Let the lines $Re(a+ib)=0$ and $Re(c+id)=0$ be perpendicular.
From section 1.2: Let $a = A+iB$ and $c= C+iD$. Then the lines are $Ax-By+Re(b)=0$ and $Cx-Dy+Re(d)=0$
Setting the slope of the first equal to the negative reciprocal of the other we get: $\frac{A}{B} = - \frac{D}{C} \iff AC=-BD$
Finally, $Re(a \bar{c}) = Re[(A+iB)(C-iD)]=AC+BD=-BD+BD=0$