If $z$ is any complex number satisfying $|z - 3 - 2i| \leq 2$,then the minimum value of $|2z - 6 + 5i|$ is

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

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\arg(z_{1}-z_{2})=\frac{\pi}{4}$ and $z_{1}, z_{2}$ satisfy the equation $|z-3|=\operatorname{Re}(z)$. Then the imaginary part of $z_{1}+z_{2}$ is equal to ..... .

If ${\tan ^{ - 1}}(\alpha + i\beta ) = x + iy,$ then $x =$

For any two non-zero complex numbers $z_1$ and $z_2$, if $|z_1+z_2|^2=|z_1|^2+|z_2|^2$, then