I want to prove this statement:
Let $f,g$ be analytic on $D(0,2)$. If $f(z)g(z) = 0$ when $z = 1/n$ for $n \in \mathbb{N}$, then either $f \equiv 0$ or $g \equiv 0$ in $D(0,2)$.
My attempt:
Define $z_n = 1/n$. Then for each $n$, $f(z_n)g(z_n) = 0$ and as $n \to \infty$, $z_n \to 0$ so $0$ is a limit point of the set $Z_{fg} = \{z \in D(0,2) : f(z)g(z) = 0\}$. But $Z_{fg} = Z_f \cup Z_g$ where $Z_{f} = \{z \in D(0,2) : f(z) = 0\}$ and $Z_{g} = \{z \in D(0,2) : g(z) = 0\}$. So $0$ is a limit point of either $Z_f$ or $Z_g$. Therefore, by the Identity Theorem, either $f \equiv 0$ or $g \equiv 0$ in $D(0,2)$.
