If $\omega$ is a complex cube root of unity whose imaginary part is positive and $|z - \omega| = |z + \omega|$,then $arg(z)$ can be-

The number of points $z$ on the Argand plane which satisfy the conditions $\operatorname{Re}\left(\frac{z-2}{z-4i}\right)=0$ and $\operatorname{Im}\left(\frac{z-2}{z-4i}\right)=1$ simultaneously 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 upwards parallel to the positive $y$-axis by $3$ units to reach a point $Z_1$. From $Z_1$,the particle moves $\sqrt{2}$ units in the direction of vector $\hat{i} + \hat{j}$ and then it moves through an angle $\frac{\pi}{2}$ in the anticlockwise direction on a circle with the centre at the origin to reach point $Z_2$. Then $Z_2 =$

If $z = x + iy$,then the area of the triangle whose vertices are the points $z$,$iz$,and $z + iz$ is

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

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