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

Oct 23, 2014

De Moivre's Theorem

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos \left(n \theta\right) + i \sin \left(n \theta\right)$

Let us now look at the posted problem.

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

by the exponential property ${\left(a b\right)}^{n} = {a}^{n} {b}^{n}$,

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

by De Moivre's Theorem,

$= 32 \left(\cos {600}^{\circ} + i \sin {600}^{\circ}\right)$

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

$= 32 \left(\cos {240}^{\circ} + i \sin {240}^{\circ}\right)$

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

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

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

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

I hope that this was helpful.