If $\{a,b,c\}\subset\mathbb C$ such that $a^2+b^2+c^2=ab +bc +ca$ prove that they lie on the vertices of an equilateral triangle in the complex plane

Question

Exercise 1.11 (from Friendly Approach to Complex Analysis).

If $a$, $b$, $c$ are real numbers such that $a^2+b^2+c^2=ab+bc+ca$, then they must be equal. Indeed, doubling both sides and rearranging gives $(a-b)^2+(b-c)^2+(c-a)^2=0$, and since each summand is nonnegative, it must be that case that each is $0$. On the other hand, now show that if $a$, $b$, $c$ are complex numbers such that $a^2+b^2+c^2=ab+bc+ca$, then they must lie on the vertices of an equilateral triangle in the complex plane. Explain the real case result in the light of this fact.

Hint: Calculate $\big((b-a)\omega+(b-c)\big)\big((b-a)\omega^2+(b-c)\big)$, where $\omega$ is a nonreal cube root of unity.

It is a solved exercise but I don't understand the solution (or the hint). More than that I don't get why the author shared this exercise.

I understand the cube roots of unity are $\{ (1,0), (-\frac{1}{2}, \frac{\sqrt{3}}{2}), (-\frac{1}{2}, -\frac{\sqrt{3}}{2}) \}$

So I have $a := (1,0)$ and $b := (-\frac{1}{2}, \frac{\sqrt{3}}{2}) $ and solved for $c$, and got two values. One of them was the last cube root of unity. However I have observed that the points (solutions for $c$ considered one at a time) lie on the vertices of an equilateral triangle.

Isn't this enough? What is the thought process behind the hint?

Answer 1

The idea behind the hint is the following: For $a,b\in\mathbb C$, the complex number $b-a$ can be interpreted as the vector from $a$ to $b$ in the complex plane. Multiplying such a vector with a non-real cube root of unity $\omega$ results in a rotation by $+120^\circ$ or $-120^\circ$ depending on which root you chose. Multiplication by $\omega^2=\frac{1}{\omega}$ is then a rotation in the opposite direction, so by $-120^\circ$ or $+120^\circ$, respectively.

Hence, the set $\{(b-a)\omega, (b-a)\omega^2\}$ contains the two vectors obtained from the vector from $a$ to $b$ by $\pm 120^\circ$ rotations, regardless of the choice of $\omega$. Now $a,b,c$ is an equilateral triangle if and only if $c$ is positioned at $b+(b-a)\omega$ or $b+(b-a)\omega^2$.

We may write this as $$ c=b+(b-a)\omega \qquad\text{or}\qquad c=b+(b-a)\omega^2, $$ or equivalently $$ (b-a)\omega+(b-c)=0 \qquad\text{or}\qquad (b-a)\omega^2+(b-c)=0. $$

Finally, since $\mathbb C$ is a field, we have $z=0$ or $w=0$ if and only if $zw=0$ so we may rewrite the condition of $a,b,c$ being an equilateral triangle as $$ \big((b-a)\omega+(b-c)\big)\cdot\big((b-a)\omega^2+(b-c)\big)=0. $$

Hence, calculating that product and checking whether it is zero tells you whether $a,b,c$ form an equilateral triangle.

Answer 2

For a different hint, let $\alpha=a-t, \beta=b-t, \gamma=c-t$, then simple calculations show that:

$$ a^2+b^2+c^2=ab+bc+ca \;\;\iff\;\; \alpha^2 + \beta^2 + \gamma^2 = \alpha \beta + \beta\gamma + \gamma \alpha $$

This means the relation is invariant to translations, and we can choose $t = \frac{a+b+c}{3}$ so that:

$$\alpha + \beta + \gamma = 0$$

Then:

$$\require{cancel} \alpha\beta + \beta\gamma + \gamma\alpha = \alpha^2+\beta^2+\gamma^2 = \cancel{(\alpha+\beta+\gamma)^2} - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \\ \;\implies\;\; \alpha\beta + \beta\gamma + \gamma\alpha = 0 $$

It follows by Vieta's relations that $\alpha, \beta, \gamma$ are the roots of $z^3 - p = 0$ for some $p \in \mathbb C$.

Answer 3

Going beyond the hint and setting $x_1=a-b$, $x_2=b-c$, $x_3=c-a$ you have the relations: $$ S_1:=x_1+x_2+x_3=0 \ \ {\rm and} \ \ S_2:=x_1^2+x_2^2+x_3^2=0.$$ Then also $x_1x_2+x_2x_3+x_3x_1= \frac12 (S_2-S_1^2)=0$. From this, we get $$ \prod_{j=1}^3 (z-x_j) = z^3 - x_1x_2x_3.$$ Thus, $x_1,x_2,x_3$ are precisely the 3 cube roots of whatever their product is.

The above generalizes to the following: If $x_1$, $x_2, \ldots,x_n$ are complex numbers so that $S_p:=x_1^p + \cdots x_n^p = 0$ for every $p=1,\ldots,n-1$ then $P(z)=\prod_{j=1}^n(z-x_j) = z^n + (-1)^n\prod_{j=1}^n x_j$ so again the numbers are precisely the $n$'th roots of $(-1)^{n-1}$ times their product. The reason behind is that $S_1,...,S_{n-1}$ form a basis for the symmetric polynomials of degree 1 to $n-1$ in $x_1,...,x_n$. In particular, any other symmetric polynomial of degree 1 to $n-1$ must vanish. This implies the above mentioned form for $P(z)$.

Answer 4

Hint.

We have $(b-a)^2+(c-a)^2 = (b-a)(c-a)$ so

$$ c-a = (b-a)\left(\frac{1\pm\sqrt{3}i}{2}\right) $$

then

$|b-a| = |c-a|$ and $\angle(b-a,c-a) = \pm\frac{\pi}{3}$

NOTE

The property $a^2+b^2+c^2 - a b- a c- b c = 0$ is translation invariant.