A problem related to Schwarz reflection principle

Question

Let $\Omega$ be a bounded domain in $\mathbb{C}$. Suppose there is a function $f$ which is analytic in $\Omega$ except a simple pole at $a\in\Omega$, such that $(z-a)f(z)$ is continuous on $\bar{\Omega}$ with $f(z)=\bar{z}$ on $\partial\Omega$. Prove that the function $g(z)=(z-a)f(z)-(z-a)\bar{a}$ is constant. What is $\Omega$?

It should be consistent with the problem if we assume $\Omega$ to be the unit disc $\mathbb{D}$, then we may use the reflection principle with respect to $\mathbb{D}$ to extend $g$ to an entire function and then conclude using the Liouville theorem that $g$ is a constant. But how can we see that this is the most general case that can happen?

Answer 1

The function $g$ is holomorphic on $\Omega$ and continuous on $\overline{\Omega}$. Its boundary values are real and positive:

$$f(z)(z-a) - \overline{a}(z-a) = \overline{z}(z-a) - \overline{a}(z-a) = \overline{z-a}(z-a) =\lvert z-a\rvert^2.$$

If $g$ weren't constant, $g(\Omega)$ would be an open subset of $\mathbb{C}$, hence $v := \operatorname{Im} g$ would attain nonzero values in $\Omega$. Since $v$ is continuous on the compact set $\overline{\Omega}$, it would attain a strictly positive maximum or a strictly negative minimum, necessarily at a point $z_0 \in \Omega$. But $v$ is a real-valued harmonic function, by assumption non-constant, hence $v$ has no local maxima or minima, contradicting the assumption.

Thus $g$ is constant, $g(z) \equiv r^2$. Thus $\Omega$ is a bounded domain whose boundary is contained in $C_r(a) = \{ z\in \mathbb{C} : \lvert z-a\rvert = r\}$, and that means $\Omega = D_r(a) = \{ z \in \mathbb{C} : \lvert z\rvert < r\}$.