11. (20 points) Solve the system of equations in the complex number range
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3 .
\end{array}\right.
$$
11. It is known that, $x y+y z+z x=3$.

Let $a_{n}=x^{n}+y^{n}+z^{n}(n \in \mathbf{N})$. Then
$$
\begin{aligned}
a_{n+3}= & (x+y+z) a_{n+2}- \\
& (x y+y z+z x) a_{n+1}+(x y z) a_{n} .
\end{aligned}
$$

Thus, $a_{3}=3 a_{2}-3 a_{1}+(x y z) a_{0}$,
$$
\begin{array}{l}
a_{4}=3 a_{3}-3 a_{2}+(x y z) a_{1}, \\
a_{5}=3 a_{4}-3 a_{3}+(x y z) a_{2} .
\end{array}
$$

From the conditions, we know $a_{0}=3, a_{1}=3, a_{2}=3, a_{5}=3$.
Therefore, $x y z=1$.
By Vieta's formulas, we know that $x, y, z$ are the roots of the equation
$$
t^{3}-3 t^{2}+3 t-1=0
$$

Thus, the solution to the system of equations is $(x, y, z)=(1,1,1)$.