I am being asked to prove the following:

$$(1+\omega)(1+\omega^2)(1+\omega^3)\cdots(1+\omega^{n-1})=\begin{cases}1,&\text{ if}\; n \;\text{is odd}\\{}\\ 0 ,&\text{ if}\; n\;\text{ is even}\end{cases}$$

I understand that $\omega_n$=$\cos(2\pi/n)+i\sin(2\pi/n)$.

I also understand that ($\omega_n)^n = 1$ .

I am not sure how to move forward with this proof.

  • 1
    $\begingroup$ Half hint: A product is zero if and only if at least one of the terms is zero. $1-1=0$ $\endgroup$ – Henry Feb 1 '17 at 16:55
  • $\begingroup$ Yes I edited and added the "....". Thanks. $\endgroup$ – NUG Feb 1 '17 at 16:57

Hint: Show that $(x-w)(x-w^2)(x-w^3)\cdots(x-w^{n-1})=1+x+x^2+\cdots +x^{n-1}$ for all $x$. Then evaluate when $x=-1$.


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