If $z=1+i \sqrt{3}$ then $|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|$ is equal to

If $z=1+i \sqrt{3}$ then $|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|$ is equal to
  1. 0
  2. $\frac{\pi}{3}$
  3. $\frac{\pi}{2}$
  4. $\frac{2 \pi}{3}$

Solution

$z=1+i \sqrt{3}$ Let $1+i \sqrt{3}=r(\cos \theta+i \sin \theta)$ Then, $\quad r \cos \theta=1$ $\ldots$ (i) $r \sin \theta=\sqrt{3}$ $\ldots$ (ii) Eq. (i) + Eq. (ii), we get $r^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=3+1$ $\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)$ $r^2=4$ $r=2$ Eq. (ii) $\div$ Eq. (i), we get $\tan \theta=\sqrt{3}$ $\Rightarrow \quad \tan \theta=\tan \frac{\pi}{3}$ $\Rightarrow \quad \theta=\frac{\pi}{3}$ So, $z=(1+i \sqrt{3})=2\left\{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right\}$ Then $\bar{z}=(1-i \sqrt{3})=2\left\{\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right\}$ $=2\left\{\cos \left(\frac{-\pi}{3}\right)+i \sin \left(\frac{-\pi}{3}\right)\right\}$ So, $\quad \operatorname{Arg} z=\frac{\pi}{3}$ $\operatorname{Arg} \bar{z}=-\frac{\pi}{3}$ $|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|=\left|\frac{\pi}{3}\right|+\left|\frac{-\pi}{3}\right|$ $=\frac{\pi}{3}+\frac{\pi}{3}$ $=\frac{2 \pi}{3}$

Asked in: AP EAMCET 2010

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