Question

If $\alpha$ is a non real root of $z=(1{)}^{1/5},$ then the value of $z^{(1+\alpha +{\alpha }^2+{\alpha }^{-2}+{\alpha }^{-1})}$ is

Answer 1

$1+\alpha +{\alpha }^2+{\alpha }^{-2}+{\alpha }^{-1}$
$=1+\alpha +{\alpha }^2+{\alpha }^3+{\alpha }^4(\because {\alpha }^5=1)$
$=0(\because \text{sum of roots is zero})$
$\therefore z^{(1+\alpha +{\alpha }^2+{\alpha }^{-2}+{\alpha }^{-1})}=z^0=1$