The [complex number](https://artofproblemsolving.com/wiki/index.php/Complex_number) $z$ is equal to $9+bi$, where $b$ is a [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [real number](https://artofproblemsolving.com/wiki/index.php/Real_number) and $i^{2}=-1$.  Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to?
Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$. Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i$.
Setting these two equal, we get that $18bi = 243bi - b^3i$, so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$. Since $b > 0$, the solution is $\boxed{015}$.