User:OdedSchramm/sb3

The Koebe 1/4 theorem states that the image of an injective analytic function $f:\mathbb {D} \to \mathbb {C}$ from the unit disk $\mathbb {D}$ onto a subset of the complex plane contains the disk whose center is $f(0)$ and whose radius is $|f\,'(0)|/4$. The theorem is named after Paul Koebe, who conjectured the result in 1907. The theorem was proven by Ludwig Bieberbach in 1914. The Koebe function $f(z)=z/(1-z)^{2}$ shows that the constant $1/4$ in the theorem cannot be improved.

Proof

There is a proof based on the area theorem and some power series calculations. Following is a proof based on the notion and properties of extremal length.

We start by assuming that $f(0)=1$ and $0\notin f(\mathbb {D} )$. Since every point $z_{0}\neq 0$ has a neighborhood in which ${\sqrt {z}}$ can be defined as an analytic function, the monodromy theorem implies that there is an analytic function $g:\mathbb {D} \to \mathbb {C}$ such that $g(z)^{2}=f(z)$ for every $z\in \mathbb {D}$. Fix such a $g$ satisfying $g(0)=1$. Note that since $f$ is injective, also $g$ must be injective, and moreover, $g(\mathbb {D} )\cap -g(\mathbb {D} )=\emptyset$. This implies that for all $r>0$ sufficiently small so that $g(\mathbb {D} )\supset B(1,r)$, the extremal distance in $\mathbb {C}$ from $B(1,r)$ to $B(-1,r)$ is at least twice the extremal distance from $B(1,r)$ to the boundary of $g(\mathbb {D} )$.