Let $z$ be a complex number such that $|z|=1$. If $\frac{2+\mathrm{k}^2 z}{\mathrm{k}+\bar{z}}=\mathrm{k} z, \mathrm{k} \in \mathbf{R}$, then the maximum distance of $\mathrm{k}+i \mathrm{k}^2$ from the circle $|z-(1+2 i)|=1$ is :

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 :

If $\alpha + i\beta$ and $\gamma + i\delta$ are the roots of $x^2 - (3 - 2i)x - (2i - 2) = 0$, $i = \sqrt{-1}$, then $\alpha \gamma + \beta \delta$ is equal to:

Let $O$ be the origin, the point $A$ be $z_1=\sqrt{3}+2 \sqrt{2} i$, the point $B\left(z_2\right)$ be such that $\sqrt{3}\left|z_2\right|=\left|z_1\right|$ and $\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6}$. Then