If $z_1, z_2, z_3 \in \mathbb{C}$ are the vertices of an equilateral triangle,whose centroid is $z_0$,then $\sum_{k=1}^3 (z_k - z_0)^2$ is equal to

$PQ$ and $PR$ are two infinite rays. $QAR$ is an arc. $A$ point lying in the shaded region,excluding the boundary,satisfies:

If ${\tan ^{ - 1}}(\alpha + i\beta ) = x + iy,$ then $x =$

Let $a = \text{Im}\left( \frac{1 + z^2}{2iz} \right)$,where $z$ is any non-zero complex number. The set $A = \{ a : |z| = 1 \text{ and } z \ne \pm 1 \}$ is equal to

$A$ complex number $z$ is such that $arg\left( \frac{z - 2}{z + 2} \right) = \frac{\pi}{3}$. The points representing this complex number will lie on

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