Complex Exercise

Exercise:

What is the Geometric representation of $\left( 3+7i \right)z+\left( 10-2i \right)\bar{z}+100=0$

Solution: Let $z=x+iy\Leftrightarrow \bar{z}=x-iy$

So $\left( 3+7i \right)z+\left( 10-2i \right)\bar{z}+100=0\Leftrightarrow \left( 3+7i \right)\left( x+iy \right)+\left( 10-2i \right)\left( x-iy \right)+100=0$

$\Leftrightarrow 3x+3iy+7ix+7{{i}^{2}}y+10x-10iy-2xi+2{{i}^{2}}y+100=0$

$\Leftrightarrow 3x+3yi+7xi-7y+10x-10yi-2xi-2y+100=0$

$\Leftrightarrow \left( 3x-7y+10x-2y+100 \right)+i\left( 3y+7x-10y-2x \right)=0$

$\Leftrightarrow \left( 13x-9y+100 \right)+i\left( -7y+5x \right)=0$

$\Leftrightarrow 13x-9y+100=0\,\,\,\,\And \,\,-7y+5x=0$

So the Geometric representation is two intersecting lines at

$x=-\frac{350}{23}\,\,,\,\,y=\frac{-250}{23}$

Thus $z=\frac{1}{23}\left( -350-i250 \right)$