§9.2 Schlicht Functions

9.2 Schlicht Functions

Definition 9.2.1.

A holomorphic function

𝑓 : 𝔻 \to ℂ

is called schlicht iff it is univalent and satisfies

𝑓 (0) = 0

and

𝑓^{'} (0) = 1

The power series expansion of schlicht $𝑓$ around $0$ is of the form

𝑓 (𝑧) = \sum_{𝑛 = 0}^{\infty} \frac{𝑓^{(𝑛)} (0)}{𝑛!} 𝑧^{𝑛}, 𝑧 \in 𝔻,

which by assumption, simplifies to

𝑓 (𝑧) = 𝑧 + \sum_{𝑛 = 2}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}, 𝑧 \in 𝔻,

where $𝑎_{𝑛} = \frac{𝑓^{(𝑛)} (0)}{𝑛!}$ for $𝑛 \geq 2$ . Two historic results regarding schlicht functions state that

The only schlicht functions satisfying $\exists 𝑛 \in ℕ$ with $| 𝑎_{𝑛} | = 𝑛$ are the Koebe functions in the form of
$𝑧 \mapsto \frac{𝑧}{{(1 - e^{i 𝜃} 𝑧)}^{2}} .$
The Koebe functions are more extremal in the sense that it satisfies the conclusion $| 𝑎_{𝑛} | = 𝑛$ for all $𝑛 \in ℕ$ ; its power series is given by
$𝑧 \mapsto \sum_{𝑛 = 1}^{\infty} 𝑛 e^{i 𝜃 (𝑛 - 1)} 𝑧^{𝑛} .$
For any schlicht $𝑓$ and $\forall 𝑛 \in ℕ$ , it holds that $| 𝑎_{𝑛} | \leq 𝑛$ .

Both were solved in the 20th century by Louis de Branges in the affirmative. The second statement is known as the Bieberbach Conjecture as it was originally hypothesized by Ludwig Bieberbach.

We will introduce two of the oldest results regarding schlicht functions, namely $| 𝑎_{2} | \leq 2$ and the Koebe Quarter Theorem (Theorem 9.2.6 $Theorem 9.2.6 Theorem 9.2.6$ ). There are many applications of the latter, for instance, we used it in the proof of Mergelyan’s Theorem (Theorem 6.2.7 $Theorem 6.2.7 Theorem 6.2.7$ ).

Lemma 9.2.2.

Let $𝑓 : 𝔻 \to ℂ$ be schlicht and define $ℎ (𝑧) = \frac{1}{𝑓 (𝑧)} = \frac{1}{𝑧} + \sum_{𝑛 = 0}^{\infty} 𝑏_{𝑛} 𝑧^{𝑛}$ . Let

𝐴_{𝑟} = {𝑧 \in ℂ : 𝑟 < | 𝑧 | < 1}

be an annulus for $0 < 𝑟 < 1$ . Then $\exists 𝜂 > 0$ such that $\forall 𝑟 > 0$ sufficiently small ( $𝜂$ independent of $𝑟$ ), $ℎ (𝐴_{𝑟})$ lies in an ellipse with a semi-major axis

𝛼 = (\frac{1}{𝑟} + | 𝑏_{1} | 𝜂) \sqrt{1 + 𝜂 𝑟^{3}}

and a semi-minor axis

𝛽 = (\frac{1}{𝑟} - | 𝑏_{1} | 𝜂) \sqrt{1 + 𝜂 𝑟^{3}} .

Proof.

Let

\tilde{ℎ} (𝑧) = \sqrt{\frac{\overline{𝑏_{1}}}{| 𝑏_{1} |}} ℎ (𝑧 \sqrt{\frac{\overline{𝑏_{1}}}{| 𝑏_{1} |}}) - \sqrt{\frac{\overline{𝑏_{1}}}{| 𝑏_{1} |}} 𝑏_{0}

(where the branch or value of the square root is the same across), so that

\tilde{ℎ} (𝑧) = \frac{1}{𝑧} + \sum_{𝑛 = 1}^{\infty} {(\frac{\overline{𝑏_{1}}}{| 𝑏_{1} |})}^{\frac{1 + 𝑛}{2}} 𝑏_{𝑛} 𝑧^{𝑛} = \frac{1}{𝑧} + \sum_{𝑛 = 0}^{\infty} \tilde{𝑏_{𝑛}} 𝑧^{𝑛}, where \tilde{𝑏_{0}} = 0 \land \tilde{𝑏_{1}} = | 𝑏_{1} | .

