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

Let $z$ be a complex number such that $\left|\frac{z-2i}{z+i}\right|=2$,where $z \neq -i$. Then $z$ lies on a circle of radius $2$ and center:

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 $.......$.

Let $C$ denote the set of all complex numbers. Define $A = \{(z, w) \mid z, w \in C \text{ and } |z| = |w|\}$ and $B = \{(z, w) \mid z, w \in C \text{ and } z^2 = w^2\}$. Then:

If $a$ is a complex number and $b$ is a real number,then the equation $\bar{a}+a+b=0$ represents $a$ as a:

The area of $\Delta$ whose vertices are $z, \omega z, z + \omega z$ is (where $\omega$ is a complex cube root of unity):