One of the values of (sqrt{3}-i)^{frac{1}{6}} | vedclass

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(C) Let $Z = (\sqrt{3} - i)^{\frac{1}{6}}$,so $Z^6 = \sqrt{3} - i$. Express $\sqrt{3} - i$ in polar form: $Z^6 = 2 \left( \frac{\sqrt{3}}{2} - i \frac

Vedclass · Accepted Answer

$2^{\frac{1}{6}} \operatorname{cis} \frac{59 \pi}{36}$

Vedclass · Answer

(C) Let $Z = (\sqrt{3} - i)^{\frac{1}{6}}$,so $Z^6 = \sqrt{3} - i$. Express $\sqrt{3} - i$ in polar form: $Z^6 = 2 \left( \frac{\sqrt{3}}{2} - i \frac{1}{2} \right) = 2 \left( \cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}) \right) = 2 \operatorname{cis}(-\frac{\pi}{6})$. Using the general form,$Z^6 = 2 \operatorname{cis}(2K\pi - \frac{\pi}{6})$ for $K = 0, 1, 2, 3, 4, 5$. Taking the $6^{th}$ root,$Z = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{2K\pi - \frac{\pi}{6}}{6} \right) = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{K\pi}{3} - \frac{\pi}{36} \right)$. For $K = 5$: $Z = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{5\pi}{3} - \frac{\pi}{36} \right) = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{60\pi - \pi}{36} \right) = 2^{\frac{1}{6}} \operatorname{cis} \left( \frac{59\pi}{36} \right)$.

One of the values of $(\sqrt{3}-i)^{\frac{1}{6}}$ is

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