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It is known that for $\Gamma_0(N)$, a weakly holomorhpic modular form is a harmonic maass form. Here is the definitions.

A weakly modular form $f$ for $\Gamma_0(N)$ is a meromorphic function on the half-plane such that $$ f(\gamma z)=j(\gamma,z)^{-k}f(z) $$ and the poles are supported in the cusps.

A harmonic Maass form $f$ for $\Gamma_0(N)$ is a real analytic on the half-plane such that $$ f(\gamma z)=j(\gamma,z)^{-k}f(z) $$ and $$ \Delta_k(f)=0, $$ where $\Delta_k=-y^2\left(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial x^2}\right) + iky\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)$, and there exists a polynomial $P(z) \in \mathbb C [q^{-1}]$ such that $$ f(z)-P(z)=O(e^{-\epsilon y}) $$ as $y \to \infty$ for some $\epsilon>0$. Analogous conditions are required at all cusps.

Since a weakly holomorphic modular form is holomorphic on the half-plane, it vanishes if we takes the hyperbolic Laplacian $\Delta_k$. But how we know that a weakly holomorphic modular form $f$ has a polynomial $P$ with the third condition of the definition of harmonic Maass form. Namely, is there a polynomial $P_x$ for the cusp $x$ of $\Gamma_0(N)$ such that $f|_k \sigma^{-1} (q)-P_x(q) = O(e^{-\epsilon y})$, where $\sigma \in \mathrm{SL}_2(\mathbb R)$ that sends $x$ to $\infty$? Why?

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A choice of polynomial $P_x$ is the "extended principal part" of $f$ at the cusp $x$.

For instance, at the cusp at infinity, we have the Fourier expansion $$ f = \frac{a_{-N}}{q^N} + \cdots + \frac{a_{-1}}{q} + a_0 + \sum_{n\ge 1}a_nq^n $$ and we can take $$ P = \frac{a_{-N}}{q^N} + \cdots + \frac{a_{-1}}{q} + a_0 \in \mathbb{C}[q^{-1}] $$

The required estimate follows from the fact that $$ |e^{2\pi i z}| = (e^{2\pi i z}e^{-2\pi i \bar{z}})^{\frac{1}{2}} = e^{-2\pi y} $$ where $y$ is the imaginary part of $z$.

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