Consider the line with equation (2-i)z + (2+i)\overline(z) = 20. Where does this line intersect the real axis?

\((2-i)z + (2+i)\bar{z} = 20\\ 2(z+\bar{z})-i(z-\bar{z}) = 20\\ 4Re(z) - 2iIm(z)= 20\\ \text{On the real axis }Im(z)=0\\ 4Re(z) = 20\\ Re(z) = 5\\ z=Re(z)+i Im(z)= 5 + i0 = 5\)

Consider...

\(\bar{r} = 2-i\\ r = 2+i\\ \bar{r}z+r\bar{z}=20 \quad | \quad \bar{r}z=r\bar{z}\\ 2\bar{r}z = 20 \\ \bar{r}z = 10 \\ (2-i)z = 10\\ 2z-\underbrace{iz}_{=0} = 20 \\ 2z=10\\ \mathbf{z=5}\)