Let \(R\) be the intersection of the right half-plane and the outside of the circle \(\abs{z - {1\over 2}} = {1\over 2}\) with the line segment \([1, 2]\) removed, i.e. \[ R = \ts{z\in \CC\st \Re(z) > 0,\,\, \abs{z-{1\over 2}} > {1\over 2} } \sm \ts{z \da x+iy \st 1\leq x\leq 2,\,\, y=0} .\] Find a conformal map from \(R\) to \(\HH\) the upper half-plane.
- Blow up the point of tangency: inverting through a circle sends inner circles to lines, fixes the real line, and preserves regions between curves. E.g. the image of \(\abs{z-i/2} =2\) is \(\ts{ \Im(z) = 2}\)
- So \(z\to 1/z\) maps the region into a half-strip.
Solution
Note: this seems unusually difficult for a UGA question! This is a bigon with one vertex \(z_2 = 0\), so send it to infinity and keep track of the slit.
In steps:
- Use \(z\mapsto 1/z\) to get a vertical strip \(0<\Re(z) < 1\), where the slit maps to \([1/2, 1]\).
Shift and dilate with \(z\mapsto \pi(z-1/2)\) to get \(-\pi/2<\Re(z) < \pi/2\), where the slit is now at \([0, \pi/2]\).
Apply \(z\mapsto \sin(z)\) to get \(\CC\sm[0, 1]\).
Move the slit with \(z\mapsto 4(z-1/2)\) to get \(\CC\sm[-2, 2]\).
Apply \(z\mapsto z+\inverseof{z}\) to get \(\CC\sm\DD\), where the slit is now eliminated.
Use \(z\mapsto 1/z\) to get \(\DD\)
Use the inverse Cayley map \(z\mapsto i{1-z\over 1+z}\) to get \(\HH\).