-3
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, Parcly Taxel, Lord Shark the Unknown, Noah Schweber, Sahiba Arora
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ What did you try? $\endgroup$ – Harald Hanche-Olsen Jul 18 '17 at 14:45
  • $\begingroup$ I just solved it. Thanks $\endgroup$ – Rishav Jul 18 '17 at 14:48
  • $\begingroup$ Why do people down vote the questions! $\endgroup$ – Rishav Jul 18 '17 at 14:49
  • 5
    $\begingroup$ Because questions are supposed to give some context / show some attempt from the OP. You may easily avoid downvotes by meeting this policy. $\endgroup$ – Jack D'Aurizio Jul 18 '17 at 14:50
  • $\begingroup$ Ok..thanks for the tip $\endgroup$ – Rishav Jul 18 '17 at 14:51
1
$\begingroup$

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 $$\frac{z_3-z_2}{z_1-z_2}=e^{i\pi/3}.$$

$\endgroup$
1
$\begingroup$

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 $(1)$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.