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= Koebe one-quarter theorem
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If $f$ is a <univalent function> on the <unit disc>, then $B(f(0),|f'(0)|/4)\subseteq f(\mathbb D)$. The constant $1/4$ is sharp, as shown by the <Koebe function>. This gives useful comparisons between the <derivative> of a <conformal map>, its <conformal radius>, and its distance to the <domain boundary>.

Koebe quarter theorem

The <Koebe quarter theorem> says that if $F:\mathbb D\to\mathbb C$ is a <univalent function> with $F(0)=0$ and $F'(0)=1$, then
$$
\boxed{\{w:|w|<1/4\}\subseteq F(\mathbb D).}
$$
Equivalently, an arbitrary <univalent function> $f$ on the <unit disc> has $B(f(0),|f'(0)|/4)\subseteq f(\mathbb D)$. The constant cannot be increased: the <Koebe function> $F(w)=w/(1-w)^2$ omits the ray $(-\infty,-1/4]$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 203 / 3 / a / Solution

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{title2=$k(z)=z/(1-z)^2$}

The Koebe function $k(z)=z/(1-z)^2$ is a <univalent function> from the <unit disc> onto $\mathbb C\setminus(-\infty,-1/4]$. To see the image, $(1+z)/(1-z)$ maps the disc onto the right half-plane and $k(z)=(((1+z)/(1-z))^2-1)/4$. Its normalization $k(0)=0$, $k'(0)=1$ proves sharpness of the <Koebe quarter theorem>.

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 Incoming links: Koebe function