# system of equations-complex number

I have a task: Prove that if for each complex number u,w,z we have:

$$uwz=1$$ and $$u+w+z=u^{-1}+w^{-1}+z^{-1}$$

then at least one of them is equal 1.

I tried substituting $$u=(uw)^{-1}$$ to the second equality and prove that by denying, but I can't solve it.

Thanks in advance.

## 3 Answers

$$w = {1\over zu} \implies z+u+{1\over zu} = {1\over z} + {1\over u} +zu$$

so $$z^2u+u^2z+1= u+z+z^2u^2$$

so $$z^2u(u-1)-z(u^2-1)+u-1=0$$

so $$(u-1)(z^2u-z(u+1)+1)=0$$

so if $$u =1$$ we are done else:

$$z^2u-zu-z+1=0\implies zu(z-1)-(z-1)=0$$

so if $$z=1$$ we are done else $$zu = 1$$ so $$w=1$$.

If $$uwz=1$$ then $$u^{-1}+w^{-1}+z^{-1}=uw+uz+wz$$, and $$(X-u)(X-w)(X-z)=X^3-(u+w+z)X^2+(uw+uz+wz)X-uwz=\cdots$$ etc.

Let $$S=u+w+z$$As observed in another answer, the equalities $$uvw=1$$ and $$S=u^{+1}+w^{-1}+w^{-1}$$ imply $$uw+wz+zu=S$$.

Now let's use Vieta's relations: $$u,w$$ and $$z$$ are the roots of the cubic polynomial $$Z^3-SZ^2+SZ-1=Z^3-1-SZ(Z-1)=(Z-1)(Z^2+Z+1-SZ),$$ which has $$Z=1$$ as a root.