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I'm asking for further clarification on two presented ways for the following problem. The problem:

Given three complex number $z_1, z_2, z_3$, prove that the the points $z_1, z_2, z_3$ are vertices of an equilateral triangle in $\mathbb{C}$ iff $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_1z_3 + z_2z_3$

The first way is the one provided by @mathlove: https://math.stackexchange.com/a/953144/820472. Here I would like to know why $z_1, z_2, z_3$ being three vertices of an equilateral triangle are equivalent with i.) $\frac{z_3 - z_1}{z_2 - z_1}$ ii.) $\cos(\pm 60^\circ) + i\sin(\pm 60^\circ)$

The second way is given by@311411 https://math.stackexchange.com/a/4150859/820472, where I would like to know

1.) to what the $\zeta$s stand for. That is, how the rotational invariance of the equilateral property is apparent in

$((z_j\,+\,\zeta )\,-\,(z_k\,+\,\zeta ))^2\,=\,(z_j\,-\,z_k)^2.$

2.) Why the solutions of the quadratics finish the proof, i.e. what is their meaning for the equilateral property?

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The author wanted to show that the expression $$(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0 \tag{1}\label{one}$$ is both translation invariant and rotation invariant.

Let us first take translation invariant. What it means is if we shift every point in the complex plane by a fixed vector $\zeta$, then too the expression (given above) doesn't change. This means if $z_i \mapsto z_i+\zeta$ and so on, then too \eqref{one} will not change. So $\zeta$ can be any (fixed) complex number. And $((z_j\,+\,\zeta )\,-\,(z_k\,+\,\zeta ))^2\,=\,(z_j\,-\,z_k)^2$ is for translational invariance not rotational.

For rotational invariance we want to see what will happen if we rotate every point in $\Bbb{C}$ by an angle $\theta$ in the counterclockwise direction, i.e. what happens when we $z_k \mapsto z_k \cdot e^{i \theta}$ (note that multiplication by $e^{i \theta}$ causes the vector represented by $z_k$ to be rotated by an angle $\theta$). As shown by the author $(z_je^{i\theta}\,-\,z_ke^{i\theta})^2\,=\,e^{i2\theta}(z_j\,-\,z_k)^2$, so every term in \eqref{one} will have the same factor $e^{2i\theta}$ so it will get canceled out and hence the expression will be invariant under rotation as well.

Once we have established that the expression \eqref{one} is invariant under both rotation and translation, then without the loss of generality we can work with values of $z_1,z_2,z_3$ that satisfy \eqref{one} but are also convenient to use. So with that the author chose $z_1=0$ and $z_2$ to be on the positive real axis (because we can translate and rotate the segment $z_1z_2$ such that $z_1$ comes to the origin and $z_2$ can be on the positive real axis and still our expression \eqref{one} will remain intact). With $z_1=0$ and $z_2=\xi >0$ we can find $z_3$ by solving the quadratic $z_3^2-z_3\xi+\xi^2=0$. From this we get $z_3=\xi e^{i\frac{\pm\pi}{3}}$ (note: this is simply rotating $z_2=\xi$ by $60^{\circ}$ in either clockwise or counterclockwise direction).

Now you realize that these points are the vertices of an equilateral triangle, then that means our general $z_1,z_2,z_3$ that satisfy \eqref{one} will also be forming an equilateral triangle.

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  • $\begingroup$ As a side note, the invariance to translation suffices to complete the proof, since it means we can translate the triangle so that its centroid moves to the origin, in other words we can assume WLOG that $\,z_1+z_2+z_3=0\,$. Then $z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_1z_3 + z_2z_3 \implies z_1z_2 + z_1z_3 + z_2z_3 = 0$ so $z_1,z_2,z_3$ are the roots of an equation of the form $z^3+c=0$ for some $c \in \mathbb C\,$. $\endgroup$
    – dxiv
    Aug 1 at 0:17

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