I've been studying from Basic complex analysis by Mardsen and Hofmann, and I came across the proposition and corollary on page 212:
Proposition 3.2.9 Suppose $f: \Omega \rightarrow \mathbb{C}$ is analytic on an open set $\Omega$ in $\mathbb{C}$ and that $c \in \Omega$. Let $D(c;r)$ be an open disk centred at c and contained in $\Omega$ and suppose $f(c)=0$. Then exactly one of the two things must occur:
$f(z)=0$ for every z in $D(c;r)$
There is an integer n such that $$ f(c)= f'(c) = f''(c)=...=f^{n-1}(c)=0 \; and \;f^{n}(c)\ne 0 $$
Corollary 3.2.9: Suppose $f: \Omega \rightarrow \mathbb{C}$ is analytic on an open set $\Omega$ in $\mathbb{C}$ and that $c\in\Omega$. If there is a sequence $z_1$,$z_2$,$z_3$,... of distinct points in $\Omega$ such that $z_k \rightarrow c $ as $k \rightarrow \infty $ and $f(z_k)=0$ for each k, then $f(z) = 0$ for each z in the largest open disk centred at c and contained in $\Omega$
I don't quite understand how to begin proving the corollary. Could I get some hints?
Suppose the conclusion of the corollary is false. Then by the Proposition we can write $f(z)=f^{(n)} (c)(z-c)^{n}/n!+f^{(n+1)} (c)(z-c)^{n+1}/(n+1)!+...$. This gives $f(z)=(z-c)^{n} g(z)$ where $g$ is continuous and $g(c) \neq 0$. [ $g$ is continuous because it is the sum of a convergent power series]. It follows that $g(z) \neq 0$ for $|z-c|$ sufficiently small. But then $f(z_k)$ cannot be $0$ for $k$ sufficiently large.