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 $$z_0\ne 0$$ 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 $$g(\mathbb D)$$.