jmc

Since $z^2+z+1=0,$ $(z-1)(z^2+z+1)=0,$ which simplifies to $z^3=1.$ Therefore, $$z^{49}=(z^3{)}^{16}\cdot z=z.$$Then $$\begin{array}{rl}{z}^{49}+z^{50}+z^{51}+z^{52}+z^{53}& =z+z^2+z^3+z^4+z^5\\ & =z+z^2+1+z+z^2\\ & =z+z^2=\boxed{-1}.\end{array}$$