[Proof] Let the complex numbers $\alpha=x(\cos A+i \sin A), \beta=y(\cos B+i \sin B)$, $\gamma=z(\cos C+i \sin C)$. Suppose $\alpha, \beta, \gamma$ are the roots of the cubic equation
$$u^{3}-a u^{2}+b u-c=0$$

By Vieta's formulas, we have:
$$\begin{aligned} a & =\alpha+\beta+\gamma=x \cos A+y \cos B+z \cos C+i F_{1}=\text { real, } \\ b & =\alpha \beta+\beta \gamma+\gamma \alpha \\ & =\frac{1}{2}\left[(\alpha+\beta+\gamma)^{2}-\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)\right] \\ & =\frac{1}{2}\left[a^{2}-\left(x^{2} \cos 2 A+y^{2} \cos 2 B+z^{2} \cos 2 C+i F_{2}\right)\right] \\ & =\frac{1}{2}\left[a^{2}-x^{2} \cos 2 A-y^{2} \cos 2 B-z^{2} \cos 2 C\right]=\text { real } \\ c & =\alpha \beta \gamma=x y z[\cos (A+B+C)+i \sin (A+B+C)] \\ & = \pm x y z=\text { real, } \end{aligned}$$

It is evident that equation (1) is a cubic equation with real coefficients and roots $\alpha, \beta, \gamma$.
Now, let $S_{r}=\alpha^{r}+\beta^{r}+\gamma^{r}$ ( $r$ is a positive integer), then to prove that for all positive integers $r$, $F_{r}=0$, it suffices to prove that for all positive integers $r$, $S_{r}$ is real. The following proof uses mathematical induction to show that $S_{r}$ is real.

Given $F_{1}=F_{2}=0$, so $S_{1}=\alpha+\beta+\gamma, S_{2}=\alpha^{2}+\beta^{2}+\gamma^{2}$ are both real, and $S_{0}=\alpha^{0}+\beta^{0}+\gamma^{0}=3$ is also real. If $S_{r-2}, S_{r-1}, S_{r}$ are all real, then for $S_{r+1}$, since $\alpha, \beta, \gamma$ satisfy equation (1), we have the equation
$$S_{r+1}-a S_{r}+b S_{r-1}-c S_{r-2}=0$$

which implies $S_{r+1}=a S_{r}-b S_{r-1}+c S_{r-2}$.
Since $a, b, c$ are all real, and $S_{r}, S_{r-1}, S_{r-2}$ are all real, it follows from the above equation that $S_{r+1}$ is also real.

Therefore, for all positive integers $r$, $F_{r}=0$.