Heuristically, we applied a rigid transformation to $ℎ$ so that the result has no constant term and whose linear coefficient is real. Trivially, if the conclusions of the statement are satisfied for $\tilde{ℎ}$ , then they naturally follow for $ℎ$ . Hence, without loss of generality, we consider only $ℎ (𝑧)$ such that $𝑏_{0} = 0$ and $𝑏_{1} \in ℝ$ , or when $ℎ (𝑧) = \frac{1}{𝑧} + 𝑏_{1} 𝑧 + 𝜙 (𝑧)$ where $𝜙 (𝑧) = 𝒪︀ (𝑧^{2})$ . For $𝑧 = 𝑟 e^{i 𝜃}$ ,

\begin{aligned} ℎ (𝑧) & = \frac{1}{𝑟} (cos 𝜃 - i sin 𝜃) + 𝑏_{1} 𝑟 (cos 𝜃 + i sin 𝜃) + 𝜙 (𝑟 e^{i 𝜃}) \\ = ((\frac{1}{𝑟} + 𝑏_{1} 𝑟) cos 𝜃 + ℜ𝔢 𝜙) + i ((𝑏_{1} 𝑟 - \frac{1}{𝑟}) sin 𝜃 + ℑ𝔪 𝜙) . \end{aligned}

Let $𝛼^{'} = \frac{1}{𝑟} + 𝑏_{1} 𝑟$ and let $𝛽^{'} = \frac{1}{𝑟} - 𝑏_{1} 𝑟$ . Then

\begin{aligned} \frac{{[ℜ𝔢 ℎ (𝑟 e^{i 𝜃})]}^{2}}{𝛼^{' 2}} + \frac{{[ℑ𝔪 ℎ (𝑟 𝑒^{i 𝜃})]}^{2}}{𝛽^{' 2}} & = {cos}^{2} 𝜃 + \frac{{[ℜ𝔢 𝜙]}^{2}}{𝛼^{' 2}} + \frac{2 cos 𝜃 ℜ𝔢 𝜑}{𝛼^{'}} \\ \begin{matrix}  \end{matrix} + {sin}^{2} 𝜃 + \frac{{[ℑ𝔪 𝜙]}^{2}}{𝛽^{' 2}} - \frac{2 sin 𝜃 ℑ𝔪 𝜙}{𝛽^{'}} . \end{aligned}

Since for all $\frac{𝛼^{'}}{\frac{1}{2 𝑟}} = 2 + 2 𝑏_{1} 𝑟^{2} \to 2$ , and $\frac{𝛽^{'}}{\frac{1}{2 𝑟}} = 2 - 2 𝑏_{1} 𝑟^{2} \to 2$ as $𝑟 \to 0^{+}$ , for sufficiently small $𝑟$ , $𝛼^{'} > \frac{1}{2 𝑟}$ and $𝛽^{'} > \frac{1}{2 𝑟}$ . Similarly, for sufficiently small $𝑟$

| \frac{ℜ𝔢 𝜙 (𝑟 e^{i 𝜃})}{𝛼^{'}} |, | \frac{ℑ𝔪 𝜙 (𝑟 e^{i 𝜃})}{𝛽^{'}} | < \frac{𝒪︀ (𝑟^{2})}{\frac{1}{2 𝑟}} = 𝒪︀ (𝑟^{3}) < 𝜂^{'} 𝑟^{3},

for some constant $𝜂^{'} > 0$ , so that

\begin{aligned} \frac{{[ℜ𝔢 ℎ (𝑟 e^{i 𝜃})]}^{2}}{𝛼^{' 2}} + \frac{{[ℑ𝔪 ℎ (𝑟 e^{i 𝜃})]}^{2}}{𝛽^{' 2}} & < 1 + 2 𝜂^{' 2} 𝑟^{6} + \frac{2 cos 𝜃 ℜ𝔢 𝜙}{𝛼^{'}} - \frac{2 sin 𝜃 ℑ𝔪 𝜙}{𝛽^{'}} \\ < 1 + ℴ︀ (𝑟^{3}) + 4 𝜂^{'} 𝑟^{3} < 1 + 𝜂 𝑟^{3} \end{aligned}

for some constant $𝜂 > 0$ and sufficiently small $𝑟$ . From rearrangement,

\frac{{[ℜ𝔢 ℎ (𝑟 𝑒^{i 𝜃})]}^{2}}{{(𝛼^{'} \sqrt{1 + 𝜂 𝑟^{3}})}^{2}} + \frac{{[ℑ𝔪 ℎ (𝑟 𝑒^{i 𝜃})]}^{2}}{{(𝛽^{'} \sqrt{1 + 𝜂 𝑟^{3}})}^{2}} < 1 .

