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.

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- Aug 19th 2008, 09:38 PMHappy Dancerroots 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 AMSean12345
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 PMHappy 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 PMmr fantastic