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?
Best Answer
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.