If $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$ where $i = \sqrt{-1}$,then $(1 + iz + z^5 + iz^8)^9$ is equal to:

If $2 + 3i$ is one of the roots of the equation $2x^3 - 9x^2 + kx - 13 = 0,$ where $k \in R,$ then the real root of this equation:

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,

If $a \pm ib$ and $b \pm ai$ are the roots of $x^4-10x^3+50x^2-130x+169=0$,then $\frac{a}{b}+\frac{b}{a}=$

Find real $\theta$ such that $\frac{3+2 i \sin \theta}{1-2 i \sin \theta}$ is purely real.

If $a, b, c$ are non-zero real numbers with $c \neq 1$ such that $a^2+b^2+c^2=c$ and if $\alpha=\frac{a+i b}{1-c}$,then $a^2+b^2=$