User:OdedSchramm/sb3

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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.

[edit] 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).