Spring 2012 - Problem 8

Solution.

Notice that $\p{x + iy}^2 = \p{x^2 - y^2} + 2xyi$. Thus, for $z \in \Omega$, we see that $0 \leq \Im{z} < 2$ and $\Re{z} \in \R$. Conversely, given $a + ib$ such that $0 \leq b < 2$, we can try to solve $z^2 = a + ib$, which means

By inspection, we see that

Since $b \geq 0$, we see that $x$ and $y$ satisfying the above will have the same sign. Since $b < 2$, the second equation implies $xy < 1$, so $z^2 = a + ib$ has a unique solution in $\Omega$ for any $a + ib$ in the strip $S = \set{0 \leq \Im{z} < 2}$. Thus, the map $\func{\phi}{\Omega}{S}$, $z \mapsto z^2$ is a conformal map.

Notice that $\phi$ extends continuously to the boundary. Moreover, if $x = 0$, then $\phi\p{x + iy} = -y^2$ and if $y = 0$, then $\phi\p{x} = x^2$. In either case, $\phi$ maps the the axes to the real line. Also, if $xy = 1$, then $\phi\p{x + iy} = x^2 - y^2 + 2i$, so the curve $xy = 1$ in the upper-half plane is mapped into the line $y = 2i$.

In light of $\phi$, it suffices to find an unbounded harmonic function which vanishes on $y = 0$ and $y = 2i$. One such choice is $v\p{z} = \Im{e^{\pi z}}$, since $e^{\pi z}$ is purely real on these lines. It is also unbounded, since $v\p{x + \frac{i}{4}} = e^{\pi x}\sin\frac{\pi}{4} \xrightarrow{x\to\infty} \infty$. Thus,

works.