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Complex Numbers

Find the value of (3i)23


Let (3i)=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]