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, $\sum _{z \in A \cap B}(Re z-Im z)=4-(-3)+(0-(-7))=14$