How do you find the square root of #16(cos60^@+isin60^@)#?

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Answer 1 · 2016-12-22T19:58:49

#2sqrt3+2i#

#z=16(cos60^0+isin60^0)#

#sqrtz=z^(1/2)=sqrt(16(cos60^0+isin60^0)#

#z^(1/2)=4(cos60+isin60)^(1/2)#

now using DeMoivre's theorem

#(costheta+isintheta)^n=cosntheta+isinntheta, AA n in RR#

#z^(1/2)=4(cos(60/2)+isin(60/2))#

#z^(1/2)=4(cos30^0+isin30^0)#

#z^(1/2)=4(sqrt3/2+i1/2)#

#z^(1/2)=2sqrt3+2i#