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Apologies if this is a stupid question but I'm pretty confused. So, a modular form $f(t) \in M_k $ is usually given by it's expansion about $\infty$ expressed in the variable $q=e^{2\pi i t} $ as:

$ \sum\limits_{n=0}^{\infty} a_n q^n $

where $n$ must start at zero here for $f(t)$ to be holomorphic at $\infty$ and therefore a modular form,

if instead we have:

$ \sum\limits_{n=-N}^{\infty} a_n q^n $, then the function is 'meromophic at $\infty$ ' with the order of the pole at $\infty$ being $N$.

We use the notation $S_k$ if the constant coeffient vanishes, that is instead we have (for a modular form so holomorhphic, i.e. cusp form):

$ \sum\limits_{n=1}^{\infty} a_n q^n $

MY QUESTION:

how from a modular form function expressed as the fourier expansion above, can we deduce the order of a zero? The last negative coeffient gives the order of a pole, and since we set $i\infty \to 0 $ do we not conclude that every cusp form has a zero of order infinity at infinity?

EXAMPLE 2

My notes say the following:

on the subject of $f$ a mermorphic modular form of weight $k$:

$ord_{\infty} (f) $= index of first non-zero coeffient in the q-expansion of $ f= ord_{q=0}(\hat{f}) $

where $ \hat{f}(q)=f(t) $ and where the order of a point has been defined as:

$ord_p(f) $ = order of vanishing of $f$ at $P$ minus the order of the pole of $f$ at $P$.

So, the first non-zero coefficient of the expansion, i know, gives the order of the pole, and the negative sign has also been taken into account by the wordings of my notes to take 'the index' , so this is implying that a meromorphic modular form of weight $k$ never has any zeros 'at infinity'. should this be obvious?

i dont understand. many thanks.

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