# Conformal map to unit disk

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.

• 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? – John Cataldo May 25 at 15:40
• See this answer. – 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.