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Fall 2020 - Problem 9

conformal mappings

Consider the following region in the complex plane:

\Omega = \set{x + iy \st 0 < x < \infty \text{ and } 0 < y < \frac{1}{x}}.

Exhibit an explicit conformal mapping $f$ of $\Omega$ onto $\D$ .

Solution.

Let $g\p{z} = z^2$ so that $g\p{x + iy} = x^2 - y^2 + 2ixy$ . Since $\Omega$ is a subset of the upper half-plane, it follows that $g$ is a conformal map onto its image. We claim that $g\p{\Omega} = \set{x + iy \st 0 < y < 2}$ . If $x + iy \in \Omega$ , then by definition,

0 < xy < 1 \implies 0 < \Im{g\p{x + iy}} = 2xy < 2.

Conversely, given $x + iy$ with $0 < y < 2$ , we need to solve

a^2 - b^2 = x \quad\text{and}\quad 2ab = y

for $a + ib \in \Omega$ . Observe that

\begin{aligned} x = a^2 - b^2 = a^2 - \frac{y^2}{4a^2} &\implies 4a^4 - 4a^2x - y^2 = 0 \\ &\implies a^2 = \frac{4x \pm \sqrt{16x^2 + 16y^2}}{8}. \end{aligned}

Since $a > 0$ , we have

a^2 = \frac{x + \sqrt{x^2 + y^2}}{2} > 0,

so such an $a + ib$ exists in $\Omega$ , and so $g$ maps $\Omega$ to a strip. Thus, $h\p{z} = e^{\pi z^2/2}$ maps $\Omega$ to the upper half-plane, so

f\p{z} = \frac{h\p{z} - i}{h\p{z} + i} = \frac{e^{\pi z^2/2} - i}{e^{\pi z^2/2} + i}

is the map we want.