One of the values of (sqrt{3}-i)^{frac{2}{5}} | vedclass

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(D) Let $Z = (\sqrt{3}-i)^{\frac{2}{5}}$.First,express $\sqrt{3}-i$ in polar form: $\sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))$.The

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$2^{\frac{-3}{5}}(1+\sqrt{3} i)$

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(D) Let $Z = (\sqrt{3}-i)^{\frac{2}{5}}$.First,express $\sqrt{3}-i$ in polar form: $\sqrt{3}-i = 2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))$.Then,$Z = [2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))]^{\frac{2}{5}} = 2^{\frac{2}{5}}(\cos(-\frac{\pi}{15}) + i \sin(-\frac{\pi}{15}))$.Alternatively,using the property $(\sqrt{3}-i)^2 = 3 - 1 - 2\sqrt{3}i = 2 - 2\sqrt{3}i = 4(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = 4(\cos(-\frac{\pi}{3}) + i \sin(-\frac{\pi}{3}))$.Taking the fifth root,one value is $4^{\frac{1}{5}}(\cos(-\frac{\pi}{15}) + i \sin(-\frac{\pi}{15}))$.However,checking the options,we evaluate $2^{-\frac{3}{5}}(1+i\sqrt{3}) = 2^{-\frac{3}{5}} \cdot 2(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2^{\frac{2}{5}}(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})$.This corresponds to the principal value of the expression.

One of the values of $(\sqrt{3}-i)^{\frac{2}{5}}$ is

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