Translation/Rotation properties for equations in the complex plane

Question

Prove that complex numbers ($z_1,z_2,z_3$) that satisfy the relation below form an equilateral triangle in the complex plane.

$$z_1^2 + z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$$

This answer first shows that the geometrical concepts of translation (introduction of an origin), and rotation are applicable, and then applies those concepts to the given relation.

I understand the answer. My question is on translation/rotation. If I am working with equations in the complex plane, do I need to prove translation/rotation for every equation? Or can I assume it works for all equations and proceed? Are there some places where it is not applicable? If so, counterexamples would help.

Note 1: This is math at AMC/AIME/JEE level, so concepts/understanding take priority over rigour and formal notation.

Note 2: I don't need the answer to the question. I need the answer to the metaquestion: does rotation/translation need to be proved for each equation separately, or is it valid for all equations in general.

Answer 1

The underlying fact is that "your" equilateral triangle can be mapped onto cubic roots of unity, considered as the "archetype" of an equilateral triangle using a "conformal" (= angle-preserving) mapping.

Let us see how (it is a different approach than the one in the reference you give). In this way, you will see that translation, rotation and homothety are naturaly involved in this kind of issues.

Relationship :

$$z_1^2 + z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$$

can be written under the form :

$$(\underbrace{z_1+z_2+z_3}_S)^2-3\underbrace{(z_1z_2+z_2z_3+z_3z_1)}_R=0$$

Otherwise said :

$$R=\frac13 S^2\tag{1}$$

Now consider the expansion of the following 3rd degree polynomial whose roots are $z_1,z_2,z_3$ :

$$P(z):=(z-z_1)(z-z_2)(z-z_3)=z^3-Sz^2+Rz-P=0 \ \text{where} \ P=z_1z_2z_3.$$

Using (1), $P(z)$ can be written :

$$z^3-Sz^2+\tfrac13S^2 z-P=0 \ \iff \ (z-\tfrac13 S)^3+\tfrac{1}{27}S^3+P=0\tag{2}$$

Setting

$$Z=z-\tfrac13 S,\tag{3}$$

and choosing an $a \in \mathbb{C}$ such that $a^3=-(\tfrac{1}{27}S^3+P)$, (2) can be written :

$$Z^3=a^3 \ \iff \ Z=a,a\omega,a\omega^2 \ \ \text{where} \ \ \omega:=e^{2i \pi/3}$$

(cubic roots of unity)

Remark about $a$: If $-(\tfrac{1}{27}S^3+P)=re^{i \theta}$, we can take : $a=\sqrt[3]{r}e^{i \theta/3}$.

Taking (3) into consideration :

$$z \to Z=az+\tfrac13 S$$

maps "your" equilateral triangle onto the equilateral triangle of the cubic roots of unity, I would say in a unavoidable way, answering (so I wish) your question, i.e., involving, in this order, a rotation by angle $\theta/3$, an homothety with ratio $\sqrt[3]{r}$ and a translation by $\tfrac13 S$.

Answer 2

Here's a couple concise ways to answer this question: Note that to prove the triangle $\overset{\triangle}{z_1z_2z_3}$ is equilateral, it suffices to show

$$|z_1-z_2|=|z_2-z_3|=|z_3-z_1|$$

First notice that the relationship given can be rewritten in the equivalent form

$$z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1\iff(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0$$

So if we set $A=z_1-z_2, B=z_2-z_3, C=z_3-z_1$ these quantities obey the relations

$$A^2+B^2+C^2=0~~,~~ A+B+C=0~~~~~~~~~~~~~~~(1)$$

Here you can proceed in at least two different ways:

a)

$C=-A-B$ and by direct substitution into the first relation in (1), obtain

$$A^2+B^2+AB=\frac{A^3-B^3}{A-B}=0$$

We see there are two cases: if $A=B$ then $A=B=C=0$ and we are done. If, however, $A\neq B$ it is true that

$$A^3=B^3\Rightarrow |A|=|B| $$

The equality $|B|=|C|$ follows from performing the previous step again but solving for $A$ instead, and the triangle is equilateral.

b)

It follows from (1) that the first and second Vieta symmetric polynomials are zero:

$$A+B+C=AB+BC+CA=0$$

This means that $A,B,C$ are roots of the cubic polynomial

$$P(x)=x^3-ABC$$

However this requires $P(A)=P(B)=P(C)=0$ which in turn requires

$$A^3=B^3=C^3=ABC$$

and hence the magnitudes of $A,B,C$ have to be equal.