# Vertices of equilateral triangle on complex plane

I come cross a problem when I read a book of complex analysis:

If three complex number $a,b,c$ satisfy the relation of

$a^2+b^2+c^2=ab+ac+bc$.

Prove that: these numbers must be three vertices of an equilateral triangle on the complex plane.

if $a,b,c$ are real numbers, we have $a=b=c$. but I’m not sure how to prove it with complex number. The hint I got is:

Calculate $((b-a)w+(b-c))*((b-a)w^2+(b-c))$, where $w$ is nonreal cube root of unity.

• What do you mean by “it’s obvious if $a,b,c$ are real numbers”? I’m not sure how the question can be made sense of in that case – User May 15 '18 at 23:06
• If they're all real numbers, then we have $a=b=c$, which are vertices of a degenerate equilateral triangle? – G Tony Jacobs May 15 '18 at 23:08
• @GTonyJacobs, yes - a degenerate, equilateral triangle on the real axis to be precise. – dezdichado May 15 '18 at 23:09
• Of course it's on the real axis; we said they're all real numbers. It's still not clear that's what OP meant. – G Tony Jacobs May 15 '18 at 23:10
• By obvious I mean, if a,b,c are real, we can show that (a-b)^2+(a-c)^2+(b-c)^2=0,from which we know a=b=c. Sorry for the confusion. Question rephrased.@User – Sizhe May 15 '18 at 23:11

What is the geometric meaning of

$((b-a)\omega + (c-a))((b-a)\omega^2 + (c-a))$

If we subtract $a$ from each point we translate them such that $a$ is on the origin.

$(b-a)\omega$ rotates $(b-a)$ 120 degrees clockwise $(b-a)\omega^2$rotates $(b-a)$ 120 degrees counter-clockwise.

If $a,b,c$ form an equilateral triangle, one of these will be exactly the negative of $(c-a)$

In which case one of the factors of $((b-a)\omega + (c-a))((b-a)\omega^2 + (c-a))$ equals $0.$

• very intuitive explanation ! – G Cab May 15 '18 at 23:46
• Fantastic, thank you! – Sizhe May 15 '18 at 23:58

The hint is one way to write it. Another is $$(x + y \omega + z \omega^2)(x + y \omega^2 + z \omega) = x^2 + y^2 + z^2 - yz-zx-xy$$

• Thanks for your reply. Would you mind to explain more? – Sizhe May 15 '18 at 23:15
• @Sizhe one of the factors must be zero. In either case, you need to relate the pairwise differences $x-y, y-z, z-x$ and show $|x-y| = |y-z| = |z-x|.$ I suppose this is why they gave the hint in that style. – Will Jagy May 15 '18 at 23:19
• @Sizhe now that I think about it, the requirement really is that $y-z$ is equal to one of these four: $(x-y)\omega, (x-y)\omega^2, -(x-y)\omega, -(x-y)\omega^2.$ You need to draw some pictures, I think two out of four are suitable. – Will Jagy May 15 '18 at 23:23
• Thank you, now I think I get it.@WillJagy – Sizhe May 15 '18 at 23:59