On definition of modular forms I've been reading Complex Analysis by E. Freitag as alternative material of A First Course in Modular Forms by F. Diamond and I am struggling to comprehend some definitions.
In the page 326, he made a remark about the terminology that he is introducing for the $q$-expansions of real periodic functions (defined in $\mathbb{H}$) which are holomorphic in some neighborhood of $i\infty$. To be precise, he says:

Let  \begin{align} f:U_C\to\mathbb{C}  \end{align} be an analytic
  function on an upper half-plane  \begin{align} U_C=\{z\in\mathbb{H}:
 \mathrm{Im} z >C\},\quad C>0. \end{align} We assume that $f$ is
  periodic, i.e. there exists a suitable $N$ with  \begin{align}
 f(z+N)=f(z),\quad N\neq 0,\quad N\in\mathbb{R}. \end{align} The
  periodicity allows a Fourier expansion ($q$-expansion) \begin{align}
 f(z)=\sum_{n=-\infty}^{\infty}a_ne^{2\pi i nz/N}, \end{align} which in
  fact corresponds to a Luarent expansion \begin{align}
 \tilde{f}(z)=\sum_{n=-\infty}^{\infty}a_nq^n\quad\big(q=e^{\frac{2\pi
 i z}{N}}\big) \end{align} in the punctured disk around the origin
  having radius $e^{-2\pi C/N}$
Terminology. The function $f$ is 
(a) non-essentially singular at $i\infty$, iff $\tilde{f}$ is non-essentially singular at the origin.
(b) regular at $i\infty$, iff $\tilde{f}$ has a removable
  singularity at the origin.
In the case of regularity, one defines 
  \begin{align}
f(i\infty):=\tilde{f}(0)\quad (=a_0). \end{align}
The notions do not depend on the choice of the period $N$, (If $f$ is
  non-constant, the set of all periods is a cyclic group.)

I can't see why do not depend of the choice of the period. I'm aware that if $f$ is not a constant, the set of periods form a discrete group of $(\mathbb{R},+)$ and is generated by a rational number in the case of modular forms.
 A: Suppose that $N$ is the smallest period of $f$. Then the other periods of $f$ are of the form $mN$ for $m \in \mathbb N$.
Let $q = e^{2\pi i z/N}$ and $q_{\rm other} = e^{2\pi i z / m N}$ be the Fourier variables used when considering $f$ as of period $N$ or $mN$ respectively.
If
$$ \widetilde f(q) = \sum_{n = -\infty}^\infty a_n q^n$$
is the Laurent expansion written in terms of $q$, then the Laurent expansion written in terms of $q_{\rm other}$ must be
$$ \widetilde f_{\rm other}(q_{\rm other}) = \sum_{n = - \infty}^\infty b_n q_{\rm other}^{n},$$
where
$$ b_n = \begin{cases} a_{n/m} & \ \ \ \ \ \ \ \ \ n \equiv 0 \mod m \\ 0 &  \ \ \ \ \ \ \ \ \  n \not\equiv 0 \mod m\end{cases}$$
$\widetilde f(q)$ is non-essentially singular at $q = 0$ iff the coefficients in its Laurent series vanish for sufficiently negative powers of $q$, i.e. if there exists an $N \in \mathbb N$ such that $n < - N \implies a_n = 0$. But this condition holds precisely iff $n < - mN \implies b_n = 0$. So $\widetilde f(q)$ is non-essentially singular at $q = 0$ iff $\widetilde f_{\rm other} (q_{\rm other})$ is non-essentially singular at $q_{\rm other} = 0$
$\widetilde f(q)$ is regular at $q = 0$ iff the coefficients in its Laurent series vanish for all negative powers of $q$, i.e. if $n < 0 \implies a_n = 0$. This condition holds precisely iff $n < 0 \implies b_n = 0$. So $\widetilde f(q)$ is regular at $q = 0$ iff $\widetilde f_{\rm other} (q_{\rm other})$ is regular at $q_{\rm other} = 0$.
If $\widetilde f(q)$ is regular, then we say that $f(i\infty) = \widetilde f(q = 0) = a_0$. But $a_0 = b_0$, so this is the same as $\widetilde f_{\rm other} (q_{\rm other} = 0)$. Thus the value that the definition assigns to $f$ at $ i\infty$ when $f$ is regular at $i \infty$ is also independent of whether we use $\widetilde f(q)$ or $\widetilde f_{\rm other}(q_{\rm other})$ in the definition.
