My complex analysis is decades in the rear view mirror. Perhaps someone here can help. I am looking for necessary and sufficient conditions on the coefficients of of a real polynomial of one complex variable (i.e. an element of ℝ[z]) so that all of its zeros will be on the unit circle. Clearly it is necessary that the polynomial is palindromic, but that is not sufficient. My question arose, after I discovered that if r,s are both real numbers then the polynomial p(z) =

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has all its roots on the unit circle. Since I had not seen anything like this before, I wondered if there were perhaps just some conditions on the coefficients one could check.

  • $\begingroup$ The first Google result for "polynomial with all roots on the unit circle" is mathpages.com/home/kmath294.htm . Please look harder for answers next time. $\endgroup$ – Qiaochu Yuan Oct 27 '09 at 16:24
  • $\begingroup$ I saw that link before posting here. It has interesting information, but so far as I can tell, does not answer the question. I suspect the answer to the question is: no simple conditions on the coefficients exist. I would be happy to learn otherwise. $\endgroup$ – rita the dog Oct 27 '09 at 17:04

Edited answer (see below for my original, less useful reply): Since your polynomial p(z) of degree 2d is palindromic, rewrite it as z^d q(z+1/z) for some polynomial q(x) of degree d. Then p(z) has 2d roots on the circle if and only if q(x) has d real roots in the interval [-2,2]. (Equivalently, q(x-2) should have d nonnegative real roots, while q(x+2) should have no positive roots.) Now you can try to extract information using e.g. Sturm sequences to try to count real roots in that interval.

I had previously posted: Since the real line parameterizes the circle (minus a point), you can transform your problem into counting the real zeros of an associated real polynomial. (See p. 182 of Rodriguez Villegas's book "Experimental Number Theory" for details; this is viewable in Google Books.)

There certainly have to exist necessary & sufficient criteria for all the zeros of a real polynomial to be real, involving rather complicated inequalities in the coefficients of the polynomial, generalizing the condition b^2-4ac >= 0 for quadratics; or for a specific polynomial you can try a real-root-counting algorithm (see Sturm's theorem on Wikipedia).

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    $\begingroup$ Yes. Your first paragraph seems way to go when coefficients are numeric. Symbolic coefficients is another story. For p(z) above, the deg 4 poly is q(t) = a_4 t^4 + a_3 t^3 + a_2 t^2 + a_1 t + a_0, where a_4 = r^8 + s^8, a_3 = -(r s^7 + s r^7), a_2 = s^2 r^6 + s^6 r^2 - 4r^8 - 4 s^8, a_1 = -s^5 r^3 - s^3 r^5 + 3s r^7 + 3s^7 r, a_0 = 2r^4 s^4 - 2s^2 r^6 - 2s^6 r^2 + 2r^8 + 2s^8. To show that for all r,s real, all roots of q(t) are in [-2,2] seems tough. My proof uses Möbius map swapping inside and out of unit circle. Coeff conditions seem unlikely. $\endgroup$ – rita the dog Oct 27 '09 at 20:30
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    $\begingroup$ There do exist coefficient conditions - indeed, this paper: arxiv.org/abs/math/9803097 finds a homeomorphism from a simplex to the set in "coefficient space" which parametrizes the set of polynomials whose roots are all on the unit circle. $\endgroup$ – j.c. Oct 28 '09 at 0:36
  • $\begingroup$ Many thanks, jc, for that reference. $\endgroup$ – rita the dog Oct 29 '09 at 15:23

See paper Palindrome-Polynomials with Roots on the Unit Circle by John Konvalina and Valentin Matache


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