University Mathematics — Year 3 · Bachelor Year 3

18Conformal Maps and the Riemann Mapping Theorem

A holomorphic 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 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 families (Montel), and proves the deepest existence theorem of the subject: every simply connected proper subdomain of $\C$ , however jagged its boundary, is conformally equivalent to the unit disc. We close with harmonic functions and the Poisson kernel, 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 is a holomorphic bijection; its inverse is automatically holomorphic (Corollary 17.10). 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’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 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).

Example 18.3 (The Joukowski map)

Beyond Möbius, the most useful conformal map 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 the exterior of a disc: corners are no obstacle to conformal equivalence, only to boundary smoothness. Images of circles through $\pm1$ but off-center 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, 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 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 to $\mathbb D$ (the singularity at $0$ is removable: $g$ is bounded near $0$ , Theorem 17.4; its value at $0$ is $f'(0)$ ). On $\abs z = r < 1$ : $\abs g \leq \frac1r$ , so by the maximum principle (Theorem 16.14) $\abs g \leq \frac1r$ on $\bar D(0,r)$ ; let $r \to 1$ : $\abs g \leq 1$ on $\mathbb D$ , which is both inequalities. Equality at an interior point makes $\abs g$ attain an 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 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.

Proof. Let $f \in \operatorname{Aut}(\C)$ and consider $g(z) = f(1/z)$ on $\C^*$ : a holomorphic function with an isolated singularity at $0$ . If it were essential, Casorati–Weierstrass (Theorem 17.4) 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 and a nonempty open set. 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(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); 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 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 subsets of $\Omega$ (to a holomorphic limit).

Proof. Local equicontinuity: 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 $K_m$ ; each $K_m$ is covered by finitely many such discs, so $\mathcal F$ is uniformly bounded and equicontinuous on $K_m$ : Arzelà–Ascoli (Theorem 7.11) extracts a subsequence converging uniformly on $K_m$ ; diagonalize over $m$ . The limit is holomorphic by Theorem 16.15. ∎

18.4 The Riemann mapping theorem

Definition 18.8

An open 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) and Proposition 16.5, on such $\Omega$ every holomorphic function has a primitive; hence every zero-free $f \in \mathcal H(\Omega)$ has a holomorphic logarithm ( $\exp\circ$ (primitive of $f'/f$ ), adjusted by a constant, as $(f\eu^{-L})' = 0$ ) and holomorphic $n$ -th roots $\eu^{L/n}$ .

Theorem 18.9 (Riemann mapping theorem)

Every simply connected open $\Omega \subsetneq \C$ , $\Omega \neq \varnothing$ , is conformally equivalent 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 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, injective (a Möbius map composed with the injective $h$ ), with $\abs g \leq \frac12 < 1$ . Composing with a Blaschke factor (Theorem 18.5) 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) extracts $f_n \to f$ uniformly on compacts; $f$ is holomorphic, $f(z_0) = 0$ , $\abs{f'(z_0)} = s$ (Theorem 16.15 for the derivatives), in particular $f$ is nonconstant; $f$ is injective by Hurwitz (Exercise 17.8(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 $\Omega$ : it has a holomorphic 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 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 $\psi$ 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 identical to $\mathbb D$ . What it does not give: any formula (explicit maps are the exception: Exercise 18.5), boundary behavior (a deeper theory — Carathéodory’s theorem — handles it), or uniqueness of extension to $\C$ or multiply connected domains: the annulus $\{1 < \abs z < 2\}$ is not conformally 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 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 $F$ , 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 extremum unless constant).

Proof. $g = u_x - \iu u_y$ satisfies the Cauchy–Riemann equations ( $P = u_x$ , $Q = -u_y$ : $P_x = u_{xx} = -u_{yy} = Q_y$ and $P_y = u_{xy} = u_{yx} = -Q_x$ ) with continuous partials: $g$ is holomorphic (Proposition 16.2; the $\R$ -differentiability follows from $\mathcal C^1$ ). Let $F_0$ be a primitive (simple connectivity, Definition 18.8); 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): adjust $F = F_0 + c$ . The properties transfer from Theorem 16.14 and Exercise 16.10 (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, 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 to $\bar{\mathbb D}$ with boundary values $g$ : the unique such harmonic function.

