2016/036) Prove that $\arg[(a+bi)(c+di)]$

$=\arg(a+bi)+\arg(c+di) \pmod \pi $
Proof:
we have $(a+bi)(c+di) = (ac - db) + (bc + ad)i$
hence $\arg[(a+bi)(c+di)] = \arg (ac - db) + (bc + ad)i$
or  $\arg[(a+bi)(c+di)] = \arctan (\frac{bc + ad}{ac-db})$
$= \arctan (\frac{\frac{b}{a}+ \frac{d}{c}}{1 - \frac{b}{a}\frac{d}{c}})$
$= \arctan (\tan ( \arctan(\frac{b}{a}) + \arctan(\frac{d}{c}))$
$= \arctan(\frac{b}{a}) + \arctan(\frac{d}{c})\pmod \pi $
$= \arg(a+bi)+\arg(c+di) \pmod \pi $