Isolated singularity is removable iff $\lim\limits_{z\to z_0} (z-z_0)f(z)=0$ Could someone explain a step in the following proof?

Theorem
An isolated singularity $z_0$ of $f$ is removable if and only if $\lim\limits_{z\to z_0} (z-z_0)f(z)=0$.
Proof ($\Leftarrow$)
Take $z_0=0$ which implies $\lim_{z\to 0} zf(z) = 0$. Since:
$$f(z) = \color{red}{\sum_{n=0}^\infty a_nz^n} + \color{blue}{\sum_{n=1}^\infty \frac{b_n}{z^n}}$$
then $z\color{red}{A(z)}\to 0$ (the holomorhic part), which implies $z\color{blue}{B(z)}\to 0$ as $z\to 0$.
This means that $\displaystyle g(z) := B\left(\frac{1}{z}\right) = \sum_{n=1}^\infty b_nz^n$ is a entire function for which $|g(z)| \leqslant |z|$ $\color{green}{(1)}$ if $|z|$ is big enough. By Liouville's theorem $g(z)$ is itself a polynomial of degree at most 1 which implies $g(z) = b_1z$ etc

Question regarding step (1)
Why is $|g(z)|$ bounded by $|z|$? Couldn't $g(z)$ be something like $g(z) = z+z^2$?

$zB(z)$ wouldn't go to zero in that situation. In fact, $zB(z)\to 0$ if and only if $b_n=0$ for all $n>1$. This seems obvious, but needs proof in case infinitely many $b_n$ are nonzero, which is why the proof is written this way. So to clarify its argument, if $|g(z)|>|z|$ held for arbitrarily large $z$, then $|g(1/z)|>|1/z|$ for arbitrarily small $1/z$ and $zB(z)=zg(1/z)$ would be bigger than $1$ at the same points, in particular, could not converge to zero.
