I need some help proving this, I've seen it proven in the other direction (prove the formula if it is an equilateral) but cant figure out how to prove it this way around.

Given three complex numbers $z_1, z_2, z_3$ prove that the points $z_1, z_2, z_3$ are vertices of an equilateral triangle in $\Bbb C$, if $$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_1z_3 + z_2z_3$$


Three points $z_1,z_2,z_3$ are vertices of an equilateral triangle $$ \begin{alignat}{10} &\iff& \frac{z_3-z_1}{z_2-z_1}&=\cos(\pm 60^\circ)+i\sin(\pm 60^\circ)=\frac{1\pm\sqrt 3i}{2}\\ &\iff& 2z_3-z_1-z_2&=\pm\sqrt 3i(z_2-z_1)\\ &\iff\quad& (2z_3-z_1-z_2)^2&=(\pm\sqrt 3i(z_2-z_1))^2\\ &\iff& z_1^2+z_2^2+z_3^2&=z_1z_2+z_2z_3+z_3z_1. \end{alignat} $$

  • $\begingroup$ I'm a bit confused about a step in your solution. When you do the step $(2z_3-z_1-z_2)^2 \implies 2z_3-z_1-z_2$, shouldn't you take into account both square roots of the number? Or why isn't the implication $(2z_3-z_1-z_2)^2 \implies \pm(2z_3-z_1-z_2)$? $\endgroup$
    – Robert Lee
    Jul 23 '20 at 19:04
  • $\begingroup$ @Robert Lee : We have $X^2=Y^2\iff X=\pm Y\iff \pm X=Y$. Does this answer your question? $\endgroup$
    – mathlove
    Jul 24 '20 at 5:05
  • $\begingroup$ Ahhh, I think I understand. So all the cases are already accounted for with the $\pm$ of the RHS, right? $\endgroup$
    – Robert Lee
    Jul 24 '20 at 5:45
  • 1
    $\begingroup$ @Robert Lee : Yes, that is what I meant. $\endgroup$
    – mathlove
    Jul 24 '20 at 5:49

Assume $z_1,z_2,z_3 \in\Bbb C$ satisfy

$$(*)\quad\quad\quad\,\,\,\, z_1^2 \,+\, z_2^2 \,+\, z_3^2 \,\,=\,\, z_1z_2 \,+\, z_2z_3 \,+\, z_3z_1.$$

It is easy to see, by opening all parentheses, that $(*)$ holds iff

$$\,\,\,(**)\quad\,\, (z_1\,-\,z_2)^2 \,+\, (z_2\,-\,z_3)^2 \,+\, (z_3\,-\,z_1)^2 \,\,=\,\, 0.$$

First assume $z_j=z_k,$ for some $j\not=k.$ Then $(**)$ implies $z_1=z_2=z_3.$ This situation we accept as a degenerate equilateral triangle. Otherwise, the three $z_j$ are distinct points and we have a proper triangle.

The property of being equilateral is invariant under any rotation about the origin, and under any translation. It is easy to see that the property $(**)$ is also so invariant, in the first case via $(z_je^{i\theta}\,-\,z_ke^{i\theta})^2\,=\,e^{i2\theta}(z_j\,-\,z_k)^2,$ and in the second case via $((z_j\,+\,\zeta )\,-\,(z_k\,+\,\zeta ))^2\,=\,(z_j\,-\,z_k)^2.$

Therefore we may assume $z_1=0$ and $z_2=\xi>0.$ Our given $(*)$ is the quadratic $\,\xi^2\,+\,z_3^2\,=\,\xi z_3.$ This quadratic admits two solutions: $\,\xi(\frac{1}{2}\,+\,i\frac{\sqrt 3}{2})\,$ and $\,\xi(\frac{1}{2}\,-\,i\frac{\sqrt 3}{2}).$

The triangle is equilateral.


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