The following question is from the book Complex Variables written by Levinson and Redheffer. It is the question 8.7 in additional problems in Chapter 3.
Let $f$ be a complex function that is analytic in $\Omega - \{\alpha\}$, where $\Omega$ is a region, and has a isolated singularity $\alpha$. If Re$f(z) \leq -m\ln$|$ z - \alpha $| for any $z \in \Omega - \{\alpha\}$, show that the singularity is removable.
I am trying to prove that $f$ is uniformly bounded in $\Omega - \{\alpha\}$ by showing that |$e^{f(z)}$| is bounded. However, I can only show that |$e^{f(z)}$|$\leq $|$\frac{1}{|z - \alpha|^m}$ but the right hand side is not bounded as $z$ approach $\alpha$.
Any hints will be appreciated.
We may assume that $m$ is a positive integer. We shall work completely in small disk $D$ around $\alpha$. There is a holomorphic function $h$ such that $(z-\alpha)^{m} e^{f(z)}=h(z)$. We can write $h(z)=(z-\alpha)^{k} \phi (z)$ where $\phi$ is holomorphic with no zeros in smaller disk $D'$ and $k \geq 0$. We can write $\phi(z) =e^{\psi(z)}$ in $D'$ for some holomorphic function $\psi$. We now get $(z-\alpha)^{m} e^{f(z)}=(z-\alpha)^{k} e^{\psi(z)}$. Now Consider a small circle $|z-\alpha| =r$ contained in $D'$. On this circle $(z-\alpha )^{m-k}$ has a continuous logarithm. As is well known this implies that $m=k$. Thus $e^{f(z)}=e^{\psi(z)}$. It is now easy to conclude that $f$ has a removable singularity at $\alpha$.