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  "body": "\n## Example\n\nThe *Koebe function* is\n$$k(z) = \\frac{z}{(1 - z)^2} = \\sum_{n=1}^{\\infty} n z^n, \\qquad \\abs{z} \u003c 1.$$\nIt is a [[univalent function|R91E83SDBD]] on the unit disc $\\mathbb{D}$ in the class $S$, and $k(\\mathbb{D}) = \\C \\setminus (-\\infty, -1/4]$.\n\nCheck: one has\n$$k(z) = \\frac{1}{4}\\left( \\left(\\frac{1 + z}{1 - z}\\right)^2 - 1 \\right).$$\nThe Möbius map $z \\mapsto (1+z)/(1-z)$ maps $\\mathbb{D}$ bijectively onto the right half-plane $\\set{\\operatorname{Re} w \u003e 0}$.\nSquaring maps the right half-plane bijectively onto $\\C \\setminus (-\\infty, 0]$.\nFinally $w \\mapsto (w - 1)/4$ maps $\\C \\setminus (-\\infty, 0]$ onto $\\C \\setminus (-\\infty, -1/4]$.\nA composition of injective holomorphic maps is injective and holomorphic, and $k(0) = 0$, $k'(0) = 1$.\n\n## Remarks\n\nThe point $-1/4$ is not in the image, so the radius $1/4$ in the [[Koebe quarter theorem|WX0HMW4GT7]] cannot be improved.\nThe coefficients are $a_n = n$, so the bound $\\abs{a_n} \\le n$ of the [[Bieberbach conjecture|ZMRFSYVWQT]] is attained for every $n$.\nThe rotations $e^{-i\\theta} k(e^{i\\theta} z)$ with $\\theta \\in \\R$ are also in $S$ and are the extremal functions of many problems on $S$.\n\n## Explanations\n\n### L1 human\n\nThe Koebe function maps the unit disc onto the whole plane with one slit cut out along the negative real axis, from $-1/4$ to $-\\infty$.\nIt is the extreme case of the class of normalised univalent functions: it stretches the disc as much as possible.\nIts boundary slit comes as close as $1/4$ to the origin, and its power series coefficients $1, 2, 3, \\ldots$ are as large as allowed.\n",
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