# A weakly holomorphic modular form is a harmonic maass form

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?

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 $$z$$.