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I am learning Laurent series, and I've learnt residue theorem before, but I am trying to solve a problem and get stuck here:

Suppose $f:\Bbb C \ (p_1, ..., p_n)$ → $\Bbb C$ is holomorphic (i.e. it only has a finitely many isolated singularities). Then how to show that $\sum^n_{j=1} Res_{p_j}f(z)+Res_\infty f(z)=0$

Paul Sinclair
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J.doe
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  • Every small circle is the boundary of two domains in $\mathbb{C}\cup{\infty}$. Apply the residue theorem to both domains. – Daniel Fischer Oct 25 '16 at 20:54
  • So I'll get that each of them is 0, so the sum is gonna be zero. Is that correct? – J.doe Oct 25 '16 at 20:59
  • No. If all residues were $0$, residues wouldn't be useful at all. With the given condition, the sum of the residues is $0$, but any individual residue would rarely be $0$. – Daniel Fischer Oct 25 '16 at 21:03

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The residue at infinity is defined by $$Res(f(z),\infty) = \lim_{R \to \infty} \frac{-1}{2i\pi}\int_{|z| = R} f(z)dz \overset{s = 1/z}= \lim_{R \to \infty} \frac{-1}{2i\pi} \int_{|s| = 1/R} \frac{f(1/s)}{s^2}ds = Res(\frac{-f(1/z)}{z^2},0)$$ Now by the residue theorem$$ \frac{1}{2i\pi}\int_{|z| = R} f(z)dz = \sum_{|p_j| < R} Res(f(z),p_j)$$ You get that $$\sum_j Res(f(z),p_j)+Res(f(z),\infty) = 0$$

reuns
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  • Could you also look at http://math.stackexchange.com/questions/1984937/integrals-and-residue-theorem?noredirect=1#comment4075075_1984937, I still don't get it. – J.doe Oct 25 '16 at 22:56
  • @J.doe What don't you understand in what I wrote ? The rest is how to compute a residue. – reuns Oct 25 '16 at 23:21
  • Nope, I got this problem. I have another question about residue theorem, and I hope maybe you can take a look at it and offer some help, if it is possible. Thank you very much. – J.doe Oct 25 '16 at 23:35
  • Could you explain why there is a s^2 in the denominator? It's must be something with dz=d(1/s) I suppose. – Algebear Apr 26 '18 at 14:54
  • I'm not entirely sure but I think in one of those steps there should not be a minus sign (I think after the first equal-sign). – Algebear Apr 26 '18 at 15:08