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The answer is #=1#
Let #z=-1/2+isqrt3/2#
We have to write #z# in trigonometric form in order to apply DeMoivre's theorem
#z=r(costheta+isintheta)#
#r=1#
#costheta=-1/2#
#sintheta=sqrt3/2#
So, #theta# is located in the 2nd quadrant
#theta=2pi/3#
Therefore, #z=cos((2pi)/3)+isin((2pi)/3)#
We need #z^3#
We use DeMoivre 's theorem, #(costheta+isintheta)^n=cosntheta+isinntheta#
#z^3=(cos((2pi)/3)+isin((2pi)/3))^3#
#=cos(2pi/3*3)+isin(2pi/3*3)#
#=cos2pi+isin2pi#
#=1#
#(-1/2+sqrt3/2i)^3=1#
According to DeMoivre's theorem, if #a+ib=rcostheta+irsintheta#
then #(a+ib)^n=r^n(cosntheta+isinntheta)#
Writing #-1/2+sqrt3/2i# in trigonometric form
#-1/2+sqrt3/2i=cos((2pi)/3)+isin((2pi)/3)#
Hence #(-1/2+sqrt3/2i)^3=1^3(cos(3xx(2pi)/3)+isin(3xx(2pi)/3))#
= #cos(2pi)+isin(2pi)#
= #1+i0#
= #1#
How do you use DeMoivre's theorem to simplify #(-1/2+sqrt3/2i)^3#?
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