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I'm trying to find a conformal map from the upper half plane to the region $\{z| |z|<1$ and $Re(z) + Im(z) >1\}$.

I know how to map the upper half plane to the unit disc, so I was hoping to then map the unit disc to this region... But I'm completely stuck. Any help would be appreciated.

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I'd rather start from the region and map to the half plane. First, $z\mapsto \frac1{z-1}$ moves one of the vertices away to infinity while the two parts of the boundary become straight lines, intersecting at $\frac1{i-1}$ with an angle of $45^\circ$. Thus appending the map $z\mapsto \left(z-\frac1{i-1}\right)^4$ "flattens" this other vertex and you at least have some half plane. Rotate to get the right one.

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  • $\begingroup$ thanks! one question: how did you know you wanted to raise $(z - \frac{1}{i-1})$ to the fourth power? $\endgroup$ – user227345 Mar 29 '15 at 20:34
  • $\begingroup$ Because $z\mapsto z^4$ multiplies the vertex angle (at $0$) by $4$. $\endgroup$ – Lubin Mar 29 '15 at 20:40
  • $\begingroup$ Why the downvote? Anything totally wrong? $\endgroup$ – Hagen von Eitzen Mar 31 '15 at 21:31

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