[proofplan] We prove that a [holomorphic function](/page/Holomorphic%20Function) $f: \Omega \to \mathbb{C}$ on a connected [open set](/page/Open%20Set) $\Omega \subset \mathbb{C}^n$ with $n \geq 2$ cannot have an isolated zero. Suppose $f(a) = 0$. We restrict $f$ to a complex line through $a$ and apply the one-variable [identity principle](/theorems/3357): on this line, $f$ is either identically zero (producing a continuum of zeros) or has an isolated zero. In the latter case, every nearby complex line through $a$ also has an isolated zero near $a$. We then use the connectivity of $\mathbb{CP}^{n-1}$ (the space of complex lines through $a$) to show that if the zero at $a$ were isolated in $\mathbb{C}^n$, the restriction to every complex line would have a zero only at $a$ with some finite order, but the zero set of a [holomorphic function](/page/Holomorphic%20Function) in several variables has real codimension $2$ (not $2n$), producing zeros in every neighbourhood of $a$. [/proofplan]

[step:Restrict $f$ to complex lines through $a$ and show at least one line carries a non-isolated zero]Suppose for contradiction that $a$ is an isolated zero of $f$, so there exists $r > 0$ with $B(a, r) \subset \Omega$ and $f(z) \neq 0$ for all $z \in B(a, r) \setminus \{a\}$. For each $v \in \mathbb{C}^n \setminus \{0\}$, define the one-variable [holomorphic function](/page/Holomorphic%20Function) \begin{align*} g_v: D_v &\to \mathbb{C} \\ t &\mapsto f(a + tv), \end{align*} where $D_v = \{t \in \mathbb{C} : a + tv \in \Omega\}$ is open in $\mathbb{C}$. Since $f(a) = 0$, we have $g_v(0) = 0$.[/step]

[guided]The idea is to probe the zero set of $f$ by slicing through $a$ along all possible complex lines. Each such line is parametrised by a direction $v \in \mathbb{C}^n \setminus \{0\}$, and the restriction $g_v(t) = f(a + tv)$ is a [holomorphic function](/page/Holomorphic%20Function) of one complex variable. By the isolated zero assumption, $g_v$ vanishes only at $t = 0$ within $\{|t| < r/|v|\}$. For each $v \in \mathbb{C}^n \setminus \{0\}$, define the one-variable [holomorphic function](/page/Holomorphic%20Function) \begin{align*} g_v: D_v &\to \mathbb{C} \\ t &\mapsto f(a + tv), \end{align*} where $D_v = \{t \in \mathbb{C} : a + tv \in \Omega\}$ is open in $\mathbb{C}$. Since $f(a) = 0$, we have $g_v(0) = 0$. If $a$ were an isolated zero, then for every $v$ with $|v| = 1$, the function $g_v$ would satisfy $g_v(t) \neq 0$ for $0 < |t| < r$ (since $|a + tv - a| = |t| < r$ implies $a + tv \in B(a, r) \setminus \{a\}$). In one complex variable, this means either $g_v \equiv 0$ on its connected component containing $0$, or $t = 0$ is a zero of some finite order $m_v \geq 1$. But $g_v \equiv 0$ is impossible under the isolated zero assumption (it would give zeros at all $t$), so each $g_v$ has a zero of finite order $m_v \geq 1$ at $t = 0$, i.e., $g_v(t) = t^{m_v} h_v(t)$ with $h_v(0) \neq 0$.[/guided]

