Find the complex number $z$ satisfying the equations $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$.

Consider the following two statements:
Statement $I$: For any two non-zero complex numbers $z_1, z_2$,
$(\left|z_1\right|+\left|z_2\right|)\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2(\left|z_1\right|+\left|z_2\right|)$
Statement $II$: If $x, y, z$ are three distinct complex numbers and $a, b, c$ are three positive real numbers such that $\frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|}$,then
$\frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=1$
Between the above two statements,

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.

Let $S$ be the set of all complex numbers $z$ satisfying $|z^2+z+1|=1$. Then which of the following statements is/are $TRUE$?
$(A) |z+\frac{1}{2}| \leq \frac{1}{2}$ for all $z \in S$
$(B) |z| \leq 2$ for all $z \in S$
$(C) |z+\frac{1}{2}| \geq \frac{1}{2}$ for all $z \in S$
$(D)$ The set $S$ has exactly four elements

If $z = x - iy$ and $z^{1/3} = p + iq$ $(x, y, p, q \in R)$, then $\frac{(\frac{x}{p} + \frac{y}{q})}{(p^2 + q^2)}$ is equal to

If $\cos (u + iv) = \alpha + i\beta ,$ then ${\alpha ^2} + {\beta ^2} + 1$ equals