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.


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.

  • $\begingroup$ 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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