Why $\bar{z_1}z_2+\bar{z_2}z_3+\bar{z_3}z_1 \in \mathbb R \iff z_1, z_2, z_3 \text{ are along the same line}$

Question

Prove:

$$\bar{z_1}z_2+\bar{z_2}z_3+\bar{z_3}z_1 \in \mathbb R \iff z_1, z_2, z_3 \text{ are along the same line}$$

My attempt:

Since $z+\bar{z} =2 \operatorname{Re}z \in \mathbb R$, so we can transform like that: $$\bar{z_1}z_2+\bar{z_3}z_2+\bar{z_3}z_1 + \bar{z_1}z_1 \in \mathbb R$$ next $$(\bar{z_1}+\bar{z_3})(z_1+z_2) \in \mathbb R$$ But we should have something like $$\frac{z_1+z_3}{z_1+z_2} \in \mathbb R$$

So, what is missing?

Thanks a lot!

Answer 1

Multiply your last fraction top and bottom by $\overline{z}_1+\overline{z}_2$