I'm working on the following past paper question: In each of the following cases, $f$ has a singularity at 0 . Classify these singularities, and, in the case of the isolated singularities, calculate the residues:

(i) $f(z)=\frac{\sin z}{1-\mathrm{e}^{-z}}$;

(ii) $f(z)=\frac{\sin z}{\left(1-\mathrm{e}^{-z}\right)^3}$;

(iii) $f(z)=\frac{\sin z}{1-\mathrm{e}^{1 / z}}$.

I've completed the first two parts by computing the Laurent series and finding the $\frac{1}{z}$ coefficient. However, I'm struggling to see what to do for (iii). I believe $z=0$ is an essential singularity, but how would I compute the residue?

I've also tried it on WolframAlpha, but it doesn't seem to know what to do at $z=0$. Some help would be much appreciated!

Also more generally, does a function with an essential singularity have a residue/Laurent expansion?


1 Answer 1


In general, the residue of a complex function, being holomorphic at the neighborhood of a point $z=z_0$ is the defined as the coefficient of the term $\frac{1}{z}$ in its Laurent series expansion. For example: $$ e^\frac{a}{z}=1+\frac{a}{z}+\frac{a^2}{2z^2}+\frac{a^3}{6z^3}+\cdots \implies \text{Rez}(e^{\frac{1}{z}})|_{z=0}=a. $$ The holomorphism condition of a complex function at the neighborhood of a point is fundamental. The problem with part C of your question is that $\frac{\sin z}{1-e^\frac{1}{z}}$ is not holomorphic at any neighborhood of $z=0$. The reason is that this function, has simple singularities at $z=\frac{1}{2k\pi i}$ for $k\in \Bbb Z-\{0\}$ and any neighborhood of $z=0$ will contain infinitely many of these simple singularities. So, no residue is defined for this function at $z=0$.

Also see how to calculate residues.

  • $\begingroup$ Very well explained, thanks! Can I just check a few things: 1) Laurent series doesn't exist either then? 2) z=0 is an essential singularity (Laurent series has infinite negative powers of z) in this case? 3) Is what you said always true for essential singularities, or are sometimes the residues defined? $\endgroup$
    – yw_2003
    Commented Jan 9, 2023 at 17:15
  • 1
    $\begingroup$ You're welcome. 1) Yes. It does not exist. The reason is that the closed integrals defining the coefficients of the Laurent series, enclose infinitely many singularities. 2) Yes. $z=0$ is essential, but not isolated (i.e. it is not the only singularity of the function in at least one of its neighborhoods.) 3) The residue at a singularity is defined only if it is isolated. For example, $z=0$ is an essential singularity for both $text{Ln} z$ and $e^\frac{1}{z}$. However, it is an isolated singularity only for the second function, for which the residue is equal to $1$. $\endgroup$ Commented Jan 9, 2023 at 17:57
  • $\begingroup$ Ayas Thanks a lot! $\endgroup$
    – yw_2003
    Commented Jan 9, 2023 at 19:23
  • $\begingroup$ Good luck!$$$$$$$$ $\endgroup$ Commented Jan 9, 2023 at 19:47

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