Calculate product of $n$-th roots of unity

Question

Consider $$a_k = \cos \frac{2k\pi}{n} -2 + i \sin \frac{2k\pi}{n}, \: n, k \in \mathbb N^*, n \text{ fixed}$$

I have to calculate $$\prod_{k=1}^n a_k = \prod_{k=1}^n [(\cos \frac{2k\pi}{n} + i \sin \frac{2k\pi}{n}) - 2]$$

I know $$\cos \frac{2k\pi}{n} + i \sin \frac{2k\pi}{n}$$ are the $n$-th roots of $1$ and they satisfy the equation $$x^{n-1} + x^{n-2} + ... +x^1 + 1 = 0$$

so their product is $(-1)^n$ but I don't know what to do with that "$2$".

Answer 1

Clearly, $a_k,1\le k\le n$ are the roots of $$(y+2)^n=1$$

$$\iff y^n+\cdots+2^n-1=0$$

Vieta's formula says: $$\prod_{k=1}^na_k=(-1)^n(2^n-1)$$