, 2016

Solution:

The value of $\omega^{4}+\omega^{3}+\omega^{2}+\omega+1=\frac{\omega^{5}-1}{\omega-1}=\frac{1}{\omega-1}$. The sum of these values is therefore the sum of $\frac{1}{\omega-1}$ over the five roots $\omega$. Substituting $z=\omega-1$, we have that $(z+1)^{5}=2$, so $z^{5}+5 z^{4}+10 z^{3}+10 z^{2}+5 z-1=0$. The sum of the reciprocals of the roots of this equation is $-\frac{5}{-1}=5$ by Vieta's.