Statement of the Principle

Let $f$ be an analytic function on a region $D$. Then if $f = 0$ on a set of points $\{z_k\} \subseteq D$ with an accumulation point in $D$, then $f$ is identically zero on $D$.

Proof

We prove the contrapositive. Suppose that $f$ is not identically zero on $D$. Let $V(f) = \{z \in D \colon f(z) = 0\}$. We will show that this set cannot have any accumulation points.

Suppose that $z_0 \in V(f)$. Then since $f$ is analytic, it has a power series expansion $f(z) = f(z_0) + c_1 z + \cdots$ convergent for $|z-z_0| < r$ for some $r > 0$. There must be a first nonzero coefficient since $f$ is not constant, say $c_k$. We factor: \[f(z) = (z - z_0)^k h(z)\] where $h(z)$ is analytic and $h(z_0) \neq 0$.

By the analyticity and hence continuity of $h$, there exists some $\delta > 0$ such that $|z-z_0| < \delta$ implies $h(z) \neq 0$. We also have that for $0 < |z-z_0| < \delta$, $(z-z_0)^k \neq 0$. So \[f(z) \neq 0, \quad\quad 0 < |z-z_0| < \delta.\] Therefore the set of zeroes of $f$ cannot have an accumulation point in $D$.

$\square$