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I am preparing for a complex analysis qual and ran into this problem on an old exam.

Find all conformal isomorphisms from the unit disk $\mathbf{D}=\{z\in\mathbf{C} : |z| < 1\}$ to the semi-disk $S=\{z\in\mathbf{D} : \operatorname{Im}z>0\}$ fixing $1$ and $-1$.

I know how to map $\mathbf{D}$ conformally onto $S$, but I can not find a mapping that fixes $1$ and $-1$.

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    $\begingroup$ If we look the other way, a conformal map $S\to \mathbf{D}$ fixing $1$ and $-1$ must convert a right angle to a straight $\pi$ angle at these points. What map do you know that doubles angles at a certain point? $\endgroup$ – Daniel Fischer May 3 '15 at 21:43
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    $\begingroup$ Every conformal isomorphism $D \to S$ differs by a conformal automorphism of the disc (i.e., if $f_1, f_2$ are conformal isomorphisms $D \to S$, then there is some conformal automorphism $g: D \to D$ such that $f_2g = f_1$). So start with a given conformal isomorphism, modify it by an automorphism of the disc so that it fixes 1 and -1, and find out what other automorphisms of the disc fix these points. $\endgroup$ – user98602 May 3 '15 at 21:43
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Thanks, I think I have it now.

We begin by applying $f(z)=\frac{z+1}{z-1}$. Note that \begin{align*} f(-1)&=0 \\ f(1)&=\infty \\ f(0)&=-1 \\ f(i)&=-i \end{align*} and so $f$ maps the semi-disk to the third quadrant. Multiplication by $-1$ followed by squaring gives us a conformal bijection to the upper half-plane, $g(z)=\left(\frac{z+1}{z-1}\right)^2$.

Let $h(z)=\frac{z-i}{z+i}$ and note that then \begin{align*} h(0)&=-1 \\ h(1)&=-i \\ h(-1)&=i \\ h(i)&=0 \end{align*} and so $h$ maps the upper half-plane conformally onto the unit disk. Hence $h(g(z))$ maps the semi-disk to the unit disk. Note that \begin{align*} h(g(1))&=h(\infty)=1 \\ h(g(-1))&=h(0)=-1 \end{align*} and so we have a conformal mapping that takes the semi-disk to the unit disk and fixes $1$ and $-1$. Suppose $f$ and $g$ are two such mappings. Then $g\circ f^{-1}$ gives an automorphism of the unit disk that fixes $1$ and $-1$. However, by the Schwarz lemma the only such automorphism is the identity and so we have that the mapping above is the unique mapping satisfying the conditions.

EDIT: As Daniel Fischer pointed out, there are automorphisms of the closed unit disk that have two fixed points.

Suppose $f$ and $g$ are two mappings satisfying the conditions from before. Then $g=d\circ f$ where $d:\overline{\mathbf{D}}\to\overline{\mathbf{D}}$ is an automorphism of the unit disk that fixes $1$ and $-1$.

Note that $d(\mathbf{D})=\mathbf{D}$ by the open mapping theorem and thus every automorphism of the closed unit disc restricts to an automorphism of the open unit disk. Hence it suffices to consider automorphisms of the form \begin{align*} d(z)=e^{i\theta}\frac{z-\alpha}{1-\overline{\alpha}z} \end{align*} for $\alpha\in \mathbf{D}$. Since $d$ must fix $1$ and $-1$ we have \begin{align*} 1-\overline{\alpha}=e^{i\theta}(1-\alpha) \\ -1-\overline{\alpha}=e^{i\theta}(-1-\alpha) \\ \end{align*} which implies that $e^{i\theta}=1$ and $\overline{\alpha}=\alpha$. Hence $$ d(z)=\frac{z-r}{1-rz} $$ for some $r\in(-1,1)$.

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  • $\begingroup$ Everything is right except the last bit, there are more automorphisms of the unit disk fixing $1$ and $-1$. For $-1 < r < 1$, consider $$T_r(z) = \frac{z-r}{1-rz}.$$ These are automorphisms of the unit disk fixing $1$ and $-1$. It remains to see that these are all. $\endgroup$ – Daniel Fischer May 7 '15 at 19:57

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