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 in $z$ on $\mathbb D$ (Exercise 16.7): $u$ is the real part of a holomorphic function, 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 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. Uniqueness: the difference of two solutions is harmonic on $\mathbb D$ , continuous on the closure, 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. — 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 ( $z \mapsto z^2$ , 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 $\varphi(z) = \frac{z - \iu}{z + \iu}$ 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

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). 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. (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 to reduce to (a)).

Solution

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 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 $f\colon \mathbb D \to \mathbb D$ , 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 self-maps contract the hyperbolic metric.

Solution

Solution of Exercise 18.3.

Fix $z$ and set $g = \varphi_{f(z)}\circ f\circ\varphi_{-z}$ : holomorphic $\mathbb D \to \mathbb D$ 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 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

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 with nonvanishing derivative and holomorphic inverse (principal $\log$ ). (b) $z \mapsto z^2$ doubles arguments: the open quadrant $\{0 < \arg z < \frac\pi2\}$ maps conformally 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 $\varphi_{1/2}(z) = \frac{z - \frac12}{1 - \frac z2}$ : $\varphi_{1/2}(\tfrac12) = 0$ and $\varphi_{1/2}'(\tfrac12) = \frac{1 - \frac14}{(1 - \frac14)^2} = \frac43 > 0$ .

Exercise 18.5 ★★

(a) Show that no conformal map $\C \to \mathbb D$ or $\C \to \mathbb H$ exists, and none $\mathbb D \to \C$ . (b) Which of the following are conformally equivalent to $\mathbb D$ ? Justify via Theorem 18.9 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(1) — develop this.)

Solution

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, and proper: conformally $\mathbb D$ (Theorem 18.9). The cut plane $\C\setminus\intoc{-\infty}0$ is star-shaped about $1$ (segments from $1$ avoid the cut), hence simply connected, and proper: conformally $\mathbb D$ . 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): $g$ 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 of $0$ and $w$ have images that are open and share the value $\tilde g(0)$ , hence share other values too (open sets): $g$ 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 $\mathbb D$ , so $F = \eu^L$ for holomorphic $L$ (Definition 18.8). Let $\sigma$ be the circle $\abs z = \frac32$ in $A$ and $\gamma = F^{-1}\circ\sigma$ , a closed path 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

Solution of Exercise 18.6.

(a) $T(w) = \frac{w - 1}{w + 1}$ maps $\{\operatorname{Re}w > 0\}$ conformally 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 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 onto the right half-plane.

Exercise 18.7 ★★★

Where does the proof of Theorem 18.9 use each hypothesis? Trace: (i) simple connectivity (twice); (ii) $\Omega \neq \C$ ; (iii) connectedness. Then show that the theorem fails for $\Omega = \C$ and for the annulus, pinpointing which step of the proof breaks in each case.

Solution

Solution of Exercise 18.7.

(i) Simple connectivity enters exactly twice, through the existence of holomorphic square roots of zero-free functions (Definition 18.8): 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 is used whenever the identity theorem or Hurwitz (Exercise 17.8) 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(a)). Failure for the annulus: not simply connected — the square-root construction breaks (e.g. $z$ itself, zero-free on $A$ , has no holomorphic square root: the same index computation as in Exercise 18.5(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

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 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 between $\frac{1-r}{1+r}$ and $\frac{1+r}{1-r}$ ; apply the representation on slightly smaller discs and pass to the limit). Deduce: a harmonic function on $\C$ bounded below is constant.

Solution

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 of $\bar{\mathbb D}$ , hence equals its Poisson integral (Theorem 18.12, 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 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 — prove it via Proposition 18.11 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) with the required boundary limits at every $x \neq 0$ , and transport it to the disc by Cayley to re-derive Exercise 18.8(c).

Solution

Solution of Exercise 18.10.

Locally, $u = \operatorname{Re}F$ with $F$ holomorphic (Proposition 18.11), so $u\circ f = \operatorname{Re}(F\circ f)$ is harmonic wherever defined: harmonicity is conformally invariant. On $\mathbb H$ : the principal logarithm gives $\log z = \ln\abs z + \iu\arg z$ holomorphic 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 (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(c); evaluating at the center 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. (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 $\bar D(0, r)$ , $r < 1$ (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 of $z_0 = \frac12$ .

