I'm trying to prove (or disprove) that the roots of an even degree real symmetric coefficient polynomial are all on the unit circle. If it is not true, I will then try to find the conditions such that they do lie on the unit circle.

Here is what I have so far.

  • If $\alpha$ is a root, then $\frac{1}{\alpha}$ is also a root.
  • Non-real roots occur in conjugate pairs.

    Now the product of roots of any such polynomial will be equal to 1 due to the constant term being equal to the leading coefficient. But since the non-real roots occur in complex conjugate pairs, then the product of the moduli is also equal to 1 (using $\alpha\bar{\alpha}=1$).

So what we've got so far is

$r_1r_2r_3...r_k|c_1||c_2||c_3|...|c_j|=1$, where $r$ are the real roots and $c$ are the non-real roots.

Using the fact that the roots occur in reciprocal pairs, the product of the real roots must be equal to 1, so we have


And that for every $c_m$, there exists a $c_n$ such that $c_m=\frac{1}{c_n}$ due to the symmetric nature of the coefficients.

From here, how do I put the pieces together to finally prove or disprove (in which case, finding conditions for) the roots lying on the unit circle?

It's awfully tempting to say that since the product of their moduli is equal to 1, then they all have modulus 1, but that of course isn't always the case.

  • 2
    $\begingroup$ You have almost given a counterexample, $(x-2)(x-1/2)$. $\endgroup$ – André Nicolas Sep 28 '14 at 18:55
  • $\begingroup$ Thank you, it seems like my hypothesis is not true in general then. In which case, where should I start investigating to determine what conditions are needed for the roots to lie on the unit circle? $\endgroup$ – Trogdor Sep 28 '14 at 18:56
  • $\begingroup$ There are several known criteria for all the roots to be inside the unit circle: 1. Jury stability criterion 2. Bistritz stability criterion $\endgroup$ – Sasha Sep 28 '14 at 19:00

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