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.
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.
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.