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 $g(z) = z+z^2$?
 A: $zB(z)$ 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.
A: Since $zB(z)\to 0$ as $z \to 0$, we have that $g(z)/z \to 0$ as $|z|\to \infty$. Hence, letting $\varepsilon  = 1$, there corresponds a positive number $k$ such that $|g(z)/z| \le 1$ for all $|z| > k$, i.e., $|g(z)| \le |z|$ for all $|z| > k$.
A: $\lim\limits_{z\to z_0} (z-z_0)f(z)=0$ implies that there exists $\delta_0>0$
such that $$0<|z-z_0|<\delta_0\implies|(z-z_0)f(z)|<1.$$
Actuall, the coefficients of Laurent series of $f$ at $z_0$ are
$$b_n=\frac{1}{2\pi i}\int_{|z-z_0|=r}\frac{f(z)}{(z-z_0)^{-n+1}}\mathrm{d}z,n\geq1,$$
where $0<r<\delta_0$.
When $n\geq2$,
$$|b_n|=\left|\frac1{2\pi i}\int_{|z-z_0|=\rho}\frac{(z-z_0)f(z)}{(z-z_0)^{-n+2}}\,\mathrm{d} z\right|
\leq\frac1{2\pi}\cdot2\pi\rho\cdot\frac{1}{\rho^{-n+2}}=\rho^{n-1}\to0\quad(\rho\to0).$$
So $$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0},$$
and this implies
$$0=\lim_{z\to z_0}(z-z_0)f(z)=b_1.$$
So we have
$$b_1=b_2=\cdots=b_n=\cdots=0.$$
