Complex Geometry

There exists two reals $p,q$ such that the equation of a straight line can be written as :

$2px+2qy+b=0$ & Let $z=x+iy$ .

$p(z+\overline{z}) - iq(z-\overline{z}) + b =0$

$\implies z(p-iq) + \overline{z}(p+iq) + b=0$

This can be written as : $\overline{a}z+a\overline{z}+b=0$ where $a=p+iq\ne0$

Slope of the line = $-\frac{2p}{2q}=-\frac{\cancel{2}p}{\cancel{2}q} = -\frac{p}{q}=-\frac{Re(a)}{Im(a)}$

We are given : $\begin{cases} a\overline{z}+\overline{a}z+1=0 \to \text{(1)} \\ b\overline{z}+\overline{b}z-1=0 \to \text{(2)} \end{cases}$

Slope of (1) = $-\frac{Re(a)}{Im(a)}=-i\frac{a+\overline{a}}{a-\overline{a}}=m_1$

Slope of (2) = $-\frac{Re(b)}{Im(b)}=-i\frac{b+\overline{b}}{b-\overline{b}}=m_2$

We know since they are perpendicular , $m_1m_2=-1$

$m_1m_2=-1\implies \cancel{i^2}\frac{(a+\overline{a})(b+\overline{b})}{(a-\overline{a})(b-\overline{b})}=\cancel{-1}$

$\implies \cancel{ab}+\cancel{\overline{ab}} + a\overline{b}+\overline{a}b = \cancel{ab}+\cancel{\overline{ab}} - a\overline{b}-\overline{a}b$

$\implies \boxed{a\overline{b}+\overline{a}b=0}$