jmc

Since $z^2+z+1=0,$ $(z-1)(z^2+z+1)=0.$ This expands as $z^3-1=0,$ so $z^3=1.$ Therefore, $$z^{97}=z^{32\cdot 3+1}=(z^3{)}^{32}z=z.$$Similarly, we can reduce $z^{98},$ $z^{99},$ $z^{100},$ $z^{101},$ to $z^2,$ 1, $z,$ $z^2,$ respectively, so $$\begin{array}{rl}{z}^{97}+z^{98}+z^{99}+z^{100}+z^{101}& =z+z^2+1+z+z^2\\ & =(1+z+z^2)+(1+z+z^2)-1\\ & =\boxed{-1}.\end{array}$$