My very specific question:

Given $(z_1,z_2) \in \mathbb{C}^2 $ and $$ I = \frac{1}{z_1 z_2 +1} $$

a) Where are the poles of $I$?

b) What are the residues of $I$?

Note: $z_1$ and $z_2$ are not in the same complex plane, this is a question about $\mathbb{C}^2$.

My broader question(s):

  • What are the best sources for learning more about and computing such residues and poles?

  • Does there exist a nice treatment of multi-dimensional complex analysis?

  • Do such questions automatically lead to algebraic geometry?

Possible solution (highlighting the confusion):

Take "directed residues" (i.e. a residue in one copy of $\mathbb{C}$ followed by a residue in the other copy) : $$ I = \frac{1}{z_2(z_1+1/z_2)} \quad \Rightarrow \quad I_1 \equiv \underset{z_1 = -1/z_2}{\textrm{Res}} \ I = \frac{1}{z_2} \quad \Rightarrow \quad \underset{z_2 = 0}{\textrm{Res}}\ I_1 = 1. $$

In the full $\mathbb{C}^2$ space, the pole is at $$\{(z_1,z_2) \ |\ z_2 = 0,\ z_1 = -1/z_2 \}.$$ I guess this means the pole is at infinity in the $z_1$ plane? It's also a bit strange because at $z_2 = 0$, $ I|_{z_2=0} = 1 $, so there is naively no pole in the $(z_1,0)$ complex plane. But maybe that's fine as long as I'm careful about tracking points at infinity and order of limits?

I have seen treatments (cf. Eq. (1.27)) that show another way of computing residues, but the equation I reference appears to require as many polynomials in the denominator as variables, whereas in my example there are two variables and only one polynomial. Also, that equation says to find the solution $z_2 z_1+1 =0$; yet the poles I computed above are specific points on that solution. Why does the directed residue approach select only two points (the second from taking a residue at $z_2 = -1/z_1$ first) along the full solution?

There is also a discussion here but the example I give here is even simpler.


With loop integrals in quantum field theory, physicists care about "leading singularities" - maximum codimension residues - of integrals. This is interesting in that context because such residues completely localize the integration variables (just as $z_1$ and $z_2$ get fixed in my "Possible Solution" above). As far as I can tell, this is not common language in math. I'm curious how mathematicians approach such issues.


Poking around a little more I found this answer. I don't know yet how/if the sources mentioned there address my specific question.

  • $\begingroup$ Hi, there is a generalization of holomorphic functions in dimension $N$ by replacing $\mathbb{C}$ by $N\times N$ orthogonal matrices. en.wikipedia.org/wiki/Holomorphic_functional_calculus but you didn't really explain why you need some sort of residues so maybe it's not what you search for. $\endgroup$ – reuns Jan 4 '16 at 2:59

The singularities of your function lie on a curve $z_1 z_2 = -1$. They are not isolated. There is no such thing as the "poles and residues" for your function on $\mathbb C^2$. All you can say is that for fixed $z_1\ne 0$ you have a pole at $z_2 = -1/z_1$ with residue $1/z_1$, and for fixed $z_2 \ne 0$ you have a pole at $z_1 = -1/z_2$ with residue $1/z_2$.


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