Let $z_1$, $z_2$ and $z_3$ be complex vertices of an equilateral triangle. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

Question

Prompt:

Let $z_1$, $z_2$ and $z_3$ represent vertices of an equilateral triangle in the complex plane. Show $z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$.

Question:

I hope the following question isn't too soft. I know a solution to the above question by taking the sides of the triangle and equating them by rotations in the complex plane of $e^{i\pi/3}$.

What, if anything, is the significance of the result other than showing some clever algebraic manipulation? Does it tell us something unique about equilateral triangles defined by vertices in the complex plane?

Is there any intuition to be gained here?

Answer 1

You can make a nice, somewhat abstract argument that connects this; in particular, equilateral triangles have various nice properties - like:

• For fixed $z_1,z_2$, there are exactly two $z_3$ such that $(z_1,z_2,z_3)$ is an equilateral triangle. If $z_1$ and $z_2$ are real, then those $z_3$ are reflections around the real axis.
• Any permutation of an equilateral triangle is an equilateral triangle - so $(z_1,z_2,z_3)$ being equilateral implies $(z_1,z_3,z_2)$ and $(z_2,z_3,z_1)$ are also equilateral.
• The map $z\mapsto az+b$ preserves equilateral triangles.

If we make a leap to trying to think about an analogous polynomial $P(z_1,z_2,z_3)$ such that, at any root, $(z_1,z_2,z_3)$ is an equilateral triangle, then the above conditions translate to:

• $P(z_1,z_2,z_3)$ has degree two in each of its variables. It has only real coefficients.
• $P(z_1,z_2,z_3)$ is a symmetric polynomial.
• $P(z_1,z_2,z_3)$ is homogenous.

Together, these yield that if such a $P$ existed, $P$ would have to be a sum of the only two symmetric and homogenous polynomials of degree two, meaning, for some $\alpha$ and $\beta$: $$P(z_1,z_2,z_3)=\alpha(z_1^2+z_2^2+z_3^2)+\beta(z_1z_2+z_2z_3+z_3z_1)$$ but, since we desired $P(z,z,z)=0$ to be a solution, we can calculate that clearly $\alpha=-\beta$ and, dividing out by $\alpha$ would give that the only polynomial that could feasibly represent equilateral triangles would be: $$P(z_1,z_2,z_3)=z_1^2+z_2^2+z_3^2-z_1z_2-z_2z_3-z_3z_1.$$ Then, if you showed that if $P(z_1,z_2,z_3)=0$ implied $P(z_1+c,z_2+c,z_3+c)=0$, which could be done algebraically, or perhaps through some clever manipulation, you would have that the roots of this polynomial satisfied all the conditions we wanted of equilateral triangles - and those conditions suffice to uniquely define the set of equilateral triangles, which proves that $P$'s roots are all equilateral triangles.

The point of this answer is, of course, to draw a connection between the symmetries defining the original problem and how the carry over to symmetries in the algebra and, indeed, suffice to show the statement. You could do this the other way round to learn geometric symmetries from the algebraic form of $P$ as well.

Answer 2

Here's an example of what I mean in my comment. The fact that all sides are equal, and that the angles between them are equal can be expressed quite elegantly as $$\frac{z_3-z_2}{z_2-z_1}=\frac{z_1-z_3}{z_3-z_2}=\frac{z_2-z_1}{z_1-z_3}.$$

The desired equation now follows from some elementary algebra. I believe this result can be generalized to any regular n-gon.

Answer 3

If $z_1=w_1+x$, $z_2=w_2+x$ and $z_3=w_3+x$, the identity becomes $$ w_1^2+2w_1x+x^2+ w_2^2+2w_2x+x^2+ w_3^2+2w_3x+x^2=\\ w_1w_2+w_1x+w_2x+x^2+ w_2w_3+w_2x+w_3x+x^2+ w_3w_1+w_3x+w_1x+x^2 $$ that becomes $$ w_1^2+w_2^2+w_3^2=w_1w_2+w_2w_3+w_3w_1 $$ so, taking $x=-z_3$, we can assume $z_3=0$ and we need to prove that if $z_1$, $z_2$ and $0$ are the vertices of an equilateral triangle, then $$ z_1^2+z_2^2=z_1z_2. $$

If $z_1=w_1u$ and $z_2=w_2u$ (with $u\ne0$), the identity above becomes $w_1^2+w_2^2=w_1w_2$, so we can assume $z_1=e^{i\pi/6}$, which means that $z_2=e^{i(\pi/6+\pi/3)}$ or $z_2=e^{i(\pi/6-\pi/3)}$.

If $z_2=e^{i(\pi/6+\pi/3)}=e^{i\pi/2}=i$, then a rotation brings us in the situation with $z_1=e^{-i\pi/6}$ and $z_2=e^{i\pi/6}$. So we are reduced to proving that $$ e^{i\pi/3}+e^{-i\pi/3}=1 $$ which is obvious.

