The value of i^{1/3} is | vedclass

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Question

(A) We know that $i = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})$.Using De Moivre's theorem,$i^{1/3} = (\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}))^{1

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$\frac{\sqrt{3} + i}{2}$

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(A) We know that $i = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})$.Using De Moivre's theorem,$i^{1/3} = (\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}))^{1/3} = \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})$.Evaluating the trigonometric values,we get $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$.Thus,$i^{1/3} = \frac{\sqrt{3}}{2} + i\frac{1}{2} = \frac{\sqrt{3} + i}{2}$.

The value of $i^{1/3}$ is

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