something with the Laurent series is confusing me, first I'll give a background of what I think I know.

If $z_0$ is an isolated singularity of a function $f$ we can find the $Res(f, z_0)$ by finding the coefficient of the $\frac{1}{z-z_0}$ of the laurent series expansion. So for example in $$f(z) = z + \frac{i}{z-1}$$ If we want to find $Res(f, 1)$ we can see that the residue is $i$ from the above definition.

However, if instead we expand the series in $z_0 = 0$ we get a Taylor Series valid for $|z| < 1$ that goes like $$f(z) = z - i\sum_{n=0}^\infty z^n$$ and a Laurent series expansion valid for $|z| > 1$ which is $$f(z) = z + \sum_{n=0}^\infty \frac{i}{z^{n+1}}$$ and if I wanted I could just make $n = 0$ and I would get the same $i$, which is the $Res(f, 1)$, and it doesn't happen just on this exercise. Is this valid or to get the $Res(f, z_0)$ I do really need powers of $z-z_0$ and this is all just a coincidence? Everything I read says that to get the residue at $z_0$ I need to have powers of $z - z_0$, however most of the times using a series of $z^n$ works as I get the same value, by using the expansion valid for $|z| > z_0$ and it gives me the correct value of $Res(f, z_0)$.

Would appreciate clarification on this subject as I feel I'm missing something here.

  • $\begingroup$ But is it valid for any expansion? I was under the impression that it had to be powers of (z-z0) since it's what I've read on most places. $\endgroup$ Commented Jan 14, 2019 at 18:34
  • $\begingroup$ There is a single simple pole at $z=1$ and a corresponding residue associated with it. There is no pole at $z=0$. Note that for $|z|<1$ and $|z|>1$ you have two series representations. For the latter, $z=0$ is NOT included in the domain. For the former, $z=0$ IS included in the domain and the residue at $z=0$ is $0$. $\endgroup$
    – Mark Viola
    Commented Jan 14, 2019 at 18:37

1 Answer 1


That will fail in most cases. For instance,$$\operatorname{res}_{z=1}\left(\frac1{1-z^2}\right)=-\frac12,$$but if you apply your method, then you will get $0$, of course, sine $\dfrac1{1-z^2}$ is an even function.

  • $\begingroup$ I checked wolfram since I couldn't calculate the expansions on that one myself, and it seems to check out, thanks for giving an example where this clearly fails! I guess the ones that were given to me were all similar and I was getting that result. Just as a curiosity, in the example I've given, would there even be any reason to calculate Laurent's series with powers of simply z? Since I couldn't get the residue on 1 that way. $\endgroup$ Commented Jan 14, 2019 at 18:48

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