# Equilateral triangle and complex numbers [closed]

Prove that $z_1, z_2, z_3 \in \mathbb{C}$ will represent the vertices of an equilateral triangle if and only if $$\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0$$

## closed as off-topic by TheGeekGreek, Parcly Taxel, Lord Shark the Unknown, Noah Schweber, Sahiba AroraJul 18 '17 at 18:43

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• What did you try? – Harald Hanche-Olsen Jul 18 '17 at 14:45
• I just solved it. Thanks – Rishav Jul 18 '17 at 14:48
• Why do people down vote the questions! – Rishav Jul 18 '17 at 14:49
• Because questions are supposed to give some context / show some attempt from the OP. You may easily avoid downvotes by meeting this policy. – Jack D'Aurizio Jul 18 '17 at 14:50
• Ok..thanks for the tip – Rishav Jul 18 '17 at 14:51

Hint. If $z_1$, $z_2$, $z_3$ (clockwise) are the vertices an equilateral triangle then $$\frac{z_3-z_1}{z_2-z_1}=\frac{|z_3-z_1|e^{i(t+\pi/3)}}{|z_2-z_1|e^{it}}=e^{i\pi/3}$$ Similarly $$\frac{z_3-z_2}{z_1-z_2}=e^{i\pi/3}.$$
Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. $A,B,C$ (counter-clockwise) are the vertices of an equilateral triangle iff $$(B-A)(\omega+1)=(C-A)$$ or $$A+\omega B+\omega^2 C = 0.\tag{1}$$ If $a,b,c$ are three complex numbers such that $a+b+c=a^{-1}+b^{-1}+c^{-1}=0$, by defining $p(z)$ as $(z-a)(z-b)(z-c)$ we get $p(z)=z^3-abc$ by Vieta's theorem, hence $a,b,c$ are given by $\sqrt[3]{abc},\omega\sqrt[3]{abc},\omega^2\sqrt[3]{abc}$. In particular $C=a-b,A=b-c,B=c-a$ fulfill $(1)$.