If $z$ is a purely real number such that ${\mathop{\rm Re}\nolimits} (z) < 0$, then $arg(z)$ is equal to

If $z=x+i y, x y \neq 0$, satisfies the equation $z^2+i \bar{z}=0$, then $\left|z^2\right|$ is equal to:

$\left| {\frac{1}{2}({z_1} + {z_2}) + \sqrt {{z_1}{z_2}} } \right| + \left| {\frac{1}{2}({z_1} + {z_2}) - \sqrt {{z_1}{z_2}} } \right|$ =

For $a \in C$, let $A =\{z \in C: \operatorname{Re}( a +\overline{ z }) > \operatorname{Im}(\bar{a}+z)\}$ and $B=\{z \in C: \operatorname{Re}(a+\bar{z}) < \operatorname{Im}(\bar{a}+z)\}$. Then among the two statements :

$(S 1)$ : If $\operatorname{Re}(A), \operatorname{Im}(A) > 0$, then the set $A$ contains all the real numbers

$(S2)$: If $\operatorname{Re}(A), \operatorname{Im}(A) < 0$, then the set $B$ contains all the real numbers,

The argument of the complex number $\frac{{13 - 5i}}{{4 - 9i}}$is

The amplitude of $\sin \frac{\pi }{5} + i\,\left( {1 - \cos \frac{\pi }{5}} \right)$