$p$ is polynomial, set bounded open set with at most $n$ components Assume $p$ is a non constant polynomial of degree $n$. Prove that the set $\{z:|(p(z))| \lt 1\}$ is a bounded open set with at-most $n$ connected components. Give example to show number of components can be less than $n$.
thanks.
EDIT:Thanks,I meant connected components.

Fact 1 : If $f : G \to \mathbb{C} $ is analytic, $G$ open and connected and $f(G)$ is a subset of a circle (or a line) then $f$ is constant.
Fact 2 : $f:\overline{G} \to \mathbb{C}$ analytic on a bounded region $G$ and $|f|(\partial G) = \{c\}$, then either $f$ has a root in $G$ or $f$ is a constant.
To prove fact 2, say $c = 0$, then $ |f| \leq 0 $ by the maximum principle, but then $f = 0$ a constant.
Say $c \neq 0$ and $f \neq 0$ in $G$, then $\frac{1}{f} : \overline{G} \to \mathbb{C}$ is analytic in the interior and then assumes maximum on $\partial G$. We need then $|f| \leq c$ and $1/|f| \leq 1/|c|$ in $G$, then $ |f| = |c|$ in $G$.
From this, we have that $f$ maps connected $G$ on the circle $\{z \, : \, |z| = |c|\}$, then $f$ is constant by Fact 1.
$\square$
Suppose now we have $f$ a polynomial. Say $G \subset |f|^{-1}([0,c)) \doteq A$ a connected component of the bounded(Liouville) open $A$. 
Note that $|f| = c$ at $\partial G$, by Fact 2 $f$ is either a constant polynomial or have a zero in the interior of $G$.
Say we have $\gamma(t) = (1-t)a + tb$ where $a$ and $b$ are zeros, $\{\gamma\}$ is compact, say then $|f| \leq c$ on this curve, then the entire curve $\{\gamma\}\subset |f|^{-1}([0,c+1)) = A_{c+1}$.
Note thar $a$ and $b$ will be two roots of $f$ lying on the same connected component of $A_{c+1}$.
[Pretty much Conway's Chapter VI]
