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 $z_1$ and $z_2$ be two complex numbers and roots of the equation $z^2 + az + b = 0$. If $O$ is the origin such that $OA = OB$ and $a^2 = \lambda b \cos^2 \frac{\alpha}{2}$,where $\alpha$ is the angle $\angle AOB$,then $\lambda$ 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

If the point $\left(\frac{k-1}{k}, \frac{k-2}{k}\right)$ lies on the locus of $z$ satisfying the inequality $\left|\frac{z+3i}{3z+i}\right| < 1$,then the interval in which $k$ lies is

Let the complex numbers $z_1, z_2$ and $z_3$ be the vertices of an equilateral triangle. Let $z_0$ be the circumcentre of the triangle,then $z_1^2 + z_2^2 + z_3^2 = $

If ${z_1}, {z_2}, {z_3}$ are affixes of the vertices $A, B$ and $C$ respectively of a triangle $ABC$ having centroid at $G$ such that $z = 0$ is the midpoint of $AG$,then: