Show that factorization of a $n - 1$ degree polynomial is $P(z) = \prod_{k=1}^{n - 1} (z^2 - 2z\cos(\frac{k\pi}{n}) + 1)$

Question

Let be $P(z) = \prod\limits_{k = 1}^{n - 1} z - z_k$, we know that $z_k = e^{i\frac{k}{\pi}}$ for $k \in [0, n - 1]$.

We want to show that $P(z)$ can be written as: $P(z) = \prod\limits_{k=1}^{n - 1} \left(z^2 - 2z\cos\left(\frac{k\pi}{n}\right) + 1\right)$.

So far, I have been able to prove (through $e^{i\theta}$ form) :

$\prod\limits_{k=1}^{n - 1} (z - z_k)^2 = \prod\limits_{k=1}^{n - 1} z^2 - 2z\cos\left(\frac{k\pi}{n}\right) + i\left[\sin^2 \frac{k\pi}{n} - 2z\sin \frac{k\pi}{n}\right] + \cos^2 \left(\frac{k\pi}{n}\right)$

I felt that at some point, some expression would cancel out and only $1$ would remain, but I am out of ideas.

Answer 1

Hint:

Group together linear factors with conjugate $z_k$s (you'll have to separate the cases $n$ odd and $n$ even). Remember that $$(z-z_k)(z-\bar z_k)=z^2-2\operatorname{Re}(z_k)z+1.$$