The given equation can be written as \[\frac{z^9 + 1}{z + 1} = 0.\]Then $ z^9 + 1 = 0, $ or $ z^9 = -1 $. Since $ z = e^{i \theta}, $ \[e^{9i \theta} = -1.\]This means $ 9 \theta = \pi + 2 \pi k $ for some integer $ k $. Since $ 0 \le \theta < 2 \pi, $ the possible values of $ k $ are 0, 1, 2, 3, 5, 6, 7, and 8. (We omit $ k = 4, $ because if $ k = 4, $ then $ \theta = \pi, $ so $ z = -1, $ which makes $ z + 1 = 0 $.) Therefore, the sum of all possible values of $ \theta $ is \[\frac{\pi}{9} + \frac{3 \pi}{9} + \frac{5 \pi}{9} + \frac{7 \pi}{9} + \frac{11 \pi}{9} + \frac{13 \pi}{9} + \frac{15 \pi}{9} + \frac{17 \pi}{9} = \boxed{8 \pi}.\]