How do you convert 4 (sqrt(3) +i) to polar form?

Apr 18, 2016

$8 {e}^{i \left(\frac{\pi}{3}\right)} = 8 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)$

Explanation:

$x = 4 \sqrt{3} = y$
$x + i y = r \left(\cos \theta + i \sin \theta\right)$

$r = \sqrt{{x}^{2} + {y}^{2}} = 8$
$\cos \theta = \frac{x}{r} = \frac{\sqrt{3}}{2} , \sin \theta = \frac{y}{r} = \frac{1}{2}$, Both are positive. $\theta$ is in the 1st quadrant.
$\theta = \frac{\pi}{6}$
So, $4 \left(\sqrt{3} + i\right) = 8 \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}\right)$