I am trying to find a way to calculate the inverse of a planar flow.
In general, I understand that with normalizing flows, one can simply go from one distribution to the other with the change of variable formula: $$p_1(\textbf{z'}) = p_0 (\textbf{z})\left\lvert\det (\frac{df(\textbf{z})}{d\textbf{z}})\right\rvert^{-1},$$ where $f$ is a planar flow and $\textbf{z'} = f(\textbf{z})$.
However, it is unclear for me how to calculate the inverse of $f$, such that $f^{-1}(\textbf{z'}) = \textbf{z}$, when $f$ can be written like this: $\textbf{z'} = f(\textbf{z}) = \textbf{z} + \textbf{A}\tanh(\textbf{B}\textbf{z} + \textbf{b})$.
I guess the answer is pretty straight forward since normalizing flows are made to be invertible (see this introduction), but I could not find it written anywhere explicitly.
Could someone help me out?