# How do you find z, z^2, z^3, z^4 given z=1/2(1+sqrt3i)?

Aug 13, 2016

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

${z}^{2} = \cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right) = \frac{1}{2} \left(- 1 + \sqrt{3} i\right)$

${z}^{3} = \cos \left(3 \frac{\pi}{3}\right) + i \sin \left(3 \frac{\pi}{3}\right) = - 1$

${z}^{4} = \cos \left(4 \frac{\pi}{3}\right) + i \sin \left(4 \frac{\pi}{3}\right) = - \frac{1}{2} \left(1 + \sqrt{3} i\right)$

#### Explanation:

The easiest method is to use De Moivre's theorem. For complex number $z$

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

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

So we want to convert our complex number to polar form. The modulus $r$ of a complex number $a + b i$ is given by

$r = \sqrt{{a}^{2} + {b}^{2}}$

$r = \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$

The complex number will be in the first quadrant of an Argand diagram so the argument is given by:

$\theta = {\tan}^{- 1} \left(\frac{b}{a}\right)$

$\theta = {\tan}^{- 1} \left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = {\tan}^{- 1} \left(\sqrt{3}\right) = \frac{\pi}{3}$

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

${z}^{2} = \cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right) = \frac{1}{2} \left(- 1 + \sqrt{3} i\right)$

${z}^{3} = \cos \left(3 \frac{\pi}{3}\right) + i \sin \left(3 \frac{\pi}{3}\right) = - 1$

${z}^{4} = \cos \left(4 \frac{\pi}{3}\right) + i \sin \left(4 \frac{\pi}{3}\right) = - \frac{1}{2} \left(1 + \sqrt{3} i\right)$