Classify the singularity of the function $$f(z) = e^{−1/z}$$ at $z = 0$ in the complex plane, and write the first three terms of the Laurent series expansion of $f(z)$ about $z = 0$.

I pull out my definition of Laurent Series:

$$f(z) = \sum_{-\infty}^{+\infty} a_n(z-z_0)^n$$

$$a_n = \frac{1}{2\pi i} \oint \frac{f(z') dz'}{(z' - z_0)^{n+1}}$$

Finding $a_{-1}$ seems like a reasonable thing to try first:

$$a_{-1} = \frac{1}{2\pi i} \oint e^{-1/z} dz$$

After about half an hour of fruitless u-substitution, I realized it wasn't solvable in terms of elementary functions, so I was on the wrong track. Not knowing what else to do, I looked at the answer, which tells me:

At $z=0$ there is an essential singularity: the Laurent series can be written as

$$f(z) = 1 + \sum_{n=1}^{\infty} \frac{1}{n!} \frac{1}{z^n} = 1 + \sum_{n=-1}^{-\infty} \frac{z^n}{n!}$$

I don't understand at all how this result was arrived at. It looks roughly like the Taylor expansion, but a) Taylor expansions aren't valid around the singular point (that's why we're using a Laurent series in the first place) and b) I'm sure it needs a $(-1)^n$ in there somewhere, because we're dealing with powers of a negative, so the terms should have alternating signs.

Can someone explain simply how this answer works, and why we don't need to compute the integrals for $a_{-n}$?

  • $\begingroup$ $f(z)$ doesn't have a Taylor expansion about $z=0$... For starters, $f(0)$ doesn't exist. However, you can take the Taylor series of $f(1/z)$, which is well-known. $\endgroup$ – Simply Beautiful Art May 27 '17 at 19:41
  • $\begingroup$ You are effectively using the fact that the Taylor expansion of $e^x=\sum\frac{x^n}{n!}$. You can use the transformation $z\mapsto\frac1z$ to move the singularity of $e^{-1/z}$ to $\infty$ and then using the expansion of $e^{-z}$, in which case you can utilize the Taylor expansion. $\endgroup$ – Clayton May 27 '17 at 19:46
  • $\begingroup$ @Clayton "Move the singularity" -- interesting, how general is that technique? Also, I still end up with $\sum_{n=0}^{+\infty} (-1)^n \frac{1}{n! z^n}$ -- I can't get the $(-1)^n$ to disappear. Might it be a typo in the answer? $\endgroup$ – dain May 27 '17 at 19:54
  • $\begingroup$ @dain let me call $w=1/z$. If $z\neq 0$ then $w$ is a point in the complex plane. Since the exponential function is an entire function is completely licit to write $\exp(-1/z)=\exp(-w)= 1-w+w^2/2!-w^3/3!+...$ (now should be clear from where $(-1)^n$ comes from) which is exactly a Laurent expansion of $f(z)$. Since the coefficients are unique you got the Laurent expansion on the annulus $\{z\in \mathbb{C}:0<|z|\}$. $\endgroup$ – Leandro May 27 '17 at 20:02
  • $\begingroup$ @Leandro That's precisely what I did, but in the answer I quoted for the problem the $(-1)^n$ isn't present -- is just an oversight, or is there some clever way of getting rid of it? $\endgroup$ – dain May 27 '17 at 20:07

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