In the connected component of $\{z \in \mathbb{C} : |p(z)| \le 1\}$, $p$ must vanish at least once. 
If $p$ is a non-constant polynomial, and. $G$ is an open connected component of $\{z \in \mathbb{C} : |p(z)| \le 1\}$, then $p$ has at least one zero in $G$

My thoughts so far.  Suppose that $p$ is nonzero in $G$.  By the minimum modulus principle, $p(z)$ then achieves its minimum (say $m$) on the boundary of $G$.  From here, I either want to show that $G$ cannot be connected, or maybe that $p$ must be constant, and hence we have a contradiction?  Any thoughts?
 A: Hint: Show that $m$ has to be equal to $1$, so that $1$ is then both the minimum of $p$ on $G$ and an upper bound, so that $p$ is constant (equal to $1$) on the non-empty open set $G$, a contradiction.
A: Let $G$ be an open connected component of $A=\{z:|p(z)| \leq 1\}$ such that $G \subsetneq A$
Assume that $p(z)$ does not have a root in $G$ thus $g(z)=\frac{1}{p(z)}$ is  holomorphic in $G$ and $|g(z)| \geq 1,\forall z \in G$
From the minimum modulus principle exists $z_0 \in G \cup \partial G$ such that $|g(z_0)|=1$
and $|g(z)| \geq |g(z_0)|=1,\forall z \in G$
Now assume that $z_0 \in \partial G$ thus $\exists \epsilon>0$ such that  $B(z_0,\epsilon) \cap G \neq \emptyset$ and $B(z_0,\epsilon) \cap G^c \neq \emptyset$.
Then we can choose $\epsilon$ small enough such that $B(z,\epsilon) \subset A$
But $B(z_0,\epsilon)$ is a connected set so because of the fact that $G$ is a connected component of $A$,we must have that $B(z,\epsilon) \subset G$ which is a contradiction.
Thus $z_0$ is in the open region $G$ and $|p(z_0)|=1$ thus $z_0$ is a maximum of $p$ and it is attained in $G$
Thus $p$ is constant which is again a contradiction.
