If $p(z,w)=a_0(z)+a_1(z)w+\dots +a_k(z)w^k$ are non constant polynomial.

$p(z,w)=a_0(z)+a_1(z)w+\dots +a_k(z)w^k$ where $a_i(z)$ are non constant polynomials in complex variables with $k\ge 1$.

I need know if $$\{(z,w):p(z,w)=0\}$$

which of these are true or false:

$1$. bounded with empty interior

$2$. unbounded with empty interior

$3$. bounded with non empty interior

$4$. unbounded with non empty interior

I think only $1$ is true as the set consists of roots of a polynomial in two variables with some specific degree.

Am I right?

$z^{k+1}+z^kw+...+z^2w^{k-1}+zw^k$ has unbounded zero set. So 1 and 3 are ruled out.

Also, if the interior were not empty, there would've existed $U\times V$ in $\mathbb C^2$ where $p(z,w)$ is identically zero. Fix $z$ in $U$ such that it is not a root of $a_k$. You get a non constant polynomial in one variable $w$ vanishing on $V$, a contradiction. So the interior must be empty.

I only eliminated boundedness possibility. To prove it is unbounded explicitly, just put $z=N$ large enough that $a_k(z)$ is not zero. For each $n\ge N$, we have $p(n,w_n)=0$ for some $w_n$ because $p(n,w)$ is a polynomial in one variable and it is non-constant.