The roots of (2 - 2i)^{1/3} are | vedclass

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(A) Let $z = 2 - 2i$. The modulus $r = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.The argument $	heta = 	an^{-1}(\frac{-2}{2}) = -\frac{\pi}{4}$.So

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$\sqrt{2} \left( \cos \frac{\pi}{12} - i\sin \frac{\pi}{12} \right), \sqrt{2} \left( -\sin \frac{\pi}{12} + i\cos \frac{\pi}{12} \right), -1 - i$

Vedclass · Answer

(A) Let $z = 2 - 2i$. The modulus $r = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.The argument $	heta = 	an^{-1}(\frac{-2}{2}) = -\frac{\pi}{4}$.So,$z = 2\sqrt{2} \left( \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) ight) = 2\sqrt{2} \left( \cos(\frac{\pi}{4}) - i\sin(\frac{\pi}{4}) ight)$.Using De Moivre's theorem,the roots are $z_k = (2\sqrt{2})^{1/3} \left( \cos \frac{2k\pi - \pi/4}{3} + i\sin \frac{2k\pi - \pi/4}{3} ight)$ for $k = 0, 1, 2$.For $k=0$: $\sqrt{2} \left( \cos \frac{\pi}{12} - i\sin \frac{\pi}{12} ight)$.For $k=1$: $\sqrt{2} \left( \cos \frac{7\pi}{12} + i\sin \frac{7\pi}{12} ight) = \sqrt{2} \left( -\sin \frac{\pi}{12} + i\cos \frac{\pi}{12} ight)$.For $k=2$: $\sqrt{2} \left( \cos \frac{15\pi}{12} + i\sin \frac{15\pi}{12} ight) = \sqrt{2} \left( \cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4} ight) = \sqrt{2} \left( -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} ight) = -1 - i$.

The roots of $(2 - 2i)^{1/3}$ are

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