MHB Find all possible values of |x+y+z|

[anemone]

Let $x,\,y,\,z$ be complex numbers such that $|x|=|y|=|z|=1$.

If $\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}=1$, find the product of all possible values of $|x+y+z|$.

 

[Opalg]

Let $x,\,y,\,z$ be complex numbers such that $|x|=|y|=|z|=1$.

If $\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}=1$, find the product of all possible values of $|x+y+z|$.

[sp]Write the equation as $x^3 + y^3 + z^3 = xyz$. If $\lambda$ is any complex number with $|\lambda|=1$ then the equation is unaltered if we replace $x,y,z$ by $\lambda x,\lambda y,\lambda z$, and $|x+y+z|$ is then also unaltered. So choose $\lambda$ to be a cube root of $(xyz)^{-1}$. The equation then becomes $x^3 + y^3 + z^3 = xyz = 1.$

Now let $a=x^3$, $b=y^3$, $c=z^3$. Then $|a| = |b| = |c| = 1$ and $a+b+c = abc = 1$, from which $\frac12(a+b) = \frac12(1-c).$ So the points $1,a,b,c$ all lie on the unit circle, and the midpoint of the chord from $a$ to $b$ is the same as the midpoint of the chord from $1$ to $-c$. But the midpoint of a chord of a circle uniquely determines its endpoints. So $a$ and $b$ must be the same as $1$ and $-c$. Therefore the three points are $\{1,c,-c\}.$ But their product is $1$, so that $-c^2 = 1$, $c = \pm i.$

Therefore $x,y,z$ are cube roots of $1,i,-i$. That leaves only finitely many possible choices for $x,y,z$. You can check that all of them are obtained (by rotation of the circle) from two basic solutions $\{1,\omega,\omega^{-1}\}$ and $\{1,\omega^5,\omega^{-5}\}$, where $\omega = e^{i\pi/6}$. The corresponding values of $|x+y+z|$ are $\sqrt3+1$ and $\sqrt3-1$, with product $(\sqrt3+1)( \sqrt3-1) = \boxed2.$[/sp]

 

[anemone]

[sp]Write the equation as $x^3 + y^3 + z^3 = xyz$. If $\lambda$ is any complex number with $|\lambda|=1$ then the equation is unaltered if we replace $x,y,z$ by $\lambda x,\lambda y,\lambda z$, and $|x+y+z|$ is then also unaltered. So choose $\lambda$ to be a cube root of $(xyz)^{-1}$. The equation then becomes $x^3 + y^3 + z^3 = xyz = 1.$

Now let $a=x^3$, $b=y^3$, $c=z^3$. Then $|a| = |b| = |c| = 1$ and $a+b+c = abc = 1$, from which $\frac12(a+b) = \frac12(1-c).$ So the points $1,a,b,c$ all lie on the unit circle, and the midpoint of the chord from $a$ to $b$ is the same as the midpoint of the chord from $1$ to $-c$. But the midpoint of a chord of a circle uniquely determines its endpoints. So $a$ and $b$ must be the same as $1$ and $-c$. Therefore the three points are $\{1,c,-c\}.$ But their product is $1$, so that $-c^2 = 1$, $c = \pm i.$

Therefore $x,y,z$ are cube roots of $1,i,-i$. That leaves only finitely many possible choices for $x,y,z$. You can check that all of them are obtained (by rotation of the circle) from two basic solutions $\{1,\omega,\omega^{-1}\}$ and $\{1,\omega^5,\omega^{-5}\}$, where $\omega = e^{i\pi/6}$. The corresponding values of $|x+y+z|$ are $\sqrt3+1$ and $\sqrt3-1$, with product $(\sqrt3+1)( \sqrt3-1) = \boxed2.$[/sp]

What a neat solution, bravo, Opalg! And thanks for participating.

I would also like to share the solution that comes along with this good problem, here goes:

Spoiler

Let $s=x+y+z$

Then

$\begin{align*}s^3&=x^3+y^3+z^3+3(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)+6xyz\\&=xyz\left(\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}+3\left(\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\right)+6\right)\\&=xyz\left(1+\left(3(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-9\right)+6\right)\\&=xyz\left(3s\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-2\right)\\&=xyz(3s\overline{s}-2)\,\,\text{because $\overline{s}=\overline{x}+\overline{y}+\overline{z}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$}\\&=xyz(3\left| s \right|^2-2)\end{align*}$

Taking absolute values, we find $\left| s \right|^3=\left| \left| 3s \right|^2-2 \right|$. It follows that $\left| s \right|$ must be a positive real root of $a^3-3a^2+2=0$ or $a^3+3a^2-2=0$. However, since the negative real roots of $a^3-3a^2+2=0$ are exactly the additive inverses of the positive real roots of $a^3-3a^2+2=0$, and all three roots of $a^3-2a^2+2=0$ are real ($a^3-3a^2+2=0$ maybe factored as $(a-1)(a^2-2a-2)=0$, and the discriminant of $a^2-2a-2$ is positive), the product of all possible values of $\left| s \right|$ is $(-2)\cdot(-1)^n$, where $n$ denotes the number of negative real roos of $a^3-3a^2+2=0$. By Descartes's Rule of Signs, we see that $n$ is odd, so the answer is 2, as desired.