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If $z=1+i \sqrt{3}$ then $|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|$ is equal to
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The correct answer is:
$\frac{2 \pi}{3}$
$z=1+i \sqrt{3}$
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, $\quad \operatorname{Arg} z=\frac{\pi}{3}$
$\operatorname{Arg} \bar{z}=-\frac{\pi}{3}$
$|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|=\left|\frac{\pi}{3}\right|+\left|\frac{-\pi}{3}\right|$
$=\frac{\pi}{3}+\frac{\pi}{3}$
$=\frac{2 \pi}{3}$
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, $\quad \operatorname{Arg} z=\frac{\pi}{3}$
$\operatorname{Arg} \bar{z}=-\frac{\pi}{3}$
$|\operatorname{Arg} z|+|\operatorname{Arg} \bar{z}|=\left|\frac{\pi}{3}\right|+\left|\frac{-\pi}{3}\right|$
$=\frac{\pi}{3}+\frac{\pi}{3}$
$=\frac{2 \pi}{3}$
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