Let $S = \{z \in \mathbb{C} : \left|\frac{z-6i}{z-2i}\right| = 1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}\}$. Then $\sum_{z \in S} |z|^2$ is equal to

If $z$ is a complex number such that $\frac{z-1}{z-i}$ is purely imaginary and the locus of $z$ represents a circle with centre $(\alpha, \beta)$ and radius $r$,then $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=$

The roots of the cubic equation $(z + ab)^3 = a^3$,where $a \neq 0$,represent the vertices of a triangle of sides of length:

If $|z - 2|/|z - 3| = 2$ represents a circle,then its radius is equal to

$z_1, z_2, z_3$ represent the vertices $A, B, C$ of a triangle $ABC$ respectively in the Argand plane. If $|z_1-z_2|=\sqrt{25-12\sqrt{3}}$,$|\frac{z_1-z_3}{z_2-z_3}|=\frac{3}{4}$ and $\angle ACB=30^{\circ}$,then the area (in sq. units) of that triangle is

Let $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ and $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$. Then the maximum value of $\{|z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B\}$ is