Prove that a non-constant harmonic function is an open map.

Question

I'm trying to solve the following exercise of the book Functions of one complex variable, John B. Conway on page 255:

4. Prove that a harmonic function is an open map. (Hint: Use the fact that the connected subsets of $\mathbb{R}$ are intervals.)

I assumed the harmonic functions $u: U \rightarrow \mathbb{R}\ (U \subset \mathbb{C} $ is open) of the exercise are not constant. If U is connected using the hint, the solution is relatively simple by Maximum Principle (or Minimum).

Maximum principle: Be $U$ open, connected and $ u: U \rightarrow \mathbb{R} $ harmonic. If exists $ a \in U $ such that $u(z)\leq u(a),\ \forall z\in U$, then u is constant.

But the case where $U$ is not connected I could not solve. This exercise is correct? If not, is there any counterexample?

Thank you

Answer 1

Hint: Every open subset of $\mathbb{C}$ is a union of (connected) balls.

Answer 2

If $U$ in a connected open set and $u\colon U\to \mathbb R$ is harmonic, take an open disc $D\subseteq U$. So on $D$ there exists an holomorphic function $F\colon D\to \mathbb C$ such that $Re(F)=u$.

Since $F$ is holomorphic with $D$ open and connected, if $u$ is not constant, then $F$ is not constant, so $F$ is an open map, (Open Map Theorem in complex analysis). Note that the projection $p\colon \mathbb R\times\mathbb R\to \mathbb R$ is a continuous open surjection, so if $V$ is an open set un $D$ (and hence in $U$) , then $u(V)=p(F(V))$ is open.