One of the values of (-64 i)^{5 / 6} is | vedclass

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(B) We have $(-64 i)^{5 / 6} = (64)^{5 / 6} 	imes (-i)^{5 / 6}$.Since $-i = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}) = e^{i(\frac{3\pi}{2} + 2k\

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$16 \sqrt{2}(1+i)$

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(B) We have $(-64 i)^{5 / 6} = (64)^{5 / 6} 	imes (-i)^{5 / 6}$.Since $-i = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}) = e^{i(\frac{3\pi}{2} + 2k\pi)}$,we have:$(-64 i)^{5 / 6} = 32 	imes (e^{i(\frac{3\pi}{2} + 2k\pi)})^{5 / 6} = 32 	imes e^{i(\frac{15\pi}{12} + \frac{10k\pi}{6})}$.For $k = 3$,the exponent becomes $i(\frac{15\pi}{12} + 5\pi) = i(\frac{5\pi}{4} + 5\pi) = i(\frac{25\pi}{4})$.Since $\frac{25\pi}{4} = 6\pi + \frac{\pi}{4}$,we have $e^{i(6\pi + \pi/4)} = e^{i\pi/4} = \cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4})$.Thus,the value is $32(\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) = 16\sqrt{2}(1+i)$.

One of the values of $(-64 i)^{5 / 6}$ is

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