# How do you convert  -2 + 2i to polar form?

Jul 3, 2016

$- 2 + 2 i = \sqrt{8} \left[\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right]$

#### Explanation:

First we want to draw an Argand diagram.

The modulus (or length) of the complex number is simply given by using Pythagoras' theorem to find the hypotenuse, ie

$r = \sqrt{{x}^{2} + {y}^{2}} = \sqrt{{\left(- 2\right)}^{2} + {2}^{2}} = \sqrt{8}$

The argument is defined as the angle from the x-axis in the counter-clockwise direction. In this case, the angle we find is in the second quadrant, so we find the 'positive' version and then need to take it away from $\pi$.

$\theta = \pi - \arctan \left(\frac{y}{x}\right) = \pi - \arctan \left(\frac{2}{2}\right) = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$