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The Koebe function realizes the quarter-disc bound

Example

The Koebe function

$k (z) : = \frac{z}{(1 - z)^{2}}$

maps $D$ biholomorphically onto $C ∖ (- \infty, - 1 / 4]$ , so the radius $1 / 4$ in Koebe's theorem is sharp.

Facts & Assumptions

Given: The function $k (z) = z / (1 - z)^{2}$ .

[L1]

Every normalized univalent disc map contains $D (0, 1 / 4)$ (Every normalized univalent disc map contains the quarter disc).

Verification

technique · direct

1.1givencasesalgebra

The function $k$ is holomorphic on $D$ , with $k (0) = 0$ and $k^{'} (0) = 1$ . The value $w = 0$ has the unique preimage $z = 0$ . If $w \neq 0$ , solving $w = z / (1 - z)^{2}$ gives $w z^{2} - (2 w + 1) z + w = 0,$ so $z = \frac{2 w + 1 \pm \sqrt{1 + 4 w}}{2 w} .$ For $w \notin (- \infty, - 1 / 4]$ , choose the square-root branch that equals $1$ at $w = 0$ ; then the minus sign gives the unique solution with $∣ z ∣ < 1$ . Hence $k$ maps $D$ bijectively onto $C ∖ (- \infty, - 1 / 4]$ .

2.1L1step 1.1∎

The omitted point closest to $0$ is $- 1 / 4$ , so no larger disc centered at $0$ can lie in $k (D)$ . Since [L1] guarantees the quarter disc for every normalized univalent map, the constant $1 / 4$ is sharp.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Matthias Weber, Complex Analysis, Example 7.5.1 (standard reference, not scraped)