If a sequence of distinct points in a bounded connected open $\Omega$ doesn't converge in $\overline\Omega$, why can we apply analytic continuation?

Let $$\gamma\subset\Bbb C$$ be a closed, simple (if $$\gamma:[a,b]\mapsto\Bbb C$$, $$\gamma$$ is injective on $$(a,b)$$, so the curve doesn't intersect except for $$\gamma(a)=\gamma(b)$$) and piece-wise regular curve and $$z_1,z_2,...$$ is an infinite set of distinct points all inside the domain (call it $$\Omega$$) defined by $$\gamma$$ ($$\partial\Omega=\gamma$$) such that $$z_n$$ does not converge towards any limit in $$\overline\Omega$$.

The analytic continuation principle requires a set a distinct points that converge to a point in the open $$\Omega$$ such that a certain holomorphic function $$f(z_n)=0~\forall n\in\Bbb N$$ and $$z_n\ne\lim z_n~\forall n$$.

And in an exercise about the formula $${1\over 2\pi i}\int_{\gamma}\frac{f'}{f}=\{\text{the number of zeros of } f$$ inside $$\Omega$$ counted with their multiplicity (order)$$\}$$, the following explentation confused me a little bit:

Let $$z_1,...,z_m$$ be in $$\Omega$$ the zeros of $$f$$ of order $$k_1,...,k_m \ge 1$$ respectively. Then $$m$$ is finite because otherwise, by the analytic continuation principle, $$f$$ would be identically zero, which is not the case because $$f$$ is not constant.

• $\Omega$ is not countable. – Kavi Rama Murthy Jan 9 at 10:35
• You don't need analytic continuation for that. If the sequence $(f(z_i))_i$ admits an accumulation point, $f$ is constant and equal to the limit. – James Jan 9 at 13:27
• Is that a theorem? What is it called? – John Cataldo Jan 9 at 13:28
• My advice would be to prove those things assuming $f$ is analytic. If the $a_n$ are distinct $a_n \to a, f(a_n) = 0$ and $f$ is analytic at $a$ then from the values of $f(a_n)$ you can find $f^{(k)}(a)$ for every $k$, so $\forall n, f(a_n) =0 \implies \forall k, f^{(k)}(a) = 0$ and hence $f=0$. The alternative is to show show directly that $f$ analytic implies $f(z) = f(z_0) + C(z-z_0)^k+O((z-z_0)^{k+1})$ so the zeros of $f$ are isolated. Finally the same result applies for $f$ holomorphic once you showed that holomorphic $\implies$ analytic. – reuns Jan 9 at 16:38