# Find the curve defined by the equation $\frac{z-z_2}{z_1-z_2}=\Big(\overline{\frac{z-z_2}{z_1-z_2}}\Big)$

I am trying to find the curve defined by $$\frac{z-z_2}{z_1-z_2}=\Big(\overline{\frac{z-z_2}{z_1-z_2}}\Big)$$

Letting $z=x+iy,\,\, z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ and using $z\bar{z}=|z|^2$ I got after some calculations $$\frac{x-x_2}{x_1-x_2}=\frac{y-y_2}{y_1-y_2}$$ which gives a straight line passing through $z_1$ and $z_2$.

My query is that Am I right about the answer and is there any shorter version of finding this. Thanks in advance.

• This equation simply means that, if $P=z$ is a point on the curve, then the directed angle $\angle PBA$ is the same as the directed angle $\angle P'B'A'$, where $X'$ denotes the reflection of $X$ about the real axis. – Batominovski Sep 19 '16 at 7:49

Yes, there is a slightly shorter proof, based on a different approach.

By using the fact that $Z=\bar{Z} \Longleftrightarrow Z \in \mathbb{R}$, the constraint can be written :

$$\exists r \in \mathbb{R} \ \ \text{such that} \ \ \dfrac{z-z_2}{z_1-z_2}=r \ \Longleftrightarrow \ z-z_2=r(z_1-z_2)$$

$$\Longleftrightarrow \ \exists r \in \mathbb{R} \ \ \text{such that} \ \ \ z=(1-r)z_2+rz_1$$

this last expression being a "barycentrical" ("weighted sum") description of the line passing through $z_1$ and $z_2$.

You are correct (assuming $z_1 \ne z_2$ of course, otherwise the problem is ill posed).

For a shorter proof, let $\lambda = \frac{z-z_2}{z_1-z_2}$ and note that the premise is $\lambda=\overline{\lambda}$ which is equivalent to $\lambda \in \mathbb{R}$.

Then it follows that $z = z_2 + \lambda(z_1 - z_2)$ with $\lambda \in \mathbb{R}$ which is the line passing through $z_2$ (at $\lambda = 0$) and $z_1$ (at $\lambda = 1$).