[Math] How to practically classify singularities in complex analysis

complex numberscomplex-analysislaurent series

I am having trouble developing an intuition around the different types of singularity in complex analysis.

The types of singularity that I am aware of are:

  1. Poles – These arise at $a_{0}$ when $\lim_{a\to a_{0}}f(a)$ does not exist, but $\lim_{a\to a_{0}}\frac{1}{f(a)}$ does.
  2. Removable singularities – These arise when both limits exist and are well-defined.
  3. Essential Singularities – Neither limit exists

However, I am having real trouble understanding this in practice. For instance, Wikipedia states that $e^{1/z}$ has an essential singularity at $z=0$. But, surely:

$$\lim_{z\to0} \frac{1}{e^{1/z}}=\lim_{z\to 0}e^{-\frac{1}{z}}=0$$

So I don't understand why it doesn't have a pole at $e^{1/z}$?

Moreover, how do we determine the Laurent Series of a function at an essential singularity? Ordinarily, if we have a pole of order $N$ at $z_{0}$, we have that:

$$(z-z_{0})^{N}f(z)=\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$$

As $(z-z_{0})^{N}f(z)$ would now be holomorphic at $z=z_{0}$, but with an essential singularity, I'm not sure how this works?

Best Answer

Your understanding of singularities is not correct. Isolated singularities are classified as

  1. Removable: $\lim_{z\to a}f(z)$ exists; $\lim_{z\to a}\dfrac{1}{f(z)}$ may $\infty$ (example: $f(z)=\sin^2z/z$.) Defining $f(a)=\lim_{z\to a}f(z)$ makes $f$ analytic at $a$.
  2. Pole: $\lim_{z\to a}f(z)=\infty$ (and $\lim_{z\to a}\dfrac{1}{f(z)}=0$.)
  3. Essential: none of the above, which is in fact equivalent to $\lim_{z\to a}f(z)$ does not exist, either as a finite complex number or $\infty$.

As explained in the comments, $\lim_{z\to0}e^{1/z}$ dos not exist.

An example of the Laurent series at an essential singularity: $$ e^{1/z}=\sum_{n=0}^\infty\frac{1}{n!}\,\frac{1}{z^n}!+\frac1z+\frac1{2\,z^2}+\dots $$

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