Composition of an entire function with the principal part of a Laurent series $\def\bbC{\mathbb{C}}
\def\bbR{\mathbb{R}}
$For $r,R\in\bbR$, denote
$$
A(r,R)=\{z\in\bbC:r<|z|<R\}
$$
to the open annulus in $\bbC$ of inner and outer radii $r$ and $R$, respectively. Suppose that
$$
f(z)=\sum_{k=-\infty}^{+\infty}a_{k}z^{k}
$$
is a convergent Laurent series in $A(r,R)$. We call $\sum_{k\geq0}a_kz^k$ and $\sum_{k<0}a_kz^k$ the Taylor part and principal part of the Laurent series, respectively. Let $g$ be an entire function such that $g(0)=0$. We have that $g\circ f$ is a holomorphic function in $A(r,R)$ and hence it has a convergent Laurent series development in this annulus. My question is: if the Taylor part of $f$ vanishes, then the Taylor part of $g\circ f$ vanishes as well?
A physicist approach tells me that yes, since if $g(z)=\sum_{k=1}^{+\infty}b_kz^k$, then we have
$$
g(f(z))=b_1f(z)+b_2f(z)^2+b_3f(z)^3+\cdots
$$
and by expanding $f$ and “reordering the terms,” we must end up with a Laurent series with vanishing Taylor part.
 A: The “Taylor part” $\sum_{k\geq0}a_kz^k$ of $f$ vanishes if and only if $f(z) = \sum_{k<0}a_kz^k$ in $A(r, R)$, and that happens if and only if $f$ is the restriction of a function $F$ which is holomorphic in $|z| > r$ with $\lim_{z \to \infty} F(z) = 0$.
In that case is $g \circ f$ the restriction of $g \circ F$, which is also holomorphic in $|z| > r$ with $\lim_{z \to \infty} g(F(z)) = 0$, so that the Taylor part of $g \circ f$ vanishes as well.
A: Okay, here is a rigorous proof. Write the Laurent series of the composition
$$
g(f(z))=\sum_{k=-\infty}^{+\infty}c_kz^k,\quad z\in A(r,R).
$$
We want to show that $c_k=0$ for all $k\geq 0$. Fix $\rho$ between $r$ and $R$. For any $k$, we have
\begin{align}
2\pi ic_k&=\int_{|z|=\rho}\frac{g(f(z))}{z^{k+1}}dz\\
&=\int_{|z|=\rho}\sum_{n=1}^{+\infty}b_k\frac{f(z)}{z^{k+1}}dz\\
&=\sum_{n=1}^{+\infty}b_k\int_{|z|=\rho}\frac{f(z)}{z^{k+1}}dz,
\end{align}
where we used the Fubini-Tonelli theorem, seeing the sum as an integral with the counting measure.
By the same argument, we have
\begin{align}
\int_{|z|=\rho}\frac{f(z)}{z^{k+1}}dz
&=\int_{|z|=\rho}\sum_{m=-1}^{-\infty}a_m\frac{z^m}{z^{k+1}}dz\\
&=\sum_{m=-1}^{-\infty}a_m\int_{|z|=\rho}\frac{z^m}{z^{k+1}}dz.
\end{align}
Thus, if $k\geq 0$, the integral $\int_{|z|=\rho}\frac{z^m}{z^{k+1}}dz$ vanishes and so does $c_k$.
