If $z $ is a complex number of unit modulus and  argument $\theta$, then ${\rm{arg}}\left( {\frac{{1 + z}}{{1 + (\bar z)}}} \right)$ equals.

For any complex number $w = c + id$, let $\arg ( w ) \in(-\pi, \pi]$, where $i =\sqrt{-1}$. Let $\alpha$ and $\beta$ be real numbers such that for all complex numbers $z=x+$ iy satisfying arg $\left(\frac{z+\alpha}{z+\beta}\right)=\frac{\pi}{4}$, the ordered pair $( x , y )$ lies on the circle

$x^2+y^2+5 x-3 y+4=0 .$

Then which of the following statements is (are) TRUE?

$(A)$ $\alpha=-1$  $(B)$ $\alpha \beta=4$   $(C)$ $\alpha \beta=-4$   $(D)$ $\beta=4$

The solutions of equation in $z$, $| z |^2 -(z + \bar{z}) + i(z - \bar{z})$ + $2$ = $0$ are $(i = \sqrt{-1})$

$arg\,(5 - \sqrt 3 i) = $

If $z_1, z_2  $ are any two complex numbers, then $|{z_1} + \sqrt {z_1^2 - z_2^2} |$ $ + |{z_1} - \sqrt {z_1^2 - z_2^2} |$ is equal to

If the set $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right): z \in C , \operatorname{Re}(z)=3\right\}$ is equal to the interval $(\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to