Proving two lines are perpendicular Given $z_1,z_2,z_3 $ and $z_4$ are complex numbers, prove that the line joining $z_1,z_2$ and the line joining from $z_3,z_4$ are perpendicular iff $Re\{(z_1-z_2)(\bar z_3-\bar z_4)\}=0$. Try not to use polar form. 
I try to start with writing $Re\{(z_1-z_2)(\bar z_3-\bar z_4)\}=Re\{z_1\bar z_3\}-Re\{z_1\bar z_4\}-Re\{z_2\bar z_3\}+Re\{z_2\bar z_4\}$ (I'm not sure if it's right)
Then how can I make use of the perpendicular condition? Any hints for the reverse direction, or I just have to reverse the argument?
Thank you!

Let $z_1-z_2=ae^{i\theta}$ and $z_3-z_4=be^{i\phi}$, then
$$(z_1-z_2)\overline{(z_3-z_4)} = abe^{i(\theta-\phi)}$$
$$\operatorname{Re}[(z_1-z_2)\overline{(z_3-z_4)}] = ab\cos (\theta-\phi)$$
\begin{align*}
  \operatorname{Re}[(z_1-z_2)\overline{(z_3-z_4)}]=0
  & \iff \cos (\theta-\phi) =0 \\
  & \iff \theta-\phi=\left(n+\frac{1}{2} \right) \pi \\
  & \iff (z_1-z_2)\perp (z_3-z_4)
\end{align*}
where $a$, $b> 0$
