Identity theorem in several variables

Remarks

Put $h = f - g$ and let $E$ be the set of points of $D$ near which $h$ vanishes identically. The set $E$ is open by definition and nonempty by hypothesis. It is also closed in $D$. At a limit point $a \in D$ of $E$, all partial derivatives of $h$ vanish by continuity, so the power series of $h$ at $a$ is zero and $h = 0$ near $a$. Since $D$ is connected, $E = D$. The one-variable condition "agreement on a set with a limit point" is not enough when $n \ge 2$: $f(z_1, z_2) = z_1$ vanishes on the whole line $\set{z_1 = 0}$ but is not zero. Connectedness is needed. Let $D = \set{\abs{z_1} < 1} \cup \set{\abs{z_1 - 3} < 1}$ in $\C$, or its product with $\C^{n-1}$. The function equal to $0$ on the first piece and $1$ on the second is holomorphic and vanishes on an open set without being zero.

Going deeper

For a student

This is the several-variable form of the identity theorem. The proof uses only the local power series expansion and connectedness, exactly as in one variable. What changes is the size of the set needed. Zero sets of holomorphic functions in $\C^n$, $n \ge 2$, are never isolated points, so agreement on a discrete set or even on a complex hypersurface says little. It is enough that $f$ and $g$ agree on a set that is not contained in any proper analytic subset of $D$. An example of such a set is a nonempty open subset of $\R^n \cap D$, when $D$ meets $\R^n$. The theorem underlies uniqueness of analytic continuation in several variables.