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So the excercise says: Being $z,w,a,b\in \mathit C$ such that $z\neq w$ and $a\neq b$. Show that the segment $zw$ is parallel to $ab$ iff: $$\frac{z-w}{\bar z-\bar w}=\frac{a-b}{\bar a-\bar b}$$ My thoughts on this are that I can express any of them as $z=(x_1,y_1)=x_1+y_1i$. So for $z-w$ to be parallel to $a-b$ : $(x_1-x_2,y_1-y_2)=r(a_1-b_1,a_2-b_2)$ for some integer $r$. I also thought it would be possible if try to use the dot product here, but first if $\bar z-\bar w$ then $(x_1,-y_1)-(x_2,-y_2)$, and because of the dot poduct, this is $x^2-y^2$wich is like trying to find the square root of a complex, wich would give me the real part of a complex. So I really don't know how to proceed, can you help me? Any comment or observation mean a lot.

Roma_Rayado
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    Hint: $z-w$ represents the segment from $w$ to $z$ in the complex plane, and $\arg(z-w)$ represents the oriented angle it makes with the positive real axis. Think at what $\arg\left(\frac{z-w}{\bar z - \bar w}\right)$ is. – dxiv May 15 '23 at 22:16
  • $r$ can be any non-zero real number, and is not necessarily an integer. – aschepler May 15 '23 at 22:24

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We have that by $z-w=r_1e^{i\theta_1}$and $a-b=r_2e^{i\theta_2}$

$$\frac{z-w}{\bar z-\bar w}=\frac{a-b}{\bar a-\bar b} \iff \frac{r_1e^{i\theta_1}}{r_1e^{-i\theta_1}}=\frac{r_2e^{i\theta_2}}{r_2e^{-i\theta_2}} \iff e^{i2\theta_1}=e^{i2\theta_2} \iff \theta_1=\theta_2+k\pi$$

using that $\bar z-\bar w=\overline{z-w}$.

user
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