JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Complex Numbers

Note that $z=\omega =e^{i2\pi /3}$ is a primitive cube root of unity, so ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$.





For $P^2=-I$:

1. ${\omega }^{2r}+{\omega }^{4s}=-1$
2. ${\omega }^{2s}[(-\omega {)}^r+{\omega }^r]=0$

From condition 2: either ${\omega }^{2s}=0$ (impossible) or $(-\omega {)}^r+{\omega }^r=0$.

If $r$ is odd: $(-\omega {)}^r=-{\omega }^r$, so $-{\omega }^r+{\omega }^r=0$. ✓

If $r$ is even: $(-\omega {)}^r={\omega }^r$, so $2{\omega }^r=0$. ✗

So $r\in \{1,3\}$.

From condition 1 with $r$ odd: ${\omega }^{2r}+{\omega }^{4s}=-1$, which means $1+{\omega }^{2r}+{\omega }^{4s}=0$.

This is satisfied when $\{1,{\omega }^{2r},{\omega }^{4s}\}=\{1,\omega ,{\omega }^2\}$.

Testing $(r,s)=(1,1)$: ${\omega }^2+{\omega }^2=2{\omega }^2\ne -1$. ✗

Actually, re-examining: for $(r,s)=(1,1)$: ${\omega }^2+{\omega }^2=2{\omega }^2$. We need ${\omega }^{2r}+{\omega }^s=-1$ (reducing exponents mod 3).

Testing systematically with $r\in \{1,3\}$ and $s\in \{1,2,3\}$, only $(r,s)=(1,1)$ works.

Answer: 1.