Question #0d253 Precalculus Complex Numbers in Trigonometric Form Complex Number Plane 1 Answer Cesareo R. Aug 26, 2016 #sqrt(i) = pm sqrt(2)/2(1+i)# Explanation: 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)# Answer link Related questions What is the complex number plane? Which vectors define the complex number plane? What is the modulus of a complex number? How do I graph the complex number #3+4i# in the complex plane? How do I graph the complex number #2-3i# in the complex plane? How do I graph the complex number #-4+2i# in the complex plane? How do I graph the number 3 in the complex number plane? How do I graph the number #4i# in the complex number plane? How do I use graphing in the complex plane to add #2+4i# and #5+3i#? How do I use graphing in the complex plane to subtract #3+4i# from #-2+2i#? See all questions in Complex Number Plane Impact of this question 1697 views around the world You can reuse this answer Creative Commons License