If $z$ is a complex number satisfying $|\operatorname{Re}(z)|+|\operatorname{Im}(z)|=4,$ then $|z|$ cannot be

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1$, then the origin and the points represented by $z_{1}$ and $z_{2}$:

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

Let $A(3-i)$ and $B(2+i)$ be two points in the Argand plane. If the point $P$ represents the complex number $z=x+iy$,which satisfies $|z-3+i|=|z-2-i|$,then the locus of the point $P$ 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