Hartogs's Theorem Let $f$ be a holomorphic function on a set $G \setminus K$, where $G$ is an open subset of $\mathbb{C}^n$ ($n \ge 2$) and $K$ is a compact subset of $G$. If the complement $G\setminus K$ is connected, then $f$ can be extended to a unique to a unique holomorphic function on $G$.
This theorem can be used to show the following result about the zeros of analytic functions of several variables.
Suppose that $f$ is an analytic function on some open set $U$ and that $f$ is not identically zero on $U \subset \mathbb{C}^n$ with $n \ge 2$. Then, the set of zeros of $f$ (i.e. $\Lambda(f)=\{ z: f(z)=0\}$) is not compact.
Since $\Lambda(f)$ is not compact we can have the following three possibilities:
- $\Lambda(f)$ is closed but is not bound
- $\Lambda(f)$ is not closed but bounded
- $\Lambda(f)$ is not closed and not bounded
My question is the following: Can we come up with examples of $f$ for each of the three cases?
Here is an example of the function that satisfies the first case. Let $f_1(z_1,z_2)=z_1 \cdot z_2$ then \begin{align} \Lambda(f_1)= \{ (z_1,z_2) : z_1=0 \} \cup \{ (z_1,z_2) : z_2=0 \}. \end{align} where $\Lambda(f_1)$ is closed but not bounded.
The cases 2 and 3 cannot exist by continuity. Since $f\colon U \rightarrow \mathbb{C}$ is holomorphic, it is continuous. Hence $\{0\}$ being a closed subset implies that $f^{-1}(0) = \Lambda(f)$ is closed in $U$.