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.