How do you find ${[2 ({cos 120}^{\circ} + i {sin 120}^{\circ})]}^{5}$ $[2(\cos 120^\circ + i \sin 120^\circ)]^5$ using the De Moivre's theorem?

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Answer 1 · 2014-10-23T22:03:04

De Moivre's Theorem

${(cos θ + i sin θ)}^{n} = cos (n θ) + i sin (n θ)$ $(cos theta+i sin theta)^n=cos(n theta)+isin(n theta)$

Let us now look at the posted problem.

${[2 ({cos 120}^{\circ} + i {sin 120}^{\circ})]}^{5}$ $[2(cos120^{circ}+isin120^{circ})]^5$

by the exponential property ${(a b)}^{n} = a^{n} b^{n}$ $(ab)^n=a^nb^n$ ,

$= 2^{5} {({cos 120}^{\circ} + i {sin 120}^{\circ})}^{5}$ $=2^5(cos120^{circ}+i sin120^{circ})^5$

by De Moivre's Theorem,

$= 32 ({cos 600}^{\circ} + i {sin 600}^{\circ})$ $=32(cos600^{circ}+i sin600^{circ})$

since $600^{\circ} = 360^{\circ} + 240^{\circ}$ $600^{circ}=360^{circ}+240^{circ}$ ,

$= 32 ({cos 240}^{\circ} + i {sin 240}^{\circ})$ $=32(cos240^{circ}+isin240^{circ})$

since ${sin 240}^{\circ} = - \frac{1}{2}$ $sin240^circ=-1/2$ and ${cos 240}^{\circ} = - \frac{\sqrt{3}}{2}$ $cos240^circ=-sqrt{3}/2$ ,

$= 32 (- \frac{1}{2} - \frac{\sqrt{3}}{2} i)$ $=32(-1/2-sqrt{3}/2i)$

by factoring out $- \frac{1}{2}$ $-1/2$ ,

$= - 16 (1 + \sqrt{3} i)$ $=-16(1+sqrt{3}i)$

I hope that this was helpful.

How do you find [2(\cos 120^\circ + i \sin 120^\circ)]^5[2(cos120∘+isin120∘)]5[2(\cos 120^\circ + i \sin 120^\circ)]^5 using the De Moivre's theorem?