$\textrm{Re}(f'(z)) > 0$ but $f$ is not injective.

Question

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.

How about $f(z) = z^{3/2}$ on the domain $–\pi < \arg(z) < \pi$? — Parthiv Basu, Jul 19 '19 at 11:58
@ParthivBasu Nice, you should write an answer with a few more details. — zhw., Jul 19 '19 at 20:05

score 4 · Accepted Answer · answered Jul 19 '19 at 21:26

4

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 $\text{Re}(f'(z)) > 0$.

answered Jul 19 '19 at 21:26

Parthiv Basu

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I think there are theorems that say for every nonconvex domain you could find such functions. And that if a domain satisfies the univalent function property then it is almost convex or something. – Parthiv Basu Jul 19 '19 at 21:28

$\textrm{Re}(f'(z)) > 0$ but $f$ is not injective.

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