Question

$\arg \bar{z}+\arg z;z\ne 0$ is equal to :

• A. $\frac{\pi }{4}$
• B. $\pi$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

Let $z=r(\cos \theta +i\sin \theta )$
Then $r=|z|$ and $\theta =\arg (z)$
Now $z=r(\cos \theta +i\sin \theta )$
$\Rightarrow \bar{z}=r(\cos \theta -i\sin \theta )$
$=r[\cos (-\theta )+i\sin (-\theta )]$
$\therefore \arg (\bar{z})=-\theta$
$\Rightarrow \arg (\bar{z})=-\arg (z)$
$\Rightarrow \arg (\bar{z})+\arg (z)=0$