If $z_1$ and $z_2$ are two roots of the equation $z^2+az+b=0$ with $a^2 < 4b$, then the origin, $z_1$, and $z_2$ form an equilateral triangle if

The region represented by $\{z=x+iy \in \mathbb{C} : |z|-\operatorname{Re}(z) \leq 1\}$ is also given by the inequality

The reflection of the complex number $(3 + 2i)$ in the straight line $z = -i \bar{z}$ is-

For any two complex numbers $z_1, z_2$,if $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$,then:

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

Let $z_1$ and $z_2$ be two complex numbers such that $\frac{z_1}{z_2} + \frac{z_2}{z_1} = 1$. Then