Solution

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 at $0$ ) is holomorphic on $\mathbb D$ with $\abs h \leq 1$ (Schwarz); $\abs h < 1$ everywhere, else the maximum principle (interior maximum of $\abs h$ ) would make $h$ a unimodular constant, i.e. $f$ a rotation — excluded. On the 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 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 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) by integrating the Cauchy–Riemann equations; identify $f(z) = u + \iu v$ as a polynomial in $z$ . (b) Show that on a star-shaped open set, every harmonic function 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’s primitive machinery, or Chapter 21’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).

Solution

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 the Poincaré lemma (Theorem 21.15; or the primitive construction of Theorem 16.8 applied to the holomorphic $u_x - \iu u_y$ ) provides $v$ with $\dd v = \omega$ , i.e. the Cauchy–Riemann system: $u + \iu v$ is holomorphic. Two conjugates differ by a function with vanishing gradient: a constant (connectedness).

(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), whose integral along the unit circle is $2\pi \neq 0$ : not exact, so no global conjugate exists — a conjugate would be a continuous 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 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 and injective on $\{\abs w > 1\}$ .

For $\rho > 1$ , let $A_\rho$ be the area (Lebesgue measure) of the 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 of $\bar g$ and $g'$ on $C_\rho$ and integrate term by term (normal convergence; only the frequency-zero products survive).
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 $f = z + a_2z^2 + \cdots \in \mathcal S$ .

Show that $f(z^2)/z^2$ is holomorphic and zero-free on $\mathbb D$ , and admits a holomorphic 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.)
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 $\abs{a_2} \leq 2$ .
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.

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: $A_2 = a_2 + \frac1c$ .
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.
Deduce a quantitative Riemann-map estimate: if $\varphi \colon \Omega \to \mathbb D$ is the Riemann map of Theorem 18.9 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.

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 $f \in \mathcal S$ .

(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) 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).
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:

$\Bigl|\,z\,\frac{f''(z)}{f'(z)} - \frac{2r^2}{1 - r^2}\Bigr| \;\leq\; \frac{4r}{1 - r^2} .$
Extract the real part:

$\frac{2r^2 - 4r}{1 - r^2} \;\leq\; \operatorname{Re}\Bigl(z\,\frac{f''(z)}{f'(z)}\Bigr) \;\leq\; \frac{2r^2 + 4r}{1 - r^2} .$
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:

$\frac{1 - r}{(1 + r)^3} \;\leq\; \abs{f'(z)} \;\leq\; \frac{1 + r}{(1 - r)^3}, \qquad \abs z = r .$
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 by Corollary 17.10 — 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 $\abs{\dd\zeta} \geq \dd\abs\zeta$ ).
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.

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 $\abs{a_2} = 2$ .
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.
(Coefficients on the cheap) Combine the growth theorem with the Cauchy estimates (Theorem 16.10) 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.)
(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.)
(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) 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?
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 $\intoo01$ .
(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: in both, one interior inequality rigidifies the whole geometry, and the extremals are unique up to rotation.

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

Show that the class $\mathcal S$ is compact for locally uniform convergence: it is locally bounded by the growth theorem, hence normal (Theorem 18.7); 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, the limit being nonconstant). Why does this matter for extremal problems like Bieberbach’s?
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.
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

Solution of Problem 18.1.

1. $g$ is injective and holomorphic; 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 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 square root $\varphi$ with $\varphi(0) = 1$ (Definition 18.8; $\mathbb D$ 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 of $\mathbb D$ onto the right half-plane; $w^2$ maps that half-plane conformally 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, 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): $F$ 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 from $0$ to $z$ in $\mathbb D$ ( $f^{-1}$ holomorphic, Corollary 17.10). 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 on the circle $\abs z = r$ (Theorem 16.10), 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) 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, 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. 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 $f(\mathbb D)$ — 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 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) 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: 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 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). If $f_n \in \mathcal S \to f$ locally uniformly: $f$ is holomorphic 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) makes the limit of injective maps injective: $f \in \mathcal S$ . A continuous functional (such as $f \mapsto \abs{a_2} = \frac{\abs{f''(0)}}2$ ) on a 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.