<s>[INST] Writing a number in polar form (help converting $\theta$ to $\pi$) 
Write $w = \sqrt{3}  - i$ in polar form. 
  

How is $\theta = \frac{-1}{\sqrt3} \textrm{ converted into } \frac{-\pi}{6}$? I understand that $w$ lies in the fourth quadrant of the unit circle, but that's about all I understand. I tried to convert degrees to radians, but I don't think that was the smart decision (I failed). 
 [/INST] The first thing to realize is that if you ever have $x^2 + y^2 = 1$ then there is a $\theta$ so that $x = \cos \theta; y = \sin \theta$ and $\tan \theta = \frac yx$.
And second thing to realise is that if you have any $a,b$ where they don't both equal $0$, you can rewrite $a = (\sqrt{a^2+b^2})a'$ and $b = (\sqrt{a^2 + b^2})b'$ where $a' = \frac a{\sqrt{a^2 + b^2}}$ and $b' = \frac b{\sqrt{a^2 + b^2}}$ and you will then have $a'^2 + b'^2 = 1$.
Which means there is $\theta$ so that $\tan \theta = \frac {b'}{a'} = \frac ba$, $a' = \cos \theta$ and $a = (\sqrt{a^2 + b^2})\cos \theta$ and $b' = \sin \theta$ and $b = (\sqrt{a^2 + b^2})\sin \theta$.
That means for any $a + bi \ne 0$ so that $a+bi$ is a complex number and $a$ is its "real" part and $bi$ is its "imaginary part", we can write:
$a+bi = (\sqrt{a^2 + b^2})(\frac a{\sqrt{a^2 + b^2}} + \frac b{\sqrt{a^2+b^2}}i)$ 
and that means there exist a $\theta$ so that 
$a + bi = (\sqrt{a^2 + b^2})(\cos \theta + i\sin \theta)$.
If we imagine $a+bi$ as the point $(a,b)$ on a complex plane, this should not be surpising.  $\sqrt{a^2 + b^2}$ is simply the distance from $(a,b)$ to $(0,0)$ and $\theta$ is simply the angle between $(a,b)$ and $(0,0)$ and the real axis.
And we can always find this $\theta$ by solving $\tan \theta = \frac ba$ and determining the proper quadrant by the positive and negative values of $a,b$.
So if $w = \sqrt 3 - i$ then if $r = \sqrt {\sqrt 3^2 + 1} = 2$ we have
$w = 2(\frac {\sqrt 3}2 - \frac 12 i)$. 
And we can solve $\cos \theta = \frac{\sqrt{3}}2$ and $\sin \theta = -\frac 12$ and $\tan \theta = \frac 1 {\sqrt{3}}$.  That theta is $-\frac {\pi}6$.
So
$w = \sqrt 3 - i$
$=2(\frac {\sqrt 3}2 - \frac 12 i)$
$= 2(\cos (-\frac \pi 6) + \sin (-\frac \pi 6) i)$.
And .... that's that.
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Oh!  Important post-script!
If you have $w = a + bi$ and you want to find the right $\theta$ and you think,  "Okay, I've got to figure out $\sqrt {a^2 + b^2}$ and I have to divide $a$ and $b$ by it..."
You don't.  Note:  $\tan \theta = \frac {\frac a{\sqrt{a^2+b^2}}}{\frac b{\sqrt{a^2 + b^2}}} = \frac ab$.  So $\theta = \arctan \frac ab$ or $\theta = \arctan \frac ab +\pi$. 
So given $w = \sqrt 3 - i$ we need to simply solve $\tan \theta =  \frac {-1}{\sqrt 3}$ with $\theta$ in the fourth quadrant. 
As to WHY that works is because $w = 2(\frac {\sqrt 3}2 - \frac 12 i)$ but we don't have to do the (somewhat) tedious calculation $a' = \frac {\sqrt 3}{\sqrt{(\sqrt 3)^2 + (-1)^2}}$ and $ b' = \frac {-1} {\sqrt{(\sqrt 3)^2 + (-1)^2}}$
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