I'm interested in how residue at a point operation complies with algebraic operations:

$$\underset{z_0}{\operatorname{Res}}(f + g) = \, ?$$

$$\underset{z_0}{\operatorname{Res}}(f g) = \, ?$$

my guess is that

$$\underset{z_0}{\operatorname{Res}}(f + g) = \underset{z_0}{\operatorname{Res}} f + \underset{z_0}{\operatorname{Res}} g$$

whereas nothing can be said about multiplication. If that's true, what about

$$\underset{z_0}{\operatorname{Res}} \sum_{n=0}^\infty f_n = \, ?$$

Can I pull the residue operator under the summation sign?

Probably there might be some algebraic properties involving compositions of elementary functions.

  • 3
    $\begingroup$ Hint: Write Laurent expansions of $f,g.$ $\endgroup$ – Ehsan M. Kermani Apr 3 '13 at 7:53
  • 1
    $\begingroup$ A bit basic, but note you also have $\mathrm{Res}(fg,z_0) = f(z_0) \mathrm{Res}(g,z_0)$ in the case where $f$ has no pole at $z_0$ (i.e. $f$ is actually holomorphic at $z_0$). $\endgroup$ – Mike F Jun 7 '15 at 18:50

(i) Assume that $f$ and $g$ are analytic in a punctured neighborhood of $z_0$. Then by definition of the residue one has $${\rm Res}_{z_0}(f+g)={1\over2\pi i}\int_{\gamma}\bigl(f(z)+g(z)\bigr)\ dz\ ,$$ where $\gamma$ is a sufficiently small circle with center $z_0$. It is then obvious that $${\rm Res}_{z_0}(f+g)={\rm Res}_{z_0}(f)+{\rm Res}_{z_0}(g)\ .$$ (ii) There is no such formula for ${\rm Res}_{z_0}(f\cdot g)$. However, when $f$ and $g$ just have poles at $z_0$ it is possible to compute ${\rm Res}_{z_0}(f\cdot g)$ in a "finitary" fashion from the Laurent expansions of $f$ and $g$ as follows: One has (I'm assuming $z_0=0$ for simplicity) $$f(z)=\sum_{k=-m}^\infty a_kz^k,\quad g(z)=\sum_{l=-n}^\infty b_lz^l\ .$$ Therefore $$f(z)g(z)=\sum_{k=-m}^{n-1} a_kz^k\ \sum_{l=-n}^{m-1} b_lz^l+ h(z)\ ,$$ where $h$ is analytic at $0$. The residue of $f\cdot g$ at $0$ can now be extracted: $${\rm Res}_0(f\cdot g)=\sum_{k=-m}^{n-1}a_k b_{-k-1}\ .$$ (iii) When the series $\sum_{n=0}^\infty f_n$ converges uniformly in annuli $\epsilon\leq |z-z_0|\leq2\epsilon$ then one has $${\rm Res}_{z_0}(\sum_n f_n)={1\over 2\pi i}\int_\gamma \sum f_n(z)\ dz=\sum_n {1\over 2\pi i}\int_\gamma f_n(z)\ dz=\sum_n {\rm Res}_{z_0}(f_n)\ .$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.