Question

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

Answer 1

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