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Fall 2019 - Problem 11

conformal mappings

Find a conformal map of the domain

D = \set{z \in \C \mid \abs{z - 1} < \sqrt{2},\ \abs{z + 1} < \sqrt{2}}

onto the open unit disc centered at the origin. It suffices to write this map as a composition of explicit conformal maps.

Solution.

$D$ is the unit of two circles which intersect at $i$ and $-i$ at an angle of $\frac{\pi}{2}$ . First, we translate so one of the intersection points is at the origin:

\func{\phi_1}{D}{\Omega_1 = D + i}, \quad \phi_1\p{z} = z + i.

Next, we send the other intersection point $2i$ to $\infty$ in a symmetric way. More specifically, we want $0 \mapsto 0$ , $i \mapsto i$ , and $2i \mapsto \infty$ , so we get

\func{\phi_2}{\Omega_1}{\Omega_2}, \quad \phi_2\p{z} = \frac{iz}{2i - z}.

Observe that

\phi_2\p{z} = \frac{iz\p{-2i - \conj{z}}}{\abs{2i - z}^2} = \frac{2z - i\abs{z}^2}{\abs{2i - z}^2} = \frac{2\Re{z}}{\abs{2i - z}^2} + i\p{\frac{2\Im{z} - \abs{z}^2}{\abs{2i - z}^2}}.

In other words, because $D + i$ was symmetric about $\Re{z} = 0$ , it follows that $\Omega_2$ is also symmetric about $\Re{z} = 0$ . Since angles are preserved by conformal mappings, we see that the circles are mapped to half-lines containing $0$ which intersect at an angle of $\frac{\pi}{2}$ . Again by symmetry, we see that $\Omega_2 = \set{\frac{\pi}{4} < \Arg{z} < \frac{3\pi}{4}}$ , where we take the standard branch $\p{0, 2\pi}$ . After the rotation

\func{\phi_3}{\Omega_2}{\Omega_3 = \set{0 < \Arg{z} < \frac{\pi}{2}}}, \quad \phi_3\p{z} = e^{-i\pi/4}z,

we get the first quadrant. Thus, we can complete the problem with

\begin{aligned} \func{\phi_4&}{\Omega_3}{\Omega_4 = \H}, \quad &&\phi_4\p{z} = z^2 \\ \func{\phi_5&}{\Omega_4}{\Omega_5 = \D}, \quad &&\phi_5\p{z} = \frac{z - i}{z + i}, \end{aligned}

so the desired conformal mapping is $\phi_5 \circ \cdots \circ \phi_1$ .