Let =8-14i, A=z C: z- zz^2- z^2-112i=1 and B=z C: |z+3i|=4.

Question

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

Answer: 14

Accepted Answer

With $z=x+iy$: $\alpha z-\bar\alpha\bar z = 2i\operatorname{Im}(\alpha z)$, and after simplification the condition for set $A$ reduces to
$$\frac{2i(-14x+8y)}{i(4xy-112)}=1 \implies (x-4)(y+7)=0 \implies x=4 \text{ or } y=-7.$$

Set $B$: $x^2+(y+3)^2=16$.

When $x=4$: $16+(y+3)^2=16 \implies y=-3$, giving $z=4-3i$.

When $y=-7$: $x^2+16=16 \implies x=0$, giving $z=-7i$.

$$A\cap B=\{4-3i,\ -7i\}.$$

$$\sum(\operatorname{Re}z-\operatorname{Im}z) = \big(4-(-3)\big)+\big(0-(-7)\big) = 7+7=14.$$

Solutions · 1