Let $A = \{z \in \mathbb{C} : 1 \leq |z - (1 + i)| \leq 2\}$ and $B = \{z \in A : |z - (1 - i)| = 1\}$. Then,$B$ is:

Let $\alpha = 8 - 14i$,$A = \{ z \in \mathbb{C} : \frac{\alpha z - \bar{\alpha} \bar{z}}{z^2 - (\bar{z})^2 - 112i} = 1 \}$,and $B = \{ z \in \mathbb{C} : |z + 3i| = 4 \}$. Then $\sum_{z \in A \cap B} (\operatorname{Re}(z) - \operatorname{Im}(z))$ is equal to $...............$.

Let complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $(x-x_0)^2+(y-y_0)^2=r^2$ and $(x-x_0)^2+(y-y_0)^2=4r^2$,respectively. If $z_0=x_0+iy_0$ satisfies the equation $2|z_0|^2=r^2+2$,then $|\alpha|=$

Let $u = \frac{2z + i}{z - ki}$, where $z = x + iy$ and $k > 0$. If the curve represented by $\operatorname{Re}(u) + \operatorname{Im}(u) = 1$ intersects the $y$-axis at the points $P$ and $Q$ such that $PQ = 5$, then the value of $k$ is:

If $S = \{z \in \mathbb{C} : |z - i| = |z + i| = |z - 1|\}$,then $n(S)$ is:

If $z=x+iy$ is a complex number satisfying $\left|z+\frac{i}{2}\right|^2=\left|z-\frac{i}{2}\right|^2$,then the locus of $z$ is