=\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
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