Suppose $f(z)$ is analytic on $0<|z-z_0|<R$. And we find a Laurent's series for $f(z)$ on annulus $r<|z-z_0|<R$ 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 $0<|z-z_0|<R$). I find this hard to understand. In particular, fix an enclosed curve $\gamma'$ contained in $r<|z-z_0|<R$ 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 $0<|z-z_0|<R$, 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 $0 < |z-z_0|<R$ 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 $z_0$."

  • $\begingroup$ 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? $\endgroup$ – mathworker21 Aug 12 '19 at 23:52
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    $\begingroup$ @mathworker21 Certainly not. How about $\frac 1 {z-z_0}$? $\endgroup$ – Kavi Rama Murthy Aug 12 '19 at 23:53
  • $\begingroup$ @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). $\endgroup$ – Daniel Li Aug 12 '19 at 23:53
  • $\begingroup$ @KaviRamaMurthy thanks. I forgot $0$ was a number $\endgroup$ – mathworker21 Aug 12 '19 at 23:54
  • $\begingroup$ As far as I can see what you are saying is correct and $a_{-1}$ is the residue. Where did you find this statement? $\endgroup$ – Kavi Rama Murthy Aug 12 '19 at 23:58

Thank you for your comment on my answer here


I believe that you are absolutely correct.

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.


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