I've been thinking about this problem for a while but with little success.
Assume $n \in \mathbb{N}$ and $a \in [-1, 1] \subseteq \mathbb{R}$. Let $z_0$ be an arbitrary root of the polynomial $P$ defined as $P(z) = z^{n + 1} - az^n - az + 1$ where $z \in \mathbb{C}$. Prove that $\lvert z_0 \rvert = 1$.
I tried applying $e^{i \varphi}$ assuming the argument is one of the roots, then representing the output using the trigonometric representation and obtaining the following system of equations
$$ \begin{cases} \cos{(n + 1) \varphi} - a\cos{n \varphi} - a\cos{\varphi} + 1 = 0 \\ \sin{(n + 1) \varphi} - a\sin{n \varphi} - a\sin{\varphi} = 0\end{cases} $$ which comes directly from comparing the real and imaginary part of $P(e^{i \varphi})$ with zero. If I proved $n + 1$ numbers of this form satisfy the system, then that would be enough. Unfortunately, I've only been able to prove that the system is equivalent to the equation $\cos{[(n + 1) \varphi / 2]} = a \cos{[(n - 1) \varphi / 2]}$.
This is a problem from my linear algebra course, so using analytic tools shouldn't be necessary. On the contrary, it should be provable using only properties of complex numbers and general properties of polynomials over $\mathbb{C}$ (e.g. there being a unique factorisation or every polynomial having a complex root).
Thank you for the help in advance.