It says we define $w={\normalsize e}^{\Large{\frac{2\pi i}{N}}}$, the $N^{th}$ root of unity (that is $w^N=1$ )
also, $\bar{w}$ is the complex conjugate of $w$ that is since $w={\normalsize e}^{\Large{\frac{2\pi i}{N}}}$ then $\bar{w}$ $={\normalsize e}^{\Large{\frac{-2\pi i}{N}}}$
I am wondering should the complex conjugate of the root of unity be also 1. It is the only thing that makes sense to me to solve the below problem
so what i means is
$\bar{w}^N$=1=${w}^N$
also, as you will see below. is it also legal to do
$\bar{w}^{\textbf{-N}}$=1=${w}^N$ where that N is negative in the first part if you cannot see it. If that is true than the calculation below would be correct.
I have read that in the root of unity it is always 1, so the complex conjugate would also have to be 1 in order for it to be a root of unity.
My overall question is I am trying to do the below
$(\bar{w}^{(N-2)})^{N-2}$ needs to turn into $\bar{w}^4$
$(\bar{w}^{(N-1)})^{N-2}$ needs to turn into $\bar{w}^2$
$(\bar{w}^{(N-1)})^{N-1}$ needs to turn into $\bar{w}$
here is what i did for each of them
$(\bar{w}^{(N-2)})^{N-2}$ =$\bar{w}^{(N^2-4N+4)}$=$(\bar{w}^N)^N(\bar{w}^{-4N})(\bar{w}^4)$=(1)(1)($\bar{w}^4$)=$\bar{w}^4$
$(\bar{w}^{(N-1)})^{N-2}$=$\bar{w}^{N^2-3N+2}$ =$(\bar{w}^N)^N$ $(\bar{w}^{-3N})$ $\bar{w}^2$ = (1)(1)($\bar{w}^2$)=$(\bar{w}^2)$
$(\bar{w}^{(N-1)})^{N-1}$=$\bar{w}^{N^2-2N+1}$=$(\bar{w}^N)^N$ $(\bar{w}^{-2N})$ $(\bar{w})$=(1)(1)($\bar{w}$)=$(\bar{w})$
since we have the root of unity I am thinking i have completed the above part correctly.
This is all being used for a Vandermonde Matrix
