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 $...............$.

The points in the set $\{z \in \mathbb{C} : \arg \left(\frac{z-2}{z-6i}\right) = \frac{\pi}{2}\}$ (where $\mathbb{C}$ denotes the set of all complex numbers) lie on the curve which is a

If $|8 + z| + |z - 8| = 16$,where $z$ is a complex number,then the point $z$ will lie on

Let $C$ be the circle in the complex plane with centre $z_0 = \frac{1}{2}(1 + 3i)$ and radius $r = 1$. Let $z_1 = 1 + i$ and the complex number $z_2$ be outside the circle $C$ such that $|z_1 - z_0| |z_2 - z_0| = 1$. If $z_0, z_1$ and $z_2$ are collinear,then the smaller value of $|z_2|^2$ is equal to $.............$.

Convert the given complex number into polar form: $-3$.

If $z = x + iy$ represents a point in the Argand plane,then a point which is not in the region represented by $|z - 1 + i| \leq 2$ is