# How can I classify the singularities of a function, without calculating the Laurent series?

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...

• Actually it is very easy to compute the Laurent expansion of $(1-e^{iz})/z^2$. You can see that $0$ is a pole of order 2. – Crostul May 3 '17 at 15:50
• Okay, poor example maybe (since the singularity is at $z=0$). What about for the function $\frac{e^{iz}}{z^2 + 4}$ which has singularities at $\pm 2i$? – M Smith May 3 '17 at 15:52
• $\dfrac{1 - e^{iz}}{z^2}$ has a simple pole at $0$. If you have your function given in the form $f(z) = \frac{g(z)}{h(z)}$ where $g$ and $h$ are holomorphic at $z_0$, the singularity is removable if the order with which $g$ vanishes at $z_0$ is not smaller than the order with which $h$ vanishes there, and if $\mathscr{o}(g) < \mathscr{o}(h)$, then it's a pole of order $\mathscr{o}(h) - \mathscr{o}(g)$. If $g$ and/or $h$ has a pole at $z_0$ it's similar. If one and only one has an essential singularity at $z_0$ and $g\not\equiv 0$, then $f$ has an essential singularity at $z_0$. – Daniel Fischer May 3 '17 at 16:03
• If $g$ and $h$ both have an essential singularity at $z_0$, it can be anything, removable, a pole, an essential isolated singularity or a non-isolated singularity. – Daniel Fischer May 3 '17 at 16:03
• Thank you for your response. To clarify, when you say 'the order with which [the function] vanishes', do you mean the value at which it tends to 0? – M Smith May 3 '17 at 17:28