If $arg\,(z) = \theta $, then $arg\,(\overline z ) = $

Argument of $ - 1 - i\sqrt 3 $ is

If $\frac{{z - i}}{{z + i}}(z \ne - i)$ is a purely imaginary number, then $z.\bar z$ is equal to

Consider the following two statements :

Statement $I$ : For any two non-zero complex numbers $\mathrm{z}_1, \mathrm{z}_2$

$\left(\left|z_1\right|+\left|z_2\right|\right)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ and

Statement $II$ : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and a, b, c are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$, then

$\frac{\mathrm{a}^2}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^2}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^2}{\mathrm{x}-\mathrm{y}}=1$

Between the above two statements,

If$z = \frac{{1 - i\sqrt 3 }}{{1 + i\sqrt 3 }},$then $arg(z) = $ ............. $^\circ$

If $z$ is a complex number such that $\frac{{z - 1}}{{z + 1}}$ is purely imaginary, then