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Let $z_1,z_2,z_3$ be complex numbers of module 1. Prove that they form an equilateral triangle if, and only if, $z_1+z_2+z_3=0$.

I know that if they form am equilateral triangle, then $$ \frac{z_3-z_2}{z_2-z_1}= \frac{z_1-z_3}{z_3-z_2}. $$ I would appreciate any hint or idea to show that the sum is 0. Thank you.

Senna
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2 Answers2

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HINT

The numbers $z_1,\,z_2,\,z_3\,$ represent vertices of a triangle, and also vectors which start at $0$.
If the triangle is equilateral, then $0$ is its center of gravity and $z_1+z_2+z_3=0.$

EDIT

enter image description here

user376343
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  • Can you explain that? I started assuming, without loss of generality, that $z_1 =1$. Now I've to prove that $z_3 + z_3 + 1 = 0$. And I want to do it using algebra and basics complex numbers. How can I proceed? – Senna Nov 29 '20 at 17:44
  • I added a picture, I hope it can help. – user376343 Nov 29 '20 at 19:00
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    thank you!. It was of help. +1 for you – Senna Nov 29 '20 at 19:02
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Hint: $z_1+z_2=-z_3$, so $|z_1+z_2|=1$.

Arthur
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  • thanks, that was a good hint. Can you help me now for the other implicance? – Senna Nov 29 '20 at 17:45
  • @Senna If the three numbers make an equilateral triangle, then by mirror symmetry, $z_1+z_2$ must lie on the line that contains $z_3$ and the origin. And by basic trigonometry their sum has absolute value $1$. And $z_1+z_2\neq z_3$. There's really only one way to make that happen. – Arthur Nov 29 '20 at 17:59
  • Problem solved, I dont know why I got stucked in such a thing. Thanks for your explanation! +1 – Senna Nov 29 '20 at 19:03