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Question

(C) We know that $i = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})$.Applying De Moivre's theorem for the square root:$\sqrt{i} = (i)^{1/2} = [\cos(\frac

Vedclass · Accepted Answer

$\pm \frac{1 + i}{\sqrt{2}}$

Vedclass · Answer

(C) We know that $i = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})$.Applying De Moivre's theorem for the square root:$\sqrt{i} = (i)^{1/2} = [\cos(\frac{\pi}{2} + 2n\pi) + i\sin(\frac{\pi}{2} + 2n\pi)]^{1/2}$ for $n = 0, 1$.For $n = 0$:$\cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} = \frac{1 + i}{\sqrt{2}}$.For $n = 1$:$\cos(\frac{5\pi}{4}) + i\sin(\frac{5\pi}{4}) = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} = -\frac{1 + i}{\sqrt{2}}$.Thus,$\sqrt{i} = \pm \frac{1 + i}{\sqrt{2}}$.Trick: Squaring the option $\pm \frac{1 + i}{\sqrt{2}}$ gives $\frac{(1 + i)^2}{2} = \frac{1 + 2i + i^2}{2} = \frac{1 + 2i - 1}{2} = i$.

$\sqrt{i} = $

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