Question: I have the following complex valued function 
$$
f(z) = z + \sqrt{z^2 - 2}.
$$
I would like to find a function $g$ such that
$$
g(f(z)) = f(g(z)) = z
$$ 
on some domain of $\mathbb{C}$.

Answer: 
**HINT:** Note that $f(z)^2=z^2+2z\sqrt{z^2-2}+z^2-2=2zf(z)-2$.
$\begin{align} f(z) & = z+\sqrt{z^2-1} \\
(f(z)-z)^2 &= z^2-2 \\
f(z)^2 -2zf(z) + z^2 &= z^2-2 \\
-2zf(z) &= -2-f(z)^2 \\
z &= \frac{2+f(z)^2}{2f(z)} \\
\end{align}$

In order to know wether a transformation I apply at each step is injective or not I think of complex numbers either in polar or cartesian coordinate. 

$g(z)=\sqrt z$ appears injective (rotation + homothety) if picture it in polar coordinate for example. While $g(z) = z - 2$ appears injective (translation) if you picture it in cartesian coordinates.