If the amplitude of $(Z-2)$ is $\frac{\pi}{2}$,then the locus of $Z$ is:

All the points in the set $S = \left\{ \frac{\alpha + i}{\alpha - i} : \alpha \in R \right\} (i = \sqrt{-1})$ lie on a

Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$. Then

Let $\left|\frac{\bar{z}-i}{2 \bar{z}+i}\right|=\frac{1}{3}$,where $z \in \mathbb{C}$,be the equation of a circle with center at $C$. If the area of the triangle,whose vertices are at the points $(0,0)$,$C$,and $(\alpha, 0)$,is $11$ square units,then $\alpha^2$ equals

Let $S_{1}, S_{2}$ and $S_{3}$ be three sets defined as:
$S_{1} = \{ z \in C : |z - 1| \leq \sqrt{2} \}$
$S_{2} = \{ z \in C : \operatorname{Re}((1 - i)z) \geq 1 \}$
$S_{3} = \{ z \in C : \operatorname{Im}(z) \leq 1 \}$
Then the set $S_{1} \cap S_{2} \cap S_{3}$

The set of all $\alpha \in R$,for which $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$ is a purely imaginary number,for all $z \in C$ satisfying $|z| = 1$ and $\text{Re}(z) \neq 1$,is