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 it follows that $x^{n+3}-a x^{n+2}+b x^{n+1}-c x^{n}=0$ (where $n$ is a natural number), which also holds for $y$ and $z$. That is,
$$
\begin{array}{l}
x^{n+3}-a x^{n+2}+b x^{n+1}-c x^{n}=0, \\
y^{n+3}-a y^{n+2}+b y^{n+1}-c y^{n}=0, \\
z^{n+3}-a z^{n+2}+b z^{n+1}-c z^{n}=0,
\end{array}
$$

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

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