The locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$,is

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

$z_1$ 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 $z$ is a complex number such that $|z| \geq 1$,then the minimum value of $\left|z+\frac{1}{2}(3+4 i)\right|$ is:

Given $z$ is a complex number such that $|z| < 2$,then the maximum value of $|iz + 6 - 8i|$ is equal to-

Let $z = x + iy$ be a complex number,$A = \{z : |z| \leq 2\}$ and $B = \{z : (1-i)z + (1+i)\bar{z} \geq 4\}$. Then which one of the following options belongs to $A \cap B$?