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

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

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

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

= $-8$

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