How do you use DeMoivre's theorem to simplify #(-1/2+sqrt3/2i)^3#?

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Answer 1 · 2016-11-23T14:03:44

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#

Answer 2 · 2016-11-23T14:05:03

#(-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#