Considered as functions $\Bbb C \to \Bbb C$, are polynomials open maps? If $p$ is a polynomial, is it true that $p(A)$ is open for every open $A\subseteq\Bbb C$?
I really don't know how to approach this. I'm fairly certain that they are closed maps, though.
 A: The answer is yes, as long as the polynomial is not constant.  A slightly more general result than this is the open mapping theorem.
A: Here is a direct proof that requires no complex analysis beyond the fundamental theorem of algebra.  Suppose $p(z)$ is a polynomial of degree $d>0$, $U\subseteq\mathbb{C}$ is open, and $p(U)$ is not open. Then there is some $a\in U$ such that $p(a)$ is not in the interior of $p(U)$.  Let $(w_n)$ be a sequence such that $w_n\to p(a)$ and $w_n\not\in p(U)$ for each $n$.  For each $n$, let $z_{n,1},\dots,z_{n,d}$ be the $d$ roots of the polynomial $p(z)-w_n$, in some order.  Note that the set $\{w_n\}$ is bounded, and $p(z)\to \infty$ as $z\to\infty$; it follows that the set $\{z_{n,k}\}$ is also bounded.  By Bolzano-Weierstrass, we can thus replace $(w_n)$ with a subsequence such that the sequences $(z_{n,1}),(z_{n,2}),\dots,(z_{n,d})$ all converge; write $b_k=\lim z_{n,k}$.  Since $w_n\not\in p(U)$ for all $n$, $z_{n,k}\not\in U$ for all $n$ and $k$, and hence $b_k\not\in U$ for all $k$ since $U$ is open.
However, I claim that the $b_k$ are exactly the roots of $p(z)-p(a)$.  Indeed, for each $n$, we have $p(z)-w_n=c\prod_{k=1}^d (z-z_{n,k})$, where $c$ is the coefficient of $z^d$ in $p(z)$.   Taking the limit as $n\to\infty$, we find that $p(z)-p(a)=c\prod_{k=1}^d (z-b_k)$.  In particular, this means one of the $b_k$ must be equal to $a$.  Since $b_k\not\in U$ and $a\in U$, this is a contradiction.
