Find a conformal map \[ \DD\sm \left[ {1\over 2}, 1\right) \to \DD .\]
Solution
The picture:
In steps:
\(f_1\): send \(1/2\to 0\) and \(1\to 1\) in order to lengthen the slit. Mobius transformations preserve lines, so take \(f_1(z) = {z-1/2 \over 1-1/2}\). New domain: \(\DD\sm [0, 1)\).
\(f_2\): send \(\DD\to \HH\) and keep track of the slit. Use the standard inverse to the Cayley transform, \(f_2(z) = i{1-z\over 1+z}\). New domain: \(\HH\sm [i, i\infty)\), noting that \(\HH\) is the open half-plane.
\(f_3\): rotate the slit so it becomes a line segment from \(-1\) to \(1\) passing through \(\infty\). Use \(f_3(z) = z^2\). New domain: \(\CC\sm\qty{(-\infty, 1] \union [1, \infty) }\)
\(f_4\): move the slit off to infinity to be left with a ray, so send \(-1\to \infty\) and \(0\to 0\). Take \(f_4(z) = {z\over z+1}\), the new domain is \(\CC\sm [0, \infty)\).
\(f_5\): fold it back. Branch cut log along \([0, \infty)\) to define \(f_5(z) = z^{1\over 2}\), so the new domain is \(\HH\).
\(f_6\): apply the standard Cayley transform \(f_6(z) = {i-z\over i+z}\)