Let $z_{1}$ be a fixed point on the circle of radius $1$ centered at the origin in the Argand plane and $z_{1} \neq \pm 1$. Consider an equilateral triangle inscribed in the circle with $z_{1}, z_{2}, z_{3}$ as the vertices. Then, $z_{1} z_{2} z_{3}$ is equal to

For $a \in \mathbb{C}$, let $A = \{z \in \mathbb{C} : \operatorname{Re}(a + \bar{z}) > \operatorname{Im}(\bar{a} + z)\}$ and $B = \{z \in \mathbb{C} : \operatorname{Re}(a + \bar{z}) < \operatorname{Im}(\bar{a} + z)\}$. Then among the two statements:
$(S1) : \text{If } \operatorname{Re}(a), \operatorname{Im}(a) > 0, \text{ then the set } A \text{ contains all the real numbers.}$
$(S2) : \text{If } \operatorname{Re}(a), \operatorname{Im}(a) < 0, \text{ then the set } B \text{ contains all the real numbers.}$

Let $S$ be the set of all complex numbers $z$ satisfying $|z-2+i| \geq \sqrt{5}$. If the complex number $z_0$ is such that $\frac{1}{|z_0-1|}$ is the maximum of the set $\left\{\frac{1}{|z-1|}: z \in S\right\}$,then the principal argument of $\frac{4-z_0-\bar{z}_0}{z_0-\bar{z}_0+2i}$ is

Let $A, B, C$ be three sets of complex numbers defined as $A = \{z : \text{Im}(z) \ge 1\}$,$B = \{z : |z - 2 - i| = 3\}$,and $C = \{z : \text{Re}((1 - i)z) = \sqrt{2}\}$. If $z$ is any point in $A \cap B \cap C$,then $|z + 1 - i|^2 + |z - 5 - i|^2$ lies between:

$\omega$ 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

The points $z_1, z_2, z_3, z_4$ in the complex plane are the vertices of a parallelogram taken in order,if and only if