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I'm trying to find a conformal map from $\{z : |z|<1, |z-i|<\sqrt{2}\}$ to the upper half plane $\{z\in\mathbb{C}: \text{Im}(z)>0\}$.

Following the advice from this answer, I've been able to do the following: On the top this region is bounded by the top arc of the unit circle, and on the bottom by part of the bottom arc of the circle $|z-i|<\sqrt{2}$. The intersection is at $-1$ and $1$. The fractional linear transformation $\frac{az-z}{cz+c}$ maps $1$ to $0$ and $-1$ to $\infty$. Now I need to map to some power $z\mapsto z^\beta$ and this is where I'm stuck...

J. Bishop
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The inverse Cayley transform $f(z) =\frac{iz+i}{-z+1}$ maps the unit disk to the upper half plane. Moreover, $f(-1) = 0$, $f(1) = \infty$ and $f((1-\sqrt{2})i) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$. As Möbius transforms map circles to circles or lines, the circle $\{z : |z-i| = \sqrt{2}\}$ is mapped to the line $\{z : \mathrm{Re}(z) = \mathrm{Im}(z)\}$. As $f(0) = i$, $\{z : |z| < 1, |z-i| < \sqrt{2}\}$ is mapped to the sector $\{z = re^{it}: t \in \left(\frac{\pi}{4},\pi\right)\}$. It is now easy to choose the right exponent to map this to a half plane and then rotate it to the upper half plane.

More generally, if you have a set $S$ that is the intersection of two disks, choose a Möbius transform that maps one intersection of the circles to $\infty$ and the other to $0$. This forces $S$ to be mapped to a sector. From there it is then easy to get to the upper half plane.

Klaus
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