# How do you use DeMoivre's theorem to simplify (5(cos(pi/9)+isin(pi/9)))^3?

Aug 6, 2016

$= 125 \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$

Could also write as $125 {e}^{\frac{i \pi}{3}}$ using Euler's formula if you so desired.

#### Explanation:

De Moivre's theorem states that for complex number

$z = r \left(\cos \theta + i \sin \theta\right)$

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

So here,

$z = 5 \left(\cos \left(\frac{\pi}{9}\right) + i \sin \left(\frac{\pi}{9}\right)\right)$

${z}^{3} = {5}^{3} \left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$

$= 125 \left(\frac{1}{2} + \frac{\sqrt{3}}{2} i\right)$