# How do you convert 4-4i to polar form?

Apr 18, 2016

$4 \sqrt{2} {e}^{i \frac{7 o u}{4}} = 4 \sqrt{2} \left(\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right)\right)$

#### Explanation:

$x = 4 \mathmr{and} y = - 4$.
$x + i y = r \left(\cos \theta + i \sin \theta\right)$

$r = \sqrt{{x}^{2} + {y}^{2}} = 4 \sqrt{2}$.

$\cos \theta = \frac{x}{r} = \frac{1}{\sqrt{2}} , \sin \theta = \frac{y}{r} = - \frac{1}{\sqrt{2}}$

sine is negative and cosine is positive.
So, $\theta$ is in the fourth quadrant,
$\theta = 2 \pi - \frac{\pi}{4} = \frac{7 \pi}{4}$.