Let Z0 be a root of the quadratic equation,

$\because z_{0}$ is a root of quadratic equation

$x^{2}+x+1=0$

$\therefore \quad z_{0}=\omega$ or $\omega^{2} \Rightarrow z_{0}^{3}=1$

$\therefore \quad z=3+6 i z_{0}^{81}-3 i z_{0}^{93}$

$=3+6 i\left(\left(z_{0}\right)^{3}\right)^{27}-3 i\left(\left(z_{0}\right)^{3}\right)^{31}$

$=3+6 i-3 i$

$=3+3 i$

$\therefore \quad \arg (z) \tan ^{-1}\left(\frac{3}{3}\right)=\frac{\pi}{4}$