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

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.

• The product of $n-1$ quadratic polynomials cannot have degree $n-1$ – Hagen von Eitzen Sep 18 '16 at 13:31
• @HagenvonEitzen You're definitely right, my bad. So what I have been proving is false. – Raito Sep 18 '16 at 13:33
• There are some nice answers for special case $n=10$ here math.stackexchange.com/questions/1909362/… – mathreadler Sep 18 '16 at 13:55
• It will still be true if you take product over half a revolution instead of a whole as the roots come in pairs in the nice way Bernard explains. – mathreadler Sep 18 '16 at 14:01

Group together linear factors with conjugate $z_k$s (you'll have to separate the cases $n$ odd and $n$ even). Remember that $$(z-z_k)(z-\bar z_k)=z^2-2\operatorname{Re}(z_k)z+1.$$
• Does it means that $\deg P(z) = 2(n - 1)$, I was supposing that my equation had $n - 1$ roots, but I guess I'm wrong? I don't understand really why, could you explain a bit more about this point, please? – Raito Sep 18 '16 at 14:16