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Koebe's $1/4$ theorem claims that if $f:\mathbb{D}\to \mathbb{C}$ is an injective holomorphic function defined on a unit disk such that $f(0) =0$ and $f'(0)= 1$, then the image $f(\mathbb{D})$ of $\mathbb{D}$ contains an open disk of radius $1/4$ centered at the origin. There are some known proofs of the theorem, but all of them seems to be nonintuitive. For example, one proof uses Bieberbach's coefficient theorem. Is there any possible geometric explanation or even proof of Koebe's theorem? Thanks in advance.

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    $\begingroup$ The intuitive concept is just: If we consider $f(z)=z$, the image contains a disk of radius $1$. In order to fail to contain even a smaller disk, say to have $r\notin f(\Bbb D)$ with $0<r<1$, the function must thus be more or less distorted. For small enough $r$, the distortion is so great that $f(\Bbb D)$ must wrap around the point $r$ enough to overlap with itself and thus $f$ cannot be injective. This does not directly explain though that $r=\frac14$ plays a special role ... $\endgroup$ – Hagen von Eitzen Feb 4 at 6:57
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    $\begingroup$ Hagen's explanation is spot on. Re the value $1/4$ itself, I remember a joke to the effect that the maximal $r$ fitting Koebe's theorem, which was at first known to exist without knowing its value, once known to equal $1/4$, could not be named Koebe's constant since this constant already had a name (that is, one fourth...). $\endgroup$ – Did Feb 4 at 7:44
  • $\begingroup$ @Did It will be good if there's a geometric intuition for the existence of Koebe's constant, although it is hard to show that the constant is exactly $1/4$. $\endgroup$ – Seewoo Lee Feb 4 at 8:04

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