How do you use DeMoivre's theorem to simplify ${(\sqrt{3} + i)}^{8}$ $(sqrt3+i)^8$ ?

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Answer 1 · 2016-11-21T09:07:30

${(\sqrt{3} + i)}^{8} = - 128 - 128 \sqrt{3} \cdot i$ $(sqrt3+i)^8=-128-128sqrt3*i$

DeMoivre's theorem states that given a complex number in its trigonometric form $r (cos θ + i sin θ)$ $r(costheta+isintheta)$ ,

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

Now let $\sqrt{3} + i = r cos θ + i r sin θ$ $sqrt3+i=rcostheta+irsintheta$ , which means

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

$r = \sqrt{{(\sqrt{3})}^{2} + 1^{2}} = \sqrt{3 + 1} = \sqrt{4} = 2$ $r=sqrt((sqrt3)^2+1^2)=sqrt(3+1)=sqrt4=2$

as such $cos θ = \frac{\sqrt{3}}{2}$ $costheta=sqrt3/2$ and $sin θ = \frac{1}{2}$ $sintheta=1/2$ i.e. $θ = \frac{π}{6}$ $theta=pi/6$

Hence, ${(\sqrt{3} + i)}^{8} = {(2 (cos (\frac{π}{6}) + i sin (\frac{π}{6})))}^{8}$ $(sqrt3+i)^8=(2(cos(pi/6)+isin(pi/6)))^8$

and using DEMoivre's Theorem, this is equal to

$2^{8} (cos (\frac{8 π}{6}) + i sin (\frac{8 π}{6}))$ $2^8(cos((8pi)/6)+isin((8pi)/6))$

= $256 (- \frac{1}{2} - i \cdot \frac{\sqrt{3}}{2})$ $256(-1/2-i*sqrt3/2)$

= $- 128 - 128 \sqrt{3} \cdot i$ $-128-128sqrt3*i$

How do you use DeMoivre's theorem to simplify (√3+i)8(sqrt3+i)^8?