# Can one compute series expansions or complex residues at essential singularities?

Title says it all.

I'm noticing a trend of failure on Mathematica/WolframAlpha's parts when trying to compute either the Laurent expansions or the residues of functions like $\sin\dfrac{1}{z}$ about $z=0$. (Specifying $z=0$ for the residue query returns nothing.)

Is there something about essential singularities $z_0$ that prevents one from computing a series expansion about or residue at $z_0$? Is this "something" explicitly mentioned in the usual definitions of series/residues that I'm missing? Admittedly, my complex analysis knowledge has started to rust.

• Just plug $u=1/z$ into $\sin u = u -u^3/3! + \cdots$ for the Larent expansion; the residue is $1.$ – zhw. Oct 28 '15 at 19:46
• @zhw. Okay, that approach works fine for $\sin\dfrac{1}{z}$, but what if I made a slight adjustment? Say, $\sin\dfrac{z}{z-1}$. Does the same process still work? – user170231 Oct 28 '15 at 19:51

Recall the definition of an essential singularity. If a function $f$ has an essential singularity at $z=z_0$, then

$$\lim_{z \to z_0} (z-z_0)^m f(z) = \infty$$

That is, an essential singularity is "more singular" than any order pole.

That all said, the definition of a residue still applies as the coefficient of $(z-z_0)^{-1}$ in the Laurent expansion of $f$.

In your example of $f(z) = \sin{\left ( \frac{z}{z-1} \right )}$, there is an essential singularity at $z=1$. The Laurent expansion of $f$ about $z=1$ may be easily determined by expressing $f$ as follows:

\begin{align}f(z) &= \sin{\left (1+\frac1{z-1} \right )} = \sin{1} \cos{\left ( \frac{1}{z-1} \right )}+ \cos{1} \sin{\left ( \frac{1}{z-1} \right )}\\ &= \sin{1}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n)!(z-1)^{2 n}} + \cos{1}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)!(z-1)^{2 n+1}}\end{align}

Hopefully it is clear that the residue of $f$ at $z=1$ is $\cos{1}$.

• Are there any functions with essential singularities for which identities and "tricks" like these don't help? I'm just trying to wrap my mind around why WA and Mma are having trouble with these computations. – user170231 Oct 28 '15 at 21:00
• @user170231: plenty. What I would do is make use of Mathematica's expansion about Infinity, i.e., Series[f[z],{z,Infinity,n}] for an n-term expansion in $1/z$. – Ron Gordon Oct 28 '15 at 21:08