Define $𝛼, 𝛽$ as in the theorem statement. It follows that $ℎ (𝑟 𝑒^{i 𝜃})$ lies in the region bounded by the ellipse centered at $0$ with semi-major and semi-minor axes $𝛼, 𝛽$ , respectively.

Since $ℎ (0) = \infty$ , it follows that $ℎ$ (which is injective by the univalence of $𝑓$ ) maps $𝐷 (0, 𝑟)$ to a neighborhood of $\infty$ and maps $𝜕 𝐷 (0, 𝑟)$ to the interior of the previous ellipse.

By the Jordan Curve Theorem (Theorem 1.2.5 $Theorem 1.2.5 Theorem 1.2.5$ ), $ℂ ∖ ℎ (𝜕 𝐷 (0, 𝑟))$ contains two connected regions. Since $ℎ$ maps a neighborhood $𝑈$ of $𝜕 𝐷 (0, 𝑟)$ to a neighborhood of $ℎ (𝜕 𝐷 (0, 𝑟))$ by the Open Mapping Theorem (Theorem 5.1.1 $Theorem 5.1.1 Theorem 5.1.1$ ), for a point $𝑤 \in ℎ (𝜕 𝐷 (0, 𝑟))$ , there exists a disk $𝐷 (𝑤, 𝜀) \subseteq ℎ (𝑈)$ , which intersects both $int ℎ (𝜕 𝐷 (0, 𝑟))$ and $ext ℎ (𝜕 𝐷 (0, 𝑟))$ (because $𝑤$ is an accumulation point of both sets as it lies in the boundary).

Thus, there exist points in $𝑈$ which map into $int ℎ (𝜕 𝐷 (0, 𝑟))$ , (hence map to points in the ellipse). Because $ℎ$ maps connected sets to connected sets, and no point in $ℎ (𝐷 (0, 𝑟))$ can map to $int ℎ (𝜕 𝐷 (0, 𝑟))$ (as otherwise the set would be disconnected), therefore all points inside $𝐴_{𝑟}$ map into $int ℎ (𝜕 𝐷 (0, 𝑟))$ . ⁠ $□$

Theorem 9.2.3 (Area Principle).

If $𝑓$ is schlicht and if

ℎ (𝑧) = \frac{1}{𝑓 (𝑧)} = \frac{1}{𝑧} + \sum_{𝑗 = 0}^{\infty} 𝑏_{𝑗} 𝑧^{𝑗},

then

\sum_{𝑗 = 0}^{\infty} 𝑗 {| 𝑏_{𝑗} |}^{2} \leq 1

Proof.

By Lemma 9.2.2 $Lemma 9.2.2 Lemma 9.2.2$ , for sufficiently small $𝑟 > 0$ , $ℎ (𝐴_{𝑟})$ lies in an ellipse $𝐸_{𝑟}$ with a semi-major axis

𝛼 = (\frac{1}{𝑟} + | 𝑏_{1} | 𝜂) \sqrt{1 + 𝜂 𝑟^{3}}

and a semi-minor axis

𝛽 = (\frac{1}{𝑟} - | 𝑏_{1} | 𝜂) \sqrt{1 + 𝜂 𝑟^{3}} .

The area of the ellipse is given by $π 𝛼 𝛽 = π (\frac{1}{𝑟^{2}} - {| 𝑏_{1} |}^{2} 𝜂^{2}) (1 + 𝜂 𝑟^{3})$ . On the other hand,

\begin{aligned} \frac{π}{𝑟^{2}} (1 + 𝜂 𝑟^{3}) \geq π (\frac{1}{𝑟^{2}} - {| 𝑏_{1} |}^{2} 𝜂^{2}) (1 + 𝜂 𝑟^{3}) & = area (𝐸_{𝑟}) \\ \geq area (ℎ (𝐴_{𝑟})), \end{aligned}

