Find the modulus and principle argument for $\sqrt{3} - i$ (a) $z = \sqrt{3} - i$
solution: 
$z = \sqrt{3} - i, |z| = \sqrt{3 + 1} = 2$. $Arg(z) = \tan^{-1}\left(\frac{-1}{\sqrt{3}} \right) = -\frac{\pi}{6}$ (z is in the 4th quadrant)
(b) $z = -\sqrt{3} - i$
Solution: 
$z = -\sqrt{3} - i, |z| = \sqrt{3 + 1} = 2$. $Arg(z) = \tan^{-1} \left(\frac{1}{\sqrt{3}} \right) + (- \pi) = \frac{-5 \pi}{6}$  (z is in the 3rd quadrant) 
Could someone please explain the solution, I really do not get how they get $Arg(z)$.
 A: For complex numbers, we can write them as $$z=|z|(\cos(Arg(z))+i\sin(Arg(z)))$$
So one starts with calculating the absolute value, and divide $z$ by that. What you get is a complex number $$\frac{z}{|z|}=\cos(Arg(z))+i\sin(Arg(z))$$
This can be written as $$\cos(Arg(z))=\Re\left(\frac{z}{|z|}\right)\\\sin(Arg(z))=\Im\left(\frac{z}{|z|}\right)$$
You can then use $$\tan(Arg(z))=\frac{\sin(Arg(z))}{\cos(Arg(z))}=\frac{\Re (z)}{\Im(z)}$$
Note however that when you do this division, the $\arctan$ or $\tan^{-1}$ function yields a value between $-\pi/2$ and $\pi/2$. So you need to be careful with it. If both $\Re(z)$ and $\Im(z)$ are positive, then $Arg(z)$ is in the first quadrant. If both are negative, you need to add $\pi$ to the value that you get from $\arctan$. Similarly, if $\Re(z)$ is positive and $\Im(z)$ is negative, $Arg(z)$ will be between in the fourth quadrant. If $\Re(z)$ is negative and $\Im(z)$ is positive, $Arg(z)$ is in the second quadrant, so you need to add (or subtract) $\pi$  
