Let $D$ be a star domain and $u: D \to \mathbb{R}$ harmonic. Assume there exists a $\zeta \in D$ s.t. $u(z) \leq u(\zeta)\; \forall z \in D$. Show that $u$ is constant.
Harmonic functions on star domains always have harmonic conjugates. So there exists a harmonic $v:D\to \mathbb{R}$ s.t. $f = u+iv$ is holomorphic on $D$. Now because $u(z) \leq u(\zeta)$ we have $\text{Re}(g) \geq 0$ where $g(z) := f(\zeta) - f(z)$. Now I'd like to show that $\text{Re}(g) > 0$ so that there exists a holomorphic $h:D\to \mathbb{C}$ s.t. $g(z) = \exp(h(z))$. Am I going the wrong direction or is there an easier way to show this?
The exponential function can be used, but not as you tried but by considering the function $h(z) = e^{f(z)}$. $h$ is holomorphic in $D$, and $$ |h(z)| = e^{\operatorname{Re}f(z)} = e^{u(z)} \le e^{u(\xi)} \, , $$ with equality for $z = \xi$. The maximum modules principle for holomorphic functions implies that $h$ is constant. It follows that $f$, and consequently $u$, are constant.