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.

  • $\begingroup$ 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 $C$? $\endgroup$ – Biswarup Saha May 1 at 8:48
  • $\begingroup$ No, because $C\cup\operatorname{int}C$ is compact and, in a compact set, every infinite set has a accumulation point. $\endgroup$ – José Carlos Santos May 1 at 8:53
  • $\begingroup$ 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 $\endgroup$ – Biswarup Saha May 1 at 9:02
  • $\begingroup$ 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 $C$. $\endgroup$ – José Carlos Santos May 1 at 9:09
  • $\begingroup$ Yes Got it, thanks $\endgroup$ – Biswarup Saha May 1 at 9:12

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