I have a homework question that I'm stuck on. It asks
Let $\Omega$ be a bounded domain whose boundary consists of two disjoint continua $C_1$ and $C_2$. Let $u(z)$ be the harmonic function on $\Omega$ such that $u(z)=0$ on $C_1$ and u(z)=1 on $C_2$, and let $v(z)$ be a locally defined harmonic conjugate of $u(z)$. Prove that there exists a constant $\lambda>0$ such that $F(z)=e^{\lambda(u(z)+iv(z))}$ is single-valued, and that $F(z)$ maps $\Omega$ conformally to the annulus $\{z \in \Bbb C:1<|z|<e^\lambda\}$.
I have a hint for the problem to first assume without loss in generality that $C_1$ is the unit circle, and $C_2$ is some simple closed analytic curve in $\{z\in \Bbb C:|z|>1\}$. I do understand this part, since we can use the Riemann mapping theorem on the complement of the bounded component of $\Omega^c$. Next, the hint says that $v(z)$ is strictly increasing on $C_1$, and that we should choose $\lambda$ such that the increase of $\lambda v(z)$ around $C_1$ is $2\pi$. I don't see why this is true. Even if I assume this part, I don't see how that shows $F(z)$ is single-valued.
Perhaps it would be better to assume $C_1$ and $C_2$ are both circles and use the ides presented in this problem.
However, I'm not sure if I'm allowed to do that.
To show that $F(z)$ maps $\Omega$ conformally into an annulus, I believe Ahlfors does this in theorem 10 on page 255 of his complex analysis book. However, I don't understand his proof, and I think it might be too complicated for my situation since I only have two components. He talks about conjugate harmonic differentials, and I don't know what that is. Any help would be extremely appreciated!
Best Answer
The function $v$ is strictly increasing along the circle
The Cauchy-Riemann equations say that $\nabla v$ is the vector $\nabla u$ rotated by 90 degrees counterclockwise. So, to show that $\dfrac{\partial v}{\partial\theta }>0$ (tangential derivative) is equivalent to showing $\dfrac{\partial u}{\partial r }>0$ (normal derivative). The latter follows from a strong form of maximum principle (Hopf Lemma), because $u$ attains its minimum everywhere on the unit circle. If you don't know the Hopf lemma, it can be proved from scratch. Indeed, for sufficiently small $\epsilon>0$ the function $u_\epsilon(z) = u(z)-\epsilon\log|z| $ satisfies $u_\epsilon\ge 0$ on both boundary components, and therefore everywhere in $\Omega$. Since $u_\epsilon=0$ on the unit circle, it follows that $\dfrac{\partial u_\epsilon}{\partial r }\ge 0$. Hence, $\dfrac{\partial u}{\partial r }\ge \epsilon$.
Choosing $\lambda$
Now we know that tracing $v$ along the unit circle results in its increase by some $c>0$. Let $\lambda = 2\pi/c$.
The function $F(z)=e^{\lambda(u(z)+iv(z))}$ is single valued
The choice of $\lambda$ ensures that $F$ is singly-defined on the unit circle. Every closed loop in $\Omega$ is homotopic to a circle, and analytic continuation along homotopic loops has the same result. Hence, $F$ is single valued in $\Omega$. (There is a part to be filled in since the unit circle is not a part of $\Omega$ but rather its boundary. One way is to notice that $u$, and subsequently $F$, can be reflected across the unit circle, since $u=0$ on the circle.)