Question #0d253

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Question #0d253

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Answer 1 · 2016-08-26T23:14:22

$\sqrt{i} = \pm \frac{\sqrt{2}}{2} (1 + i)$ $sqrt(i) = pm sqrt(2)/2(1+i)$

Explanation:

Using de Moivre's identity

$e^{i x} = cos x + i sin x$ $e^(ix) = cosx + i sin x$

we have

$e^{i (\frac{π}{2} + 2 k π)} = i, k = 0, \pm 1, \pm 2, \dots$ $e^(i (pi/2+2kpi))=i, k=0,pm1,pm2,cdots$ so

$\sqrt{i} = \sqrt{e^{i (\frac{π}{2} + 2 k π)}} = e^{i (\frac{π}{4} + k π)} = e^{i \frac{π}{4}} e^{i k π}$ $sqrt(i) = sqrt(e^(i (pi/2+2kpi)))=e^(i( pi/4+kpi))=e^(i(pi)/4)e^(i kpi)$

but

$e^{i \frac{π}{4}} = cos (\frac{π}{4}) + i sin (\frac{π}{4})$ $e^(i(pi)/4)=cos(pi/4)+isin(pi/4)$ and $e^{i k π} = {(- 1)}^{k}$ $e^(ikpi) = (-1)^k$

Finally

$\sqrt{i} = \pm \frac{\sqrt{2}}{2} (1 + i)$ $sqrt(i) =pmsqrt(2)/2(1+i)$

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Question #0d253

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