8. If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ .
8. -1 .

Let $z$ be a root satisfying the condition. Then the original equation is equivalent to
$$
z^{2008}(z+1)=-1 \text {. }
$$

Taking the modulus on both sides, we get $|z+1|=1$.
Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle I: with center $(-1,0)$ and radius 1, and the circle with center $(0,0)$ and radius 1 in the complex plane. Therefore, $z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
: Upon verification, both roots are roots of the original equation. Thus, the sum of all roots with modulus 1 is -1.