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Suppose that $z_1, z_2, z_3 \in \mathbb{C}$ such that $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_1z_3 + z_2z_3$. In order to conclude that the triangle formed by $z_1, z_2, z_3$ is equilateral, it is sufficient to show that all angles of the triangle are equal. In ProofWiki this is done by showing that $\frac{z_2 - z_1}{z_3 - z_2} = \frac{z_3 - z_1}{z_1 - z_2}$, after which the article concludes that: "Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_3 - z_2$ and $z_1 - z_2$". Link to the article.

What known theorem/property yields the result between the angles from the equality $\frac{z_2 - z_1}{z_3 - z_2} = \frac{z_3 - z_1}{z_1 - z_2}$? I have done barely any geometrical arguments for the past few years, so I am quite rusty with the machinery.

Epsilon Away
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  • $\frac{z_2-z_1}{z_3-z_2}=-\frac{z_1-z_2}{z_3-z_2}=-a$. $|a|=\frac{|z_1-z_2|}{|z_3-z_2|}$ and $\arg a$ is angle of rotation from $z_3$ to $z_1$ with center in $z_2$. – Ivan Kaznacheyeu Jun 13 '22 at 14:06
  • @IvanKaznacheyeu Could you further elaborate on this answer? I don't see how it establishes the claimed connection with $\frac{z_3 - z_1}{z_1 - z_2}$. – Epsilon Away Jun 15 '22 at 17:31
  • @EpsilonAway The wiki page explains that in the links to Complex Multiplication. If there is something in there which is not clear, you should ask about that specifically. – dxiv Jun 15 '22 at 17:55
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    @dxiv Yeah, it just occurred to me why it holds true: If we just consider the arguments, then the fractions read as: "The difference between the arguments of $z_2 - z_1$ and $z_3 - z_2$ is equal to that of the difference between $z_3 - z_1$ and $z_1 - z_2$". – Epsilon Away Jun 15 '22 at 18:15
  • @EpsilonAway Right, that's all there is to it. About the problem itself, see also 1, 2, 3. – dxiv Jun 16 '22 at 02:14

3 Answers3

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The argument of a complex number is the counterclockwise angle made by the vector joining the point representing the complex number (in the Argand plane) to the origin, with the positive X axis.
Any angle measured clockwise, starting from the positive direction of the X axis, is taken as negative by convention.

More specifically, $arg(z_1-z_2)$ is the counterclockwise angle made by the vector joining the point representing $z_2$ to the point representing $z_1$, with the positive X axis.

We must also note that the following property holds given that $z,\omega≠0:$ $$arg(\frac zω)=arg(z)-arg(\omega)$$
In the image, let $\alpha$ be divided into $\theta_1$ and $\theta_2$. Since $\theta_2$ is measured clockwise wrt the red dotted reference line, we will always write it as $-\theta_2$.
Then, $$arg(z_1-z_2)-arg(z_3-z_2)= \theta_1-(-\theta_2)= \theta_1+\theta_2.$$ Thus, $$arg\left(\frac{z_1-z_2}{z_3-z_2}\right)=\alpha.$$
Thus we have the result: $arg\left(\frac{z_1-z_2}{z_3-z_2}\right)$ represents the angle made by the vectors joining $(z_2,z_1)$ and $(z_2,z_3)$ with $z_2$ as the vertex.
Note, however, that the $z_2$ (that is, The complex number representing the vertex) must have the same sign in both the numerator and denominator (either both positive or both negative ).

insipidintegrator
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The answer to my question occurred to me right after I posted the bounty: If we just consider the arguments, then the fractions read as: "The difference between the arguments of $z_2 - z_1$ and $z_3 - z_2$ is equal to that of the difference between $z_3 - z_1$ and $z_1 - z_2$", which means exactly that $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_2$ and $z_3 - z_2$.

Epsilon Away
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We can also go from modulus. If you call the lengths of the sides of the triangle, $a,b,c$, then $a^2=bc$, $b^2=ca$, $c^2=ab$.

Bob Dobbs
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