If $|z|=1$ and $z \neq \pm 1$,then all the values of $\frac{z}{1-z^2}$ lie on

If $z, iz$ and $z+iz$ are the vertices of a triangle and if $|z|=4$,then the area (in sq. units) of that triangle is:

Let $a = \text{Im}\left( \frac{1 + z^2}{2iz} \right)$,where $z$ is any non-zero complex number. The set $A = \{ a : |z| = 1 \text{ and } z \ne \pm 1 \}$ is equal to

If $z_1, z_2, z_3$ are the vertices of an equilateral triangle and $z$ is its circumcentre,then

Let $C$ be the set of all complex numbers. Let $S_{1} = \{z \in C : |z-3-2i|^{2}=8\}$,$S_{2} = \{z \in C : \operatorname{Re}(z) \geq 5\}$,and $S_{3} = \{z \in C : |z-\bar{z}| \geq 8\}$. Then the number of elements in $S_{1} \cap S_{2} \cap S_{3}$ is equal to:

If $n$ is a positive integer greater than unity and $z$ is a complex number satisfying the equation $z^n = (z + 1)^n$,then