If $z$ is a non-real complex number, then the minimum value of $\frac{\operatorname{Im}(z^5)}{(\operatorname{Im} z)^5}$ is

Let $z$ be a complex number (not lying on the $X$-axis) of maximum modulus such that $\left| z + \frac{1}{z} \right| = 1$. 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,

If the point $(x, y)$ satisfies the equation $\frac{x+i(x-2)}{3+i}-i=\frac{2y+i(1-3y)}{i-3}$,then $x+y=$

Let $z = a + ib, b \neq 0$ be a complex number satisfying $z^{2} = \overline{z} \cdot 2^{1-|z|}$. Then the least value of $n \in N$ such that $z^{n} = (z + 1)^{n}$ is equal to:

The complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ is equal to $.....$