Consider the line with equation \((2-i)z + (2+i)\overline{z} = 20\). Where does this line intersect the real axis?
\((2-i)z+(2+i)\bar{z} = 20\\ z = x + i y\\ (2-i)(x+iy) + (2+i)(x-i y) = 20\\ (2x+y) - i(x-2y)+(2x+y)+i(x-2y)=20\\ 2x+y=10\\ y=-2x+10\\ y=0 \Rightarrow x=5\\ \text{Intersection is at $z=5$}\)
