# Residue and Laurent Series, is this valid?

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.

• 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. Jan 14, 2019 at 18:34
• 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 $0$. Jan 14, 2019 at 18:37

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.