Vtyjkyuk

Hartogs's Theorem Let $f$ be a holomorphic function on a set $G setminus K$, where $G$ is an open subset of $mathbbC^n$ ($n ge
2$) and $K$ is a compact subset of $G$. If the complement $Gsetminus
K$ 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 $U subset mathbbC^n$ 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:

1. $Lambda(f)$ is closed but is not bound


2. $Lambda(f)$ is not closed but bounded


3. $Lambda(f)$ 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
beginalign
Lambda(f_1)= (z_1,z_2) : z_1=0 cup (z_1,z_2) : z_2=0 .
endalign
where $Lambda(f_1)$ is closed but not bounded.

The cases 2 and 3 cannot exist by continuity. Since $fcolon U rightarrow mathbbC$ is holomorphic, it is continuous. Hence $0$ being a closed subset implies that $f^-1(0) = Lambda(f)$ is closed in $U$.