If $|z^2 - 1| = |z|^2 + 1$,then $z$ lies on

Let $C$ be the set of all complex numbers. Let $S_{1}=\{z \in C:|z-2| \leq 1\}$ and $S_{2}=\{z \in C: z(1+i)+\overline{z}(1-i) \geq 4\}$. Then,the maximum value of $\left|z-\frac{5}{2}\right|^{2}$ for $z \in S_{1} \cap S_{2}$ is equal to:

Let $S$ be the set of all complex numbers $z$ satisfying $|z-2+i| \geq \sqrt{5}$. If the complex number $z_0$ is such that $\frac{1}{|z_0-1|}$ is the maximum of the set $\left\{\frac{1}{|z-1|}: z \in S\right\}$,then the principal argument of $\frac{4-z_0-\bar{z}_0}{z_0-\bar{z}_0+2i}$ is

If ${\tan ^{ - 1}}(\alpha + i\beta ) = x + iy,$ then $x =$

The point $P$ denotes the complex number $z=x+iy$ in the Argand plane. If $\frac{2z-i}{z-2}$ is a purely real number,then the equation of the locus of $P$ is

If $|Z_1|=|Z_2|=|Z_3|=1$ and $Z_1+Z_2+Z_3=0$,then the area of the triangle whose vertices are $Z_1, Z_2, Z_3$ is