How to solve this system of equations for $x^2+y^2+z^2$?

Question

For the complex numbers $x,y,z$, the system of equations

$x^2-yz=i~~~~~ y^2-zx=i~~~~~ z^2-xy=i$

It is not easy for me to get $x^2+y^2+z^2$ from the above. I don't need the values of $x,y,z$

I'm stuck in what to do at first. Any advice would be helpful.

My attempt : we can get $x^2+y^2+z^2-xy-yz-zx=3i$,
so $(x-y)^2+(y-z)^2+(z-x)^2=6i$
I actually don't know whether this is a right way to solve the above.

Answer 1

Subtracting the equations in pairs we get

$$(x-y)(x+y+z) = 0$$ $$(x-z)(x+y+z) = 0$$ $$(y-z)(x+y+z) = 0$$

If $x+y+z \not = 0$ then $x=y=z$, but then $x^2-yz = 0 \not = i$ so we must have $x+y+z = 0$.

Next we can use the identity

$$x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+xz)$$

togeather with the equation system to arrive at

$$x^2 + y^2 + z^2 = 2 i + \frac{(x+y+z)^2}{3} = 2i$$

Answer 2

^{$\sqrt i$ denotes $e^{i\pi/4}$} $$(x^2-yz)-(z^2-xy)=i-i\\\implies(x-z)(x+z)-y(x+z)=0\\\implies(x+z)(x-y-z)=0$$

Case I: $x+z=0\implies z=-x$

$$x^2-xy=i,y^2+x^2=i\implies x^2-xy=x^2+y^2\implies y(x+y)=0$$

Case Ia: $x+y=0\implies y=-x$

$$2x^2=i\implies x=\sqrt\frac i2,y=z=-\sqrt\frac i2$$

Case Ib: $y=0$

$$x=\sqrt i,y=0,z=-\sqrt i$$

Case II: $x=y+z$

^continue....