Fall 2014 - Problem 7

Solution.

Let $\Omega$ be the domain in question.

First, consider the map $\phi_1\p{z} = \frac{z}{z - ih}$, which is the Möbius transformation which sends $0 \mapsto 0$, $ih \mapsto \infty$, and $\infty \mapsto 1$. Notice that

so the real line is mapped to the generalized circle containing $0, \frac{1}{1 + h^2}\p{1 + ih}, \frac{1}{1 + h^2}\p{-1 + ih}, 1$. By symmetry, this is the circle centered at $\frac{i}{2}$ of radius $\frac{1}{2}$. Notice that if $t \geq 0$, then

so the ray $\poc{0, ih}$ is mapped to $\poc{0, -\infty}$, and the ray $\poc{ih, \infty}$ is mapped to $\br{1, \infty}$. Hence, $\phi_1$ maps $\Omega$ to $\Omega_1 = \set{\abs{z - \frac{i}{2}} > \frac{1}{2}} \setminus \p{-\infty, 0}$. If we let $\phi_2\p{z} = 2\p{z - \frac{i}{2}}$, then $\phi_2 \circ \phi_1$ maps $\Omega$ to $\Omega_2 = \set{\abs{z} > 1} \setminus \p{-i\infty, -i}$, i.e., the exterior of the unit disk with the bottom ray removed.

Let $\phi_3\p{z} = \frac{1}{z}$. Then $\set{\abs{z} > 1}$ is mapped to $\set{\abs{z} < 1}$, and the ray $\p{-i\infty, -i}$ gets mapped to $\p{0, i}$, so $\phi_3$ maps $\Omega_2$ conformally to $\Omega_3 = \set{\abs{z} < 1} \setminus \p{0, i}$. Let $\phi_4\p{z} = -i$, which maps $\Omega_3$ to $\Omega_4 = \set{\abs{z} < 1} \setminus \p{0, 1}$, which is the unit disk with a segment removed.

Let $\phi_5\p{z} = \sqrt{z}$ where the argument of a number is taken in $\pco{0, 2\pi}$. This maps $\Omega_4$ to $\Omega_5 = \set{\abs{z} < 1} \cap \set{\Im{z} > 0}$, which is the upper half of the unit disk.

Let $\phi_6\p{z}$ be the Möbius transform sending $-1 \mapsto 0$, $1 \mapsto \infty$, and $i \mapsto i$, i.e.,

This maps $\p{-1, 1}$ to $\p{0, \infty}$ and the upper half of the unit circle to $\p{0, i\infty}$, so this maps $\Omega_5$ to $\Omega_6 = \set{\Im{z} > 0} \cap \set{\Re{z} > 0}$. Let $\phi_7\p{z} = z^2$, which maps $\Omega_6$ to $\Omega_7 = \set{\Im{z} > 0}$.

Finally, we can let $\phi_8\p{z} = \frac{z - i}{z + i}$, the Cayley transform, which maps the upper half-plane to the unit disk. Hence, $\phi_8 \circ \phi_7 \circ \cdots \circ \phi_1$ is a conformal mapping from $\Omega$ to the unit disk.