JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Complex Numbers

The equation $x^4+x^2+1=0$ can be factored as $(x^2+x+1)(x^2-x+1)=0$. This gives roots $x=\pm \omega ,\pm {\omega }^2$ where $\omega =e^{2\pi i/3}$ is a primitive cube root of unity (so ${\omega }^3=1$ and $1+\omega +{\omega }^2=0$). For any such root $\alpha$, we have ${\alpha }^3=\pm 1$. Computing the powers modulo 3: $1011=337\times 3$, $2022=674\times 3$, $3033=1011\times 3$. Thus ${\alpha }^{1011}=({\alpha }^3{)}^{337}=(\pm 1{)}^{337}=\pm 1$, and similarly for the other terms. Checking carefully: ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}=1+1-1=1$.