$f:A\to C$ is analytic in the open set $A$ and $f'(z_0)\ne 0$ for all $z_0$ in $A$ . Show that $\{Re f(z) | z \in A\}$ is an open set.

Question

My idea to show this is through the inverse function theorem. I know that, since A is an open set, we can use the theorem and show that all of f(A) will also be an open set (with the conditions that we have specifically) but how do I go from there to formally show that just the real part of f(A) is open as well.

Answer 1

I'll use $C = f(A)$ for convenience. We know $C$ is open, so let $x \in \Re(C)$. Then there is some $z \in C$ with $\Re(z) =x$, but $C$ open so there is an $\epsilon > 0$ with $B_\epsilon(z) \subset C$. But the real part of this ball is just an interval $(x-\epsilon, x+\epsilon)$, which we know is entirely contained in $\Re(C)$, hence $\Re(C)$ is open.

Answer 2

The topology on $\mathbb{C}$ is the product topology on $\mathbb{R}^2$ for which the projections are open.

In my opinion, you have done the hard part of showing that $f(A)$ is open using the inverse function theorem.