(0, \frac r{(1+r)^2}\bigr)
\subseteq U$. Sharpness: $k(-r) = -\frac r{(1+r)^2}$ is the image of a boundary point of $D(0,r)$, and $k$ is injective, so $k(-r) \notin k(D(0, r))$: the radius cannot be increased. As $r \to 1^-$, $\frac r{(1+r)^2} \to \frac14$: the quarter theorem for the full disc.

**20.** First, $\partial f(\mathbb D) \neq \emptyset$: otherwise $f(\mathbb D)$, open, closed, nonempty, would be all of $\C$, and $f^{-1} \colon \C \to \mathbb D$ would be a bounded nonconstant entire function, against [Corollary 16.12](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#cor-b3-holomorphic-liouville). Left inequality: the Koebe transform $F$ of question 10 lies in $\mathcal S$ and $F(\mathbb D) = \frac{f(\mathbb D) -
f(z_0)}{f'(z_0)(1-\abs{z_0}^2)}$; by question 7 it contains $D(0, \frac14)$, so

$$
f(\mathbb D) \supseteq f(z_0) + D\Bigl(0,\
\tfrac14\abs{f'(z_0)}\bigl(1 - \abs{z_0}^2\bigr)\Bigr) ,
$$

and every point of $\partial f(\mathbb D)$ — disjoint from the [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $f(\mathbb D)$ — is at distance $\geq
\frac14(1-\abs{z_0}^2)\abs{f'(z_0)}$ from $f(z_0)$. Right inequality: let $d = d(f(z_0), \partial f(\mathbb D)) <
\infty$. The disc $D(f(z_0), d)$ lies in $f(\mathbb D)$: a segment from $f(z_0)$ to any of its points stays at distance $< d$ from $f(z_0)$, so never meets $\partial f(\mathbb D)$, and the [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) argument of question 19 keeps it in $f(\mathbb D)$. Then $\hat g(w) = f^{-1}(f(z_0) + dw)$ maps $\mathbb D$ into $\mathbb D$ with $\hat g(0) = z_0$, and $\chi = \psi^{-1}\circ\hat g$, with $\psi(z) = \frac{z +
z_0}{1 + \bar z_0z}$, fixes $0$: Schwarz ([Theorem 18.4](#thm-b3-conformal-schwarz)) gives $\abs{\chi'(0)}
\leq 1$. Since $\chi'(0) = \frac{\hat g'(0)}{\psi'(0)} =
\frac{d}{f'(z_0)\,(1 - \abs{z_0}^2)}$, this is $d \leq (1 -
\abs{z_0}^2)\abs{f'(z_0)}$.

**21.** $k(\mathbb D) = \C\setminus
\intoc{-\infty}{-\frac14}$, so $\partial k(\mathbb D) =
\intoc{-\infty}{-\frac14}$, and for the positive real point $k(r) = \frac r{(1-r)^2}$ the nearest boundary point is $-\frac14$:

$$
d = k(r) + \frac14 = \frac{4r + (1-r)^2}{4(1-r)^2}
= \frac{(1+r)^2}{4(1-r)^2} .
$$

Left member of question 20: $\frac14(1 - r^2)k'(r) =
\frac14\,\frac{(1-r^2)(1+r)}{(1-r)^3} =
\frac{(1+r)^2}{4(1-r)^2} = d$: equality. (The right member equals $4d$: the full factor $4$ separates the two sides, and the Koebe function sits exactly at the bottom.)

**22.** The single principle is [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval): univalence forbids overlap, so the area of the complement of the image of $\{\abs w > \rho\}$, expanded in Fourier modes on circles, is nonnegative — the area theorem is an $L^2$ identity with a sign. Everything else is that inequality transported: a square root (question 3) turns it into $\abs{a_2} \leq 2$; a Möbius reflection off an omitted value (question 6) turns $\abs{a_2} \leq 2$ into the quarter theorem; disc automorphisms (question 10) spread $\abs{a_2}
\leq 2$ over the whole disc as the distortion theorem; radial integration converts distortion into growth, and growth into covering. At every stage the case of equality propagates too, always landing on the rotated Koebe functions — one extremal family for the whole theory, just as the rotations are the unique extremals of Schwarz’s lemma. One [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) inequality, plus rigidity of its equality case, governs the entire geometry: conformal mapping is the art of exploiting such inequalities.

**23.** The growth theorem bounds $\abs f \leq
\frac{r}{(1-r)^2}$ uniformly on each $\bar D(0, r)$, $r <
1$, for all $f \in \mathcal S$ at once: locally bounded, so $\mathcal S$ is a normal family ([Theorem 18.7](#thm-b3-conformal-montel)). If $f_n \in \mathcal S \to
f$ locally uniformly: $f$ is [holomorphic](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#def-b3-holomorphic-holo) with $f_n' \to f'$ locally uniformly (Weierstrass), so $f(0) = 0$, $f'(0) = 1$ — in particular $f$ is nonconstant — and Hurwitz ([Exercise 17.8](https://one-course.com/books/math/5/en/chapter/17-laurent-series-and-the-residue-theorem#exo-b3-residues-8)) makes the limit of injective maps injective: $f \in \mathcal S$. A [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functional (such as $f \mapsto \abs{a_2} = \frac{\abs{f''(0)}}2$) on a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) class *attains* its supremum: extremal functions exist before one knows what they are — the starting point of every variational attack on coefficient problems, Bieberbach’s included.

**24.** Write $g(w) = w + A_2w^2 + O(w^3)$ and compose:

$$
z = g(f(z)) = f(z) + A_2f(z)^2 + O(z^3)
= z + (a_2 + A_2)z^2 + O(z^3),
$$

so $A_2 = -a_2$ and $\abs{A_2} \leq 2$ (Bieberbach), with equality iff $\abs{a_2} = 2$, i.e. iff $f$ is a rotated Koebe function (question 16) — and then $g$ is the corresponding inverse, defined on the slit plane.

**25.** $f(z_1) - f(z_2) = (z_1 - z_2)\bigl(1 + a(z_1 +
z_2)\bigr)$. If $\abs a \leq \frac12$: for $z_1 \neq z_2$ in $\mathbb D$, $\abs{a(z_1 + z_2)} < 2\abs a \leq 1$ (strict: $\abs{z_1 + z_2} < 2$), so the second factor cannot vanish: injective, and $f \in \mathcal S$ (normalizations are built in). If $\abs a > \frac12$: the point $s =
-\frac1a$ has $\abs s < 2$, so $z_{1,2} = \frac s2 \pm
\varepsilon$ lie in $\mathbb D$ for small $\varepsilon > 0$, are distinct, and $z_1 + z_2 = s$ kills the factor: $f(z_1)
= f(z_2)$, not injective. So $\mathcal S$ contains $z +
az^2$ exactly for $\abs a \leq \frac12$. Bieberbach’s bound $\abs{a_2} \leq 2$ is thus wildly unsaturated by polynomials of degree $2$ — the Koebe function’s coefficients $a_n = n$ come from an infinite series conspiring along the omitted ray, a behavior no polynomial (which belongs to $\mathcal S$ only with tiny coefficients) can imitate.