Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3|z_1| = 4|z_2|$. If $z = \frac{3z_1}{2z_2} + \frac{2z_2}{3z_1}$,then:

Given that the equation $z^2 + (p + iq)z + r + is = 0$,where $p, q, r, s$ are real and non-zero,has a real root,then:

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,

$\omega$ is a complex cube root of unity and if $Z$ is a complex number satisfying $|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$,then the set of possible values of $a$ is

Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0$,where $z \in \mathbb{C}$. Then $4(\alpha^2+\beta^2)$ is equal to:

The expression $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ equals