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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.

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Hint: Write Laurent expansions of $f,g.$ –  Ehsan M. Kermani Apr 3 '13 at 7:53
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up vote 2 down vote accepted

(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 $${\rm Res}_{z_0}(\sum_n f_n)={1\over 2\pi i}\int_\gamma \sum f_n(z)\ dz=\sum_n {1\over 2\pi i}\int_\gamma f_n(z)\ dz=\sum_n {\rm Res}_{z_0}(f_n)\ .$$

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