# roots of unity

• Aug 19th 2008, 09:38 PM
Happy Dancer
roots of unity
The question asks, find the n solutions of z^n=1, which I got to be z=e^(2*i*k*Pi)/n, but where n is an odd positive integer. I now don't understand what the question is asking, is my answer wrong then? Any help would be much appreciated.
• Aug 20th 2008, 03:42 AM
Sean12345
Hi Happy Dancer,

Your answer is correct: it doesn't matter whether n is odd or even the result still follows.

You could say;

The roots of $\displaystyle z^n~=~1~$ where $\displaystyle n~=~2u+1$ for some $\displaystyle u~\in~\mathbb{Z}$ and $\displaystyle u~\geq~1$ are

$\displaystyle 1,~w,~w^2,....,~w^{2u}$ where $\displaystyle ~w~=~e^{\frac{2\pi i}{2u+1}}$

But that seems a tedious thing to do.
• Aug 20th 2008, 10:10 PM
Happy Dancer
Thanks, also whats the best way to prove that their sum is zero, is it just to state the formula for the geometric series?
• Aug 20th 2008, 10:53 PM
mr fantastic
Quote:

Originally Posted by Happy Dancer
Thanks, also whats the best way to prove that their sum is zero, is it just to state the formula for the geometric series?

Read this: Root of unity - Wikipedia, the free encyclopedia