JEE Advanced 2016 · Paper 1, Mathematics question 54

Simplify. $z=\omega$, a cube root of unity. Then
$$P^2=\begin{bmatrix}\omega^{2r}+\omega^{4s}&\omega^{2s}\left((-\omega)^r+\omega^r\right)\\\omega^{2s}\left((-\omega)^r+\omega^r\right)&\omega^{4s}+\omega^{2r}\end{bmatrix}.$$

Off-diagonal zero. $(-1)^r=-1$, so $r$ is odd: $r=1$ or $3$.

Diagonal $=-1$.

• $r=1$: $\omega^2+\omega^{4s}=-1$ needs $\omega^{4s}=\omega$, so $s=1$.
• $r=3$: $1+\omega^{4s}=-1$ is impossible.

Only $(r,s)=(1,1)$ works.

Answer: 1