Question

$\arg \bar{ z }+\arg z ; z \neq 0$ 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)$
$\Rightarrow \arg (\bar{z})+\arg (z)=0$