Mathematics - Complex Number Question with Solution | TestHub

Given:

$A=\left\{\mathrm{z}\in \mathrm{ℂ}:\frac{\alpha \mathrm{z}-\overline{\alpha }\overline{\mathrm{z}}}{{\mathrm{z}}^2-(\overline{\mathrm{z}}{)}^2-112\mathrm{i}}=1\right\}$

$B=\left\{z\in ℂ:\left|z+3i\right|=4\right\}$

$\alpha =8-14\mathrm{i}$

Let

$z=x+iy$

So,

$\alpha z=\left(8-14i\right)\left(x+iy\right)$

$\Rightarrow \alpha z=(8x+14y)+i(-14x+8y)$

Also,

$z+\overline{z}=2x$ and $z-\overline{z}=2iy$

Set A:

$\frac{\alpha \mathrm{z}-\overline{\alpha }\overline{\mathrm{z}}}{{\mathrm{z}}^2-(\overline{\mathrm{z}}{)}^2-112i}=1$

$\Rightarrow \frac{2i\left(-14x+8y\right)}{\left(z+\overline{z}\right)\left(z-\overline{z}\right)-112i}=1$

$\Rightarrow \frac{2i\left(-14x+8y\right)}{4ixy-112i}=1$

 $\Rightarrow \frac{-14x+8y}{\left(2xy-56\right)}=1$

$\Rightarrow xy+7x-4y-28=0$

$\Rightarrow \left(x-4\right)\left(y+7\right)=0$

$\Rightarrow x=4$ or $y=-7$

Set $B$:

$\left|z+3i\right|=4$

$\Rightarrow \left|x+i\left(y+3\right)\right|=4$

$\Rightarrow x^2+{\left(y+3\right)}^2=16$

When $x=4\Rightarrow y=-3$

When $y=-7\Rightarrow x=0$

$\therefore A\cap B=\left\{4-3i,0-7i\right\}$

So, $\sum _{z\in A\cap B}\left(Rez-Imz\right)=4-(-3)+(0-(-7\left)\right)=14$