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

If $z$ and $\omega$ are two non-zero complex numbers such that $|z \omega|=1$ and $\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,then $\bar{z} \omega =$

For all complex numbers $z_1, z_2$ satisfying $|z_1| = 12$ and $|z_2 - 3 - 4i| = 5$,the minimum value of $|z_1 - z_2|$ is

Let $z_{1}$ and $z_{2}$ be two imaginary roots of $z^{2}+pz+q=0$, where $p$ and $q$ are real. The points $z_{1}, z_{2}$ and the origin form an equilateral triangle if

The inequality $|z - 4| < |z - 2|$ represents the region given by

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