Find the square root of $- i$

\pm \frac{1}{\sqrt{(2)}} (1 + i) .

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\pm \frac{1}{\sqrt{(2)}} (1 - i) .

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\pm \frac{1}{2} (1 + i) .

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\pm \frac{1}{2} (1 - i) .

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Solution

The correct option is B $\pm \frac{1}{\sqrt{(2)}} (1 - i) .$
Let $\sqrt{- i} = x + i y .$ Then $- i = x^{2} - y^{2} + 2 i x y,$
$∴ x^{2} - y^{2} = 0$ and $2 x y = - 1$ .
Solving these, we get $x = - y = \pm \frac{1}{\sqrt{(2)}} .$
Hence $\sqrt{i} = \pm (\frac{1}{\sqrt{(2)}} - \frac{1}{\sqrt{(2)}} i) = \pm \frac{1}{\sqrt{(2)}} (1 - i) .$
Ans: B

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