If $\frac{z-1}{2z+1}$ is a purely imaginary number,then the locus of $z$ represents a circle. Find its radius.

Let $A = \{z \in \mathbb{C} : |z - 2 - i| = 3\}$, $B = \{z \in \mathbb{C} : \operatorname{Re}(z - iz) = 2\}$ and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to . . . . . . .

The locus of the complex number $z$ such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{3}$ is:

$A$ particle $P$ starts from the point $z_0 = 1 + 2i$,where $i = \sqrt{-1}$. It moves first horizontally away from the origin by $5$ units and then vertically away from the origin by $3$ units to reach a point $z_1$. From $z_1$,the particle moves $\sqrt{2}$ units in the direction of the vector $\hat{i} + \hat{j}$ and then it moves through an angle $\frac{\pi}{2}$ in the anticlockwise direction on a circle with the center at the origin,to reach a point $z_2$. The point $z_2$ is given by:

The locus of $z$ satisfying the inequality $\left|\frac{z+2 i}{2 z+i}\right| < 1$,where $z=x+i y$,is

The minimum value of $|z-1|+|z-5|$ is