Let $z$ be a complex number satisfying $z^2 + z + 1 = 0.$  Compute
\[z^{97} + z^{98} + z^{99} + z^{100} + z^{101}.\]
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{align*}
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{align*}