Question:
Prove that $\arg (z)+\arg ^{(\bar{z})}=0$
Solution:
Let $z=r(\cos \theta+i \sin \theta)$
$\Rightarrow \arg (z)=\theta$
Now, $\bar{z}=r(\cos \theta-i \sin \theta)=r(\cos (-\theta)+i \sin (-\theta))$
$\Rightarrow \arg (\bar{z})=-\theta$
Thus, $\arg (z)+\arg (\bar{z})=\theta-\theta=0$
Hence proved.