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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.

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What makes you so sure of the truth of the statement? – José Figueroa-O'Farrill Mar 4 '10 at 17:51
is there any relation to Mikio Sato hyperfunctions? – kakaz Mar 4 '10 at 18:24
I don't know whether there is a relation with hyperfunctions – senti_today Mar 4 '10 at 20:35
It looks like $x^n + (1-\epsilon)x$ has $(n-1)^2$ self-intersections. – Douglas Zare Mar 5 '10 at 6:57
I previously posted an answer that was wrong with no redeeming qualities. Then I did some Googling. Douglas Zare found an example which reaches the upper bound proved in Quine's paper. – Jonas Meyer Mar 5 '10 at 7:04
up vote 20 down vote accepted

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.

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Thank you so much! (My own attempts at Googling this have failed earlier.) – senti_today Mar 5 '10 at 16:00

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!).

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The problem is that f(z)=z^2 has the image going over itself, so this abstract topological approach doesn't seem to work. – Kevin Buzzard Mar 5 '10 at 7:27
This does prove that, if the number of intersections is not finite, then the unit circle must be a branched cover of the image curve. I've been thinking about how to turn that into a proof and making some progress, but Jonas apparently found a relevant reference first. – David Speyer Mar 5 '10 at 11:59
David, It would still be interesting to see a new proof if you're interested in completing it. – Jonas Meyer Mar 5 '10 at 18:32

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