How do you use DeMoivre's theorem to simplify #(sqrt3+i)^8#?

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Answer 1 · 2016-11-21T09:07:30

#(sqrt3+i)^8=-128-128sqrt3*i#

DeMoivre's theorem states that given a complex number in its trigonometric form #r(costheta+isintheta)#,

#(r(costheta+isintheta))^n=r^n(cosntheta+isinntheta)#

Now let #sqrt3+i=rcostheta+irsintheta#, which means

#rcostheta=sqrt3# and #rsintheta=1# and hence

#r=sqrt((sqrt3)^2+1^2)=sqrt(3+1)=sqrt4=2#

as such #costheta=sqrt3/2# and #sintheta=1/2# i.e. #theta=pi/6#

Hence, #(sqrt3+i)^8=(2(cos(pi/6)+isin(pi/6)))^8#

and using DEMoivre's Theorem, this is equal to

#2^8(cos((8pi)/6)+isin((8pi)/6))#

= #256(-1/2-i*sqrt3/2)#

= #-128-128sqrt3*i#