If $z \neq 1$ and $\frac{z^2}{z-1}$ is real,then the point represented by the complex number $z$ lies:

The complex numbers $z_1, z_2, z_3$ are the vertices of a triangle. Then the complex numbers $z$ which make the triangle into a parallelogram are

Let $S = \{ z \in \mathbb{C} : |z - 2| \leq 1, z(1 + i) + \overline{z}(1 - i) \leq 2 \}$. Let $|z - 4i|$ attain minimum and maximum values,respectively,at $z_1 \in S$ and $z_2 \in S$. If $5(|z_1|^2 + |z_2|^2) = \alpha + \beta \sqrt{5}$,where $\alpha$ and $\beta$ are integers,then the value of $\alpha + \beta$ is equal to

The area (in sq. units) of the region $S = \{z \in \mathbb{C} : |z-1| \leq 2, (z+\overline{z}) + i(z-\overline{z}) \leq 2, \operatorname{Im}(z) \geq 0\}$ is

$\alpha$ is the real root and $\beta, \gamma$ are the other roots of the equation $x^3-a^3=0$ $(a>0)$. Then the number of common points of the curves given by $|z-\beta|=\frac{\sqrt{3} a}{2}$ and $|z-\gamma|=\frac{\sqrt{3} a}{2}$ is

If $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,$c = \cos \gamma + i\sin \gamma$ and $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1$,then $\cos (\beta - \gamma ) + \cos (\gamma - \alpha ) + \cos (\alpha - \beta )$ is equal to