Restriction of a complex polynomial to the unit circle 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.
 A: The image of the unit circle is a real-algebraic curve, so the number of self-intersections should be finite. 
Addendum:
I'm not sure how to complete the argument, but here's a heuristic (following Speyer's suggestion). The curve $x^2+y^2=1$ is a rational curve (i.e., birational to $\mathbb{CP}^1$, the Riemann sphere), so it's image is also a rational curve (here, I'm thinking of the map $p:\mathbb{C}\to \mathbb{C}$ as a polynomial map $\mathbb{R}^2 \to \mathbb{R}^2$, and its extension to $\mathbb{C}^2\to \mathbb{C}^2$ and $\mathbb{CP}^2\to \mathbb{CP}^2$). Complex conjugation gives an antiholomorphic involution of $x^2+y^2=1$, fixing the circle (on the Riemann sphere, this must be conjugate to complex conjugation). The image (in $\mathbb{CP}^2$) is a singular sphere, and again complex conjugation should be an anti-holomorphic involution fixing the image of the unit circle. So the map should be a composition of a polynomial map with a map of the Riemann sphere sending the circle to the circle.  All such maps of the Riemann sphere are products of Mobius transformations of the form $\frac{z-\varphi}{1-\overline{\varphi}z}$ (this is an exercise in Ahlfors, making use of the Schwarz lemma). If the composition is to be a polynomial map, then $\varphi$ must $=0$ in each factor, and the map is of the form $z^k$. My algebraic geometry is quite weak, so I'm not sure if this argument can be made rigorous (and I probably shouldn't have posted an answer in the first place!). 
A: You're right.  Quine proved in "On the self-intersections of the image of the unit circle under a polynomial mapping" 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 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.

