Find the value of (−√3−i)23
Let (−√3−i)=r(cosθ+isinθ)
Then, rcosθ=−√3; rsinθ=−1
∴ \; 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]