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.