If $i = \sqrt{-1}$, then $\frac{e^{xi} + e^{-xi}}{2} = $

Let $Z_1, Z_2, Z_3$ be three non-zero complex numbers such that $a = |Z_1|, b = |Z_2|, c = |Z_3|$. If the determinant $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$,then:

Let $A = \{\theta \in [0, 2\pi] : 1 + 10 \operatorname{Re}\left(\frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta}\right) = 0\}$. Then $\sum_{\theta \in A} \theta^2$ is equal to

$z = \frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}, \quad (i = \sqrt{-1})$ will be purely imaginary if $\theta =$

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} = $