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.
Best Answer
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.