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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.

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  • $\begingroup$ u are correct... $\endgroup$
    – user476275
    Dec 2, 2017 at 5:29
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    $\begingroup$ 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. $\endgroup$
    – Sam Wong
    Jun 5, 2020 at 11:10

1 Answer 1

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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.


$$\tag*{$\blacksquare$}$$

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