# Why $a_{-1}$ term of Laurent series may not be residue?

Suppose $$f(z)$$ is analytic on $$0<|z-z_0|. And we find a Laurent's series for $$f(z)$$ on annulus $$r<|z-z_0| where $$r$$ may not be $$0$$. Then it is said that $$a_{-1}$$ of such Laurent's series may not be residue unless $$r=0$$ (Residue is defined as $$Res(f,z_0)=\frac{1}{2 \pi i}\int_\gamma f(z)dz$$ for any enclosed curve $$\gamma$$ on $$0<|z-z_0|). I find this hard to understand. In particular, fix an enclosed curve $$\gamma'$$ contained in $$r<|z-z_0| where the Laurent's series applies. Substitute $$f$$ with this Laurent series into $$Res(f,z_0)=\frac{1}{2 \pi i}\int_{\gamma'} f(z)dz$$. Then integral of all except the $$a_{-1}/(z-z_0)$$ terms should evaluate to $$0$$ as we have finished substitution and are just evaluating the integral with Cauchy formula (and thus are no longer concerned by where Laurent's series apply). In the end since $$Res(f,z_0)=\frac{1}{2 \pi i}\int_\gamma f(z)dz$$ takes same value for all $$\gamma$$ in $$0<|z-z_0|, our result based on $$\gamma'$$ applies in general. Where is the mistake in this proof?

Note 1. I have checked explicit expression for $$a_{-1}$$ in a Laurent expansion where $$r>0$$. I think $$a_{-1}$$ should be residue. The statement in Note 2 may be false. I hope the author or somebody with expertise in complex analysis could confirm.

Note 2. The original statement I was referring to can be found in Simon's answer in this post: Calculate residue at essential singularity:

"In fact the residue of $$f(z)$$ at an isolated singularity $$z_0$$ of $$f$$ is defined as the coefficient of the $$(z-z_0)^{-1}$$ term in the Laurent Series expansion of $$f(z)$$ in an annulus of the form $$0 < |z-z_0| for some $$R > 0$$ or $$R = \infty$$.

If you have another Laurent Series for $$f(z)$$ which is valid in an annulus $$r < |z-z_0|< R$$ where $$r > 0$$, then it might differ from the first Laurent Series, and in particular the coefficient of $$(z-z_0)^{-1}$$ might be different, and hence not equal to the residue of $$f(z)$$ at $$z_0$$."

• i'm confused. if $f$ is analytic on $0 < |z-z_0| < R$, then isn't the laurent series just composed of non-negative powers? – mathworker21 Aug 12 '19 at 23:52
• @mathworker21 Certainly not. How about $\frac 1 {z-z_0}$? – Kavi Rama Murthy Aug 12 '19 at 23:53
• @mathworker21 I'm not an expert on complex analysis but I think you are referring to the case where $f$ is analytic on $|z-z_0|<R$. I'm thinking $z_0$ is a singularity (essential or pole). – Daniel Li Aug 12 '19 at 23:53
• @KaviRamaMurthy thanks. I forgot $0$ was a number – mathworker21 Aug 12 '19 at 23:54
• As far as I can see what you are saying is correct and $a_{-1}$ is the residue. Where did you find this statement? – Kavi Rama Murthy Aug 12 '19 at 23:58

It was my fault for using the same letter $$R$$ in my first and second paragraphs there. I did not intend it to be the same $$R$$, so certainly I should have used two different symbols ! Please see the edit that I made to that answer, where I have gone into a little more detail.
Don't confuse $$\int_{|z| = r+\epsilon} f(z)dz= 2i\pi a_{-1}, \qquad \int_{|z| = \epsilon} f(z)dz = 2i\pi b_{-1}$$ where $$f$$ is assumed to be analytic on $$|z| \in (0,2\epsilon)$$ and $$|z| \in (r,R)$$ and $$a_n,b_n$$ are the Laurent coefficients of the expansion on each annulus.
If $$f$$ is analytic on $$|z|\in (0, R)$$ then $$a_n = b_n$$.
If $$f$$ has a pole on $$|z| \in (2\epsilon,r)$$ then both contour integrals won't give the same result.