In my book, Rouche's theorem is stated as follow:
$f,g$ are meromorphic in a simply connected domain $D$,
$C$ is a simple closed contour in $D$ such that $f$ and $g$ have no zeros or poles on $C$,
and $|f(z)+g(z)| < |f(z)|+|g(z)|$ for all $z \in C$.
Then $Z_f-P_f=Z_g-P_g$. Here, $Z_f$ is the number of zeros of $f$ that lie inside $C$ and $P_f$ is the number of poles that lie inside $C$.
Usually, I see the different type of Rouche's theorem as follows:
If $f$ and $h$ are functions that are analytic inside and on a simple closed contour C and if the strict inequality $|h(z)|<|f(z)|$ holds at each point on C, then $f$ and $f+h$ must have the same total number of zeros inside C.
Don't we need this kind of assumption "$f(z) \neq 0$ and/or $h(z)\neq 0$ for $z \in C$" in order to prove the theorem?
Would you give me any comment about it? Thanks in advance!
If $|f(z)+h(z)-f(z)| = |h(z)| < |f(z)| \le |f(z)|+ |f(z)+h(z)|$ for $z \in C$ then we see that $f$ has no zeros in $C$.
If $f(z)+h(z)=0$ then $|h(z) = |f(z)|$, so we see that $f+h$, $f$ have no zeros in $C$.