A problem regarding removable singularity in complex analysis.

Question

I am reading complex analysis from Stein and Shakarchi.I am struggling with a problem given in the exercise.The problem is given below:

Suppose $f(z)$ is holomorphic in a punctured disc $D_r(z_0)-\{z_0\}$.Suppose also that $|f(z)|\leq A|z-z_0|^{-1+\epsilon}$ for some $\epsilon>0$ and all $z$ near $z_0$.Show that the singularity of $f$ at $z_0$ is removable.

I have no idea how to start the problem.I think that somehow I have to use Riemann theorem on removable singularity,which states that if $f$ is bounded in a neighborhood of $z_0$,then $f$ has removable singularity.That means I have to show that $f$ is bounded.Can someone help me to show that provided I am on the right track.

Addendum

I saw similar question on stack exchange but that too had no satisfactory answer.

Answer 1

Note that $|f(z) (z-z_0)|\le A|z-z_0|^{\epsilon}$, hence $f(z)(z-z_0)$ is bounded near $z_0$. Therefore $z_0$ is a removable singularity of $f(z)(z-z_0)$. And from the inequality again, $|f(z)(z-z_0)|\rightarrow 0$ as $z\rightarrow z_0$. Now we have the expansion $f(z)(z-z_0)=\sum_{n=1}^\infty a_n(z-z_0)^n$ where $n$ starts from $1$, as the LHS is $0$ at $z=z_0$. Hence $f(z) = \sum_{n=1}^\infty a_n (z-z_0)^{n-1}$ has analytic continuation to $z=z_0$ as well.

Answer 2

In the lavender region, $$ \begin{align} f(a) &=\frac1{2\pi i}\int_\gamma\frac{f(z)}{z-a}\,\mathrm{d}z\tag{1a}\\ &=\frac1{2\pi i}\int_{\partial B(z_0,r)}\frac{f(z)}{z-a}\,\mathrm{d}z\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: Cauchy's Integral Formula
$\text{(1b)}$: the integral along the oppositely directed linear segments cancel
$\phantom{\text{(1b): }}$the integral along $B(z_0,\delta)$ is at most $\frac{2\pi\delta A\delta^{-1+\epsilon}}{|z_0-a|-\delta}=\frac{2\pi A\delta^{\epsilon}}{|z_0-a|-\delta}\to0$

The function defined by $\text{(1b)}$ is equal to $f$ in $B(z_0,r)\setminus\{z_0\}$ and is analytic in $B(z_0,r)$.