[proofplan] We prove the [identity principle](/theorems/3357) by an open-and-closed argument. Define $U = \{z \in \Omega : f \text{ vanishes identically in a neighbourhood of } z\}$. By definition $U$ is open. We show $U$ is closed relative to $\Omega$ by proving that if $z_0 \in \Omega$ is a limit point of $U$, then all partial derivatives of $f$ at $z_0$ vanish, forcing the Taylor series of $f$ at $z_0$ to be identically zero, so $f$ vanishes in a polydisc around $z_0$. Since $\Omega$ is connected and $U$ is non-empty, open, and closed, $U = \Omega$. [/proofplan] [step:Define the set $U$ where $f$ vanishes locally and verify it is open] Define \begin{align*} U &:= \{z \in \Omega : \text{there exists } \varepsilon > 0 \text{ such that } f \equiv 0 \text{ on } B(z, \varepsilon) \cap \Omega\}. \end{align*} By hypothesis, $f$ vanishes on a non-empty open subset $W \subset \Omega$, so $W \subset U$ and $U \neq \varnothing$. The set $U$ is open: if $z_0 \in U$, then $f \equiv 0$ on $B(z_0, \varepsilon)$ for some $\varepsilon > 0$ with $B(z_0, \varepsilon) \subset \Omega$. For any $z \in B(z_0, \varepsilon/2)$, the ball $B(z, \varepsilon/2) \subset B(z_0, \varepsilon) \subset \Omega$, and $f \equiv 0$ on $B(z, \varepsilon/2)$, so $z \in U$. [/step] [step:Show that $U$ is closed in $\Omega$ by proving all Taylor coefficients vanish at limit points] Let $z_0 \in \Omega$ be a limit point of $U$ (i.e., $z_0 \in \overline{U} \cap \Omega$). We claim $z_0 \in U$. Since $f \in \mathcal{O}(\Omega)$, there exists a polydisc $\mathbb{D}^n(z_0, r) \subset \Omega$ on which $f$ is represented by its Taylor series: \begin{align*} f(z) &= \sum_{\alpha \in \mathbb{N}_0^n} c_\alpha (z - z_0)^\alpha, \quad z \in \mathbb{D}^n(z_0, r), \end{align*} where $c_\alpha = \frac{1}{\alpha!} \partial^\alpha f(z_0)$ and $\partial^\alpha = \partial_{z_1}^{\alpha_1} \cdots \partial_{z_n}^{\alpha_n}$. We show $c_\alpha = 0$ for every multi-index $\alpha$. Fix $\alpha \in \mathbb{N}_0^n$. The function $\partial^\alpha f: \Omega \to \mathbb{C}$ is holomorphic on $\Omega$ (since $f \in \mathcal{O}(\Omega)$ and holomorphic functions can be differentiated to all orders), hence continuous. On $U$, we have $f \equiv 0$ locally, so all partial derivatives of $f$ vanish on $U$: $\partial^\alpha f(z) = 0$ for all $z \in U$. Since $z_0$ is a limit point of $U$ and $\partial^\alpha f$ is continuous at $z_0$: \begin{align*} c_\alpha = \frac{1}{\alpha!} \partial^\alpha f(z_0) = \frac{1}{\alpha!} \lim_{\substack{z \to z_0 \\ z \in U}} \partial^\alpha f(z) = 0. \end{align*} Therefore $f(z) = \sum_\alpha 0 \cdot (z - z_0)^\alpha = 0$ for all $z \in \mathbb{D}^n(z_0, r)$, which shows $f \equiv 0$ on the polydisc $\mathbb{D}^n(z_0, r)$. Hence $z_0 \in U$. [guided] The core idea is that holomorphic functions in several variables are analytic (representable by convergent [power series](/page/Power%20Series)), and a [power series](/page/Power%20Series) is determined by its coefficients, which are given by partial derivatives at the centre. If all coefficients vanish, the function vanishes identically in a neighbourhood. Why do all coefficients vanish? The Taylor coefficient $c_\alpha = \frac{1}{\alpha!}\partial^\alpha f(z_0)$ is the value at $z_0$ of the [holomorphic function](/page/Holomorphic%20Function) $\partial^\alpha f$. On the set $U$, $f$ vanishes identically in neighbourhoods, so all its partial derivatives vanish on $U$. Concretely, if $z \in U$, then $f \equiv 0$ on a ball $B(z, \varepsilon)$, so $\partial^\alpha f(z) = 0$ by differentiating the zero function. Since $z_0$ is a limit point of $U$, there exists a sequence $(z_k)_{k \geq 1}$ in $U$ with $z_k \to z_0$. By continuity of $\partial^\alpha f$: \begin{align*} \partial^\alpha f(z_0) = \lim_{k \to \infty} \partial^\alpha f(z_k) = \lim_{k \to \infty} 0 = 0. \end{align*} This gives $c_\alpha = 0$ for every $\alpha$, so the Taylor series of $f$ at $z_0$ is the zero series, and $f \equiv 0$ on $\mathbb{D}^n(z_0, r)$. Note the contrast with the one-variable [identity principle](/theorems/3357): in one variable, the zeros of a non-constant [holomorphic function](/page/Holomorphic%20Function) are isolated, and the proof uses the order of vanishing at a point. In several variables, zeros are never isolated (by the [No Isolated Zeros](/theorems/3380) theorem), so the one-variable argument does not directly generalize. Instead, we use the stronger hypothesis that $f$ vanishes on an [open set](/page/Open%20Set) (not just at a point with an accumulation of zeros), and the argument becomes simpler: we only need that all Taylor coefficients vanish, which follows from continuity of derivatives. [/guided] [/step] [step:Conclude by connectedness of $\Omega$] The set $U$ is non-empty (since the [open set](/page/Open%20Set) on which $f$ vanishes is contained in $U$), open (by the first step), and closed relative to $\Omega$ (by the second step). Since $\Omega$ is connected and $U \subset \Omega$ is a non-empty clopen subset, $U = \Omega$. Therefore $f \equiv 0$ on $\Omega$. [/step]