Product of modular forms: poles, zeros expansion

[binbagsss]

Homework Statement

question concerning part c.

Homework Equations

The question is pretty simple if there is no zero of order $N$ at infinity, such that it does not cancel the pole of $f(t)$ at infinity of order $N$.

In this case it follows that $f(t) g(t) \in M^{!}_2$ and so we just set the constant coefficient to zero.

However, if there is a zero of order $N$ at infinity then $f(t)g(t) \in M_2$ instead, and the constant coefficient is not zero in general.

So my question is how do we know there is not a zero of order $N$ at infinity?

The Attempt at a Solution

From what I understand the expansion of $f(t)$ in terms of $q$ is the expansion of $f(t)$ near $\infty$ and it is holomorphic - no pole at $\infty$- if there are no negative coefficients, and if there are negative coefficients the last one gives the order of the pole.

However for positive coefficients it runs to $\infty$ , how do you deduce the order of any zeros? or

 

[binbagsss]

So my question is how do we know there is not a zero of order $N$ at infinity?

Ok, so I think this is obvious since $g(t) \in M_k$ not $S_k$ so there is no zero at $q=\infty$ due to the non-zero constant coefficient. HOWEVER, so in the case a function $\in S_k$ there is a zero at $\infty$ and from the Fourier expansion $\sum\limits_{n=0}^{\infty} q^n$ wouln't i conlclude it is a zero of order infinity? But I suspect this is wrong? thanks

 

[binbagsss]

Ok, so I think this is obvious since $g(t) \in M_k$ not $S_k$ so there is no zero at $q=\infty$ due to the non-zero constant coefficient. HOWEVER, so in the case a function $\in S_k$ there is a zero at $\infty$ and from the Fourier expansion $\sum\limits_{n=0}^{\infty} q^n$ wouln't i conlclude it is a zero of order infinity? But I suspect this is wrong? thanks

actually no, since $S_k$ is a sub-group of $M_k$ so we haven't necessarily been told that there is a non-vanishing constant coefficient.