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Solve the system of equations $x+y+z=1, x y z=1,|x|=|y|=|z|=1, x, y, z \in \mathbb{C}$

I tried with polar representation, letting $x=\exp{(ia)}$,$y=\exp{(ib)}$, $z=\exp{(ic)}$, with this I got $a+b+c=2 \pi n$, so the three numbers are essentially three points on a unit circle, with their centroid being the point $1/3$. But I don't know how to proceed.

math_learner
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  • $i, -i,1$ is an answer – pie Mar 30 '24 at 14:41
  • From first expression, we can say $$\sin A + \sin B + \sin C = 0$$ From $xyz=1...(ii)$, we can say $$\sin(A+B+C)=0$$ After dividing (i) by (ii),we get $$ \sin (A+B)+sin(B+C)+sin(C+A)=0$$. I found all these. Maybe these will be helpful – Gwen Mar 30 '24 at 14:57

2 Answers2

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Use $xy= 1/z = \overline{z}$ and similar to conclude that $xy+xz+yz = \overline{x+y+z} = 1$. Then $x, y, z$ are the solutions of the cubic equation $t^3 - (x+y+z)t^2 + (xy+xz+yz)t - xyz=0$, hence of $t^3-t^2+t-1=0$. That equation can be written as $(t-1)(t^2+1)=0$, hence the solutions are $i$, $-i$, and $1$.

Catalin Zara
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If $x,y,z$ are distinct then $x+y+z=1$ means that the orthocenter is $1$ and lies on the circumcircle.

Hence the triangle is a right angled triangle with one of its vertices coinciding with $1$.

So if $x=1$ we have $y+z=0$ and $yz=1$ so that the other two variables are $i,-i$

Hari Shankar
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