Find a conformal map: \[ \DD \sm \ts{\abs{z - {1\over 2}} = {1\over 2} } \to \DD .\]
Solution
The picture:
Key insight: send the point of tangency to \(\infty\) to get parallel lines. Send \(z\mapsto {1+z \over 1-z}\) to send \(-1\to 0, 0\to 1, 1\to \infty\) to get a vertical strip.
Rotate and dilate: \(z\mapsto i\pi z\) to get a strip \(\RR \cross i(0, \pi)\).
Standard exponential \(z\mapsto e^{iz}\) sends this to \(\HH\).
Standard Cayley \(z\mapsto {z-i \over z+i}\) sends this to \(\DD\).