Suppose that $p(t) = a_0 + a_1 \cos(t)+ b_1 \sin(t) + a_2 \cos(2t) + b_2 \sin(2t)$ is a quadratic trig polynomial with complex coefficients. Assume that $a_2$ and $b_2$ are not real multiples of the same number. Is it true that $p(t)$ has at most finitely many self-intersections, that is, are there at most finitely many pairs $t_1, t_2 \in [0,2\pi)$ such that $p(t_1) = p(t_2)$ and $t_1 \neq t_2$?
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1$\begingroup$ Yes, use $\tan t/2$ to reduce to a rational function. $\endgroup$– Igor RivinCommented Oct 4, 2011 at 19:13
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$\begingroup$ Hi Brian! What does it mean that $a_2$ and $b_2$ are not real multiples of the same number? For instance, $a_2/b_2$ is a real number unless $b_2 = 0$. $\endgroup$– Matt YoungCommented Oct 4, 2011 at 19:15
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$\begingroup$ Oops, misread the question since the coefficients are complex. Never mind. $\endgroup$– Matt YoungCommented Oct 4, 2011 at 19:19
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1$\begingroup$ @Igor, I know that $p(t)$ could be replaced by a rational function on the unit circle, but I don't know what that would do. There is a paper by Quine that gives a bound on the number of self-intersections for polynomials on the unit circle, but I am unaware of any similar result for rational functions. $\endgroup$– Brian LinsCommented Oct 5, 2011 at 0:56
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$\begingroup$ @Brian: I will ponder... $\endgroup$– Igor RivinCommented Oct 5, 2011 at 7:11
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