Trouble with argument in a complex number We have: 
$z_1 = 2 + 2i$ and $z_2 = -1 -\sqrt{3} i$ 
I am asked for obtaining arg$\left(\frac{z_1}{z_2}\right)$ and arg($z_1z_2$)
As we know:
$$\theta = \tan^{-1} \left(\frac{y}{x}\right)$$
The division of complex numbers is:
$$\frac{z_1}{z_2} = z_1z_2^{-1} = \frac{z_1z_2^*}{z_2z_2^*}$$
Doing so I got $\frac{\sqrt{3}-1}{-1-\sqrt{3}}$ but I do not know how to get the angle without calculator using this result.
The same happened to me in the product:
$$arg(z_1z_2) = \tan^{-1}\left(\frac{-1-\sqrt{3}}{-1+\sqrt{3}}\right) = ...$$
I know it has to be a silly thing but I am stuck here.

I do not really know if you wanted this. But, try using the following steps to find the value
$$tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{- 1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} \right)$$
$$\therefore \tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{-1 + \sqrt{3}} \right) = \tan^{-1} \left( \dfrac{\tan \left( \dfrac{\pi}{4} \right) + \tan \left( \dfrac{\pi}{3} \right)}{1 - \tan \left( \dfrac{\pi}{4} \right) \cdot \tan \left( \dfrac{\pi}{3} \right)} \right)$$
$$\therefore \tan^{-1} \left( \dfrac{-1 - \sqrt{3}}{-1 + \sqrt{3}} \right) = \tan^{-1} \left( \tan \left( \dfrac{7 \pi}{12} \right) \right) = \dfrac{7 \pi}{12}$$
You can try the same method for other as well.
