# Problem regarding zeros and poles of a meromorphic function inside a circle (Argument Principle)

Basically, I have confusion when I am going through the Argument Principle from the book of Stein & Shakarchi.

Suppose, f is meromorphic in a region $$\Omega$$ and $$C$$ be a circle inside $$\Omega$$ containing its interior. Let, $$z_1,z_2,...,z_m$$ and $$w_1,w_2,...,w_n$$ be the zeros and poles of $$f$$ inside $$C$$ respectively and $$f$$ doesn't have any zero or pole on $$C$$ (i.e. on the boundary).

In the book, the author proves that if a holomorphic and not identically zero function $$f$$ has a zero (say $$z_0$$) in $$\Omega$$. Then $$\exists$$ a neighbourhood (say $$D(z_0)$$) of $$z_0$$ in $$\Omega$$ and a unique $$p\in\Bbb{N}$$ such that $$f(z)=(z-z_0)^pg(z)\ \forall z\in D(z_0)$$ where $$g$$ is a non-vanishing holomorphic function on $$D(z_0)$$.
As a corollary, he also shows that if $$f$$ has a pole (say $$w_0$$) in $$\Omega$$. Then $$\exists$$ a neighbourhood (say $$D(w_0)$$) of $$w_0$$ in $$\Omega$$ and a unique $$q\in\Bbb{N}$$ such that $$f(z)=(z-w_0)^{-q}h(z)\ \forall z\in D(w_0)$$ where $$h$$ is a non-vanishing holomorphic function on $$D(w_0)$$.
So, these statements holds locally i.e. existance of the functions $$g$$ or $$h$$ is in a neighbourhood of the zero or pole. And even these are proved for either one pole or one zero separately.
But to prove the result of Argument principle i.e. $${1\over2\pi i}\int_{C} {f'(z)\over f(z)}dz=$$(#of zeros with multiplicity)$$-$$(#of poles with multiplicity), we need to establish the fact that
$$f(z)=\prod_{i=1}^{m}(z-z_i)^{p_i}\prod_{j=1}^{n}(z-w_j)^{-q_j}G(z)\ \forall z\in C\cup\operatorname{int}C$$ and $$G$$ is non-vanishing, holomorphic in $$C\cup\operatorname{int}C$$.
So, that I can get holomorphic $$G'/G$$ on $$C\cup\operatorname{int}C$$ and $$\int_{C}{G'(z)\over G(z)}dz=0$$
But I can't prove this. Can anybody give me explanation to remove the confussion. Thnaks for assistance in advance.

If we define$$G(z)=\prod_{j=1}^m(z-z_j)^{-p_j}\prod_{k=1}^n(z-w_k)^{q_k}f(z),$$for each $$z\in\Omega$$, then $$G$$ is a holomorphic function without zeros or poles and therefore $$\frac{G'}G$$ is a holomorphic function.
• I have another query. Number of poles of $f$ inside $C$ is finite due to meromorphic property and $C\cup\operatorname{int}C$ is compact hence BW-compact. But what about number of zeros? Can there be infinitely many zeros of meromorphic $f$ inside $C$? – Biswarup Saha May 1 at 8:48
• No, because $C\cup\operatorname{int}C$ is compact and, in a compact set, every infinite set has a accumulation point. – José Carlos Santos May 1 at 8:53
• Yes, it is right. As per the definition of Meromorphic function the set of points where $f$ had poles must not contain any accumulation point in domain. But what about zeros? I'm asking about the zeros – Biswarup Saha May 1 at 9:02
• Me too. If $z_0$ is an accumulation point of the set of zeros, then $z_0\in C$ or $z_o\in\operatorname{int}C$. If $z_0\in\operatorname{int}C$, then $f$ is the null function. And you don't have zeros in $C$. – José Carlos Santos May 1 at 9:09