# Classification of isolated singularity by limit

Let $z_0$ be an isolated singularity of $f$ so there exist a punctured ball $B'$ centered in the singularity where $f$ is holomorphic. Let $f(z)=\sum_n a_n (z-z_0)^n$ be the Laurent series of $f$ in $B'$.

1. $z_0$ is a removable singularity iff $\lim_{z \to z_0}f(z)$ exists

2. $z_0$ is a pole iff $\lim_{z \to z_0}f(z)=\infty$

3. $z_0$ is an essential singularity iff $\lim_{z \to z_0}f(z)$ does not exists (either finite or infinite)

Proof of 1. If $z_0$ is a removable singularity then $f(z)=\sum_{n=0}^{+\infty} a_n (z-z_0)^n$ so it follows $\lim_{z \to z_0}f(z)=a_0$. Conversely if the limit exists, putting $f(z_0):=\lim_{z \to z_0}f(z)$ gives an holomorphic extension of $f$.

Proof of 2. If $z_0$ is a pole of order $m$ then $g(z)=f(z)(z-z_0)^m$ is holomorphic and so $\lim_{z \to z_0}f(z)=\lim_{z \to z_0}\frac{g(z)}{(z-z_0)^m}=\infty$. Conversely by hypotesis one has that $\lim_{z \to z_0}g(z)=0$ where $g(z):=1/f(z)$; so $g$ can be extended holomorphic over all the ball and has a zero in $z_0$. It follows that $f$ has a pole in $z_0$.

Proof of 3. Let $z_0$ be an essential singularity. By Casorati-Weierstrass, $f(B')$ is dense in $\mathbb C$ and so $\forall w \in \mathbb C, \forall \epsilon>0, \exists \zeta \in B'$ such that $|f(\zeta)-w|<\epsilon$. Take $w,w' \in \mathbb C$ with $w \neq w'$, then using the previous statement one can construct two sequences $\{z_n\}$ and $\{u_n\}$ such that $f(z_n) \to w$ and $f(u_n) \to w'$. So $\lim_{z \to z_0}f(z)$ does not exist.

Can I use exclusion for the converse? I mean, if the limit of $f$ in $z_0$ does not exist then $z_0$ is not a removable singularity or a pole and so it is an essential singularity.

• u are correct...
– user476275
Dec 2, 2017 at 5:29
• Note that if a singularity is an essential singularity, then it's neither a pole nor a removable singularity. And you have proved the first two equivalent statements. So you can automatically get the third equivalent statement. But your proof for the statement $2$ is not complete. I will add some details you missed later. Jun 5, 2020 at 11:10

## 1 Answer

I am writing to add some necessary details to the proof of the statement $$\textbf{2}$$.

The OP didn't provide his definition of $$\textit{pole}$$. By inspecting his proof, I guess his definition of $$\textit{pole}$$ is based on the singular part of Laurent series. To be more specific, his definition might be that, if the Laurent series at a singularity $$z_0$$ has at least one and at most finitely many terms of negative powers, then $$z_0$$ is called a $$\textit{pole}$$.

So in the proof of the converse direction of the statement $$\textbf{2}$$, we have to show that the Laurent series of $$f$$ at $$z_0$$ has at least one and at most finitely many terms of negative powers.

Proof:

First let's suppose that $$\text{lim}_{z\to z_0}f(z)=\infty$$.

Let $$g(z):=\frac{1}{f(z)}$$. Then $$g(z)$$ is well-defined in a ball centered at $$z_0$$ and $$\text{lim}_{z\to z_0}g(z)=0.$$ This implies that $$g$$ is holomorphic in the ball.

Since $$g$$ is holomorphic in the ball and has a zero at $$z_0$$, we can find some holomorphic function $$h$$ s.t. $$g(z)=(z-z_0)^k h(z)$$ with $$h(z_0)\neq0$$, where $$k\ge 1.$$

Since $$h$$ is holomorphic in the ball and $$h(z_0)\neq0$$, we may assume $$h$$ is nonzero in the ball. (Because we can always shrink the ball to a smaller ball, it doesn't really matter.) So $$\frac{1}{h}$$ is holomorphic in the ball and hence we can write $$\frac{1}{h(z)}=\sum_{n=0}^\infty a_n(z-z_0)^n$$ for some coefficients $$\{a_n\}$$ by the Taylor theorem.

So at every point $$z$$ in the punctual ball (i.e. the ball excluding $$z_0$$), we have $$f(z)=\frac{1}{g(z)}=\frac{1}{(z-z_0)^k h(z)}=\sum_{n=0}^\infty a_n(z-z_0)^{n-k}=\sum_{n=-k}^\infty a_{n+k}(z-z_0)^n.$$

By the definition, we conclude that $$z_0$$ is indeed a pole.

$$\tag*{\blacksquare}$$