<s>[INST] Polar form from a complex number? How do I solve for the polar form of $-3\sqrt{2} - 3\sqrt{3}\,i\; ?$ 
I think I solved for $r$ which is $3\sqrt{5}$ from using $r=\sqrt{a^2+b^2}.$
When I look for theta I use $\tan\theta = \frac ab,$ but when doing so I get $\frac{\sqrt{6}}{ 2}.$ Anyway I looked in the answer key for $\theta$ and it's in degrees. Around $230.8^{\circ}$. How do I figure that out from $\frac{\sqrt{6}}{ 2}?$ Did I even do it right ?
 [/INST] Same way always:
$z = -3\sqrt{2} + 3\sqrt 3 i = re^{i\theta}$ where $r = |-3\sqrt{2} + 3\sqrt 3 i| = \sqrt{(-3\sqrt{2})^2 + (3\sqrt 3)^2} = \sqrt{9*2 + 9*3}=\sqrt {45} = 3\sqrt 5$.
And $\theta$ is so that $\cos \theta = \frac {-3\sqrt{2}}r=\frac {-3\sqrt{2}}{3\sqrt 5}=\frac {-\sqrt 2}{\sqrt 5}; \sin\theta = \frac {3\sqrt{3}}r= \frac {3\sqrt{3}}{3\sqrt 5}=\frac {\sqrt 3}{\sqrt 5}; \tan \theta = \frac {3\sqrt{3}}{-3\sqrt {2}}= -\frac{\sqrt{3}}{\sqrt{2}}$.
As $\cos \theta <0$ and $\sin \theta > 0$ we know $\frac \pi 2 < \theta < \pi$.
$\theta = \arctan \frac {-\sqrt 2}{\sqrt 5} = 2.45687$
So $z = 3\sqrt 5e^{2.45687i}$.
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