where

\begin{aligned} area (ℎ (𝐴_{𝑟})) & = \frac{1}{2 i} \iint_{𝐴_{𝑟}} {| ℎ^{'} (𝑧) |}^{2} d \overline{𝜁} \land d 𝜁 (Lusin Area Theorem) \\ = \int_{𝑟}^{1} \int_{0}^{2 π} {| ℎ^{'} (𝜌 e^{i 𝜃}) |}^{2} 𝜌 d 𝜃 d 𝜌 \\ = \int_{𝑟}^{1} \int_{0}^{2 π} {| - 𝜌^{- 2} e^{- 2 i 𝜃} + \sum_{𝑗 = 0}^{\infty} 𝑗 𝑏_{𝑗} 𝜌^{𝑗 - 1} e^{i 𝜃 (𝑗 - 1)} |}^{2} 𝜌 d 𝜃 d 𝜌 \\ = \int_{𝑟}^{1} \int_{0}^{2 π} [𝜌^{- 4} + (\sum_{𝑗 = 0}^{\infty} 𝑗 𝑏_{𝑗} 𝜌^{𝑗 - 1} e^{i 𝜃 (𝑗 - 1)}) \overline{\sum_{𝑗 = 0}^{\infty} 𝑗 𝑏_{𝑗} 𝜌^{𝑗 - 1} e^{i 𝜃 (𝑗 - 1)}} \\ \begin{matrix}  \end{matrix} + 2 ℜ𝔢 (- 𝜌^{- 2} \overline{e^{- 2 i 𝜃}} \sum_{𝑗 = 0}^{\infty} 𝑗 𝑏_{𝑗} 𝜌^{𝑗 - 1} e^{i 𝜃 (𝑗 - 1)})] 𝜌 d 𝜃 d 𝜌 \\ = \int_{𝑟}^{1} [2 π 𝜌^{- 3} + \int_{0}^{2 π} \sum_{𝑗, 𝑘 \in ℤ_{\geq 0}} 𝑗 𝑘 𝑏_{𝑗} \overline{𝑏_{𝑘}} e^{i 𝜃 (𝑗 - 𝑘)} 𝜌^{𝑗 + 𝑘 - 1} d 𝜃 (a) \\ \begin{matrix}  \end{matrix} - \frac{2}{𝜌} \int_{0}^{2 π} ℜ𝔢 (\sum_{𝑗 = 0}^{\infty} 𝑗 𝑏_{𝑗} 𝜌^{𝑗 - 1} e^{i 𝜃 (𝑗 + 1)}) d 𝜃] d 𝜌 \\ = \int_{𝑟}^{1} [2 π 𝜌^{- 3} + \sum_{𝑗, 𝑘 \in ℤ_{\geq 0}} 𝑗 𝑘 𝑏_{𝑗} \overline{𝑏_{𝑘}} (\int_{0}^{2 π} e^{i 𝜃 (𝑗 - 𝑘)} d 𝜃) 𝜌^{𝑗 + 𝑘 - 1} \\ \begin{matrix}  \end{matrix} - \frac{2}{𝜌} ℜ𝔢 (\sum_{𝑗 = 0}^{\infty} 𝑗 𝑏_{𝑗} 𝜌^{𝑗 - 1} \int_{0}^{2 π} e^{i 𝜃 (𝑗 + 1)}) d 𝜃] d 𝜌 . (b) \end{aligned}

Observe that for any $𝑛 \in ℤ$ ,

\int_{0}^{2 π} e^{i 𝑛 𝜃} d 𝜃 = {\begin{cases} 2 π & if 𝑛 = 0, \\ {\frac{1}{𝑛 i} e^{i 𝑛 𝜃} |}_{0}^{2 π} & otherwise \end{cases} = {\begin{cases} 2 π & if 𝑛 = 0, \\ 0 & otherwise . \end{cases}

Thus, in the second term, only terms with $𝑗 = 𝑘$ persist. In the third term, since $𝑗 + 1$ never vanishes, the entire third term vanishes as well.

The expansion of the summation in (a) is justified by the absolute convergence of the two factors. The switch in (b) is justified by locally uniform convergence of a power series.

