0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.