Let the complex number $z = x + iy$ be such that $\frac{2z - 3i}{2z + i}$ is purely imaginary. If $x + y^2 = 0$,then $y^4 + y^2 - y$ is equal to:

If $\alpha+i \beta$ and $\gamma+i \delta$ are the roots of $x^2-(3-2 i) x-(2 i-2)=0$,where $i=\sqrt{-1}$,then $\alpha \gamma+\beta \delta$ is equal to :

If $e^{i \theta} = \operatorname{cis} \theta$, then find the value of $\sum_{n=0}^{\infty} \frac{\cos (n \theta)}{2^n}$.

If $a = \cos (2\pi /7) + i\sin (2\pi /7)$,then the quadratic equation whose roots are $\alpha = a + a^2 + a^4$ and $\beta = a^3 + a^5 + a^6$ is

If ${z_r} = \cos \frac{{r\alpha }}{{{n^2}}} + i\sin \frac{{r\alpha }}{{{n^2}}}$,where $r = 1, 2, 3, \dots, n$,then $\mathop {\lim }\limits_{n \to \infty } {z_1}{z_2}{z_3} \dots {z_n}$ is equal to

Among the statements:
$(S1) :$ The set $\{z \in \mathbb{C} - \{-i\} : |z|=1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\}$ contains exactly two elements,and
$(S2) :$ The set $\{z \in \mathbb{C} - \{-1\} : |z|=1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\}$ contains infinitely many elements.