Answer 4

An analogous condition is $(z_1 + z_2 + z_3)^2 = 3(z_1 z_2 + z_1 z_3+z_2 z_3)$, or

$$\left(\frac{z_1+z_2+z_3}{3}\right)^2 = \frac{z_1 z_2 + z_1 z_3 + z_2 z_3}{3}$$

that is, the equation with roots $z_k$, $k=1,3$, is of the form

$$z^3 - 3 a z^2 + 3 a^2 z + b = 0$$ or

$$(z-a)^3 = c$$

This also indicates how to generalize the statement for $n$ complex numbers ( impose $n-2$ equalities).

Answer 5

@Milo Brandt is completely correct.

Now, there aren't any things left to know about the equilateral triangles, but there are a lot of amazing things with the complex numbers.

I am talking about the relationships of complex numbers and the coordinate geometry...

With the help of complex numbers, it becomes very easy for us to calculate or prove things like this. Imagine proving the same thing with the coordinates/coordinate geometry. It makes our work easier.

Now, since you want to know about the equilateral triangles, these properties,or I should say, these results may help you:

• let the circum centre of the triangle be $z_0$ , then, $3z^2_0 = z^2_1 + z^2_2 + z^2_3$
• The angle between the points must be $60°$ or $π/3^c$(most obvious). Thus, $arg((z_3 - z_1)/(z_2 - z_1)) = π/3^c$. And similar for all other angles.

Best ever property for equilateral triangle:

• The triangle with vertices as $z_1,z_2$ and $z_3$ can only be equilateral if and only if: $1/(z_1 - z_2) + 1/(z_2 - z_3) + 1/( z_3 - z_1) =0$.

I am going to prove the last point for your convenience...

Let,

$a = z_1 - z_2$

$ b = z_2 - z_3$

$ c = z_3 - z_1$

Thus,

$a + b + c = 0$

Also,

$1/a + 1/b + 1/c = 0$. [ GIVEN IN QUESTION]

Thus,

$1/a + (b+c)/bc = 0$

$1/a - a/bc = 0$ [ as $a + b + c =0$]

Thus,

$a^2 = bc$

$|a^3| = |abc|$

Similarly,

$|b^3| = |abc|$ and $|c^3| = |abc|$

Now, add these equations

$|a^3| + |b^3| + |c^3| = 3|abc|$

Which is only possible when $|a| + |b| + |c| = 0$ or $ |a| = |b| = |c| $

But since $|a|,|b|,|c|>0$ ,

Thus $|a| = |b| = |c|$.

Thus the triangle is an equilateral triangle!

Hence, Proved!

Answer 6

The significance of the expression $$z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1\tag1$$ is that it is the most concise and easiest to apply in analyzing an equilateral triangle in the complex plane.

There may be some suspicion on the statement. It is tempting, for example, to point out that the expression $$|z_1-z_2| = |z_2-z_3| = |z_3-z_1|\tag2 $$ is simpler. However, it is only so in appearance, not in complex algebraic operation. It is actually the following complex conjugate expression in disguise $$(z_1-z_2)(\bar z_1-\bar z_2 )= (z_2-z_3)(\bar z_2-\bar z_3 ) =( z_3-z_1)(\bar z_3-\bar z_1 )\tag3$$ which is more involved than (1). Besides, the expression (2) is in fact a system of two equations, which is again more involved than the single equation (1). In practical application, it is more straightforward to check the condition (1) in determining whether $z_1$, $z_2$ and $z_3$ form an equilateral triangle than, say, checking the condition (2).

Answer 7

This equation can be restated as $(z_1-z_2)^2+(z_2-z_3)^2+(z_3-z_1)^2=0$. With $z:=z_1-z_2$, some $\omega=(\exp\pm2\pi i/3)$ satisfies $z_2-z_3=\omega z,\,z_3-z_1=\omega^2z$. Since $\omega^3=1\ne\omega$, $\omega^2+\omega+1=\frac{\omega^3-1}{\omega-1}=0$. The LHS is then $z^2(1+\omega^2+\omega^4)=z^2(1+\omega^2+\omega)=0$.

Answer 8

Does it tell us something unique about equilateral triangles?

In the complex plane the sides of any triangle are represented by numbers $a,b,c$ where $$a+b+c=0.$$

The equation you are interested in is equivalent to the equation $$a^2+b^2+c^2=0$$ and therefore tells us that equilateral triangles have the very special property that all quadratic symmetric polynomials of the sides are constant.

This gives us an elegant tool for investigating these triangles with minimal algebraic manipulation. I will give just two examples.

The sides of an equilateral triangle satisfy $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

Simply divide the symmetric quadratic equation $ab+ba+cb=0$ by $abc$.

The equation $a^2+b^2+c^2=0$ only represents equilateral triangles

$abc-c^3=c(ab+bc+ca)-c^2(a+b+c)=0$. Then $a^3,b^3,c^3$ are all equal (to $abc$) and therefore $a,b,c$ all have the same length.