Equation of line in argand plane is given by $\dfrac{z - z_1}{\bar z - \bar z_1 } = \dfrac{z_1- z_2}{\bar z_1 - \bar z_2}$
Let complex slope be defined as $\omega = \dfrac{z_1- z_2}{\bar z_1 - \bar z_2}$
Now my book says that if two lines are perpendicular then $\omega_1 + \omega_2 = 0$.
I just can't figure out why this is the case.
Let the equations of two lines $\ell_1,\ell_2$ be $$\dfrac{z - z_1}{\bar z - \bar z_1 } = \dfrac{z_1- z_2}{\bar z_1 - \bar z_2},\dfrac{z - z_1'}{\bar z - \bar z_1' } = \dfrac{z_1'- z_2'}{\bar z_1' - \bar z_2'} $$ respectively. Note that $\ell_1\perp\ell_2$ iff $$ z_2-z_1=ri(z_2'-z_1')\tag{1}$$ for some real number $r\neq0$. From (1), one has $$ \bar z_2-\bar z_1=-ri(\bar z_2'-\bar z_1')\tag{2} $$ from which it is easy to check $$ \omega_1+\omega_2=\dfrac{z_1- z_2}{\bar z_1 - \bar z_2}+\dfrac{z_1'- z_2'}{\bar z_1' - \bar z_2'}=0.$$