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

In the Argand plane,the values of $z$ satisfying the equation $|z-1|=|i(z+1)|$ lie on

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:

If the vertices of a square are $z_1, z_2, z_3$ and $z_4$ taken in the anti-clockwise order, then $z_3=$

Let $z = x + iy$ be a non-zero complex number such that $z^{2} = i|z|^{2},$ where $i = \sqrt{-1}.$ Then $z$ lies on the:

In the Argand plane,the vector $z = 4 - 3i$ is turned in the clockwise sense through $180^o$ and stretched three times. The complex number represented by the new vector is