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. A related result is the Schwarz lemma, and a notion related to both is conformal radius.
Suppose that is univalent in |z| > 1. Then In fact, if r > 1, the complement of the image of the disk |z| > r is a bounded domainX. Its area is given by Since the area is positive, the result follows by lettingr 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 Grönwall.
Koebe function
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 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 with α a complex number of absolute value 1. The Koebe function and its rotations are schlicht: that is, univalent and satisfying f = 0 and f′ = 1.
Let 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. This result was proved by Ludwig Bieberbach in 1916 and provided the basis for his celebrated conjecture that |an| ≤ n, proved in 1985 by Louis de Branges.
Proof of quarter theorem
Applying an affine map, it can be assumed that so that If w is not in f, then is univalent in |z| < 1. Applying the coefficient inequality to f and h gives so that
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 be a univalent function on |z| < 1 normalized so that f = 0 and f' = 1 and let r = |z|. Then with equality if and only if f is a Koebe function