How do you find the square root of #5(cos120+isin120)#?

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Answer 1 · 2016-09-13T12:59:09

#+-sqrt 5/2(1+isqrt 3)#

Use #cos(180^o + a)=-cos a and sin (180^o + a)=-sin a#

#(5(cos 120^o + i sin 120^o))^(1/2)#

#sqrt 5(cos(2k(180)^o + 120^o)+i sin (2k(180)^o + 120^o))^(1/2#

#=sqrt 5(cos((2k(180) + 120^o)/2)+i sin(( (2k(180)^o + 120^o))/2)). k=0,1#

#=sqrt 5(cos 60^o + i sin 60^o)=sqrt 5/2(1+i sqrt 3)#, for k =0 and

#=sqrt 5(cos 240^o + i sin 240^o)#, for k = 1

#=sqrt 5(cos(180^o +60^o) + i sin(180^o + 60^o) #

#=-sqrt 5(cos 60^o + i sin 60^o) #

#=-sqrt 5/2(1+isqrt 3)#