Math 2.0

Functions of a complex variable Geometric function theory Example

Koebe function

Example

The Koebe function is k(z)=z(1−z)2=∑n=1∞nzn,∣z∣<1.k(z) = \frac{z}{(1 - z)^2} = \sum_{n=1}^{\infty} n z^n, \qquad \abs{z} < 1. It is a univalent function on the unit disc 𝔻\mathbb{D} in the class SS, and k(𝔻)=ℂ∖(−∞,−1/4]k(\mathbb{D}) = \C \setminus (-\infty, -1/4].

Check: one has k(z)=14((1+z1−z)2−1).k(z) = \frac{1}{4}\left( \left(\frac{1 + z}{1 - z}\right)^2 - 1 \right). The Möbius map z↦(1+z)/(1−z)z \mapsto (1+z)/(1-z) maps 𝔻\mathbb{D} bijectively onto the right half-plane {Re⁡w>0}\set{\operatorname{Re} w > 0}. Squaring maps the right half-plane bijectively onto ℂ∖(−∞,0]\C \setminus (-\infty, 0]. Finally w↦(w−1)/4w \mapsto (w - 1)/4 maps ℂ∖(−∞,0]\C \setminus (-\infty, 0] onto ℂ∖(−∞,−1/4]\C \setminus (-\infty, -1/4]. A composition of injective holomorphic maps is injective and holomorphic, and k(0)=0k(0) = 0, k′(0)=1k'(0) = 1.

Remarks

The point −1/4-1/4 is not in the image, so the radius 1/41/4 in the Koebe quarter theorem cannot be improved. The coefficients are an=na_n = n, so the bound ∣an∣≤n\abs{a_n} \le n of the Bieberbach conjecture is attained for every nn. The rotations e−iθk(eiθz)e^{-i\theta} k(e^{i\theta} z) with θ∈ℝ\theta \in \R are also in SS and are the extremal functions of many problems on SS.

Where this leads

2 pages build on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Koebe_quarter_theorem
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