# How do you use demoivre's theorem to simplify [2(cos((pi)/2)+isin((pi)/2))]^8?

Oct 15, 2016

${\left[2 \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)\right]}^{8} = 256$

#### Explanation:

According to de Moivre's Theorem we can calculate any integer power of a complex number given in trigonometric form.

The theorem says that:

If a complex number $z$ is given in a form:

$z = | z | \left(\cos \varphi + i \sin \varphi\right)$

Then the $n - t h$ power of $z$ is:

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

Using this formula we get:

${\left[2 \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)\right]}^{8} = {2}^{8} \left[\cos \left(\frac{8 \pi}{2}\right) + i \sin \left(\frac{8 \pi}{2}\right)\right] =$

$= 256 \left[\cos 4 \pi + i \sin 4 \pi\right] = 256$