TITLE: $\textrm{Re}(f'(z)) > 0$ but $f$ is not injective.
QUESTION [5 upvotes]: I'm looking for an example of a simply connected open set in $\mathbb{C}$ and a holomorphic function $f \colon \Omega \to \mathbb{C}$ such that
$$\textrm{Re}(f'(z)) > 0$$
for all $z \in \Omega$ but $f$ is not injective.
One can show that under these conditions, $f$ must be injective if $\Omega$ is convex. I'm not sure how to come up with a counterexample for a nonconvex set though. It seems for many elementary functions such as $z^2, e^z$ that the set where $\textrm{Re}(f'(z)) > 0$ has convex connected components.

REPLY [4 votes]: Consider $f(z)= z^{3/2}$ where the branch cut is along the negative real axis, that is, $-\pi < arg(z) < \pi$. This is clearly a nonconvex domain. $f$ is not injective as $z^3$ is not injective and we have $f'(z) = \frac{3}{2} z^{1/2}$, so $\text{Re}(f'(z)) > 0$.