Let $\varepsilon _k$ be $\cos \frac {2k \pi} {n} + i \sin \frac {2k \pi} {n}$. Find the value of the product: $$\prod _ {k=1}^n (2+\varepsilon _k-\varepsilon _k^2).$$
Asked
Active
Viewed 94 times
2
-
did you try anything? – Arnaldo Jan 13 '17 at 20:36
-
I don't know how to start. I calculated the product for $n=2,3,4$. – M. Stefan Jan 13 '17 at 20:38
1 Answers
4
For $z^n-1=0$ : $$ \prod_1^n(2+\varepsilon_k-\varepsilon_k^2)=\prod_1^n-(-1-\varepsilon_k)(2-\varepsilon_k)=(-1)^n \prod_1^n(-1-\varepsilon_k)\prod_1^n(2-\varepsilon_k)=(-1)^n((-1)^n-1)(2^n-1) $$
Nosrati
- 29,995
-
-
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. – Jyrki Lahtonen Jan 13 '17 at 22:06
-