Question

Let $|z{|}^2+(3-4i)z+(3+4i)\overline{z}-92=0$ and $(1-i)z+(1+i)\overline{z}-16=0$ intersect at $z_1$ and $z_2.$ Then the sum of the areas of two parallelograms having origin as one common vertex, and $(z_1+z_2$) as diagonal of one parallelogram and $(z_1-z_2)$ as diagonal of the other parallelogram is

Answer 1

$|z{|}^2+(3-4i)z+(3+4i)\overline{z}-92=0\dots \left(1\right)$
$(1-i)z+(1+i)\overline{z}-16=0\dots \left(2\right)$
Let $z=x+iy\Rightarrow \overline{z}=x-iy$
$\therefore z+\overline{z}=2x,z-\overline{z}=2iy\dots \left(3\right)$

From $\left(1\right)$ and $\left(3\right),$
$|z{|}^2+3(z+\overline{z})-4i(z-\overline{z})-92=0$
$x^2+y^2+6x+8y-92=0\dots \left(4\right)$
From $\left(2\right),$
$(z+\overline{z})-i(z-\overline{z})-16=0$
$\Rightarrow 2x+2y-16=0$
$\Rightarrow x+y=8$
$\Rightarrow y=8-x\dots \left(5\right)$

$\therefore$ From $\left(4\right)$ and $\left(5\right)$,
$x^2+(8-x{)}^2+6x+8(8-x)-92=0$
$\Rightarrow x^2+64-16x+x^2+6x+64-8x-92=0$
$\Rightarrow 2x^2-18x+36=0$
$\Rightarrow x^2-9x+18=0$
$\Rightarrow (x-6)(x-3)=0$
$\Rightarrow x=3,6$
From $\left(5\right),$ $y=5,2$

$\therefore z_2=3+5i\equiv (3,5)$ and $z_1=6+2i\equiv (6,2)$


Clearly, quadrilaterals $OABC$ and $ODEA$ are the parallelograms of same area.
Area of $△OAC$ is $\frac{1}{2}|6\cdot 5-3\cdot 2|=12$
$\therefore$ Area of parallelogram $OABC$ is $2\times 12=24$

Total area $=2\times$ area$\left(OABC\right)=48$ sq. units