The argument theorem states:
If f is meromorphic in an open set containing a circle C and its interior and if f has no poles and never vanishes in C, then the number of zeros of f inside C minus the number of poles of inside C is equal to $\frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz$
Rouche's Theorem:
If functions $f$ and $g$ are holomorphic in an open set containing a circle $C$ and its interior, and if $|f(z)|>|g(z)|$ for all $z \in C$, then $f(z)$ and $f(z)+g(z)$ have the same number of zeros within the circle $C$.
Then in the proof of Rouche's theorem we can assume that $f$ and $g$ are holomorphic in an open set containing a circle $C$ and its interior, and $|f(z)|>|g(z)|$ for all $z \in C$.
Now here is where we need to apply the argument principle: We know that f is meromorphic (because all holomorphic functions are meromorphic) and we know that f never vanishes on C (because $|f(z)|>|g(z)|$ for all $z \in C$), but how do we know that there are no poles on C?
Also, the next line of the proof in my textbook states that by the argument principle the number of zeros in C=$\frac{1}{2\pi i}\int_C \frac{f'(z)}{f(z)}dz$. Why is it not the number of zeros minus the number of poles? By this point we should have shown that there are no poles on C, but are there also no poles inside of C?
Any guidance is helpful.