$|x|=|y|=|z|=1$ and $x^3+y^3+z^3=-xyz$, are there infinite many values to $|x+y+z|$

Question

I was playing around with complex numbers and I tried this.

Let $x$, $y$, $z$ be complex numbers with the properties
$$|x|=|y|=|z|=1,$$ $$x^3+y^3+z^3=-xyz.$$

The question is how many values the following expression can take? $$|x+y+z|.$$ It is easy to see that $$|x+y+z|≤3,$$

but I found only $2$ solutions if $x=z=-y$ then $|x+y+z| =1$, and $$x=\frac{1}{2}+\frac{\sqrt{3}}{2}i ,\; y=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,\; z=1 $$ then $|x+y+z|=2$.

I believe there are infinitely many values but I couldn't find any other than these two solution.

I want a prove that there are infinite many values for $|x+y+z|$.

If there are finite values for$|x+y+z|$ then how many values? and also what is the greatest value of$|x+y+z|$.

Answer 1

I think there are only finitely many possible values.

The equation $$x^3+y^3+z^3=-xyz\qquad(*)$$ is homogeneous, so everything in sight is unaffected by the transformation $$(x,y,z)\mapsto (\lambda x,\lambda y, \lambda z),$$ where $|\lambda|=1$. So without loss of generality we can assume that $z=-1$ (use $\lambda=-1/z$), when $(*)$ reads $$ x^3+y^3=1+xy.\qquad(**) $$ As $x$ and $y$ are on the unit circle, so are $x^3,y^3,1$ and $xy$. Therefore the arguments of both sides of $(**)$ are the averages of the arguments of the terms. In other words. $$ \frac12(3\arg x+3\arg y)\equiv \frac12(\arg x+\arg y)\pmod{2\pi}. $$ Moving all the terms to the left hand side this can be rewritten to give $\arg x+\arg y\equiv0\pmod{2\pi}.$ In other words, $x$ and $y$ must have opposite arguments. As they are on the unit circle, it follows that $$y=\overline{x}=\frac1x.$$ Plugging this into the equation $(**)$ yields $$x^3+\frac1{x^3}=2.\qquad(***)$$ This is a degree six equation, so there are at most six solutions for $x$. Hence at most six possible values for $|x+y+z|$. Actually the solutions of $(***)$ are the third roots of unity, so we only get the two possible values for $|x+y+z|$ that you found yourself.

Answer 2

$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

$$-4xyz=(x+y+z)((x+y+z)^2-3(xy+yz+zx))$$

from $xy+yz+zx$ take $xyz$ common so you get $$-4xyz=(x+y+z)((x+y+z)^2-3xyz(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}))$$

Now $|x|=1$ so we have $x \bar{x}=1$ so $\frac{1}{x}=\bar{x}$

So, $$-4xyz=(x+y+z)((x+y+z)^2-3xyz(\bar{x}+\bar{y}+\bar{z}))$$ $$-4xyz=(x+y+z)((x+y+z)^2-3xyz(\overline{x+y+z}))$$

$$-4xyz=(x+y+z)^3-3xyz(|x+y+z|^2)$$

$$xyz( 3|x+y+z|^2-4)=(x+y+z)^3$$

Taking modulus both sides

$$|3|x+y+z|^2-4|=|x+y+z|^3$$

substitute $|x+y+z|=t$ $$|3t^2-4|=t^3$$

Now you can solve for t. There will definitely be finite number of t

Answer 3

Hint: $\;$ let $\,u=\dfrac{x}{z}\,$, $\,v=\dfrac{y}{z}\,$ where $\,z \ne 0\,$ since $\,|z|=1\,$, then the system can be written as:

$$ \begin{align} |u|=|v|=1 \\ u^3 + v^3 + 1 + uv = 0 \tag{1} \end{align} $$

Taking the conjugate of $\,(1)\,$, and using that $\,\overline u = \dfrac{|u|^2}{u} = \dfrac{1}{u}\,$, $\,\overline v = \dfrac{1}{v}\,$:

$$ \begin{align} \overline u^3 + \overline v^3 + 1 + \overline u \overline v = 0 \\ u^3 + v^3 + u^3v^3 +u^2 v^2 = 0 \tag{2} \end{align} $$

Subtracting $\,(2)-(1)\,$:

$$ u^3v^3 + u^2v^2 - uv - 1 = 0 $$

It follows that $\,uv = t\,$ is a root of $\,t^3+t^2-t-1=0\,$, so $\,t = \pm 1 \iff u = \pm \dfrac{1}{v} = \overline v\,$.
Substituting back in $\,(1)\,$:

$$ \pm \overline v^3 + v^3 + 1 \pm \overline v v = 0 \;\;\iff\;\; v^6 + v^3 \pm v^3 \pm 1 = 0 \tag{3} $$

What's left is to solve $\,(3)\,$ for $\,v\,$, then reverse the substitutions back to $\,u, x, y, z\,$. There will be infinitely many solutions $\,(x,y,z)\,$, but finitely many values of $\,|x+y+z|\,$ among those.