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 ?
 A: 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 $z = 3\sqrt 5e^{2.45687i}$.
A: A complex number $z$ can be written as:
$$z=a+bi,\,a,b\in\mathbb{R}$$
To convert it to its polar form, one must obtain the magnitude and angle first. The magnitude can be resolved by using Pythagoras' theorem like so.
$$|z|=\sqrt{\Re^2(z)+\Im^2(z)}=\sqrt{a^2+b^2}$$
Calculating the angle can be obtained through trigonometry (multiples of $\tau$ may be added due to the periodic nature of the angle):
$$\theta=\operatorname{atan2}(\Im(z),\Re(z))+\tau n=\operatorname{atan2}(b,a)+\tau n,\,n\in\mathbb{Z}$$
The polar form of $z$ is denoted as $|z|\cdot e^{\theta i}$, which can be expanded using Euler's Formula to $|z|\left(\cos(\theta)+\sin(\theta)i\right)$.
You provided the complex number (rearranged):
$$z=\left(-3\sqrt{2}\right)+\left(-3\sqrt{2}\right)i$$
$$|z|=3\sqrt{5},\,\theta\approx4.08$$
Substituting in the values to the polar form can approximate $z$:
$$z\approx3\sqrt{5}\cdot e^{4.08i}$$
