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I'm learning modular form and run into such function: $$ j(\tau)=\frac{(2\pi)^{12}+\ldots}{(2\pi)^{12}q+\ldots}=\frac{1}{q}+\ldots $$

where $q=e^{2\pi i\tau}$. (I omitted the details of definition of $j$, I will update if needed.)

Then I was told that $j$ has a simple pole at $\infty$ and the residue of it is 1. I'm confused.

First of all, how to see it is a simple pole? The formula given is in $q$, not in $\tau$, can we just say so from the leading term $\frac{1}{q}$?

Second, I'm not quite understand the residue at $\infty$. Is it just $-Res(j,0)$? Why $j$'s residue at $\infty$ is 1?

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The function $\,j\,$, which seems to be the invariant of an elliptic curve, should be written a little more explicitly in order to try to see what happens at zero to $\,1/q\,$... – DonAntonio Jan 5 '13 at 2:51
Well, at $\,1/\tau\,$...:) – DonAntonio Jan 5 '13 at 2:56
@DonAntonio I tried. But if we use $e^z=\sum\frac{z^n}{n!}$ to expand $q$ in term of $\tau$, we get no $1/\tau$ term. Because $1/q$ us just $e^{-2\pi i\tau}$... – hxhxhx88 Jan 6 '13 at 1:18
As I fixed above, it must be $\,1/\tau\,$ , so: $$j(\tau^{-1})=\frac{(2\pi)^{12}+\ldots}{({2\pi})^{12}e^{2\pi i/\tau}+\ldots}$$ – DonAntonio Jan 6 '13 at 2:54
I still cannot solve it. Now I think I need to calculate the residue of $\tau^{-2}j(\tau^{-1})$ at $0$. It is $\tau^{-2}(e^{-2\pi i/\tau}+\cdots)$. After expanding $e$, there is still no $\tau^{-1}$ term, right? – hxhxhx88 Jan 6 '13 at 9:15

If $f$ is holomorphic for $|z|>r$, then $$\mathrm{Res}(f, z=\infty)=\mathrm{Res}(w^{-2}f(w^{-1}), w=0)$$

Moreover, usually $\tau$ is considered to be in the upper half plane or in some fancier (and smaller) domain; therefore, I think that the residue is considered at $q=\infty$.

Anyway, if $f(q)=\frac{1}{q}+g(q)$ with $g$ holomorphic, $$\frac{1}{p^2}f(1/p)=\frac{1}{p}+\frac{1}{p^2}h(1/p)$$ so the residue of the latter at $p=0$ is $1$. You can also say that the residue at $\infty$ is minus the residue at $0$, but only if you know that the function is holomorphic elsewhere.

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But $j$ is a function of $\tau$, $q=e^{2\pi i\tau}$ is just a change of variable. Can we just calculate the residue using the expansion with respect to $q$ rather than $\tau$? – hxhxhx88 Jan 6 '13 at 1:15
is your expression for $j$ defined for all $\tau\in\mathbb{C}$? – wisefool Jan 6 '13 at 1:27
No, for $\tau$ in the upper half plane. – hxhxhx88 Jan 6 '13 at 2:13
So it is meaningless to ask for the residue at $\tau=\infty$. You need to have the function defined in a neighborhood of $\infty$, i.e. for $|\tau|>R$. Therefore, you go to the universal covering through the exponential map and compute the residue with respect to the new coordinate $q$. – wisefool Jan 6 '13 at 14:04
Your word sounds right. So what if we regard $j$ as being defined on $\mathbb{C}$? – hxhxhx88 Jan 6 '13 at 14:17

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