Question #0d253

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Answer 1 · 2016-08-26T23:14:22

#sqrt(i) = pm sqrt(2)/2(1+i)#

Using de Moivre's identity

#e^(ix) = cosx + i sin x#

we have

#e^(i (pi/2+2kpi))=i, k=0,pm1,pm2,cdots# so

#sqrt(i) = sqrt(e^(i (pi/2+2kpi)))=e^(i( pi/4+kpi))=e^(i(pi)/4)e^(i kpi)#

but

#e^(i(pi)/4)=cos(pi/4)+isin(pi/4)# and #e^(ikpi) = (-1)^k#

Finally

#sqrt(i) =pmsqrt(2)/2(1+i)#