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

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Answer 1 · 2014-10-23T22:03:04

De Moivre's Theorem

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

Let us now look at the posted problem.

$[2(cos120^{circ}+isin120^{circ})]^5$

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

$=2^5(cos120^{circ}+i sin120^{circ})^5$

by De Moivre's Theorem,

$=32(cos600^{circ}+i sin600^{circ})$

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

$=32(cos240^{circ}+isin240^{circ})$

since $sin240^circ=-1/2$ and $cos240^circ=-sqrt{3}/2$ ,

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

by factoring out $-1/2$ ,

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

I hope that this was helpful.

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