# conformal mapping in complex analysis

The question is to find a conformal map from $D$ onto $D\setminus\left[\frac{-1}{2},1\right)$.

I don't know how to start with. The line $\left[\frac{-1}{2},1\right)$ makes it crazy. Anyone can give me some hints?

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You don't say what $D$ is. I assume the unit disk? Would it be easier if the range was supposed to be $D\setminus(-1,0]$? – Thomas Andrews Nov 26 '12 at 21:05
D is the unit disk. – David Nov 26 '12 at 21:17

## 1 Answer

First map $D$ onto itself with a Möbius. $$L(z)=\frac{ z-\alpha}{ 1 -\bar{\alpha} z}$$ where $\alpha=-\frac{1}{2}$. Then follow with $z \to -z$ and finally with $\sqrt{z}$. That will give you the intersection of $D$ with the right half plane. From there it is easy.

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can you explain more clearly? – David Nov 26 '12 at 21:24
Which part is not clear? Your cut goes onto the interval $[-1,0]$. Then you take the square root and you get the right half of the disk. – PAD Nov 26 '12 at 21:26
thanks. I need some time to digest. Honestly, I am a new graduate student and I am not professional in this area. – David Nov 26 '12 at 21:28
I will try to follow. Thanks again. – David Nov 26 '12 at 21:28
@PantelisDamianou It seems to go backwards. You still need a conformal map from $D$ onto the right half-disk, don't you? – Thomas Andrews Nov 26 '12 at 21:29