Ambiguous function pole definition

Question

As I have seen on the Internet each definition of a complex function pole says, that a pole is a point such that $$\lim_{z\to z_0} f(z) = \infty$$
But what if this limit doesn't exist? What if
$$
\begin{cases}
\lim\limits_{z \to z_0 + 0} f(z) = \infty \\
\lim\limits_{z \to z_0 – 0} f(z) = -\infty \end{cases}
$$
Does this mean, that $z_0$ is the pole of $f(z)$?
For example when I investigate
$$
f(z)=\frac{1}{(z^2-4)(z^4-1)}
$$
All $\lim\limits_{z\to z_k} f(z)$ where $z_k \in \{ \pm i, \pm1, \pm2\}$ do not exist (as they give $\pm\infty$ depending on the way we approach those points), but doing a query on WolframAlpha says that all those points are function's poles.
Wolfram's output

Answer 1

On the complex plane it does not make sense to talk about $-\infty$ because there is no ordering on the complex numbers. Indeed, assume $i > 0$ then $i^2=-1>0$ and similarly for the other side. So in a sense $\infty$ can be seen as what any unbounded complex functions converges to.

Yes, your $f(z)$ does have a pole at your described points. In a sense having a pole of order $N$ at $z_0$ can be interpolated that the functions behaves like \begin{equation} \frac{1}{(z-z_0)^N} \end{equation} near $z_0$ which can be made precise by saying \begin{equation} f(z) = \frac{1}{(z-z_0)^N} g(z) \end{equation} where $g(z)$ is holomorphic on a small neighborhood of $z_0$. This can be done to your $f(z)$ by polynomial division.

But it seems like you're confused about isolated singularities. So I'll give a brief overview of the three types:

A removable singularity at $z_0$ is one where $\lim_{z \rightarrow z_0} (z-z_0)f(z) = 0$ In particular, this is called removable because one can redefine a $f(z)$ to be holomorphic at $z_0$ using Cauchy's Integral Formula representation. So this is the most mild type of singularity because essentially it's not one. For example $f(z) = z$ on $B(\varepsilon,0) - 0$ But defining $f(0)=0$, we see $f$ extends to be holomorphic on the entire ball.

A pole is what I described above, which can be characterized by $\lim_{z \rightarrow z_0} f(z) = \infty$ (recall $-\infty$ does not make sense on $\mathbb{C}$) An example is any rational function with a zero on the denominator that is not canceled out by the numerator's zero.

The worst type of singularity is an essential singularity which can be seen as an isolated singularity that is neither a pole or removable, which is equivalent to \begin{equation} \limsup_{z \rightarrow z_0} f(z) = \infty \end{equation} \begin{equation} \liminf_{z \rightarrow z_0} |f(z)| < \infty \end{equation} Some powerful theorems are that if $f$ is holomorphic on the punctured disk of $z_0$ then it's image on any $\delta$ neighborhood is dense in $\mathbb{C}$. A stronger refinement of this theorem is Picard's Big Theorem which states if $f$ is holomorphic on the punctured disk with an essential singularity at $z_0$ then it must touch every point on $\mathbb{C}$ except at most one infinitely often. An example is $f(z)=e^{1/z}$. In fact every holomorphic function that is entire and non-polynomial has an essential singularity at $\infty$.

Another way to look at them is their Laurent Series. A removable singularity has no $a_{-n}$ terms in its Laurent series, a pole has finitely many $a_{-n}$ terms, while an essential singularity has infinitely many negative $a_{-n}$ terms.