$p$ 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.
 A: Hints:


*

*For boundedness: Show that $|p(z)| \to \infty$ as $|z|\to \infty$

*For openness: The preimage $p^{-1}(A)$ of an open set $A$ under a continuous function $p$ is again open. Polynomials are continuous, so just write your set as a preimage.

*For the connected components: Recall fundamental theorem of algebra. What may these components have to do with polynomial roots? How many of them can a polynomial have? The example for fewer components is straightforward, just think of it geometrically (draw a curve). 
A: For bounding the number of connected components, you can show that every connected component contains at least one root of your polynomial. To obtain that, you can apply the minimum modulus principle to the connected component, just like in a proof of the fundamental theorem of algebra.
A: 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]
