Question:
Let $z_{0}$ be a root of the quadratic equation, $x^{2}+x+1=0$. If $z=3+6 i z_{0}^{81}-3 i z_{0}^{93}$, then arg $z$ is equal to:
Correct Option: 1
Solution:
$z_{0}=\omega$ or $\omega^{2}$ (where $\omega$ is a non-real cube root of unity)
$z=3+6 i(\omega)^{81}-3 i(\omega)^{93}$
$z=3+3 i$
$\Rightarrow \arg z=\frac{\pi}{4}$