Schwarz derivative and Möbius transformation. In https://en.wikipedia.org/wiki/Schwarzian_derivative#Properties ,
some properties about Schwarz derivative
\begin{align}
(Sf)(z)&=
\left(\dfrac{f''(z)}{f'(z)}\right)'
-\dfrac{1}{2} \left( \dfrac{f''(z)}{f'(z)} \right)^2 \\
&=
\dfrac{f'''(z)}{f'(z)}-\dfrac{3}{2}\left(\dfrac{f''(z)}{f'(z)}\right)^2
\cdots \ast
\end{align}
are described.
And it is said that "The Schwarzian derivative of any Möbius transformation
$g(z)=\dfrac{az+b}{cz+d}$ is zero. Conversely,
the Möbius transformations are the only functions with this property."
I could understand the fact that the Schwarzian derivative of any Mobius transformation is zero.
But I cannot understand the fact that the Möbius trasformations are the only function with this property.
If $z \in \mathbb{R}$, I could understand this fact, owing to the $2$nd answer of this site, using "log". https://mathoverflow.net/questions/173182/schwarzian-derivative-of-a-diffeomorphism-is-zero-iff-linear-fractionals
However, "log" in $\mathbb{R}$ is not equal to "log" in $\mathbb{C}$.
So if $z \in \mathbb{C}$, I think I cannot use the approach using "log".
Summarizing the above, I wonder why Möbius transformations are the only solutions of $\ \ast$.
I want you give me some ideas.
 A: Per comment I will present a simple (though using a trick) proof of the fact that $S(f)=0$ implies $f$ Mobius. We work on a domain where $f'$ doesn't vanish so $S$ is analytic and we show that $f=g_1/g_2$ where $g_{1,2}''=0$ and that implies $f$ is Mobius on the given domain.
Consider a function $g$ for which $2f'g'+f''g=0$ on a small disc of center $w$ within the given domain - for example we can take $k(z)=\int_w^z \frac{f''(y)dy}{2f'(y)}$ which is analytic on a small disc centered at $w$ since $f'(z) \ne 0$ there; then take $g(z)=e^{-k(z)}$ and an easy check shows that $g'=-k'g=-f''g/(2f')$
But now differentiating the relation above we get:
$2f'g''+2f''g'+f'''g+f''g'=0$ and dividing by $f'$ and substituting $g'=-f''g/(2f')$ we get
$2g''+(f'''/f'-3(f'')^2/(2f'))g=0$ and obviously the $g$ coefficient is $S(f)=0$ so we indeed get $2g''=0$ hence $g$ is linear on the given small disc.
But now if $h=fg$ one has $h''=f''g+fg''+2f'g'=0$ since $g''=0$ and $2f'g'+f''g=0$, so $h$ is linear too and then $f=h/g$ is Mobius on the small disc, but by analytic continuation it is so on the given domain and of course we can extend it to the complex plane meromorphically with at most one pole!
Note that this proof shows how to solve $S(f)=p$ for any analytic $p$ as we get by the same trick and same computations that $f=h/g$ where now $2g''+pg=0$ and $2h''+ph=0$, so $f$ is a ratio of linearly independent solutions of a fairly simple differential equation (and conversely it is fairly straightforward to show that any such ratio gives an $f$ with $S(f)=p$)
