Assumptions on domain of a meromorphic function

Question

I'm studying complex analysis from Ash's book:

The argument principle for meromorphic stated here is:

4.2.6 Definition

For $f$ meromorphic on $\Omega$, let $Z(f)$ denote the set of zeros of $f$, and $P(f)$ the set of poles of $f$. If $z\in Z(f)\cup P(f)$, let $m(f,z)$ be the order of the zero or pole of $f$ at $z$.

4.2.7 Argument Principle for Meromorphic Functions

Suppose $f$ is meromorphic on $\Omega$. Then for any closed path (or cycle) $\gamma$ in $\Omega \backslash(Z(f)\cup P(f))$ such that $\gamma$ is $\Omega$-homologous to $0$, we have $$n(f\circ \gamma, 0) = \sum_{z\in Z(f)}n(\gamma,z)m(f,z) - \sum_{z\in P(f)}n(\gamma, z)m(f,z).$$

How do I know the sum on $Z(f)$ is well defined? Must that set be countable? Is $\Omega$ always defined to be connected for a meromorphic function?

Answer 1

If $[a,b]$ is the domain of $\gamma$, then the set$$\left\{z\in\mathbb{C}\,|\,n(\gamma,z)\neq0\right\}\cup\gamma\bigl([a,b]\bigr)$$is compact. Therefore, by the identity theorem, you can have $n(\gamma,z)\neq0$ for finitely many zeros of $f$ only.

Answer 2

Let $\Omega$ be a bounded subset of $\Bbb{C}$: then by Bolzano-Weierstrass, an infinite number of zeroes would imply that $f$ vanishes everywhere, and thus you couldn't even define $\gamma$, which exists by hypothesis, hence the contradiction. Thus the concerned sum is actually finite.

On the other hand, if $\Omega$ is not bounded, $Z(f)$ could be infinite but then the subset on which the winding number is not zero would again be finite by the same compactness argument (remember a simple closed curve cuts two region, one of which is unbounded and formed by the points on which $n(\gamma, z)=0$).