I am asked to prove the following: Let $\Omega\subseteq\mathbb C$ a convex open subset and let $f$ be a holomorphic function in $\Omega$ such that $|f'(z)-1|<1$ for all $z\in\Omega$. Then $f$ is injective.
I proved this result, but then, I am asked to give a counterexample of a holomorphic $f$ in a non-convex connected open subset $\Omega\subseteq\mathbb C$ such that $|f'(z)-1|<1$ for all $z\in\Omega$ but $f$ is not injective.
I tried to think of $f$ in a way that involves the logarithm function, since the discontinuity of the argument may result in $f$ not being injective. I thought about the function $f(z)=\log(z)+z$, fixing the argument in $[-\pi,\pi]$. We have $$ |f'(z)-1|=|1/z+1-1|=|1/z|<1\iff |z|>1.$$
So we may choose $\Omega=\{z\in\mathbb C\,\colon |z|>1\}$. This set is open, not convex and connected. The problem is I don't know how to check that $f$ is not injective in $\Omega$; in other words, how to determine the points $z_1,z_2\in\Omega$ such that $f(z_1)=f(z_2)$. I would be very glad if someone can help me with that. Thank you.
Let's consider $f(z)=z^{3/2}$ where $z \notin (-\infty,0]$ defined as usual by the principal branch of the logarithm so in particular if $z=re^{i\theta}, |\theta| < \pi, r>0$ then $f(z)=r^{3/2}e^{3i\theta/2}$. Clearly $f(re^{2\pi i/3})=r^{3/2}e^{\pi i}=-r^{3/2}=f(re^{-2\pi i/3})$ for any $r>0$.
Now $f'(z)=\frac{3z^{1/2}}{2}$ and we notice that $|a-1|<1$ is equivalent to $2\Re a >|a|^2$ so using that $2\Re f'(z)=3\sqrt r \cos (\theta/2)$ and $|f'(z)|^2=9r/4$ we get that $|f'(z)-1|<1$ is satisfied as long as $\cos (\theta/2) >3\sqrt r /4$ so if we consider the domain $D$ given by $|\theta| < 3\pi/4, 0<r<(\frac{4 \cos 3\pi/8}{3})^2$ we satisfy the relation $|f'(z)-1|<1$ and $f$ is not injective since $re^{2\pi i/3}, re^{-2\pi i/3} \in D$ for any $r$ as above and $f(re^{2\pi i/3})=f(re^{-2\pi i/3})=-r^{3/2}$