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

$\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right) \times \left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right) \times \left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right) \times \ldots \infty =$

The number of all possible solutions of the equation $z^3+\overline{z}=0$ 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 $z = \frac{7 - i}{3 - 4i}$,then $z^{14} = $