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I previously held the conception that an essential singularity could be defined as a point $z_0$ of the function $f(z)$ for which: $$\lim_{z\rightarrow z_0}(z-z_0)^nf(z)$$ is not finite for any finite $n$. Although I don't think this definition is wrong (please correct me if it is), I am under the impression it is not the most useful. I think another definition of an essential singularity is:

The point $z_0$ is an essential singularity of the function $f(z)$ if and only if: $$\lim_{z\rightarrow z_0}f(z)$$ can be made to take at least two different values (taking the point at infinity to be only one value) when approaching from two different directions.

Would this definition/statement be correct? And if so can an equivalence be shown between these two definitions.

Quantum spaghettification
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    I like to think of Picard's Theorem for essential singularities. Namely, near an essential singularity, all complex values (except perhaps one) occur infinitely often. You're looking for a definition, but this distinction is a good way to think about it. For a pole, like $1/z$, the magnitude of the values of the function approach infinity. – Michael Burr Feb 24 '16 at 15:29
  • @MichaelBurr Would this be a necessary and sufficient condition for a point to be an essential singularity? or are there possible points where near that point all complex values occur infinitely often (except perhaps one) but which aren't essential singularities? – Quantum spaghettification Feb 24 '16 at 15:57
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    If the function is holomorphic at a point, this can obviously not happen. If it is a pole, then the magnitude, as Michael Burr said, approaches infinity and thus, this also doesn't happen. –  Feb 24 '16 at 16:12

2 Answers2

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Either the limit exists or it doesn't. There's no such thing as $\lim f(z)$ taking two or more different values.

You could instead say this: $f$ has an essential singularity at $z_0$ iff $\lim_{z\to z_0}f(z)$ fails to exist in $\mathbb C \cup\{\infty\}.$ As it turns out, this is equivalent to saying $f$ has an essential singularity at $z_0$ iff $f$ fails to have a limit at $z_0$ "in the worst possible way": for every $w \in \mathbb C \cup\{\infty\}$ there exists a sequence $z_n \to z_0$ such that $f(z_n) \to w.$

zhw.
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Let's start with the most popular definition of essential singularity:

Definition $\bigstar$: Let $U\subseteq\mathbb{C}$ be a open and convex set and $f:U \setminus \left\{z_{0}\right\} \longrightarrow \mathbb{C}$ be a holomorphic function with $z_{0} \in \mathbb{C}$. The point $z_{0}$ is essential singularity if $z_{0}$ is not a removable singularity or a pole.

In addition, we have the following facts that you can check in any book of complex analysis:

Proposition 1: Let $U\subseteq\mathbb{C}$ be a open and convex set and $f:U \setminus \left\{z_{0}\right\} \longrightarrow \mathbb{C}$ be a holomorphic function with $z_{0} \in \mathbb{C}$. The point $z_{0}$ is removable singularity if and only if $\lim_{z\rightarrow z_{0}}(z-z_{0})f(z)=0$.

And

Proposition 2: Let $U\subseteq\mathbb{C}$ be a open and convex set and $f:U \setminus \left\{z_{0}\right\} \longrightarrow \mathbb{C}$ be a holomorphic function with $z_{0} \in \mathbb{C}$. The point $z_{0}$ is a pole of degree $m$ if and only if $\lim_{z\rightarrow z_{0}}(z-z_{0})^{m}f(z)\neq 0$ and exists.

According to the last two propositions we can say that your statement:

... an essential singularity could be defined as a point $z_{0}$ of the function $f(z)$ for which: $$\lim_{z\rightarrow z_{0}}(z-z_{0})^{n}f(z)$$ is not finite for any finite $n$.

is correct long as you clarify that $n$ should be positive integer.

On the other hand, your other statement:

The point $z_{0}$ is an essential singularity of the function $f(z)$ if and only if: $$\lim_{z\rightarrow z_{0}} f(z)$$ can be made to take at least two different values (taking the point at infinity to be only one value) when approaching from two different directions.

is incorrect, the counterexample is to take the function $f(z)=\frac{1}{z}$, clearly $z_{0}=0$ is a pole of degree 1 of this function, but note that if we approach $z_{0}=0$ through numbers with null imaginary part and real part positive, then the function $f(z)$ approaches $\infty$. For other hand, if we approach $z_{0}=0$ through numbers with null imaginary part and real part negative, then the function $f(z)$ approaches $-\infty$. Therefore, $f(z)=\frac{1}{z}$ satisfies all requirements your statement with $z_{0}=0$, but $z_{0}=0$ is not an essential singularity. (see definition $\bigstar$).

But all is not lost, what you if you can say is:

If $z_{0}$ is essential singularity, then, for all $p,q\in \mathbb{C}$ there are sequences $(z^{*}_{n})_{n\in \mathbb{N}}\subseteq U$ and $(z'_{n})_{n\in \mathbb{N}}\subseteq U$ such that $\lim_{n \rightarrow \infty}z^{*}_{n}=z_{0}$, $\lim_{n \rightarrow \infty}z'_{n}=z_{0}$, $\lim_{n \rightarrow \infty}f(z^{*}_{n})=p$ and $\lim_{n \rightarrow \infty}f(z'_{n})=q$.

This is a consequence of Casorati–Weierstrass theorem.

Diego Fonseca
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