Finding the Laurent series of $f(z) = e^{1/z}$ about $z_0 = 0$

Question

Specifically, I want to do this without invoking the Taylor/power series definitions for $e^z$. I know they exist and accept they're valid methods, but I want to show them from the definition of Laurent series, mostly just to see how it might work out - if at all (which goes into my question).

I know from the definition of a Laurent series given by

$$f(z) = \sum_{n = -\infty}^{\infty} a_k (z - z_0)^k$$

we have

$$a_n = \frac{1}{2 \pi i} \int_C \frac{f(\zeta)}{(\zeta - z_0)^{n+1}}d\zeta$$

where $C$ is a closed curve in the annulus in which the series is convergent in oriented counterclockwise.

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For $f(z) = e^{1/z}$ have the discontinuity at $z = 0$ but other than that we don't have any "trouble points," so to speak. So we can define our annulus by $0 < |z| < \infty$, in effect.

We begin by trying to calculate the above integral, with a curve $C$ which is a circle of finite, positive radius. It becomes clear that

$$ \int_C \frac{e^{1/\zeta}}{\zeta^{n+1}}d \zeta$$

could be evaluated using the equality

$$\frac{1}{2\pi i} \int_C \frac{e^{1/\zeta}}{\zeta^{n+1}}d \zeta=\frac{1}{n!}f^{(n)}(0)$$

... but it won't work out, since the derivatives all have powers of $z$ in the denominator (thus running into problems with inputs of $0$). I certainly can't think of any other way (with our current skillset in the class) as to how it might work.

Is this just a case where it's like "you just have to use the power series for $e^z$, substitute in $1/z$, and invoke the uniqueness of Laurent series," or is there something I'm overlooking?

Answer 1

Full disclosure: This question was posed several months prior to this answer. Having completed my course in this subject, I feel a bit more confident in giving an answer, and so, for the sake of closing this question, I will answer my question.

Strictly speaking, you could calculate the Laurent series for the function manually, but it'd be a pain in the rear. Typically, we don't bother formally calculating the series from scratch; as was correctly posited in the OP, we typically instead manipulate known power series and such, and invoke the uniqueness of the Laurent series.

Let's elaborate. We know the power series for $e^z$ is given by the following:

$$e^z = \sum_{k=0}^\infty \frac{z^k}{k!} \tag 1$$

Equation $(1)$ is the expansion of $e^z$ about the point $z=0$. Taking $z \mapsto 1/z$, we can then get

$$e^{1/z} = \sum_{k=0}^\infty \frac{z^{-k}}{k!} = \sum_{k=0}^\infty \frac{1}{k! \cdot z^k} \tag 2$$

Yes, immediately, this just looks like a power series about $z=0$ again, and not a Laurent series, at least at first glance. However, that's fine! Let us consider the general forms of power and Laurent series:

$$f(z) = \sum_{k=0}^\infty a_nz^n \tag {Power series}$$ $$g(z) = \sum_{k=-\infty}^\infty a_nz^n \tag {Laurent series}$$

So suppose we end up with a power series, such as the form for $f$ above. Notice it starts at $a_0$ in lieu of some term of negative index (or $-\infty$). This is the cause in part of OP's confusion: it is obviously a power series. However, it is also a Laurent series: let $a_k = 0$ for all $k \leq -1$: then you could have a Laurent series in that sense, and by its uniqueness, you'd have the power series.

This same phenomenon occurs in $(2)$: let $a_n = 0$ for all negative integers $n$. Then you have the Laurent series for $e^{1/z}$!