Let $x,y,z$ be non-zero real numbers. Suppose $\alpha,\beta, \gamma$ are complex numbers such that $|\alpha|=|\beta|=|\gamma|=1$. If $x+y+z=0=\alpha x+\beta y+\gamma z,$ then prove that $\alpha=\beta=\gamma$.
I was trying to use the property that $\alpha\overline{\alpha}=|\alpha|^2=1$, $\beta\overline{\beta}=|\beta|^2=1$ and $\gamma\overline{\gamma}=|\gamma|^2=1$ in the following way:
It is given that $$x+y+z=0=\alpha x+\beta y+\gamma z$$ $$\implies \overline{\alpha}(x+y+z)=0=x+\overline{\alpha}(\beta y+\gamma z)...(1)$$ $$\implies \overline{\beta}(x+y+z)=0=y+\overline{\beta}(\gamma z+\alpha x)...(2)$$ $$\implies\overline{\gamma}(x+y+z)=0=z+\overline{\gamma}(\alpha x+\beta y)...(3)$$
Adding $(1),(2)$ and $(3)$, we have $\overline{\alpha}(\beta y+\gamma z)+\overline{\beta}(\gamma z+\alpha x)+\overline{\gamma}(\alpha x+\beta y)=0.$
This implies that $x(\alpha\overline{\beta}+\alpha\overline{\gamma})+y(\beta\overline{\alpha}+\beta\overline{\gamma})+z(\gamma\overline{\alpha}+\alpha\overline{\beta})=0.$