As a result, we are hence left with

\begin{aligned} \frac{π}{𝑟^{2}} (1 + 𝜂 𝑟^{3}) & \geq \int_{𝑟}^{1} (2 π 𝜌^{- 3} + 2 π \sum_{𝑗 = 0}^{\infty} 𝑗^{2} {| 𝑏_{𝑗} |}^{2} 𝜌^{2 𝑗 - 1}) d 𝜌 \\ = {- π 𝜌^{- 2} |}_{𝑟}^{1} + π \sum_{𝑗 = 0}^{\infty} 𝑗 {| 𝑏_{𝑗} |}^{2} {𝜌^{2 𝑗} |}_{𝑟}^{1} \\ = \frac{π}{𝑟^{2}} - π + π \sum_{𝑗 = 0}^{\infty} 𝑗 {| 𝑏_{𝑗} |}^{2} (1 - 𝑟^{2 𝑗}) . \end{aligned}

Thus,

1 + 𝜂 𝑟^{3} \geq \sum_{𝑗 = 0}^{\infty} 𝑗 {| 𝑏_{𝑗} |}^{2} (1 - 𝑟^{2 𝑗}) .

Since each term in the right-hand side is always positive, for any $𝑁 \in ℕ$ ,

1 + 𝜂 𝑟^{3} \geq \sum_{𝑗 = 0}^{𝑁} 𝑗 {| 𝑏_{𝑗} |}^{2} (1 - 𝑟^{2 𝑗}) .

Taking $𝑟 \to 0^{+}$ ,

1 \geq \sum_{𝑗 = 0}^{𝑁} 𝑗 {| 𝑏_{𝑗} |}^{2},

which holds for any $𝑁$ . Thus, letting $𝑁 \to \infty$ gives the desired result. ⁠ $□$

Theorem 9.2.4.

𝑓

is schlicht, then there exists a schlicht

𝑔

such that

{𝑔 (𝑧)}^{2} \equiv 𝑓 (𝑧^{2})

𝔻

Proof.

We aim to show $𝑓 (𝑧^{2})$ has a holomorphic square root. Let $𝜇 : 𝑧 \mapsto \frac{𝑓 (𝑧)}{𝑧}$ , satisfying that $𝜇 (0) = 1$ and that $𝜇$ vanishes nowhere (because $𝑓$ vanishes at $0$ , it may not vanish elsewhere). Therefore, $𝜇 (𝑧)$ admits a holomorphic square root $𝑟 (𝑧)$ . Let

𝑔 (𝑧) = 𝑧 𝑟 (𝑧^{2})

for $𝑧 \in 𝔻$ . Then $𝑔 (0) = 0$ , $𝑔^{'} (0) = 1$ , and $𝑔^{2} \equiv 𝑓 \circ (𝑧 \mapsto 𝑧^{2})$ . We are left to show the univalence of $𝑔$ . Let $𝑧_{1}, 𝑧_{2} \in 𝔻$ such that $𝑔 (𝑧_{1}) = 𝑔 (𝑧_{2})$ . Then $𝑓 (𝑧_{1}^{2}) = 𝑓 (𝑧_{2}^{2})$ , so $𝑧_{1}^{2} = 𝑧_{2}^{2}$ .

Then either $𝑧_{1} = 𝑧_{2}$ or $𝑧_{1} = - 𝑧_{2}$ . Assume the latter case. Since $𝑧_{1}^{2} = 𝑧_{1}^{2}$ , $𝑟 (𝑧_{1}^{2}) = 𝑟 (𝑧_{2}^{2})$ , therefore

𝑔 (𝑧_{1}) = 𝑧_{1} 𝑟 (𝑧_{1}^{2}) = - 𝑧_{2} 𝑟 (𝑧_{2}^{2}) = - 𝑔 (𝑧_{2}),

a contradiction. Thus $𝑧_{1} = 𝑧_{2}$ and $𝑔$ is schlicht. ⁠ $□$

Theorem 9.2.5.

𝑓 : 𝔻 \to ℂ

is schlicht and expands to

𝑧 + \sum_{𝑛 = 2}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}

, then

| 𝑎_{2} | \leq 2

Proof.

By Theorem 9.2.4 $Theorem 9.2.4 Theorem 9.2.4$ , there exists a schlicht function $𝑔$ such that ${𝑔 (𝑧)}^{2} = 𝑓 (𝑧^{2})$ . Then there exists a unique Laurent series

\frac{1}{𝑔 (𝑧)} = \frac{1}{𝑧} + \sum_{𝑛 = 0}^{\infty} 𝑏_{𝑛} 𝑧^{𝑛} .

By absolute convergence,

\frac{1}{{𝑔 (𝑧)}^{2}} = \frac{1}{𝑧^{2}} + \frac{2 𝑏_{0}}{𝑧} + (2 𝑏_{1} + 𝑏_{0}^{2}) + 𝒪︀ (𝑧) = \frac{1}{𝑓 (𝑧^{2})} .

