Let $f(z)$ be a modular form of weight $k$ for $\text{SL}_{2}(\mathbb{Z})$ and set $g(z) := f(Nz)$ for some $N>1$, $g$ is a modular form of weight $k$ for $\Gamma_{0}(N)$. I am trying to evaluate $g$ at the cusp $0$. Suppose $f$ has some Fourier expansion $\displaystyle f(z) = \sum_{n=0}^{\infty}a_{n} e^{2\pi i nz}$, then in essence I'm trying to compute the Fourier expansion of $z^{-k}g(-\frac{1}{z})$.
So far the only reasonable way I can see of doing this is to use the modularity of $f$ as follows:
$$ z^{-k}g\left(-\frac{1}{z}\right) = z^{-k}f\left(-\frac{N}{z}\right)$$
well if $z \in \mathbb{H}$, then $-\frac{1}{z} \in \mathbb{H}$ and thus $-\frac{N}{z} \in \mathbb{H}$. Since $f$ is modular of weight $k$, we have that for any $z\in \mathbb{H}$ that
$$ f\left(-\frac{1}{z} \right) = z^{k}f(z) \Rightarrow f\left(-\frac{N}{z} \right) = \frac{z^{k}}{N^{k}}f\left( \frac{z}{N}\right) = \frac{z^{k}}{N^{k}}\sum_{n=0}^{\infty}a_{n} e^{2\pi i n \frac{z}{N}}.$$
So it would appear that
$$ z^{-k}g\left(-\frac{1}{z}\right) = \frac{1}{N^{k}}\sum_{n=0}^{\infty}a_{n} e^{2\pi i n \frac{z}{N}}$$
So I guess my question first of all: is this right, or at least does it look right?
From this, allegedly one can deduce modular forms in $\Gamma_{0}(N)$ which take the value of $0$ at $\infty$ and $1$ at $0$ and vice versa. However, the leading constant coefficients in both the expansions of $g(z)$ and $z^{-k}g\left(-\frac{1}{z}\right)$ must both be zero if one of them is. So I feel the above must be wrong.
Any help appreciated!
Yes it is correct.
If $g\in M_k(\Gamma_0(N))$ then the value at the cusp $\Gamma_0(N)i\infty$ can be different to the value at the cusp $\Gamma_0(N)0$ (ie. $g(i\infty)\ne z^{-k}g(-1/z)(i\infty)$
But here you are taking $g$ in the subspace such that $g(z/N) = (z/N)^{-k}g(-1/z)$ so obviously it vanishes at one cusp iff it vanishes at the other.
Try with $f(z)-f(Nz)$ for $f\in M_k(SL_2(\Bbb{Z}))$ not a cusp form, it vanishes at $\Gamma_0(N)i\infty$ but not at $\Gamma_0(N)0$.