# How do you convert 3 sqrt2 - 3 sqrt 2i to polar form?

In Polar form the equation is $6 \left(\cos \left(- 0.785\right) + i \sin \left(- 0.785\right)\right)$
$Z = 3 \sqrt{2} - 3 \sqrt{2} i$ so Modulas $Z = \sqrt{{\left(3 \sqrt{2}\right)}^{2} + {\left(- 3 \sqrt{2}\right)}^{2}} = 6$
Argument $z = \theta = {\tan}^{-} 1 \left(- \frac{3 \sqrt{2}}{3 \sqrt{2}}\right) = {\tan}_{1} \left(- 1\right) = - 0.785 r a \mathrm{di} a n$ So in Polar form the equation is $6 \left(\cos \left(- 0.785\right) + i \sin \left(- 0.785\right)\right)$