Let $f$ be a nonconstant smooth function on $\mathbb C$ such that the set $\Gamma$ given by $\Gamma = \{z \in \mathbb C : |f(z)| = 7\}$ is a smooth simple closed curve in $\mathbb C$. Denote $G$ the bounded region enclosed by $\Gamma$. Assume $f$ is holomorphic in $G$.
Assume in addition that $\Gamma$ contains no zero of $f' = \partial f/\partial z$. Suppose $f$ has $m$ zeros counting multiplicities in $G$. How many zeros counting multiplicities does $f'$ have in $G$? Prove your assertion.
I assume there may be an argument with argument principle (no pun intended), and/or perhaps some clever use of Rouche's Theorem. My guess is that the answer is $m - 1$ after trying some functions, but there could be something else hidden there.
For what it's worth, we have already shown earlier that $f$ has at least one zero, which follows from the maximum princple.
Any help would be wonderful.
