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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!

user153012
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Zhen Zhang
  • 805
  • You are almost done! Note that $\frac{z_1+z_3}{z_1+z_2}\in\mathbb{R}$ is equivalent to $\frac{z_1+z_3}{z_1+z_2}=\frac{\bar{z_1}+\bar{z_3}}{\bar{z_1}+\bar{z_2}}$. – karvens Sep 06 '14 at 09:26
  • You're not using $z+\bar{z}\in\mathbb{R}$, but rather that $z\bar{z}\in\mathbb{R}$. – egreg Sep 06 '14 at 09:57
  • Take a look at another solution in https://math.stackexchange.com/q/3070084 – Jean Marie Jan 16 '19 at 23:21

1 Answers1

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

Rene Schipperus
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