Let $\alpha=8-14 i, A=\left\{z \in C: \frac{a z-\bar{\alpha} \bar{z}}{z^{2}-(\bar{z})^{2}-112 i}=1\right\}$ and $B=\{z \in C:|z+3 i|=4\}$. The $\sum_{z \in A \cap B}(\operatorname{Rez}-\operatorname{Imz})$ is equal to \_\_\_\_.

Solution

$\alpha=8-14 \mathrm{i}$, let $\mathrm{z}=\mathrm{x}+\mathrm{iy} \Rightarrow \mathrm{az}=(8 \mathrm{x}+14 \mathrm{y})+\mathrm{i}(-14 \mathrm{x}+8 \mathrm{y})$
$\mathrm{z}+\overline{\mathrm{z}}=2 \mathrm{x}, \mathrm{z}-\overline{\mathrm{z}}=2 \mathrm{iy}$ Set $\mathrm{A}: \frac{2 \mathrm{i}(-14 \mathrm{x}+8 \mathrm{y})}{\mathrm{i}(4 \mathrm{xy}-112)}=1 \Rightarrow(\mathrm{x}-4)(\mathrm{y}+7)=0 \Rightarrow \mathrm{x}=4$ or $\mathrm{y}=-7$
Set $\mathrm{B}: \mathrm{x}^{2}+(\mathrm{y}+3)^{2}=16$. When $\mathrm{x}=4, \mathrm{y}=-3$ when $\mathrm{y}=-7, \mathrm{x}=0 \therefore \mathrm{~A} \cap \mathrm{~B}=\{4-3 \mathrm{i}, 0-7 \mathrm{i}\}$
So, $\sum_{\mathrm{z} \in \mathrm{A} \cap \mathrm{B}}(\operatorname{Rez}-\operatorname{Imz})=4-(-3)+(0-(-7))=14$