Why is $f(z)y^k$ bounded for $f$ a cusp form?

Question

For $f$ is a cusp form of weight $2k, k>0$ ($f(z)=(cz+d)^{-2k}f(\frac{az+b}{cz+d}$)), then why is $f(z)y^k$ bounded?

If expanded $f$ in $\sum a_nq^n$, it's domain is a open disc, hence I can't use compactness to conclude, help please.

Answer 1

It suffice to prove it on the fundamental domain, and for $y$ large, $|y^kq(y)|\leq 1$, hence $f(z)y^k$ is bounded for $|y|\geq N$, and its complement in fundamental domain is compact, hence is bounded.