4・135 Find all real or complex roots of the system of equations
$$\left\{\begin{array}{l}
x+y+z=3 \\
x^{2}+y^{2}+z^{2}=3 \\
x^{5}+y^{5}+z^{5}=3
\end{array}\right.$$
[Solution] Let $x, y, z$ be the roots of the cubic equation $r^{3}-a r^{2}+b r-c=0$. Then, by Vieta's formulas, we have: $a=3, b=3$, and 
$$x^{n+3}-a x^{n+2}+b x^{n+1}-c x^{n}=0 \quad (n \text{ is a natural number})$$
This also holds for $y$ and $z$. Thus, we have:
$$\begin{array}{l}
x^{n+3}-a x^{n+2}+b x^{n+1}-c x^{n}=0 \\
y^{n+3}-a y^{n+2}+b y^{n+1}-c y^{n}=0 \\
z^{n+3}-a z^{n+2}+b z^{n+1}-c z^{n}=0
\end{array}$$

Adding these three equations, we get:
$$\begin{array}{l}
\left(x^{n+3}+y^{n+3}+z^{n+3}\right)-a\left(x^{n+2}+y^{n+2}+z^{n+2}\right)+b\left(x^{n+1}+y^{n+1}+\right. \\
\left.z^{n+1}\right)-c\left(x^{n}+y^{n}+z^{n}\right) \\
= 0
\end{array}$$

Setting $n=0,1,2$, we obtain:
$$\begin{array}{l}
x^{3}+y^{3}+z^{3}=3 c \\
x^{4}+y^{4}+z^{4}=12 c-9 \\
x^{5}+y^{5}+z^{5}=30 c-27
\end{array}$$
$\therefore \quad c=1$, hence we have $(r-1)^{3}=0$, which gives the roots of the equation as
$$x=y=z=1$$