2
$\begingroup$

Let $\varepsilon _k$ be $\cos \frac {2k \pi} {n} + i \sin \frac {2k \pi} {n}$. Find the value of the product: $$\prod _ {k=1}^n (2+\varepsilon _k-\varepsilon _k^2).$$

  • 0
    did you try anything? – 2017-01-13
  • 0
    I don't know how to start. I calculated the product for $n=2,3,4$. – 2017-01-13

1 Answers 1

4

For $z^n-1=0$ : $$ \prod_1^n(2+\varepsilon_k-\varepsilon_k^2)=\prod_1^n-(-1-\varepsilon_k)(2-\varepsilon_k)=(-1)^n \prod_1^n(-1-\varepsilon_k)\prod_1^n(2-\varepsilon_k)=(-1)^n((-1)^n-1)(2^n-1) $$

  • 0
    Nice! Good job! – 2017-01-13
  • 0
    @JyrkiLahtonen But it's false. help to improve. it's right for even $n$. – 2017-01-13
  • 0
    Looks correct to me. It may be simpler to say that it is zero when $n$ is even and $2(2^n-1)$ when $n$ is odd. – 2017-01-13
  • 0
    @JyrkiLahtonen Thanks. – 2017-01-13