8. Suppose the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is
Answer: $-1$.
Let $z$ be a root that satisfies 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 in the complex plane, so $z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$. Upon inspection, these two roots are indeed roots of the original equation, so the sum of all roots with modulus 1 is -1.