I'm looking for some counterexample for the following situation : let $S$ et $T$ be two power series, with respective positive radius $R_S$ and $R_T$, with $T(0)=0$. Therefore there is $\rho>0$ such that $\rho<R_T$ and for all $z\in\mathbb C$ such that $|z|<\rho$, $\left|T(z)\right|<R_S$.
We know that the power series $S\circ T$ has a radius at least $\rho$, and for all $z\in\mathbb C$ such that $|z|<\rho$, $(S\circ T)(z) = S(T(z))$.
To be more precise, $S\circ T$ is the formal power series $\sum_n s_nT^n$, and $S(T(z)) = \sum_n s_n(T(z))^n$.
What I'm looking for is some example of $S$ and $T$ such that there is $z\in\mathbb C$, $|z|<R_T$ and $(S\circ T)(z) \ne S(T(z))$.
I'm at a loss to find such an example, so any help would be appreciated. As my english is a bit poor, I'm sure you'll want some more explanation, feel free to ask me :-)
An easy way is to use a non-bijective function and its inverse defined on some neighborhood of the origin. When the function (within its radius of convergence) takes a value twice you get a contradiction.
There are many possibilities but the simplest one I could think of is to take $T(z)=\exp(z)-1$ and $S(z)=\log(1+z)$. Clearly, $S\circ T(z) =z$ but $$S(T(2\pi i))= S(1-1)=0\neq 2\pi i.$$