Let $ |z_1 − 8−2i| \leq 1 $ and $ |z_2−2+6i| \leq 2 $, $ z_1, z_2 \in \mathbb{C} $. Then the minimum value of $ |z_1 − z_2| $ is :

Let $z_1, z_2$ and $z_3$ be three complex numbers on the circle $|z|=1$ with $\arg \left(z_1\right)=\frac{-\pi}{4}, \arg \left(z_2\right)=0$ and $\arg \left(z_3\right)=\frac{\pi}{4}$. If $\left|z_1 \bar{z}_2+z_2 \bar{z}_3+z_3 \bar{z}_1\right|^2=\alpha+\beta \sqrt{2}, \alpha, \beta \in Z$, then the value of $\alpha^2+\beta^2$ is :

Let $$z$$ be a complex number such that the real part of $$\frac{z-2 i}{z+2 i}$$ is zero. Then, the maximum value of $$|z-(6+8 i)|$$ is equal to

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 :