Let $z, w \in \mathbb{C}$ satisfy $z^2 + \bar{w} = z$ and $w^2 + \bar{z} = w$. Then,the number of ordered pairs of complex numbers $(z, w)$ is equal to

If $z=x-iy$ and $z^{1/3}=a+ib$,then $\frac{(x/a+y/b)}{a^2+b^2}=$

If the moduli of two complex numbers are less than unity,then the modulus of the sum of these complex numbers is:

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 $2i$ is a root of $f(z) = z^4 + z^3 + 2z^2 + 4z - 8 = 0$,then which among the following cannot be a root of $f(z) = 0$?

If $P, Q$ and $R$ are angles of an isosceles triangle and $\angle P = \frac{\pi}{2}$, then the value of $\left(\cos \frac{P}{3} - i \sin \frac{P}{3}\right)^3 + (\cos Q + i \sin Q) (\cos R - i \sin R) + (\cos P - i \sin P) (\cos Q - i \sin Q) (\cos R - i \sin R)$ is: