Inverting the complex function $f(z)=z+\sqrt{z^2-2}$.

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 1

HINT: Note that $f(z)^2=z^2+2z\sqrt{z^2-2}+z^2-2=2zf(z)-2$.

Answer 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.