# 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.

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.