# How do you simplify z^6 = a + bi given z = sqrt 3 + i?

Jul 8, 2016

$a = - 64 , b = 0$

#### Explanation:

if you are familiar with Euler's rule $z = {e}^{i x} = \cos x + i \sin x$ we can express z as

$z = 2 \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = 2 \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right) = 2 {e}^{i \frac{\pi}{6}}$

thus ${z}^{6} = {\left(2 {e}^{i \frac{\pi}{6}}\right)}^{6}$

$= {2}^{6} \cdot {\left({e}^{i \frac{\pi}{6}}\right)}^{6}$

$= 64 \cdot {e}^{i \pi}$

and then reversing back into original form

$= 64 \left(\cos \pi + i \sin \pi\right)$

$= 64 \left(- 1 + i \left(0\right)\right)$

$= - 64$