I am pretty sure that the following statement is true. I would appreciate any references (or a proof if you know one).


Let $f(z)$ be a polynomial in one variable with complex coefficients. Then there is the following dichotomy. Either we can write $f(z)=g(z^k)$ for some other polynomial $g$ and some integer $k>1$, or the restriction of $f(z)$ to the unit circle is a loop with only finitely many self-intersections. (Which means, more concretely, that there are only finitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$.)


EDIT. Here are a couple reasons why I believe the statement is correct.


1) The statement is equivalent to the following assertion. Consider the set of all ratios $z/w$, where $|z|=1=|w|$ and $f(z)=f(w)$ (here we allow $z=w$). If $f$ is a nonconstant polynomial, then this set is finite.


[[ Here is a proof that the latter assertion implies the original statement. Suppose that there are infinitely many pairs $(z,w)$ such that $|z|=1=|w|$, $z\neq w$ and $f(z)=f(w)$. Then some number $c\neq 1$ must occur infinitely often as the corresponding ratio $z/w$. However, this would imply that $f(cz)=f(z)$ (as polynomials). It is easy to check that this forces $c$ to be a root of unity, and if $k$ is the order of $c$, then $f(z)=g(z^k)$ for some polynomial $g(z)$. ]]


Going back to the latter assertion, note that the set of all such ratios is a compact subset of the unit circle, and it is not hard to see that 1 must be an isolated point of this set. So it is plausible that the whole set is discrete (which would mean that it is finite).


2) If I am not mistaken, experiments with polynomials that involve a small number of nonzero monomials (such as 2 or 3) also confirm the original conjecture.



Restriction of a complex polynomial to the unit circle
You're right. Quine proved in "[On the self-intersections of the image of the unit circle under a polynomial mapping](http://www.jstor.org/stable/2039005)" that if the degree is $n$ and $f(z)\neq g(z^k)$ with $k>1$, then the number of points with at least 2 distinct preimage points is at most $(n-1)^2$. An example shows that this is sharp. Here's the [review](http://www.ams.org/mathscinet-getitem?mr=313485) in MR.


After the proof, there is a remark: 



> 
> As a simple consequence of this theorem we note that a polynomial $p$
>  cannot map $|z| < 1$ conformally onto a domain with a slit, for in this case
>  $p(e^{i\phi})$ would have an infinite number of vertices.
> 
> 
> 


