JEE Main 2022 (Online) 29th June Evening ShiftQuestion 1
Let $\alpha$ be a root of the equation $1+x^{2}+x^{4}=0$. Then the value of $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}$ is equal to :
Complex Numbers
Solution
$1+x^2+x^4=0\Rightarrow x^6=1$ (since $x^2\ne1$, $x^6-1=(x^2-1)(x^4+x^2+1)$). $1011=6\cdot168+3\Rightarrow\alpha^{1011}=\alpha^3$; $2022=6\cdot337\Rightarrow\alpha^{2022}=1$; $3033=6\cdot505+3\Rightarrow\alpha^{3033}=\alpha^3$. Value $=\alpha^3+1-\alpha^3=1$. Option (A).