# How do you convert 4sqrt3-4i to polar form?

$r = 8$
$\theta = - \frac{\pi}{6}$
$4 \sqrt[2]{3} - i 4$ is a vector on the Gauss plane whose length is
$r = \sqrt[2]{{4}^{2} \cdot 3 + {4}^{2}} = 8$ and the angle referred to the real positive semi axis is $\arctan \left(- \frac{1}{\sqrt[2]{3}}\right) = - \frac{\pi}{6}$