Let z = - 1 + 3 i 2 , where i = - 1 , and r, s ∈ \ 1, 2, 3\ . Let P...

\text{z} = \text{\omega}$($\text{\omega}$is complex cube root of$unity)
\text{P} =\left[$\begin{cases}{\left(- \text{\omega}\right)}^{\text{r}} & {\text{\omega}}^{2 \text{s}} \\ {\text{\omega}}^{2 \text{s}} & {\text{\omega}}^{\text{r}}\end{cases}$\right]$,${$$\text{P}$$}^{2} = - \text{I}$<br />$⇒ {\text{P}}^{2} = \left[$\begin{cases}{\text{\omega}}^{2 \text{r}} + {\text{\omega}}^{4 \text{s}} &{\text{\omega}}^{\text{r} +2\text{s}}\left({\left(- 1\right)}^{\text{r}} + 1\right) \\ {\text{\omega}}^{\text{r} +2 \text{s}}\left({\left(- 1\right)}^{\text{r}} + 1\right) &{\text{\omega}}^{4 \text{s}} + {\text{\omega}}^{2 \text{r}}\end{cases}$\right] = - $\text{I}$=$\left[\begin{cases}- 1 & 0 \\ 0 & - 1\end{cases}\right]
⇒ {\left(- 1\right)}^{$\text{r}$} + 1 = 0 ⇒ r$is odd$⇒ $\text{r}$=1, 3<br />Also {$\text{\omega}$}^{2 \text{r}} + {$\text{\omega}$}^{4 \text{s}} = - 1$<br />$∴ $\text{r}$ \ne 3$<br />By$$\text{r}$=1 ⇒ {$\text{\omega}$}^{2} + {$\text{\omega}$}^{4$\text{s}$} = - 1 ⇒ {$\text{\omega}$}^{4$\text{s}$} =$\text{\omega}$={$\text{\omega}$}^{4}
$⇒$$\text{s}$$=1$
\left($\text{r}$$,$$\text{s}$\right) = \left(1, 1\right)$
Only 1 pair