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I have the following family of polynomials

$$ (p-1)(1-x^{2(n+1)}) - x(1-x^{2n}) = 0 $$

where $p\in\mathbb{R}$. By construction this polynomial has roots on the unit circle, and this is reflected in the fact that it is anti-palindromic. I can divide out two roots, $x=1,-1$ and get the following polynomial

$$ (p-1)x^{2n} - x^{2n-1} + ... -x + (p-1) =0 $$ Which is a palindromic polynomial. What's interesting about these is that for every root $\alpha$ there is a corresponding root $\frac{1}{\alpha}$ and generically you can reduce this type of polynomial to something of the form

$$ x^m g(x+\frac{1}{x}) $$

for some value of $m$, where $g$ is a polynomial of lower order (for this case, $m=n$ and $g(x+\frac{1}{x}) = 2T_n (x + \frac{1}{x}) - \frac{2}{p-1}T_{n-1} (x+\frac{1}{x}) + ... + 1$ where $T_n$ is the $n$th Chebyshev polynomial of the first kind) which also has the information of the roots.

My question is how can I determine which, if any, roots of my original polynomial are roots of unity i.e. of the form $e^{i\frac{r}{s}\pi}, r,s\in\mathbb{Z}$, for general values of $p$.

There are special values of $p$ for which this equation is monic with all unital coefficients and the roots are all roots of unity, namely $p=0,1,2$. I expect that the roots depend continuously on the parameter $p$, with $p=1$ a singular point, but given that the polynomial is not monic for all other values of $p$ is it reasonable to expect that the roots are no longer simple roots of unity in these regions.

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In [Uber die Anzahl der Wurzeln einer algebraischen Gleichung in einem Kreise, Math. Z., 14, 1922, 110-148], A. Cohn established the following result:

Let $p_n(z) = \sum_{k=0}^n a_k z^k$ be a polynomial of degree $n$ with complex coefficient, then all zeros of $p_n(z)$ lie on the unit circle if and only if $p_n(z)$ satisfies

(a) $a_{n-k} = \mu \bar{a}_k$, $0 \leq k \leq n$, $|\mu| = 1$;

(b) all the zeros of $p'_n(z)$ lie in or on the unit circle.

Edit You can find a lot of papers dealing with the same type of problem in the literature and such kind of polynomials are called self-reciprocal polynomials.

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  • $\begingroup$ Thanks for this comment, but it doesn't really answer my question. I was aware that the roots would lie on the unit circle as they are palindromic polynomial. My question was whether it was possible to determine whether any of the roots remained roots of unity as a function of p. $\endgroup$ – Aran May 14 '18 at 14:47

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