Find the value of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$
Let $\left(- \sqrt{3} - i\right) = r \left(\cos \theta + i \sin \theta\right)$
Then, $\;$ $r \cos \theta = - \sqrt{3}$; $\;\;$ $r \sin \theta = -1$
$\therefore$ $\;$ $r = \sqrt{\left(- \sqrt{3}\right)^2 + \left(-1\right)^2} = 2$
Now, $\;$ $\cos \theta = - \dfrac{\sqrt{3}}{2}$; $\;$ $\sin \theta = - \dfrac{1}{2}$ $\implies$ $\theta = -\pi + \dfrac{\pi}{6} = \dfrac{- 5 \pi}{6}$
$\begin{aligned}
\therefore \; \left(- \sqrt{3} - i\right)^{\frac{2}{3}} & = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5\pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^{\frac{2}{3}} \\\\
& = 2 ^{\frac{2}{3}} \left\{\left[\cos \left(\dfrac{-5 \pi}{6}\right) + i \sin \left(\dfrac{-5 \pi}{6}\right)\right]^2\right\}^{\frac{1}{3}} \\\\
& = 2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{3}\right) + i \sin \left(\dfrac{-5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\
& = 2^{\frac{2}{3}} \left[\cos \left(2 k \pi - \dfrac{5 \pi}{3}\right) + i \sin \left(2 k \pi - \dfrac{5 \pi}{3}\right)\right]^{\frac{1}{3}} \\\\
& = 2^{\frac{2}{3}} \left\{\cos \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] + i \sin \left[\left(6k - 5\right) \dfrac{\pi}{9}\right] \right\} \;\;\; where \; k = 0, 1, 2
\end{aligned}$
$\therefore$ $\;$ The values of $\;$ $\left(- \sqrt{3} - i\right)^{\frac{2}{3}}$ $\;$ are
$2^{\frac{2}{3}} \left[\cos \left(\dfrac{-5 \pi}{9}\right) + i \sin \left(\dfrac{-5 \pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{\pi}{9}\right) + i \sin \left(\dfrac{\pi}{9}\right)\right]$, $\;$ $2^{\frac{2}{3}} \left[\cos \left(\dfrac{7\pi}{9}\right) + i \sin \left(\dfrac{7\pi}{9}\right)\right]$