# Identifying when a family of palindromic polynomials has roots of unity

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.

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.