Let $A = \{z : (\frac{z - \bar{z}}{2i})^2 \leqslant 2(\frac{z - \bar{z}}{2i})\}$ where $i = \sqrt{-1}$ and $B = \{z : |z| \leqslant \sqrt{5}\}$. The number of points with integral real and imaginary parts of $z$ lying in $A \cap B$ is -

Let the set of all values of $k \in R$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0, z \in C$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:

If $f(x)$ is a polynomial of degree $n$ with rational coefficients and $1+2i, 2-\sqrt{3}$ and $5$ are three roots of $f(x)=0$,then the least value of $n$ is

The sum of all possible values of $\theta \in[-\pi, 2 \pi]$, for which $\frac{1+i \cos \theta}{1-2 i \cos \theta}$ is purely imaginary, is equal to (in $\pi$)

Let $z_1, z_2 \in \mathbb{C}$ be the distinct solutions of the equation $z^2 + 4z - (1 + 12i) = 0$. Then $|z_1|^2 + |z_2|^2$ is equal to:

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,