If $|z^2 - 1| = |z|^2 + 1$,then $z$ lies on

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

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

The minimum value of the expression $|z|+|z-1|+|z-1-i|+|z-i|$,where $z$ is a complex number and $i=\sqrt{-1}$,is

If the vertices $A, B$ and $C$ of an isosceles $\triangle ABC$ are respectively $z_1, z_2$ and $z_3$ and if $\angle C=90^{\circ}$,then

Let $z \in \mathbb{C}$ be such that $|z| < 1$. If $w = \frac{5 + 3z}{5(1 - z)}$,then