TITLE: Can a holomorphic function of several variables have just one zero?
QUESTION [6 upvotes]: Of course “several” in the title means $n$ strictly greater than 1 and the function is defined on some open subset of $\mathbb{C}^n$.
I tried to use the Weierstrass preparation theorem because it’s the only result on analysis of several variables I know but I couldn’t find a contradiction (I believe the answer is “no”).

REPLY [3 votes]: Suppose for simplicity that $f$ is holomorphic in an open set $U\subset \mathbb C^2$ with $(0,0)\in U.$ Assume that $f$ has an isolated zero at $(0,0).$ Then we can choose $r>0$ such that $f$ is holomorphic in $D(0,2r)^2$ and $f\ne 0$ in $D(0,2r)^2\setminus \{(0,0)\}.$
For $w\in D(0,r)$ define
$$g(w)=\frac{1}{2\pi i}\int_{|z|=r} \frac{D_zf(z,w)}{f(z,w)}\,dz.$$
The argument principle shows that $g(0)$ is a nonzero integer $m,$ while $g(w)= 0$ for $0<|w|<r.$ Note also that $g$ is a continuous function of $w.$ Thus $g(0) = m = \lim_{z\to 0} g(w)=0,$ which is a contradiction.