The Argument Principle
Suppose a function $f$ is meromorphic on an open set that contains a circle $C$ and its interior. Further assume that $f$ has no zeroes on $C$ (but may have zeroes in the interior of $C$). Then,
$$\frac{1}{2\pi i} \oint_C \frac{f'(z)}{f(z)} dz \ =$$
the number of zeroes of $f$ in the interior of $C$ (counted with multiplicity) minus the number of poles of $f$ in the interior of $C$ (counted with multiplicity).
Rouche's Theorem
Suppose that $f$ and $g$ are holomorphic functions on a domain and let $C$ be a circle whose interior also lies in the domain. If $|f(z)-g(z)| \lt |f(z)|$ for all $z$ on $C$, then $f$ and $g$ have the same number of zeroes in the interior of $C$, counting multiplicities.
My Question
Why does the Argument Principle require $f$ to have no zeroes on $C$, whereas this requirement is relaxed for Rouche's Theorem?
Thank you.
Best Answer
In the Argument Principle: If $f$ has a zero $z_0 \in C$, then $z_0$ is a pole for the quotient $f'/f$. Therefore you cannot define $\int_C f'(z)/f(z)\,dz$.
In Rouche's Theorem: The inequality $|f-g|<|f|$ on $C$ implies that $f$ and $g$ have no zeroes on $C$. In fact, if $g$ (resp. $f$) had a zero $z_0 \in C$, then $|f(z_0)|<|f(z_0)|$ (resp. $|g(z_0)|<0$), which is a contradiction.