If z=1+isqrt{3},then |operatorname{Arg} z|+|o | vedclass

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(D) Given $z=1+i\sqrt{3}$.
We know that for a complex number $z=x+iy$,the argument $\operatorname{Arg} z$ is $	an^{-1}(\frac{y}{x})$.
Here,$x=1$ an

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

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(D) Given $z=1+i\sqrt{3}$.
We know that for a complex number $z=x+iy$,the argument $\operatorname{Arg} z$ is $	an^{-1}(\frac{y}{x})$.
Here,$x=1$ and $y=\sqrt{3}$.
Since $x>0$ and $y>0$,$z$ lies in the first quadrant.
$\operatorname{Arg} z = 	an^{-1}(\frac{\sqrt{3}}{1}) = 	an^{-1}(\sqrt{3}) = \frac{\pi}{3}$.
Now,the conjugate of $z$ is $\bar{z} = 1-i\sqrt{3}$.
Here,$x=1$ and $y=-\sqrt{3}$.
Since $x>0$ and $y < 0$,$\bar{z}$ lies in the fourth quadrant.
$\operatorname{Arg} \bar{z} = 	an^{-1}(\frac{-\sqrt{3}}{1}) = 	an^{-1}(-\sqrt{3}) = -\frac{\pi}{3}$.
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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