Complex Number Question 6
Question 6 - 29 January - Shift 2
Let $\alpha=8-14 i, A={z \in \mathbb{C}: \frac{\alpha z-\bar{\alpha} \bar{z}}{z^{2}-(\bar{z})^{2}-112 i}=1}$
and $B={z \in \mathbb{C}:|z+3 i|=4}$.
Then $\sum _{z \in A \cap B}(Re z-Im z)$ is equal to
Show Answer
Answer: 14
Solution:
Formula: Properties of conjugate, Algebra of complex numbers
$\alpha=8-14 i$
$z=x+iy$
$\alpha z=(8 x+14 y)+i(-14 x+8 y)$
$z+\overline{z}=2 x \quad \text {and} \quad z-\overline{z}=2 iy$
Set $A={z \in \mathbb{C}: \frac{\alpha z-\bar{\alpha} \bar{z}}{z^{2}-(\bar{z})^{2}-112 i}=1}$
$(x-4)(y+7)=0 \Rightarrow x=4 \text{ or } \quad y=-7$
Set $B={z \in \mathbb{C}:|z+3 i|=4}$
Set $B: x^{2}+(y+3)^{2}=16$
when $x=4$ then $y=-3$
when $y=-7$ then $x=0$
$\therefore A \cap B={4-3 i, 0-7 i}$
So, $\sum _{z \in A \cap B}(Re z-Im z)=4-(-3)+(0-(-7))=14$
