Analytical function with boundary condition has a removable singularity Let $U \subset \mathbb{C}$ with $z_0 \in \mathbb{C}$ such that $f$ is an analytic function on $U \setminus \{z_0\} $ 
$|f(z)| \leq M|z-z_0|^{-p} \quad$ where $z \in U, M \in \mathbb{R}$ and $p < 1$
I now have to prove that $z_0$ is a removable singularity but I'm not really sure how I can prove this. 
 A: Define $g:U\rightarrow\mathbb{C}$ to be $$g(z)=\begin{cases} (z-z_0)^2f(z) & z\neq z_0\\ 0 & z=z_0\end{cases}.$$ Due to your bound on $f$, we get that $g$ is continuous on $U$. In fact, it is not hard to see that $g$ is complex differentiable, with $$g'(z)=\begin{cases} 2(z-z_0)f(z)+(z-z_0)^2f'(z) & z\neq z_0\\ 0& z=z_0\end{cases},$$ where the derivative at zero is obtained using the definition and your bound (the choice of $p$ is key here). Hence, $g$ is holomorphic on $U$, and so on some neighborhood $O$ of $z_0$, $$g(z)=\sum\limits_{n=0}^\infty a_n (z-z_0)^n$$ on $O$. Since $g(z_0)=g'(z_0)=0,$ we can write $$g(z)=(z-z_0)^2h(z),$$ where $h(z)=\sum\limits_{n=0}^\infty a_{n+2}(z-z_0)^n$ on $O$. Comparing this to our definition of $g$ yields that $h(z)=f(z)$ on $O\setminus\{z_0\},$ and so we can uniquely analytically extend $f$ to $U$ via $$\tilde{f}(z)=\begin{cases} f(z) & z\neq z_0\\ h(z_0) & z=z_0\end{cases},$$ removing the singularity. 
This follows Taylor's argument in his notes on complex analysis.
