The derivative of a two-to-one complex function has no zeros.

Question

Let $U$ be open in $\mathbb{C}$ and $f \in H(U)$ such that $f$ is two-to-one on $U$. Prove that $f'$ has no zeros in $U$.

I am thinking about using Cauchy's Integral formula for derivatives, but I do not know what to do with two-to-one condition.

Answer 1

If $f'$ has a zero of order $k$ at $a$, then in some neighborhood of $a$, $f$ can be written as $f(z)=f(a)+((z-a) h(z))^{k+1}$ where $h$ is a holomorphic function with $h(a)\ne 0$.

So, the values near $f(a)$ are attained $k+1$ times in this neighborhood, while $f(a)$ itself is attained exactly once in this neighborhood.

By the 2-to-1 assumption, $k$ must be equal to $1$. Also, the value $f(a)$ must be attained the second time at some point $b$. By the open mapping property, every value near $f(a)$ is also attained in a neighborhood of $b$. But this makes them have at least three preimages, a contradiction.