Prove that

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.

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