Given $f(z) = z^2-19z$, there are complex numbers $z$ with the property that $z$, $f(z)$, and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$. There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$. Find $m+n$.
Notice that we must have \[\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .\]However, $f(t)-t=t(t-20)$, so
\begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\ &=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\ &=(z-19)(z+1)\\ &=(z-9)^2-100. \end{align*}
Then, the real part of $(z-9)^2$ is $100$. Since $\text{Im}(z-9)=\text{Im}(z)=11$, let $z-9=a+11i$. Then, \[100=\text{Re}((a+11i)^2)=a^2-121\implies a=\pm\sqrt{221}.\]It follows that $z=9+\sqrt{221}+11i$, and the requested sum is $9+221=\boxed{230}$.
(Solution by TheUltimate123)