[step:Count the zero set dimension via slicing and reach a contradiction]Under the isolated zero assumption, for every unit vector $v \in S^{2n-1} := \{v \in \mathbb{C}^n : |v| = 1\}$, the function $g_v(t) = f(a + tv)$ has a zero of finite order $m_v \geq 1$ at $t = 0$, and no other zeros in $\{|t| < r\}$. Consider two linearly independent directions $v, w \in \mathbb{C}^n$. The restriction of $f$ to the two-dimensional complex affine plane $P = \{a + sv + tw : s, t \in \mathbb{C}\}$ is a [holomorphic function](/page/Holomorphic%20Function) of two complex variables $(s, t)$ that vanishes at $(0,0)$. If the zero were isolated in $\mathbb{C}^n$, it would in particular be isolated in $P \cap B(a, r)$ — but we are working in $P \cong \mathbb{C}^2$, and the zero set of a non-constant [holomorphic function](/page/Holomorphic%20Function) of two variables has complex dimension at least $1$ (real dimension at least $2$). To see this, define \begin{align*} F: \{(s,t) \in \mathbb{C}^2 : a + sv + tw \in \Omega\} &\to \mathbb{C} \\ (s,t) &\mapsto f(a + sv + tw). \end{align*} The function $F$ is holomorphic and $F(0,0) = 0$. Since $g_v$ is not identically zero, $F$ is not identically zero on the plane $P$. By the [Weierstrass Preparation Theorem](/theorems/???), after a possible linear change of coordinates in the $(s,t)$-plane, $F$ can be written near $(0,0)$ as \begin{align*} F(s,t) = U(s,t) \cdot W(s,t), \end{align*} where $U$ is a unit (non-vanishing [holomorphic function](/page/Holomorphic%20Function) near the origin) and $W$ is a Weierstrass polynomial in $t$ of degree $m \geq 1$: \begin{align*} W(s,t) = t^m + \sum_{j=0}^{m-1} a_j(s) t^j \end{align*} with $a_j$ holomorphic near $s = 0$ and $a_j(0) = 0$. The zero set of $F$ near $(0,0)$ coincides with the zero set of $W$. For each $s$ sufficiently close to $0$, the polynomial $t \mapsto W(s,t)$ has exactly $m$ roots (counted with multiplicity) near $t = 0$. In particular, for $s \neq 0$ sufficiently small, there exist values of $t$ with $W(s,t) = 0$ and hence $F(s,t) = 0$, i.e., $f(a + sv + tw) = 0$. These zeros $(s, t(s))$ with $s \neq 0$ correspond to points $z = a + sv + t(s)w \neq a$ in $B(a, r)$ where $f$ vanishes. This contradicts the assumption that $a$ is an isolated zero of $f$.[/step]

[guided]The deeper geometric point is as follows. The zero set of a non-constant [holomorphic function](/page/Holomorphic%20Function) $F: U \subset \mathbb{C}^k \to \mathbb{C}$ on a connected [open set](/page/Open%20Set) $U$ has complex dimension exactly $k - 1$ (real dimension $2k - 2$). When $k = 1$, zeros are isolated (dimension $0$). When $k \geq 2$, the zero set has dimension at least $1$, so zeros are never isolated. Why does the [Weierstrass Preparation Theorem](/theorems/3381) give us non-isolated zeros? The key is the polynomial $W(s,t) = t^m + a_1(s)t^{m-1} + \cdots + a_m(s)$. At $s = 0$, all roots coincide at $t = 0$ (since $a_j(0) = 0$ for all $j$). As $s$ varies, these roots move continuously. For $s \neq 0$ sufficiently small, the $m$ roots $t_1(s), \ldots, t_m(s)$ are close to $0$ but not necessarily all equal to $0$. Even if some roots coincide, the point is that $(s, t_j(s))$ gives a zero of $F$ for every $s$ near $0$ — so the zero set contains a curve, not just a point. To make this completely explicit: fix $s$ small with $s \neq 0$. The polynomial $t \mapsto W(s, t)$ has degree $m \geq 1$ in $t$, so by the [fundamental theorem of algebra](/theorems/347) it has at least one root $t_0 \in \mathbb{C}$. Since the coefficients $a_j(s) \to 0$ as $s \to 0$, the root $t_0$ can be made arbitrarily small by choosing $|s|$ small. The point $z = a + sv + t_0 w$ then satisfies $f(z) = F(s, t_0) = U(s, t_0) W(s, t_0) = 0$, and $z \neq a$ since $s \neq 0$ (and if $t_0 = 0$ then $z = a + sv \neq a$). This produces zeros of $f$ in $B(a, r) \setminus \{a\}$, contradicting the assumption. This argument reveals a fundamental difference between one and several complex variables: in $\mathbb{C}^1$, the zero set of a [holomorphic function](/page/Holomorphic%20Function) is discrete, while in $\mathbb{C}^n$ with $n \geq 2$, it is a complex hypersurface of real codimension $2$. In $\mathbb{R}^{2n}$, a set of real codimension $2$ is far too large to consist of a single point when $n \geq 2$.[/guided]