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.
Best Answer
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.