If z=1+i sqrt{3} then |operatorname{Arg} z|+| | vedclass

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(D) $z=1+i \sqrt{3}$ Since $z$ is in the first quadrant,$\operatorname{Arg} z = 	an^{-1}\left(\frac{\sqrt{3}}{1}ight) = \frac{\pi}{3}$. For the con

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$\frac{2 \pi}{3}$

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(D) $z=1+i \sqrt{3}$ Since $z$ is in the first quadrant,$\operatorname{Arg} z = 	an^{-1}\left(\frac{\sqrt{3}}{1}ight) = \frac{\pi}{3}$. For the conjugate $\bar{z} = 1-i \sqrt{3}$,which is in the fourth quadrant,$\operatorname{Arg} \bar{z} = 	an^{-1}\left(\frac{-\sqrt{3}}{1}ight) = -\frac{\pi}{3}$. Therefore,$|\operatorname{Arg} z| + |\operatorname{Arg} \bar{z}| = |\frac{\pi}{3}| + |-\frac{\pi}{3}| = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2 \pi}{3}$.

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

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