So when we have an $\alpha$ with $0 < \alpha < \pi$, I need to find a conformal map from

$S_\alpha = \{z \in \mathbb{C} ~|~ |z| > \frac{1}{2}, -\alpha < \arg(z) < \alpha\}$

to the open unit disk $\mathbb{D}$ but I have no idea how I can construct this. I tried to find a conformal map to the upper half plane but I can't find anything.

  • $\begingroup$ Maybe solving $\begin{cases}f({1\over2}e^{i\alpha})=i\\f(-{1\over2}e^{i\alpha})=-i\end{cases}$ where $f$ is a Möbius transformation will help? $\endgroup$ – John Cataldo May 25 at 15:40
  • $\begingroup$ See this answer. $\endgroup$ – Maxim May 29 at 18:12

Assuming you can map a half disc to a disc, consider this: $z\to 1/z$ maps your $S_\alpha$ to the sector

$$\{|z|<2, -\alpha < \arg z < \alpha \}.$$

Follow this with the map $z\to z^{\pi/(2\alpha)}$ to obtain a half disc.


By taking logarithm, you map $S_\alpha$ to rectangular strip with one end at infinity. Scale and translate to have its vertices at $-\frac\pi2i$, $\frac\pi2i$, and $+\infty$. Now apply $z\mapsto e^{-z}$, and you have the right half of the unit disk, i..e, we have a line and a circular arc meeting at $90^\circ$ angles in $\pm i$. Apply the Möbius transform $z\mapsto\frac{z+i}{z-i}$ that sends $i$ to infinity to arrive at a quarter plane with vertex at $0$. Square to arrive at a half plane. From here, you know the way to the unit disk.enter image description here


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