Let $\omega = z \bar{z} + k_1 z + k_2 i z + \lambda(1 + i)$,where $k_1, k_2 \in R$. Let $\operatorname{Re}(\omega) = 0$ be the circle $C$ of radius $1$ in the first quadrant touching the line $y = 1$ and the $y$-axis. If the curve $\operatorname{Im}(\omega) = 0$ intersects $C$ at $A$ and $B$,then $30(AB)^2$ is equal to $.......$.

If $|z - 2 - 3i| + |z + 2 - 6i| = 4$,where $i = \sqrt{-1}$,then the locus of $P(z)$ is

The vertices $B$ and $D$ of a parallelogram are $1 - 2i$ and $4 + 2i$. If the diagonals are at right angles and $AC = 2BD$,the complex number representing $A$ is

Let $z_1$ and $z_2$ be two distinct complex numbers and let $z = (1-t)z_1 + tz_2$ for some real number $t$ with $0 < t < 1$. If $\operatorname{Arg}(w)$ denotes the principal argument of a non-zero complex number $w$,then which of the following are true?
$(A)$ $|z-z_1| + |z-z_2| = |z_1-z_2|$
$(B)$ $\operatorname{Arg}(z-z_1) = \operatorname{Arg}(z-z_2)$
$(C)$ $\left|\begin{array}{cc} z-z_1 & \bar{z}-\bar{z}_1 \\ z_2-z_1 & \bar{z}_2-\bar{z}_1 \end{array}\right| = 0$
$(D)$ $\operatorname{Arg}(z-z_1) = \operatorname{Arg}(z_2-z_1)$

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

Let $z \neq -i$ be any complex number such that $\frac{z - i}{z + i}$ is a purely imaginary number. Then $z + \frac{1}{z}$ is