# Conformal map of doubly connected domain into annulus.

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!

### 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.)