# Product of roots of unity.

Let $$x_n$$ be the $$n$$ roots of unity.

Does there exist a closed expression for

$$F_n:=\prod_{i=1}^n(1+x_i+x^{n-1}_i)?$$

Interestingly, if $$n=1$$, then $$F_n=3$$ and if $$n=2$$ then $$F_2=-3$$ and if $$n=3$$ then $$F_3=0.$$ So maybe it just oscillates between these numbers, but how could one prove something like this?

Note that $${x_i}^{n-1}=\frac{1}{x_i}$$ because $${x_i}^n=1, \forall 1 \leq i \leq n$$. So, rewriting, $$F_n = \displaystyle\prod_{i=1}^{n}\dfrac{{x_i}^2+x_i+1}{x_i}=\prod_{i=1}^n\dfrac{(x_i-\omega)(x_i-{\omega}^2)}{x_i}$$, where $$\omega$$ satisfies $$x^2+x+1=0$$ i.e. cube root of unity. Since $$x_i, 1 \leq i \leq n$$ is an $$n$$th root of unity, it is a root of the polynomial $$z^n-1=0$$. So, $$x^n-1=\displaystyle\prod_{i=1}^n(x-x_i)$$. Thus, $$F_n=\displaystyle\prod_{i=1}^n\dfrac{(x_i-\omega)(x_i-{\omega}^2)}{x_i}=\dfrac{\prod_{i=1}^n(\omega-x_i)\prod_{i=1}^n({\omega}^2-x_i)}{(-1)^n\prod_{i=1}^n(-x_i)}=\dfrac{f(\omega)f({\omega}^2)}{(-1)^nf(0)}$$, where $$f(z)=z^n-1=\displaystyle\prod_{i=1}^n(z-x_i)$$. So, $$F_n=\dfrac{((\omega)^n-1)(({\omega}^2)^n-1)}{(-1)^{n+1}}=(-1)^{n+1}(1-\omega^n)(1-\omega^{2n})$$. Can you proceed?
• Very nice! – You could mention that $\omega \ne 1$ is a third root of unity (although one can infer it easily from the context). May 31 at 11:41
Yes, a truncated form will exist. Note that if $$x_k$$ is $$n^{th}$$ root of unity, then we have: $$x_k=e^{\frac {2k\pi i}{n}}$$ Hence, using $$e^{i\theta}=\cos \theta +i\sin \theta$$, we have $$F_n=\prod_{k=1}^n (2\cos (\frac {2k\pi}{n})+1)=2^{2n}\prod_{k=1}^n \cos(\frac {\pi}{6}+ \frac {k\pi}{n})\prod_{k=1}^n \cos (-\frac {\pi}{6} +\frac {k\pi}{n})$$ Now, both the products are evaluated using a similar technique, which is as follows: We have, $$\sin nx =2^{n-1} \prod_{k=0}^{n-1} \sin (x+\frac {k\pi}{n})$$ This result comes from Euler's formula itself. Now, substitute $$x=t+\frac {\pi}{2}$$ to get a cosine product, after changing limit of product too to match the question($$k=0$$ term should be divided while $$k=n$$ term should be multiplied externally). Then put $$t=\frac {\pi}{6}$$ in the formula to obtain: $$\prod_{k=1}^n \cos(\frac {\pi}{6}+\frac {k\pi}{n})=-\frac {\sin(\frac {2n\pi}{3})}{2^{n-1}}$$ Similarly evaluate the other term, your final answer should be: $$F_n= 4\sin(\frac {n\pi}{3})\sin(\frac {2n\pi}{3})$$
• Simplified: $$F_n = \begin{cases} 0, & n \bmod 3 = 0 \\ 3, & n \bmod 6 =1,5 \\ -3, & n \bmod6 = 2,4 \end{cases}$$ May 31 at 13:28