Let $z_1, z_2 \in \mathbb{C}$ such that $|z_1 + z_2| = \sqrt{3}$ and $|z_1| = |z_2| = 1$. Then the value of $|z_1 - z_2|$ is

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

If the imaginary part of $\frac{2z + 1}{iz + 1}$ is $-2$,then the locus of the point representing $z$ in the complex plane is

$z_1$ and $z_2$ are two complex numbers such that $\left|z_1-z_2\right| < k$. If a complex number $z$ satisfies the condition $\left|z-z_1\right|+\left|z-z_2\right|=k$,then $z$ lies on:

If ${z_1}$ and ${z_2}$ are two complex numbers satisfying the equation $\left| \frac{z_1 + z_2}{z_1 - z_2} \right| = 1$,then $\frac{z_1}{z_2}$ is a number which is

If $z=x+iy$ is a complex number satisfying $\left|\frac{z-2i}{z+2i}\right|=2$ and the locus of $z$ is a circle,then its radius is