Let $|z_1 - 8 - 2i| \leq 1$ and $|z_2 - 2 + 6i| \leq 2$,where $z_1, z_2 \in \mathbb{C}$. Then the minimum value of $|z_1 - z_2|$ is:

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 $\bar{z} \omega =$

The set of all real values of $c$ for which the equation $z \bar{z} + (4 - 3i) \bar{z} + (4 + 3i) z + c = 0$ represents a circle is

The largest value of $r$ for which the region represented by the set $\{ \omega \in \mathbb{C} : |\omega - 4 - i| \le r \}$ is contained in the region represented by the set $\{ z \in \mathbb{C} : |z - 1| \le |z + i| \}$ is equal to

Let $S = \{z \in \mathbb{C} : \left|\frac{z-6i}{z-2i}\right| = 1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}\}$. Then $\sum_{z \in S} |z|^2$ is equal to

If $z_{1}, z_{2}$ are complex numbers such that $\operatorname{Re}(z_{1})=|z_{1}-1|$, $\operatorname{Re}(z_{2})=|z_{2}-1|$ and $\arg(z_{1}-z_{2})=\frac{\pi}{6}$, then $\operatorname{Im}(z_{1}+z_{2})$ is equal to