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Let $p(z,w)=\alpha_0(z)+ \alpha_1(z)w+\cdots+\alpha_k(z)w^k$ ,where $k \le 1$ and $p(z,w)=\alpha_0(z)+ \alpha_1(z)w+\cdots+\alpha_k(z)w^k_0,\cdots, p(z,w)=\alpha_0(z)+ \alpha_1(z)w+\cdots+\alpha_k(z)w^k_k$ are non-constant polynomials in the complex variable $z$. Then:

$(z,w) \in C\times C:p(z,w)=0$ is:

  • Bounded with empty interior.
  • Unbounded with empty interior.
  • bounded with nonempty interior .
  • unbounded with nonempty interior.

How can I be able to solve this problem ? I have no idea at all.

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    Your first two lines with three versions of $p(z,w)$ and undefined $w_0$ and $w_k$ don't make sense. Also, $k\leq 1$ is very strange.2012-12-16

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If a complex polynomial is locally (on an open set) equal to zero, it is globally equal to zero. Therefore the interior of the set of zeroes is empty since $p \neq 0$. Each of the terms $\alpha_m$ has only finitely many zeroes. In particular there are only finitely many $z \in \mathbb{C}$ for which $\alpha_m(z) = 0$ for all $m \in \{2, \dotsc, k\}$ simultaneously. If $|z|$ is large enough then $z$ will not be an element of this finite set and $w \mapsto p(z,w)$ is a non-constant polynomial in $w$ which therefore has a zero in $\mathbb{C}$. This shows that the set of zeroes of $p$ is unbounded.