A polynomial with two complex variables Let $p(z,w)=a(z) + b(z)w +...+ t(z)w^k$, where $k\ge1$ and $a(z),b(z),...,t(z)$ are non-constant polynomials in the complex variable $z$. Then
$\{(z,w)\in \Bbb C\times\Bbb C: p(z,w)=0\}$ is


*

*bounded with empty interior?

*unbounded with empty interior?

*bounded with nonempty interior?

*unbounded with nonempty interior?

 A: Let's call your set $V$ (for "variety").
For any fixed $z$, if we plug the value of $z$ into the functions $a(z), b(z), \ldots$, then unless they all simultaneously vanish (which only happens for finitely many values of $z$, since this would mean $z$ is a root of each polynomial!), there are between $1$ and $k$ distinct values of $w \in \mathbb{C}$ such that $p(z,w)=0$ (exactly $k$ as long as $t(z) \neq 0$ and you count multiplicities), by the fundamental theorem of algebra.
Thus, $V$ cannot be bounded, since it includes points with all but finitely many values of $z$ (and of $w$ by the same argument flipped around). 
Now, assume $(z_0,w_0)$ is in the interior of $V$. This means there is some open ball $U \subseteq V$ with $(z_0,w_0) \in U$. Since $U$ is open, so is $U \cap L$, where $L =\{ (z_0, w), w \in \mathbb{C} \}$, in the subspace topology on $L$. But an open set in $L$ must have more than finitely many points, and we've seen that $V \cap L \supseteq U \cap L$ is finite.
To be more elementary, if $(z_0,w_0)$  is an interior point of $V$, then $V$ must contain all points within distance $r$ of $(z_0,w_0)$  for some $r >0$. But clearly there are more than finitely many points within distance $r$ of $(z_0,w_0)$  such that the first coordinate is $z_0$.
A: I'm thinking unbounded with an empty interior.
First, if you had a non-empty interior you'd have a little room to wiggle around in there while keeping $p$ identically zero. Fixing a $z_0$ in the interior, you'd find that the single variable polynomial you get by plugging in $z_0$ must have infinitely many zeros, a no-no.
Secondly, take any unbounded sequence of complex numbers $z_i$. Each time I plug one of those guys into $p$ I get a new polynomial with $k$ zeros (counting with multiplicity). As my sequence of $z_i$ is unbounded, the zeros of $p$ gotta be unbounded.
