1. ## 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.

2. 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.

3. Thanks, also whats the best way to prove that their sum is zero, is it just to state the formula for the geometric series?

4. 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