One of the values of (-32 i)^{frac{2}{5}} is | vedclass

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(B) Given expression: $(-32 i)^{\frac{2}{5}}$We can write $-i$ as $\cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2}$.So,$(-32 i)^{\frac{2}{5}} = (32)^{\f

Vedclass · Accepted Answer

$4 \operatorname{cis} \frac{3 \pi}{5}$

Vedclass · Answer

(B) Given expression: $(-32 i)^{\frac{2}{5}}$We can write $-i$ as $\cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2}$.So,$(-32 i)^{\frac{2}{5}} = (32)^{\frac{2}{5}} (-i)^{\frac{2}{5}} = 4 (\cos \frac{3 \pi}{2} + i \sin \frac{3 \pi}{2})^{\frac{2}{5}}$.Using De Moivre's theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n 	heta) + i \sin(n 	heta)$,we consider the principal value:$= 4 (\cos(\frac{3 \pi}{2} 	imes \frac{2}{5}) + i \sin(\frac{3 \pi}{2} 	imes \frac{2}{5}))$$= 4 (\cos \frac{3 \pi}{5} + i \sin \frac{3 \pi}{5})$$= 4 \operatorname{cis} \frac{3 \pi}{5}$.

One of the values of $(-32 i)^{\frac{2}{5}}$ is

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