Zeroes of automorphic function $\phi (\tau) $ = $\frac {\Delta (q\tau)} { \Delta(\tau) } $

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I am self studying Tom M. Apostol, Modular functions and Dirichlet series in number theory and I could not think about this argument on page 85.

Now $\varphi$ has a zero of order $q-1$ at $\infty$ and no further zeros in $H$.

I am not able to understand how zero of this function is $\infty $ which is automorphic under $\Gamma_0 (q) $.

I think zero is at i$\infty $ and that too when q-1>0 where q is an integer.

I know I could be wrong. But if I am wrong, then please tell me how zero is $\infty $ not i$\infty $ .

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For $q $ prime the cusps are $[i\infty]=\{\gamma(i\infty),\gamma\in \Gamma_0(q)\} $ and $[0]=\{\gamma(0),\gamma\in \Gamma_0(q)\} $

(for all $\gcd(a,b)=1$ if $q| b$ then $a/b\in [i\infty]$, otherwise $a/b \in [0]$)

For each cusp $[a/b]$ we take some $\gamma\in GL_2(\Bbb{Q})$ such that $\gamma(i\infty)=a/b$ and such that $z\to z+1$ is the least translation in $h^{-1}\Gamma_0(q) h$, the order of the zero/pole of $\phi$ at $[a/b]$ is the integer $n$ such that $\phi(h(z))\sim Ce^{2i\pi nz}$ as $z\to i\infty$.

For each $\Gamma_0(q) z,z\in \Bbb{H}$ let $Stab(z) = \{\gamma \in \Gamma_0(q), \gamma(z)=z\}$, if $|Stab(z)|=1$ then $z\to \Gamma_0(q)z$ is a chart, otherwise $\Gamma_0(q) z$ is an elliptic point and the chart is $s\mapsto \Gamma_0(q)(z+s^{|S_z|})$.

Here we find, from the chart $\gamma(z)= -1/(qz)$ for $[0]$ that $$\phi(\gamma(z)) = \frac{\Delta(-1/z)}{\Delta(-1/(qz))}= \frac{z^{12}\Delta(z)}{(qz)^{12}\Delta(qz)} \sim \frac{z^{12} e^{2i\pi z}}{(qz)^{12}e^{2i\pi q z}}$$ thus $\phi$ has a pole of order $q-1$ at $[0]$.

At $[i\infty]$, $\phi(z)\sim \frac{e^{2i \pi qz}}{e^{2i\pi z}}$ has a zero of order $q-1$. Since it is given by an infinite product, it has no other poles or zeros, on the upper half-plane.

Because $\Gamma_0(q)\setminus (\Bbb{H}\cup i\infty \cup \Bbb{Q})$ is a compact Riemann surface, such a meromorphic function on a modular curve always has the same number of zeros and poles counted with multiplicity.