$|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$ is possible if

If the four complex numbers $z$,$\overline{z}$,$\overline{z}-2 \operatorname{Re}(\overline{z})$ and $z-2 \operatorname{Re}(z)$ represent the vertices of a square of side $4$ units in the Argand plane,then $|z|$ is equal to:

If a complex number $z$ satisfies $|z^2-1|=|z|^2+1$,then $z$ lies on

$A$ 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 $Z_2 =$

$\omega$ is a complex cube root of unity and $Z$ is a complex number satisfying $|Z-1| \leq 2$. The possible values of $r$ such that $|Z-1| \leq 2$ and $|\omega Z - 1 - \omega^2| = r$ have no common solution are

If $z = x + iy$ represents a point in the Argand plane,then a point which is not in the region represented by $|z - 1 + i| \leq 2$ is