Find a conformal map from \(D = \{z :\ |z| < 1,\ |z - 1/2| > 1/2\}\) to the unit disk \(\Delta=\{z: \ |z|<1\}\).
Solution
1. \(D\) is the unit disk with the disk \(\{|z - 1/2| \le 1/2\}\) removed; it is a simply connected region bounded by two circles tangent at \(z = 1\).
the two circles \(|z| = 1\) and \(|z - 1/2| = 1/2\) are tangent at \(z = 1\).
2. The Möbius map \(w = \frac{1}{1-z}\) sends \(z = 1\) to \(\infty\) and maps the two tangent circles to two parallel lines.
a Möbius map sending the tangency point to \(\infty\) turns the two tangent circles into parallel lines.
3. Under \(w = \frac{1}{1-z}\), the region \(D\) maps to a vertical strip.
3.1. The circle \(|z| = 1\) maps to the line \(\operatorname{Re} w = 1/2\).
for \(|z| = 1\), \(w = \frac{1}{1-z}\) satisfies \(\operatorname{Re} w = 1/2\) (since \(\frac{1}{1-e^{i\theta}} = \frac{1}{2} + \frac{i}{2}\cot(\theta/2)\)).
3.2. The circle \(|z - 1/2| = 1/2\) maps to the line \(\operatorname{Re} w = 1\).
for \(z = 1/2 + (1/2)e^{i\theta}\), \(w = \frac{1}{1-z} = \frac{1}{1/2 - (1/2)e^{i\theta}} = \frac{2}{1 - e^{i\theta}}\), whose real part is \(1\).
3.3. Hence \(D\) maps to the strip \(\{w : 1/2 < \operatorname{Re} w < 1\}\).
1 and 2.
4. Map the strip \(\{1/2 < \operatorname{Re} w < 1\}\) to the unit disk.
4.1. First translate and scale to the strip \(\{0 < \operatorname{Re} \zeta < 1\}\) via \(\zeta = 2w - 1\).
\(\zeta = 2w - 1\) maps \(\operatorname{Re} w = 1/2\) to \(\operatorname{Re} \zeta = 0\) and \(\operatorname{Re} w = 1\) to \(\operatorname{Re} \zeta = 1\).
4.2. The map \(\eta = e^{\pi i \zeta}\) sends the strip \(\{0 < \operatorname{Re} \zeta < 1\}\) to the upper half-plane.
If \(\zeta=x+iy\) with \(0<x<1\), then \[ e^{\pi i\zeta}=e^{-\pi y}e^{i\pi x}, \] whose argument lies in \((0,\pi)\). Conversely every point of \(\HH\) has a unique logarithm with argument in \((0,\pi)\), so this is a biholomorphism from the vertical strip onto \(\HH\).
4.3. The map \(\xi = \frac{\eta - i}{\eta + i}\) sends the upper half-plane to the unit disk.
the Cayley transform.
5. Composing, a conformal map \(D \to \Delta\) is \[z \mapsto \frac{e^{\pi i(2/(1-z) - 1)} - i}{e^{\pi i(2/(1-z) - 1)} + i}.\]
3 and 4.
6. Q.E.D.
5.