Factorization of a polynomial involving cosine into $m$ second-order factors [closed]

Question

For each $m\in\mathbb{N}$ and fixed $a>0,\theta\in\mathbb{R}$, I want to factorizate the polynomial $p_m(x) = x^{2m} - 2a^m\cos (m\theta)x^m + a^{2m}$ into $m$ polynomials of second order. Using basic algebra I get for $m=2$ the following decomposition

$$ p_2 (x)= x^4 - 2a^2\cos(2\theta)x^2 + a^4 = (x^2+2a\cos(\theta)x +a^2)(x^2 - 2a\cos(\theta)x +a^2) $$ I used the formula $\cos (2\ \cdot) = \cos^2 (\cdot) - \sin^2(\cdot)$. I want to do the general case $m\in\mathbb{N}$ and I am afraid that complex numbers are a good tool here, but I can't see how to manage this problem using complex numbers. Please, any help or comment is welcome. Thank you in advance.

Answer 1

At first, substitute $t=x^m$ in your equation to get

$t^2-2a^m\cos(m\theta)t+a^{2m}$

This is factorized as $(t-a^m e^{im\theta})(t-a^m e^{-im\theta})$. I did that using the quadratic formula, but there is probably a better way of seeing it.

But then, $x=t^{1/m}$, which is (accounting for all the roots of unity) $x=a e^{\pm i\theta} e^{i \frac {2\pi k} m}$. The quadratic factors are obtained by multiplying linear factors with conjugate roots, i.e.

$(x-ae^{im\theta + i\frac {2\pi k} m})(x-ae^{-im\theta - i\frac {2\pi k} m}) = (x^2-2a\cos(\theta + \frac {2\pi k} m)x + a^2)$

The quadratic factors are of this form for $k=0,1,\dots, m-1$. This agrees with the factorization you gave for $m=2$.