All the values of (8i)^{frac{1}{3}} are | vedclass

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(B) Let $z = 8i = 8(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = 8e^{i(\frac{\pi}{2} + 2k\pi)}$ for $k = 0, 1, 2$. Taking the cube root,$z^{\frac{1}{3

Vedclass · Accepted Answer

$\pm \sqrt{3}+i, -2i$

Vedclass · Answer

(B) Let $z = 8i = 8(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}) = 8e^{i(\frac{\pi}{2} + 2k\pi)}$ for $k = 0, 1, 2$. Taking the cube root,$z^{\frac{1}{3}} = 2e^{i(\frac{\pi}{6} + \frac{2k\pi}{3})}$. For $k=0$: $2e^{i\frac{\pi}{6}} = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2(\frac{\sqrt{3}}{2} + i \frac{1}{2}) = \sqrt{3} + i$. For $k=1$: $2e^{i(\frac{\pi}{6} + \frac{2\pi}{3})} = 2e^{i\frac{5\pi}{6}} = 2(\cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i \frac{1}{2}) = -\sqrt{3} + i$. For $k=2$: $2e^{i(\frac{\pi}{6} + \frac{4\pi}{3})} = 2e^{i\frac{9\pi}{6}} = 2e^{i\frac{3\pi}{2}} = 2(0 - i) = -2i$. Thus,the values are $\pm \sqrt{3} + i, -2i$.

All the values of $(8i)^{\frac{1}{3}}$ are

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