8. If the complex number $x$ satisfies $x+\frac{1}{x}=-1$, then $x^{2013}+\frac{1}{x^{2013}}=$ $\qquad$ .
8. 2 .

Given that $x^{2}+x+1=0$.
Since the discriminant $\Delta=-3<0$, $x$ is a complex number.
Also, $x^{3}-1=(x-1)\left(x^{2}+x+1\right)=0$
$\Rightarrow x^{3}=1$.
Therefore, $x^{2013}+\frac{1}{x^{2013}}=\left(x^{3}\right)^{671}+\frac{1}{\left(x^{3}\right)^{671}}=1+1=2$.