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!

  • $\begingroup$ If $v$ and $w$ are orthogonal vectors then $v\cdot w=0$. This is the definition of orthoganility. $\endgroup$ – Masacroso Oct 21 '16 at 8:08
  • $\begingroup$ how do we define dot product on a complex plane? And why do we have to take conjugate of $z_3$ and $z_4$? $\endgroup$ – mathshungry Oct 21 '16 at 9:28
  • 1
    $\begingroup$ The euclidean dot product on $\Bbb C^n$ is defined as $$v\cdot w:=\sum_{k=1}^n v_k \bar{w}_k$$ See the definition and characteristics of any dot product here. $\endgroup$ – Masacroso Oct 21 '16 at 9:50

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 $a$, $b> 0$


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