Identity theorem

Statement

Let $D \subseteq \mathbb{C}$ be a domain and let $f, g : D \to \mathbb{C}$ be holomorphic on $D$. Let $S = \{z \in D : f(z) = g(z)\}$. Assume that $S$ has an accumulation point in $D$. This means that there is $z_0 \in D$ such that each disc with center $z_0$ contains a point of $S$ other than $z_0$. Then $f(z) = g(z)$ for each $z \in D$.

Remarks

Proof idea: apply the isolated zeros theorem to $f - g$. The set of points near which $f - g = 0$ is open, closed in $D$ and not empty, so it is $D$ because $D$ is connected. For example, if $f$ is entire and $f(1/n) = 0$ for each $n \ge 1$, then $f = 0$. The connectedness of $D$ is necessary: on the union of two disjoint discs, a function can be $0$ on one disc and $1$ on the other. The theorem makes analytic continuation unique.

Explanations

L1 human

Two holomorphic functions on a domain that agree on a small piece, even on a convergent sequence of points with its limit, agree everywhere. So a holomorphic function is fixed everywhere by its values on a small disc.