The locus of $z=x+iy$, such that $\operatorname{Im}\left(\frac{z-3i}{iz+4}\right)=0$ is

The point $P$ denotes the complex number $z=x+iy$ in the Argand plane. If $\frac{2z-i}{z-2}$ is a purely real number,then the equation of the locus of $P$ is

The set of points on the Argand plane which satisfy both $|z| \leq 4$ and $\operatorname{Arg}(z) = \frac{\pi}{3}$ represents:

Let the point $P$ represent $z=x+iy$, where $x, y \in \mathbb{R}$, in the Argand plane. Let the curves $C_1$ and $C_2$ be the loci of $P$ satisfying the conditions $(i)$ $\frac{2z+i}{z-2}$ is purely imaginary and $(ii)$ $\operatorname{Arg}\left(\frac{z+i}{z+1}\right)=\frac{\pi}{2}$, respectively. Then the point of intersection of the curves $C_1$ and $C_2$, other than the origin, is

If $|z - 2|/|z - 3| = 2$ represents a circle,then its radius is equal to

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a