If $x+i y=\frac{a+i b}{a-i b},$ prove that $x^{2}+y^{2}=1$

Let $\mathrm{z}$ be a complex number such that $|\mathrm{z}+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is :

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?

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

Let $z$ and $w$ be two complex numbers such that $w=z \bar{z}-2 z+2,\left|\frac{z+i}{z-3 i}\right|=1$ and $\operatorname{Re}(w)$ has minimum value. Then, the minimum value of $n \in N$ for which $w ^{ n }$ is real, 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: