Constant term of weakly modular form of weight 2 vanishes

Question

I stumbled upon this fascinating statement while browsing through old exercise sheets and don't find a fruitful approach to tackle the problem.

Statement

The constant term of the Fourier expansion of a weakly holomorphic modular form of weight 2 (for the full modular group) vanishes.
I'm hoping for a hint helping me to understand that this is true.

Description

In case a detailed description involving formulae is advantageous: Let $f: \mathbb{H} \to \mathbb{C}$ be a holomorphic mapping on the upper half plane, which fulfils

$$ f\left(\frac{az+b}{cz+d}\right) = (cz+d)^2 f(z)$$

for all $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right) \in SL_2(\mathbb{Z})$. In this case, $f$ admits a Fourier expansion $f(z) = \sum_{n \in \mathbb{Z}} a(n) e^{2 \pi i n z}$, which is usually required to possess only finitely many negative terms, in which case it is called a weakly holomorphic modular form (of weight 2). The statement of concern asserts that $a(0)=0$ for such a function $f$.

Thanks for giving my request a thought!
Best, TFT

Answer 1

Thanks to @Somos first comment, the question appears to be answered.

Let $j$ denote the Klein j function, then $(j^n)'$ is a weakly holomorphic modular form of weight $2$ for all natural numbers $n \in \mathbb{N}$, has leading term $q^{-n}$ (times a nonzero constant), where $q=e^{2 \pi i z}$, and its constant term vanishes.

So let $f$ be as above (the question), then an appropriate linear combination of $f$ and different $(j^n)'$ is a holomorphic modular form of weight $2$ (by annihilating negative powers of $q$). But this function has to be identically zero, by the valence theorem.

Since the above linear combination has the same constant term as $f$ has, the constant term of $f$'s Fourier expansion has to be zero.