Since

\begin{aligned} \frac{1}{𝑓 (𝑧^{2})} & = \frac{1}{\sum_{𝑛 = 1}^{\infty} 𝑎_{𝑛} 𝑧^{2 𝑛}} \\ = \frac{1}{𝑧^{2}} \cdot \frac{1}{1 + \sum_{𝑛 = 1} 𝑎_{𝑛 + 1} 𝑧^{2 𝑛}} \\ = \frac{1}{𝑧^{2}} \cdot \frac{1}{1 + 𝑎_{2} 𝑧^{2} + 𝒪︀ (𝑧^{4})} \\ = \frac{1}{𝑧^{2}} \sum_{𝑗 = 0}^{\infty} {(- 𝑎_{2} 𝑧^{2} + 𝒪︀ (𝑧^{4}))}^{𝑗} \\ = \frac{1}{𝑧^{2}} (1 - 𝑎_{2} 𝑧^{2} + 𝒪︀ (𝑧^{4})) = \frac{1}{𝑧^{2}} - 𝑎_{2} + 𝒪︀ (𝑧^{2}), \end{aligned}

by the uniqueness of the Laurent series, the coefficients must match; therefore $2 𝑏_{1} + 𝑏_{0}^{2} = - 𝑎_{2}$ and $𝑏_{0} = 0$ . By the Area Principle (Theorem 9.2.3 $Theorem 9.2.3 Theorem 9.2.3$ ) applied to $1 / 𝑔$ , we have

\sum_{𝑛 = 0}^{\infty} 𝑛 {| 𝑏_{𝑛} |}^{2} \leq 1 ⟹ | 𝑏_{1} | \leq 1,

and hence

| 𝑎_{2} | \leq 2 | 𝑏_{1} | + {| 𝑏_{0} |}^{2} \leq 2 . □

Theorem 9.2.6 (Koebe Quarter Theorem).

𝑓 : 𝔻 \to ℂ

is schlicht with

𝑓 (𝑧) = 𝑧 + \sum_{𝑛 = 2}^{\infty} 𝑎_{𝑛} 𝑧^{𝑛}

, then the image

𝑓 (𝔻)

contains the open disk of radius

\frac{1}{4}

centered at

𝑓 (0)

Proof.

Construct a second function

𝑔 (𝑧) = 𝜙 \circ 𝑓,

where $𝜙 : 𝑤 \mapsto \frac{𝛼 𝑤}{𝛼 - 𝑤}$ is a Möbius transformation, holomorphic everywhere outside $𝑤 = 𝛼$ (which $𝑓$ does not attain). Since $𝜙$ fixes the origin, satisfies $𝜙^{'} (0) = 1$ , and is univalent, $𝑔$ is schlicht.

Moreover, the coefficient of $𝑧^{2}$ in the expansion of $𝑔$ can be obtained from

\begin{aligned} 𝑔 (𝑧) & = (𝑤 \mapsto \frac{𝑤}{1 - \frac{𝑤}{𝛼}}) \circ 𝑓 (𝑧) \\ = 𝑓 (𝑧) \sum_{𝑛 = 0}^{\infty} {(\frac{𝑓 (𝑧)}{𝛼})}^{𝑛} \\ = (𝑧 + 𝑎_{2} 𝑧^{2} + 𝒪︀ (𝑧^{3})) (1 + \frac{𝑧}{𝛼} + \frac{𝑎_{2}}{𝛼} 𝑧^{2} + 𝒪︀ (𝑧^{3})) \\ = 𝑧 + (𝑎_{2} + \frac{1}{𝛼}) 𝑧^{2} + 𝒪︀ (𝑧^{3}), \end{aligned}

where Theorem 9.2.5 $Theorem 9.2.5 Theorem 9.2.5$ applied to $𝑔$ gives $| 𝑎_{2} + \frac{1}{𝛼} | \leq 2$ and Theorem 9.2.5 $Theorem 9.2.5 Theorem 9.2.5$ applied to $𝑓$ gives $| 𝑎_{2} | \leq 2$ , therefore

| 𝑎_{2} + \frac{1}{𝛼} | + | 𝑎_{2} | \leq 4 ⟹ | \frac{1}{𝛼} | \leq 4 \Rightarrow | 𝛼 | \geq \frac{1}{4} .

Then any omitted value must lie outside $𝐷 (0, \frac{1}{4})$ , proving the result. ⁠ $□$