Koebe quarter theorem
Koebe Quarter Theorem. the image of an injective analytic function f : D → C from the unit disk 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 1916. The example of the Koebe function shows that the constant 1/4 in the theorem cannot be improved.
Gronwall's area theorem
is univalent in |z| > 1. Then
In fact, if r > 1, the complement of the image of the disk |z| > r is a bounded domain X(r). Its area is given by
Since the area is positive, the result follows by letting r decrease to 1. The above proof shows equality holds if and only if the complement of the image of g has zero area, i.e. Lebesgue measure zero.
This result was proved in 1914 by the Swedish mathematician Thomas Hakon Gronwall.
The Koebe function is defined by
Application of the theorem to this function shows that the constant 1/4 in the theorem cannot be improved, as the image domain f(D) does not contain the point z = −1/4 and so cannot contain any disk centred at 0 with radius larger than 1/4.
The rotated Koebe function is
Bieberbach's coefficient inequality for univalent functions
be univalent in |z| < 1. Then
This follows by applying Gronwall's area theorem to the odd univalent function
Equality holds if and only if g is a rotated Koebe function.
Proof of quarter theorem
Applying an affine map, it can be assumed that
If w is not in f(D), then
is univalent in |z| < 1.
Applying the coefficient inequality to f and h gives
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 f(z) be a univalent function on |z| < 1 normalized so that f(0) = 0 and f'(0) = 1 and let r = |z|. Then
with equality if and only if f is a Koebe function
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