Consider a function $f : D \rightarrow \mathbb{C}$ with Laurent Series Expansion of the form $$ f(z) = \sum_{n = 0}^\infty a_n (z - z_0 )^n $$ where $z_0 \in D$ is a singular point of $f$. We can then classify the singularity $z_0$, from the laurent expansion of $f$, as follows:
- $z_0$ is a removable singularity if $a_n = 0$ for all $n < 0$.
- $z_0$ is a simple pole if $a_n = 0$ for all $n < -1$ and $a_{-1} \neq 0$.
- $z_0$ is a pole of order $N$, where $N \in \mathbb{Z}^+$, if $a_n = 0$ for all $n < -N$ and $a_{-N} \neq 0$.
- $z_0$ is an essential singularity of $f$ if no such $N$ (from the above point) exists.
However, for some functions, such as $$ f(z) = \frac{1 - e^{iz}}{z^2} \hspace{5mm} \text{or} \hspace{5mm} f(z) = \frac{e^{iz}}{z^2+4} $$ it is not easily possible to calculate the Laurent series expansion. Thus, we must find another (easier way) to classify the singularities. How would one do this?
EDIT: I have discovered that, if $z_0$ is a singularity of $f$, then if the limit $$ \lim_{z \rightarrow z_0} (z-z_0) f(z) $$ exists and is non-zero, then $z_0$ is a Simple Pole. However, I am still unsure of how to classify a singularity of this condition is not true...