If $z = \frac{3}{2 + \cos \theta + i \sin \theta}$,then the locus of $z$ is :-

Geometrically,the set $\{z \in \mathbb{C} : |z - 2 - 2i| \leq 1\}$ represents

The area (in sq. units) of the region $S = \{z \in \mathbb{C} : |z-1| \leq 2, (z+\overline{z}) + i(z-\overline{z}) \leq 2, \operatorname{Im}(z) \geq 0\}$ is

The solutions of the equation in $z$,$|z|^2 - (z + \bar{z}) + i(z - \bar{z}) + 2 = 0$ are $(i = \sqrt{-1})$.

If a complex number $z$ is such that $\frac{z-2i}{z-2}$ is a purely imaginary number and the locus of $z$ is a closed curve,then the area of the region bounded by that closed curve and lying in the first quadrant is

Let $z_1, z_2, z_3, \omega, z_0, z'_0$ be fixed points on the complex plane such that no $3$ are collinear,satisfying the condition $Arg\left( \frac{\omega - z_1}{z_2 - z_3} \right) = Arg\left( \frac{\omega - z_2}{z_3 - z_1} \right) = Arg\left( \frac{\omega - z_3}{z_1 - z_2} \right) = \frac{\pi}{2}$. If $z_1, z_2, z_3$ satisfy the equation $|z - z_0| = R_1$ and $z_2, \omega, z_3$ satisfy the equation $|z - z'_0| = R_2$,then the ratio $\frac{R_1}{R_2}$ is equal to: