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!
Best Answer
Let $w=z_1-z_2$ and $v= z_3-z_4$. The given $Re(w\bar{v})=0$ leads to
$$w\bar{v}+\bar{w}v=0$$
Then, evaluate
$$|w-v|^2= (w-v)(\bar{w}-\bar{v}) =|w|^2+|v|^2-(w\bar{v}+\bar{w}v)= |w|^2+|v|^2$$
which means $w$, $v$ and $w-v$ form a right triangle, hence $w$ and $v$ are perpendicular to each other. The reserve also holds, i.e. $$|w-v|^2= |w|^2+|v|^2\implies w\bar{v}+\bar{w}v=0\implies Re(w\bar{v})=0$$