Let $z = x + iy$ be a complex number,$A = \{z : |z| \leq 2\}$ and $B = \{z : (1-i)z + (1+i)\bar{z} \geq 4\}$. Then which one of the following options belongs to $A \cap B$?

Let $\theta_1, \theta_2, \ldots, \theta_{10}$ be positive valued angles (in radian) such that $\theta_1+\theta_2+\ldots+\theta_{10}=2 \pi$. Define the complex numbers $z_1=e^{i \theta_1}, z_k=z_{k-1} e^{i \theta_k}$ for $k=2,3, \ldots, 10$,where $i=\sqrt{-1}$. Consider the statements $P$ and $Q$ given below:
$P: |z_2-z_1|+|z_3-z_2|+\ldots+|z_{10}-z_9|+|z_1-z_{10}| \leq 2 \pi$
$Q: |z_2^2-z_1^2|+|z_3^2-z_2^2|+\ldots+|z_{10}^2-z_9^2|+|z_1^2-z_{10}^2| \leq 4 \pi$
Then,

If $A = \{z : |\frac{z - 2}{z + 2}| = 3, z \in C\}$ and $z_1, z_2, z_3, z_4 \in A$ are $4$ complex numbers representing points $P, Q, R, S$ respectively on the complex plane such that $z_1 - z_2 = z_4 - z_3$,then the maximum value of the area of quadrilateral $PQRS$ is:

If the point $\left(\frac{k-1}{k}, \frac{k-2}{k}\right)$ lies on the locus of $z$ satisfying the inequality $\left|\frac{z+3i}{3z+i}\right| < 1$,then the interval in which $k$ lies is

$A$ complex number $z$ is such that $arg\left( \frac{z - 2}{z + 2} \right) = \frac{\pi}{3}$. The points representing this complex number will lie on

If $z$ is a complex number such that $|z| \ge 2$,then the minimum value of $|z + \frac{1}{2}|$ is: