How do you use DeMoivre's theorem to simplify #(1-sqrt3i)^3#?

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Answer 1 · 2016-11-14T16:39:05

#(1-sqrt3i)^3=-8#

According to DeMoivre's theorem

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

Now let #1-sqrt3i=r(costheta+isintheta)#

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

hence squaring and adding #r^2=(1^2+(-sqrt3)^2)=1+3=4#

and #r=2#, #costheta=1/2# and #sintheta=-sqrt3/2#

hence, #theta=-pi/3#

Therefore #1-sqrt3i=2(cos(-pi/3)+isin(-pi/3))# and using DeMoivre's theorem

#(1-sqrt3i)^3=2^3(cos(-3pi/3)+isin(-3pi/3))#

= #2^3(cos(-pi)+isin(-pi))#

= #8(-1+i*0)#

= #-8#