JEE Main 2023 (Online) 29th January Evening ShiftQuestion 1
Let $\alpha=8-14i$, $A=\left\{z\in\mathbb C:\frac{\alpha z-\overline{\alpha z}}{z^{2}-(\bar z)^{2}-112i}=1\right\}$ 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
Complex Numbers
Answer$14$
Solution
With $z=x+iy$: $\alpha z-\overline{\alpha z}=2i(8y-14x)$ and $z^2-\bar z^2=4ixy$, so $16y-28x=4xy-112$, i.e. $(x-4)(y+7)=0$. On $|z+3i|=4$: $x=4\Rightarrow y=-3$; $y=-7\Rightarrow x=0$. Sum $=(4+3)+(0+7)=14$.