Find a conformal map \(L\to \DD\) where \[ L\da \ts{\abs{z - i } < \sqrt 2} \intersect \ts{\abs{z+i} < \sqrt 2} ,\] i.e. a lune with vertices \(-1\) and \(1\).
Solution
The key insight: for lunes, map the corners to \(0\) and \(\infty\); this yields a sector. Here we want \(-1\mapsto 0\) and \(1\mapsto \infty\), so \(f(z) = {z+1\over z-1}\) gets things started.
In steps:
- \(z\mapsto {1+z\over 1-z}\) sends the lune to the sector \(\Arg(z) \in (-\pi/4, \pi/4)\), since it fixes and must be symmetric about the real axis, and preserves the right angle between the circles at \(z=-1\).
- \(z\mapsto e^{i\pi/4}z\) to rotate this into \(Q_1\).
- \(z\mapsto z^2\) to dilate into \(\HH\).
- \(z\mapsto {z-i\over z+i}\) the standard Cayley map \(\HH\to\DD\).