Question

Let $z=\frac{-1+\sqrt{3}i}{2}$, where $i=\sqrt{-1}$, and $r,s\in \{1,2,3\}$. Let $P=\left[\begin{array}{cc}(-z{)}^r& z^{2s}\\ z^{25}& z^r\end{array}\right]$ and I be the identity matrix

Answer 1

Answer: 1

$z=\omega$
$P^2=\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{25}\\ {\omega }^{2s}& {\omega }^r\end{array}\right]\left[\begin{array}{cc}(-\omega {)}^r& {\omega }^{25}\\ {\omega }^{2s}& {\omega }^T\end{array}\right]$
$P^2=\left[\begin{array}{cc}{\omega }^{2r}+{\omega }^{4s}& (-\omega {)}^T{\omega }^{2s}+{\omega }^{2s}{\omega }^r\\ {\omega }^{25}(-\omega {)}^r+{\omega }^r{\omega }^{2s}& {\omega }^{45}+{\omega }^{2r}\end{array}\right]=\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]$
$\therefore (-1{)}^r{\omega }^r{\omega }^{2s}+{\omega }^{2s}{\omega }^T=0$
So $r=1,3$
Also ${\omega }^{2r}+{\omega }^{4s}=-1$
when $r=1$
${\omega }^2+{\omega }^{4s}=-1$
$\therefore s=1$
when $r=3$
${\omega }^6+{\omega }^{4s}=-1$
$1+{\omega }^{4s}=-1$
Not possible.