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 Let $z$ be a root satisfying the condition, then the original equation is equivalent to $z^{2008}(z+1)=-1$. 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 with center $(-1,0)$ and radius 1, and the circle with center $(0,0)$ and radius 1 on the complex plane, so $z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$. Upon verification, these two roots are both roots of the original equation, thus the sum of all roots with modulus 1 is -1.