Preimage of circle under a polynomial Given a polynomial of degree $n$, is it true that $\{z: |p(z)|=r\}$ is the boundary of some domain?  
This question is motivated by the following homework problem:

Let $p$ be a polynomial of degree $n$, show that $\{z: |p(z)|=r\}$ has at most $n$ components.

This is what I have tried:
We start by a simple lemma:  If $f(z)$ is holomorphic in some connected domain $D$ which is continuous throughout the boundary, such that $|f|$ is constant at the boundary, then there exists $z_0$ such that $f(z_0)=0$.
This can be easily shown using maximum principle.
Using this lemma, if the question stated in the beginning is true--i.e $\{z: |p(z)|=r\}$ is the boundary of some domain $D$--then since $p$ has at most $n$ zeros, there must be at most $n$ components.
 A: Remark that $lim_{\mid z\mid\rightarrow +\infty}\mid p(z)\mid =+\infty$. This implies that $p^{-1}(C_r)$ is bounded, since it is closed, we deduce that $p^{-1}(C_r)$ is compact and has a finite number of connected components. 
Let $A$ be a connected connected component of $p^{-1}(C_r)$, $p(A)$ is a connected subset of $C_r$, suppose $p(A)\neq C_r$, and let $U$ be an open subset which contains $A$ and does not intersects the others connected components of $p^{-1}(C_r)$, $p(U)$ is open thus $p(A)\neq p(U)$, there exists $u\in U$ such that $p(u)\in C_r$ and $p(u)$ is not in $p(A)$, this implies that $u$ is in another component of $p^{-1}(C_r)$ contradiction, since the only connected component of $p^{-1}(C_r)$ that $U$ intersects is $A$, thus $p(A)=C_r$.
Let $v\in C_r$, $p^{-1}(v)=\{z:p(z)-v=0\}$ has  cardinal inferior to $n$ since the degree of $p$ is $n$. You can choose $v$ such that the cardinal of $p^{-1}(v)$ is $n$. Each element of $v_i$ of $p^{-1}(v)$ is contained in a unique connected component $C_i$ of $p^{-1}(C_r)$ so there exists at most $n$ such $C_i$. If $w$ is another element of $C_r$, each  connected component of $p^{-1}(C_r)$ which contains an element of $p^{-1}(w)$ contains an element in $p^{-1}(v)$ since its image is $C_r$ thus each connected component of $p^{-1}(C_r)$ which contains an element of $p^{-1}(w)$ is  a connected component of $p^{-1}(C_r)$ which contains an element of $p^{-1}(v)$. So the cardinal of the connected components of $p^{-1}(C_r)$ is inferior to $n$ since the cardinal of the connected components of $p^{-1}(C_r)$ which contains an element of $p^{-1}(v)$ is inferior to $n$. 
