Valence formula

Fix the standard fundamental domain $\mathcal{F}=\{\tau\in\mathbb{H} : |\operatorname{Re}(\tau)|\le \tfrac12,\ |\tau|\ge 1\}$, with vertices at $\rho=e^{i\pi/3}$, $\rho+1$, and $i$ on its lower boundary, extending up to a cusp at $\infty$. Truncate it at height $\operatorname{Im}(\tau)=T$ for large $T$, and indent the boundary with small circular arcs around any zeros of $f$ that happen to lie on $\partial\mathcal{F}$ itself. Call the resulting contour $C$. Since $f$ has finitely many zeros in this bounded region, the argument principle gives $\frac{1}{2\pi i}\oint_C \frac{f'(\tau)}{f(\tau)}\,d\tau$ equal to the number of zeros of $f$ strictly inside $C$.

Now evaluate each piece of $\partial\mathcal{F}$. The two vertical sides $\operatorname{Re}(\tau)=-\tfrac12$ and $\operatorname{Re}(\tau)=\tfrac12$ are identified by $T:\tau\mapsto\tau+1$, and since $f(\tau+1)=f(\tau)$, the integrand $f'/f$ is also $1$-periodic; traversed in opposite directions (one side going up, the identified side coming down), these two integrals cancel exactly. This is where the translation symmetry earns its keep: it removes two full sides of the contour for free.

The top edge at height $T$ closes up as $T\to\infty$: writing $f=\sum_{n\ge m}a_nq^n$ with $m=\operatorname{ord}_\infty(f)$ and $a_m\ne0$, we get $f'/f \to 2\pi i\,m$ as $q\to0$, so this edge contributes $-\operatorname{ord}_\infty(f)$ to the total (the minus sign from the contour's orientation, traversed leftward at the top).

What remains is the lower boundary: the arc from $\rho$ to $i$ and the arc from $i$ to $\rho+1$ along $|\tau|=1$, plus the two short vertical segments from $\rho$ up to the top-left corner region and from $\rho+1$ down. The inversion $S:\tau\mapsto-1/\tau$ maps the arc through $i$ to itself (since $S(i)=i$) and swaps the two halves of the circular arc; because $S$ fixes $i$ and rotates a neighborhood of $i$ by angle $\pi$ (the stabilizer of $i$ in $\mathrm{PSL}_2(\mathbb{Z})$ has order $2$), the small indentation around $i$ contributes only $-\tfrac12\operatorname{ord}_i(f)$ instead of a full $-\operatorname{ord}_i(f)$: only "half" of a full residue is actually swept out before $S$-symmetry identifies the rest. Likewise $\rho$ and $\rho+1=S T^{-1}(\rho)$ are identified with stabilizer of order $3$ in $\mathrm{PSL}_2(\mathbb{Z})$ (generated by $ST$, of order $3$), so their combined indentation contributes $-\tfrac13\operatorname{ord}_\rho(f)$. Any other zero $P$ on the remaining, generic part of the arc has trivial stabilizer and contributes the full $-\operatorname{ord}_P(f)$.

Adding every piece: the vertical sides cancel, the top gives $-\operatorname{ord}_\infty(f)$, the corners give $-\tfrac12\operatorname{ord}_i(f)-\tfrac13\operatorname{ord}_\rho(f)$, and any remaining boundary zero $P$ gives $-\operatorname{ord}_P(f)$. But $\frac{1}{2\pi i}\oint_C f'/f\,d\tau$ must also equal $\tfrac{k}{12}$: this constant comes from tracking how the weight-$k$ automorphy factor $(c\tau+d)^k$ forces the total turning of $\arg f$ around the two identified corner arcs (related by $S$, which sends $d\tau\mapsto \tau^{-2}d\tau$ picking up a winding governed by $k$) to contribute exactly $k/12$ net — the same constant that appears because the hyperbolic area of $\mathcal{F}$ is $\pi/3$ and $X(1)$ has exactly one cusp and elliptic points of order $2,3$, which is the Riemann–Hurwitz bookkeeping underlying the $12$ in the formula. Equating the two computations of the same contour integral and moving every zero-order term to one side gives exactly $\operatorname{ord}_\infty(f)+\tfrac12\operatorname{ord}_i(f)+\tfrac13\operatorname{ord}_\rho(f)+\sum_{P}\operatorname{ord}_P(f) = \tfrac{k}{12}$.