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All my doubts came from the following exercise:

Given

\begin{equation} f(z) = z^3 e^{-\frac{2}{z}} \end{equation}

find the singularities of the function in $\bar{\mathbb{C}}$ and specify their nature (removable, pole of order $N$, essential singularities) and the residue at these points.

Solution: $z = 0$ essential singularity, $z = \infty$ pole; $\operatorname{Res}(f(z), 0) = \frac{2}{3} = - \operatorname{Res}(f(z), \infty)$

So I started considering the expansion of $e^z$ about $0$: \begin{equation}e^z = \sum\limits_{n = 0}^{\infty}\frac{z^n}{n!}\end{equation} and then I substituted $z\to-\frac{2}{z}$ \begin{equation}e^{-\frac{2}{z}} = \sum\limits_{n = 0}^{\infty}\frac{(-1)^n2^n}{n!}\left(\frac{1}{z}\right)^n\end{equation} Multiplying everything for $z^3$ I get the function $f(z)$ expanded in Laurent series: \begin{equation}f(z) = \sum\limits_{n = 0}^{\infty}\frac{(-1)^n2^n}{n!}z^{-n+3}\end{equation}

Then considering the $c_{-1}$ coefficient I got the residues of the function in $0$: $$ \operatorname{Res}(f(z), 0) = \left(\sum\limits_{n = 0}^{\infty}\frac{(-1)^n2^n}{n!}z^{-n+3}\delta_{4}^{n}\right) = \frac{(-1)^42^4}{4!} = \frac{2}{3} $$ where $\delta_j^i$ in the Kronecker delta. Now, since a friend of mine was doing the same exercise he developed a solution expanding the function at $\infty$ and he got the same result as me, with the same sign (so wrong because $\frac{2}{3}$ should be the right in $0$, but not in $\infty$), but now I'm not sure of what I have done, since it appears to me that this expansion is not in $0$ but in $\infty$, so $\frac{2}{3}$ should be the residue of $f(z)$ at $\infty$ and not in $0$ (as my expansion is valid at $\infty$, since is $e^{-\frac{2}{z}}$.

Last I tried computing with Mathematica the series and the result and I got:

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Basically Mathematica in $0$ is not computing results, while in $\infty$ it's giving me the same series that I thought I get in $0$, but the residue computed at $\infty$ is the same that I got with my calculations.

What's going on? Am I calculating the series in $\infty$ without knowing? And if it's so, how is it possible that the residue at $\infty$ is the opposite of the $c_{-1}$ at $\infty$?

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  • $\begingroup$ I would start by looking at limits where we get singularities points which are $0$ and $\infty$ Now: a. if the limit exists so it is removable singularity b. if the limit is $\pm \infty$ it is a pole and c. if the limit does not exist it is essential singularity $\endgroup$ – gbox Sep 12 '17 at 11:34
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What you did to compute the residue at $0$ is correct.

The residue at $\infty$ is$$-\operatorname{res}_0\left(\frac1{z^2}f\left(\frac1z\right)\right)=-\operatorname{res}_0\left(\frac1{z^2}\cdot\frac1{z^3}e^{-2z}\right)=-\operatorname{res}_0\left(\frac{e^{-2z}}{z^5}\right)=-\frac23.$$

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How exactly do you define the residue at $\infty$? A definition I'm working with is: $\DeclareMathOperator{\Res}{Res}$

Let $\infty$ be an isolated singularity of $f$, meaning there exists $R > 0$ such that $f$ is holomorphic at $\mathbb{C}\setminus \overline{B(0,R)}$. Then: $$\Res(f, \infty) = -\frac{1}{2\pi i}\int\limits_{|z|=R} f(z)\,dz$$

Furthermore, $f$ has a Laurent expansion on $\mathbb{C}\setminus \overline{B(0,R)}$:

$$f(z) = \sum_{n=-\infty}^\infty c_nz^n, \quad|z| > R$$ We see that $\Res(f, \infty) = -c_{-1}$.

A result which follows from this definition is:

Let $z_1, \ldots, z_k \in \mathbb{C}$ be all isolated singularities of $f$ and $f$ is holomorphic on $\mathbb{C}\setminus\{z_1, \ldots, z_k\}$. Then: $$\sum_{i=1}^k \Res(f,z_i) + \Res(f, \infty) = 0$$

In your case, $f(z) = z^3e^{-\frac{2}{z}}$ has a Laurent series $$f(z) = \sum_{n = 0}^{\infty}\frac{(-1)^n2^n}{n!}z^{-n+3}, \quad z\ne 0 $$ on $\mathbb{C}^\times$ so using this series we get $\Res(f,0) = c_{-1} = \frac{2}{3}$ and $\Res(f, \infty) = -c_{-1} = -\frac{2}{3}$. We verify that $\Res(f,0) + \Res(f,\infty) = 0$, since $0$ is the only isolated singularity of $f$.

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Laurent series can't be about $\infty$; they are always about a finite complex number. They are power series, so that's the only thing that makes sense!

However, the same function can have multiple Laurent series that converge in different annuluses. For example, the function $f(z) = \frac{1}{1-z} + \frac{1}{2-z}$ has three different Laurent series centered on $z=0$:

$$ \sum_{n=0}^{\infty} \left(1 + 2^{-1-n} \right) z^n \qquad \qquad |z| < 1 $$ $$ \sum_{n=-\infty}^{-1} (-1) z^n + \sum_{n=0}^{\infty} 2^{-1-n} z^n \qquad \qquad 1 < |z| < 2 $$ $$ \sum_{n=-\infty}^{-1} (-1 - 2^{-1-n}) z^n \qquad \qquad 2 < |z| $$

When you ask for the series at infinity, what you're telling Mathematica to do is to give you the series for which $\infty$ is in (or on the boundary of) the domain of convergence. Mathematica seems to default to picking 0 as the center when you ask this.


Regarding the series around zero.... I don't think Mathematica explicitly documents its semantics, but presumably what's going on here is that the series command intends to return a finite approximation. For example, something of the sort

$$ \frac{1}{1-z} \approx 1 + z + z^2 + O(z^3) $$

The problem with your example is there aren't any finite approximations. The essential singularity means that you need all of the leading terms to get an approximation near $z=0$.

So, what Mathematica does is it instead factors out an analytic function expressing the essential singularity, and then gives a series approximation for what's leftover.

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  • $\begingroup$ So my expansion (the one i made with the series of $e^z$) is about $0$ and my method is right? $\endgroup$ – opisthofulax Sep 12 '17 at 12:08

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