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

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

For $a \in \mathbb{C}$, let $A = \{z \in \mathbb{C} : \operatorname{Re}(a + \bar{z}) > \operatorname{Im}(\bar{a} + z)\}$ and $B = \{z \in \mathbb{C} : \operatorname{Re}(a + \bar{z}) < \operatorname{Im}(\bar{a} + z)\}$. Then among the two statements:
$(S1) : \text{If } \operatorname{Re}(a), \operatorname{Im}(a) > 0, \text{ then the set } A \text{ contains all the real numbers.}$
$(S2) : \text{If } \operatorname{Re}(a), \operatorname{Im}(a) < 0, \text{ then the set } B \text{ contains all the real numbers.}$

Which of the following equations can represent a triangle in the complex plane?

The solutions of the equation $z^2(1-z^2)=16$,$z \in \mathbb{C}$,lie on the curve