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

The locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$,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:

If $z=x+iy$ satisfies the condition $|z+1|=1$,then $z$ lies on the

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:

The equation of any $Circle$ in the complex plane is of the form $z \bar{z} + b \bar{z} + \bar{b} z + c = 0$,where $b \in \mathbb{C}$ and $c \in \mathbb{R}$.