# How to prove conformal self map of punctured disk ${0<|z|<1}$ is rotation

My idea is to somehow show that the conformal self map (let's say $f$) of the punctured disk is actually satisfies the hypothesis of the Swartz's lemma and apply swartz's lemma to conclude. I don't know how I can prove $f$ satisfies the condition $f(0)=0$. Does someone have better and rigorous way. Please share!

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The given function $f:\ \dot D\to\dot D$ is analytic and bounded in a punctured neighborhood of $0$. Therefore it can be continued analytically to a function $\tilde f:\ D\to \bar D$. Let $\tilde f(0)=c$. A priori $|c|\leq1$, but $|c|=1$ would violate the maximum principle. If we had $c\in \dot D=f(\dot D)$ then we would have $\tilde f(0)=c$ and in addition $f(z_1)=c$ for some $z_1\in\dot D$. It is easy to see that points in a punctured neighborhood of $c$ would then be covered twice by $f$: once by images of points $z$ near $0$ and once by images of points $z$ near $z_1$. It follows that $c=0$; so the analytic continuation $\tilde f$ is actually a conformal self map of $D$ that leaves $0$ fixed. Now apply Schwarz' Lemma.

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I am confused in this line, could you please elaborate. "If we had $c\in \dot D$ then $f(z_1)=c$ for some $z_1\in\dot D$, and it is easy to see that points in a punctured neighborhood of $c$ would then be covered twice by $f$" I can follow rest of the proof very easily. Thanks –  Deepak Dec 27 '12 at 16:37
@Deepak: See my edit. –  Christian Blatter Dec 27 '12 at 16:48
cool. Thanks... –  Deepak Dec 27 '12 at 17:27
I dont understand that line which deepak has mentioned –  La Belle Noiseuse Mar 7 '13 at 17:50
@City of God: See my small edit. The sentence in question now begins with "If we had $c\in \dot D=f(\dot D)$ then $\ldots$". –  Christian Blatter Mar 7 '13 at 18:49

$f(a) = 0$ and $\phi_a$ is a Möbius transformation that takes $a$ to zero; $\phi_a : D \to D$ is defined by $\phi_a(z) = \frac{z-a}{1-\bar{a}z}$.

$\phi_a\circ f = g$, $g(0) = 0$ and $|g|\leq 1$, so by Schwartz's Lemma $g(z) = z$ now solve for $f$ by taking inverse of $\phi_a$.

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Can you be more rigorous @K.Ghosh. Thank you. –  Deepak Dec 26 '12 at 17:22