Question

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

Answer 1

$ \alpha=8-14 i$
$ z=x+i y $
$ a z=(8 x+14 y)+i(-14 x+8 y)$
$z +\overline{ z }=2 x\,\,\, z -\overline{ z }=2 iy$
Set A: $\frac{2 i(-14 x+8 y)}{i(4 x y-112)}=1$
$ (x-4)(y+7)=0 $
$ x=4 \text { or } y=-7$
Set B: $x^2+(y+3)^2=16$
when $x=4 \,\,\, y=-3$
when $y=-7 \,\,\, x=0$
$\therefore A \cap B =\{4-3 i , 0-7 i \}$
So, $\displaystyle\sum_{z \in A \cap B}(\text{Rez}-\text{Im} z )=4-(-3)+(0-(-7))=14$