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!

  • 1
    $\begingroup$ You wrote “$f$ and $g$ have no zeros or poles on $C$” in your first version. In the second version “$\lvert h(z) \rvert < \lvert f(z) \rvert$” implies that $f$ has no zero on $C$. $\endgroup$
    – k.stm
    Nov 28, 2017 at 5:52

1 Answer 1


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 $C$.


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