Let \(D\) be the domain in the complex plane consisting of the interior of the unit disk, with the subdisk of radius \(\frac12\) and center \(\frac12\) removed. Construct an analytic function that maps \(D\) conformally onto the first quadrant \(\operatorname{Re} z>0\), \(\operatorname{Im} z>0\). (A figure appears in the original exam.)
Solution
Goal. Construct a conformal map from the unit disk minus the subdisk of radius \(\tfrac12\) centered at \(\tfrac12\) onto the first quadrant.
1. The domain \(D\) is the unit disk \(\DD\) with the closed disk \(\theset{z : \abs{z - \tfrac12} \le \tfrac12}\) removed.
the subdisk of radius \(\tfrac12\) centered at \(\tfrac12\) is \(\theset{z : \abs{z - \tfrac12} < \tfrac12}\), tangent to the unit circle at \(z = 1\).
2. The Möbius map \(w = \frac{1+z}{1-z}\) sends \(\DD\) to the right half-plane and the subdisk to a half-plane.
2.1. \(w = \frac{1+z}{1-z}\) maps \(\DD\) conformally onto \(\theset{\Re w > 0}\).
this is the standard Cayley transform; it sends the unit circle to the imaginary axis and \(0\) to \(1\).
2.2. The subdisk \(\abs{z - \tfrac12} = \tfrac12\) maps to the vertical line \(\Re w = 1\).
the circle \(\abs{z - \tfrac12} = \tfrac12\) passes through \(z = 1\) (mapped to \(\infty\)) and \(z = 0\) (mapped to \(1\)); a circle through \(1\) maps to a line, and since it is tangent to the unit circle at \(1\), its image is the vertical line through \(w = 1\), i.e. \(\Re w = 1\).
2.3. The image of \(D\) is the half-strip \(\theset{w : 0 < \Re w < 1}\).
the interior of the subdisk maps to \(\theset{\Re w > 1}\), so removing it from the right half-plane leaves \(0 < \Re w < 1\).
3. Map the half-strip to the first quadrant.
3.1. \(u = \sin(\pi w)\) maps the strip \(\theset{0 < \Re w < 1}\) onto the upper half-plane.
\(\sin(\pi w)\) maps the vertical strip \(0 < \Re w < 1\) conformally onto the upper half-plane (the sine function maps the strip to the upper half-plane via the exponential).
3.2. \(v = \sqrt u\) (principal branch) maps the upper half-plane onto the first quadrant.
the principal square root maps the upper half-plane onto the first quadrant \(\theset{\Re v > 0, \Im v > 0}\).
4. The composition is the required map.
4.1. \(f(z) = \sqrt{\sin\qty(\pi \frac{1+z}{1-z})}\) maps \(D\) conformally onto the first quadrant.
compose 2.1, 3.1, and 3.2; each step is conformal.
5. Q.E.D.
4.1 is the requested conformal map.