modular form $q-$expansion not quite same as Laurent expansion to define the order of function at a point?

Question

Let $f$ be a modular form where $f:H\to \bar{C}$ is any meromorphic modular form.

Say I want to study the singularity behaviour at $i\infty$ of $f$. I can consider $z\to \frac{1}{z}$ map.

1. Then $f(\frac{1}{z})$ will give rise to laurent expansion $g(z)$ around $z=0$.

2. The other thing to get is $q-$ expansion $h(q)$ for $f$ at $q=0$ where $q=e^{2\pi iz}$.

Q1: Why order of $f$ at $i\infty$ is defined in terms of lowest degree of $q$ expansion here instead of laurent expansion $f(\frac{1}{z})$ at $z=0$'s degree?

Q2: Should not they contain the same information?

Q3: Say $f$ is weight $0$, then $f'$ is modular form of weight $2$.(This can be proven very easily.) Now I can consider either Laurent expansion's order at $i\infty$ and $q-$expansion's order of $f'$.($f'=\frac{df}{dz}$). They give me different order here for $f=j$ where $j$ is absolute modular invariant defined over the $H$ upper half plane. Was there anything wrong with my reasoning below?

Q3 a) Since $j$ has simple pole at $i\infty$, $j(w)\sim\frac{1}{w}$ where $w=\frac{1}{z}$.(I am only concerned about the order the pole here.) So $j'(w)\sim-\frac{1}{w^2}$ by taking derivative. This gives me a second order pole from laurent expansion's derivative. So order of $j'(z)$ at $z=i\infty$ is $-2$.

Q3 b) Consider $q=e^{2\pi iz}$ expansion and $j$ and one sees $j(q)\sim \frac{1}{q}$ as leading order. Take derivative to see $\frac{dj}{dz}=\frac{dj(q)}{dq}\frac{dq}{dz}\sim \frac{1}{q^2}\times -2\pi i q\sim\frac{1}{q}$. This says order of $j'$ at $i\infty$ is $-1$.

Answer 1

The inversion map $z\to -1/z$ does not send the upper half-plane to a neighborhood of $0$, unfortunately, so we cannot say that $f(-1/z)$ is a function defined on a punctured neighborhood of $0$, so we cannot talk about the nature of its "singularity" there, in the usual isolated-singularity language of complex variables.

Due to the periodicity in $z\to z+2\pi i$, modular forms and functions descend to the quotient of the upper half-plane by those translations, and are meromorphic on that quotient, and the image of the upper half-plane is a full punctured neighborhood of $0$ in that situation.

Answer 2

$f$ is only defined for positive imaginary parts, hence so is $g=f(1/z)$. To have a Laurent expansion you need to be analytic in a punctured neighborhood, not just half of one!

In Q3 a) you have $j(z)\sim \exp(-2\pi i z)$ at infinity, not $\sim z$ or $\sim1/z$. So $g(z) \sim \exp(-2\pi i /z)$ near $0$ (and note how this does not have a (partial) Laurent expansion).

Taking into account the above correction, the asymptotics of $g$ for $z\to0$ (full punctured neighborhood or not) still aren't the same as for $q\to0$ because you compare them via $$z \to 1/z \to \exp(2\pi i/z)=q(1/z)$$ whose derivative $-q(1/z)/z^2$ is not bounded near $0$, and comparing the behavior of the derivatives of the $q$-expansion and $g$ involves this factor (due to the chain rule).