$\omega$ 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

Let the locus of a point $z$ in the Argand plane satisfying the condition $\operatorname{Re}(z^2)=4$ be $C_1$ and the locus of $z$ satisfying the condition $\operatorname{Im}(z^2)=4$ be $C_2$. Then the number of common points of the two curves $C_1$ and $C_2$ are

Let $S = \{z \in \mathbb{C} - \{i, 2i\} : \frac{z^2 + 8iz - 15}{z^2 - 3iz - 2} \in \mathbb{R} \}$. If $\alpha - \frac{13}{11}i \in S$ and $\alpha \in \mathbb{R} - \{0\}$,then $242\alpha^2$ is equal to

For any real number $r$, let $A_r = \{e^{i \pi r n} : n \in \mathbb{N}\}$ be a set of complex numbers. Then,

$A$ 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:

$A$ point $z$ moves on the Argand diagram in such a way that $|z - 3i| = 2$. Then its locus will be: