How do you use DeMoivre's theorem to simplify ${(1 - \sqrt{3} i)}^{3}$ $(1-sqrt3i)^3$ ?

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Answer 1 · 2016-11-14T16:39:05

${(1 - \sqrt{3} i)}^{3} = - 8$ $(1-sqrt3i)^3=-8$

According to DeMoivre's theorem

${(r (cos θ + i sin θ))}^{n} = r^{n} (cos n θ + i sin n θ)$ $(r(costheta+isintheta))^n=r^n(cosntheta+isinntheta)$

Now let $1 - \sqrt{3} i = r (cos θ + i sin θ)$ $1-sqrt3i=r(costheta+isintheta)$

hence $r cos θ = 1$ $rcostheta=1$ and $r sin θ = - \sqrt{3}$ $rsintheta=-sqrt3$

hence squaring and adding $r^{2} = (1^{2} + {(- \sqrt{3})}^{2}) = 1 + 3 = 4$ $r^2=(1^2+(-sqrt3)^2)=1+3=4$

and $r = 2$ $r=2$ , $cos θ = \frac{1}{2}$ $costheta=1/2$ and $sin θ = - \frac{\sqrt{3}}{2}$ $sintheta=-sqrt3/2$

hence, $θ = - \frac{π}{3}$ $theta=-pi/3$

Therefore $1 - \sqrt{3} i = 2 (cos (- \frac{π}{3}) + i sin (- \frac{π}{3}))$ $1-sqrt3i=2(cos(-pi/3)+isin(-pi/3))$ and using DeMoivre's theorem

${(1 - \sqrt{3} i)}^{3} = 2^{3} (cos (- 3 \frac{π}{3}) + i sin (- 3 \frac{π}{3}))$ $(1-sqrt3i)^3=2^3(cos(-3pi/3)+isin(-3pi/3))$

= $2^{3} (cos (- π) + i sin (- π))$ $2^3(cos(-pi)+isin(-pi))$

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

= $- 8$ $-8$

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