Koebe quarter theorem

Koebe function

The Koebe function is defined by

${\displaystyle f(z)=\frac{z}{(1-z{)}^2}=\sum _{n=1}^{\mathrm{\infty }}n{z}^n}$

Application of the theorem to this function shows that the constant ${\displaystyle 1/4}$ in the theorem cannot be improved, as the image domain ${\displaystyle f(\mathbf{D})}$ does not contain the point ${\displaystyle z=-1/4}$ and so cannot contain any disk centred at ${\displaystyle 0}$ with radius larger than ${\displaystyle 1/4}$.

The rotated Koebe function is

${\displaystyle f_{\alpha }(z)=\frac{z}{(1-\alpha z{)}^2}=\sum _{n=1}^{\mathrm{\infty }}n{\alpha }^{n-1}{z}^n}$

with ${\displaystyle \alpha }$ a complex number of absolute value ${\displaystyle 1}$. The Koebe function and its rotations are schlicht: that is, univalent (analytic and one-to-one) and satisfying ${\displaystyle f(0)=0}$ and ${\displaystyle f^{\prime }(0)=1}$.

Bieberbach's coefficient inequality for univalent functions

Let

${\displaystyle g(z)=z+a_2{z}^2+a_3{z}^3+\cdots }$

be univalent in ${\displaystyle |z|<1}$. Then

${\displaystyle |a_2|\le 2.}$

This follows by applying Gronwall's area theorem to the odd univalent function

${\displaystyle g(z^{-2}{)}^{-1/2}=z-\frac{1}{2}{a}_2{z}^{-1}+\cdots .}$

Equality holds if and only if ${\displaystyle g}$ is a rotated Koebe function.

This result was proved by Ludwig Bieberbach in 1916 and provided the basis for his celebrated conjecture that ${\displaystyle |a_n|\le n}$, proved in 1985 by Louis de Branges.

Proof of quarter theorem

Applying an affine map, it can be assumed that

${\displaystyle f(0)=0,f^{\mathrm{\prime }}(0)=1,}$

so that

${\displaystyle f(z)=z+a_2{z}^2+\cdots .}$

In particular, the coefficient inequality gives that ${\displaystyle |a_2|\le 2}$. If ${\displaystyle w}$ is not in ${\displaystyle f(\mathbf{D})}$, then

${\displaystyle h(z)=\frac{wf(z)}{w-f(z)}=z+(a_2+w^{-1})z^2+\cdots }$

is univalent in ${\displaystyle |z|<1}$.

Applying the coefficient inequality to ${\displaystyle h}$ gives

${\displaystyle |w{|}^{-1}=|w^{-1}|=|-a_2+a_2+w^{-1}|\le |a_2|+|a_2+w^{-1}|\le 4,}$

so that

${\displaystyle |w|\ge \frac{1}{4}.}$

Koebe distortion theorem

The Koebe distortion theorem gives a series of bounds for a univalent function and its derivative. It is a direct consequence of Bieberbach's inequality for the second coefficient and the Koebe quarter theorem.

Let ${\displaystyle f(z)}$ be a univalent function on ${\displaystyle |z|<1}$ normalized so that ${\displaystyle f(0)=0}$ and ${\displaystyle f^{\prime }(0)=1}$ and let ${\displaystyle r=|z|}$. Then

${\displaystyle \frac{r}{(1+r{)}^2}\le |f(z)|\le \frac{r}{(1-r{)}^2}}$

${\displaystyle \frac{1-r}{(1+r{)}^3}\le |f^{\mathrm{\prime }}(z)|\le \frac{1+r}{(1-r{)}^3}}$

${\displaystyle \frac{1-r}{1+r}\le \left|z\frac{f^{\mathrm{\prime }}(z)}{f(z)}\right|\le \frac{1+r}{1-r}}$

with equality if and only if ${\displaystyle f}$ is a Koebe function

${\displaystyle f(z)=\frac{z}{(1-e^{i\theta }z{)}^2}.}$