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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 $D\setminus(-1,0]$?2012-11-26
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    D is the unit disk.2012-11-26

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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?2012-11-26
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    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.2012-11-26
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    thanks. I need some time to digest. Honestly, I am a new graduate student and I am not professional in this area.2012-11-26
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    I will try to follow. Thanks again.2012-11-26
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    @PantelisDamianou It seems to go backwards. You still need a conformal map from $D$ onto the right half-disk, don't you?2012-11-26
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    The next step is to send $i$ to $0$ and $-i$ to infinity. That will give you two perpendicular lines. The map is $$ L_4(z)=\frac{z-i}{z+i}$$. Then follow with $z^2$ to get the upper half plane. Finally, repeat $L_4$ again to get the unit disk.2012-11-26
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    @ThomasAndrews that is the easy part. I explained how you go from the half-disk onto the unit disk in my previous comment.2012-11-26
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    @ThomasAndrews You are right. I went the wrong direction. One has to invert all the maps. But I think this is probably the best way to proceed.2012-11-26