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This feels rather silly to ask, but this has been confusing me as of late. One exam question I was attempting recently was to find the contour integral of $1/\sqrt{z^2-1}$ over the contour $\Gamma$ encircling its branch points at $z=\pm1$.

First question: I am aware that $z=\pm1$ are branch points, but do they classify as other types of singularities as well?

Now I’m also familiar with computing the residue at infinity and that we can take the negative of the residue at $w=0$ for $-\frac1{w^2}f\left(\frac1{w}\right)$. The official solutions, however, are rather cryptic:

Blowing the contour $\Gamma$ up to $\infty$ we have $$\oint_\Gamma f(z)\text{ d}z=\oint_\Gamma\frac{\text{d}z}{z}\left(1+\mathcal{O}\left(\frac1{z^2}\right)\right)=2\pi i.$$

Second questions (or rather lots of): What is going on here? What is $1$ the residue of? Isn’t this a Laurent expansion about $z=0$? Where does the $\infty$ come in?

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  • $\begingroup$ The residue at infinity is $Res(f(z),\infty)=-\lim_{R \to \infty} \int_{|z|=R} f(z)dz$, if $f$ is analytic on $\Bbb{C}$ minus finitely many points it is what you need for $\sum_{a \in \Bbb{C}\cup \infty} Res(f(z),a)=0$. If $f$ vanishes as $|z| \to \infty$ then $f(1/s)$ is analytic at $0$ so $f(1/s)= Cs+O(s^2)$ and the change of variable $-\lim_{R \to \infty} \int_{|z|=R} f(z)dz=\lim_{R \to \infty} \int_{|1/s|=1/R} f(1/s)d(1/s)=\lim_{R \to \infty} \int_{|1/s|=1/R}(Cs+O(s^2))\frac{ds}{-s^2}$ gives the value $=-2i\pi$ $\endgroup$
    – reuns
    Aug 28, 2019 at 0:39
  • $\begingroup$ The question assumes that we take a branch which has a Laurent expansion around infinity (on $|z| > 1$), also see this answer. $\endgroup$
    – Maxim
    Sep 6, 2019 at 6:51

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If you imagine $z$ to be quite large, then $\frac{1}{\sqrt{z^2-1}} \approx \frac{1}{\sqrt{z^2}}$ plus extra terms. This is more or less the same as the dipole moment in physics - when there are two charges of opposite sign close to each other, from really far away it looks like there is a singularity right in the middle between them.

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  • $\begingroup$ Hi, thanks for the answer — why does that argument apply for branch points? $\endgroup$
    – user107224
    Aug 27, 2019 at 22:56
  • $\begingroup$ In the intuitive argument above, the type of singularity does not matter. If I have a singularity at $a$ and a singularity at $b$ of the same order, then from really far away it will look like a singularity at $\frac{a+b}{2}$ with double the order. The "branch" part of the points also does not matter because when you were given that function, you were probably made to assume you were dealing with the principal branch of square root. Integration over a finite number of discontinuities is okay. $\endgroup$
    – Ninad Munshi
    Aug 27, 2019 at 23:08
  • $\begingroup$ I’m not convinced why it “looks” like a singularity — I find it really hard to resolve in my head that it’s similar to a dipole -> monopole. Is there a proof I can do this? $\endgroup$
    – user107224
    Aug 28, 2019 at 13:56
  • $\begingroup$ I think it is easier to visualize in the real case. Say I had a function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x) = \frac{1}{x-d} - \frac{1}{x+d} = \frac{1}{x}\left( \frac{1}{1-\frac{d}{x}} - \frac{1}{1+\frac{d}{x}}\right)$ and suppose I was much further away from the singularities than distance between them. Then I could asymptotically use geometric series for both and we get $\frac{1}{x}(1+\frac{d}{x}+\cdots - 1 + \frac{d}{x} - \cdots) \approx \frac{2d}{x^2}$ $\endgroup$
    – Ninad Munshi
    Aug 28, 2019 at 14:08

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