The level-one valence formula

Since $f\equiv ̸0$ and $f$ has isolated zeros, and since $q^{n}g(q)$ with $g(0)\mathrm{\ne }0$ has no zeros for small $\mathrm{\mid }q\mathrm{\mid }$, there are finitely many zeros of $f$ in $\bar{D}\cap \{\mathrm{Im}\tau \le Y\}$ for each $Y$, and for large $Y$ all zero classes have a representative there. Choose such a $Y$ and $\epsilon >0$ small, let $R$ be the region obtained from $\bar{D}\cap \{\mathrm{Im}\tau \le Y\}$ by deleting the open discs of radius $\epsilon$ around each zero of $f$ in that set (together with the strip $\mathrm{Im}\tau >Y$), and let $\mathrm{\Gamma }=\mathrm{\partial }R$ with the positive orientation. Then $f$ is holomorphic on a neighbourhood of $\bar{R}$ and has no zeros on $\mathrm{\Gamma }$, so by [F2] $\frac{1}{2\pi i}{\oint }_{\mathrm{\Gamma }}\frac{f^{\mathrm{\prime }}}{f}d\tau =0$, and $\mathrm{\Gamma }$ consists of the top horizontal segment, the two vertical sides, the circular arc, cut where zeros occur, and the small circles (or circular arcs) around the zeros.

The top segment is traversed from right to left; on it $f^{\mathrm{\prime }}\mathrm{/}f=2\pi in+o(1)$ uniformly in $\mathrm{Re}\tau$ as $Y\to \mathrm{\infty }$ by [F3], so its contribution to $\frac{1}{2\pi i}\oint$ tends to $-n$. On the parts of $\mathrm{\Gamma }$ lying on the vertical sides of $\mathrm{\partial }D$ the integrand is $T$-invariant and the two sides are oppositely oriented, so they cancel exactly; the parts lying on the circular arc contribute $k\mathrm{/}12$ in the limit $\epsilon \to 0$ by [F3] (the cuts near zeros are accounted for with the small circles below).

Consider a class $P\mathrm{\ne }\mathrm{\infty }$ with ${\mathrm{ord}}_P(f)=m>0$ and all its representatives in the truncated domain. Near a zero ${\tau }_0$ of order $m$, $f^{\mathrm{\prime }}\mathrm{/}f=m\mathrm{/}(\tau -{\tau }_0)+$holomorphic by [F1], so over a circular arc of angle $\theta$ around ${\tau }_0$ inside $R$ the integral equals $im\theta +o(1)$ as $\epsilon \to 0$, contributing $-\theta m\mathrm{/}(2\pi )$ to $\frac{1}{2\pi i}{\oint }_{\mathrm{\Gamma }}$ (the boundary is traversed clockwise around the deleted disc). By [F4] the total angle of the sectors of $R$ at all representatives of $P$ is: $2\pi$ if $P$ is a non-elliptic class (one interior representative, or two boundary representatives each contributing $\pi$), $\pi =2\pi \mathrm{/}{\nu }_i$ for the class of $i$, and $\pi \mathrm{/}3+\pi \mathrm{/}3=2\pi \mathrm{/}3=2\pi \mathrm{/}{\nu }_{\omega }$ for the class of $\omega$ (represented by the two points $\omega$ and $\omega +1$). Hence each class $P\mathrm{\ne }\mathrm{\infty }$ contributes $-m\mathrm{/}{\nu }_P=-{\mathrm{ord}}_P(f)\mathrm{/}{\nu }_P$.

Summing 2.1 and 2.2 in the identity of 1.1 and letting $Y\to \mathrm{\infty }$, $\epsilon \to 0$ gives $0=\frac{k}{12}-n-{\sum }_{P\mathrm{\ne }\mathrm{\infty }}\frac{{\mathrm{ord}}_P(f)}{{\nu }_P}$, that is ${\sum }_{P\mathrm{\ne }\mathrm{\infty }}\frac{1}{{\nu }_P}{\mathrm{ord}}_P(f)+{\mathrm{ord}}_{\mathrm{\infty }}(f)=\frac{k}{12}$. All sums are finite by 1.1, and the term ${\mathrm{ord}}_{\mathrm{\infty }}(f)=n$ appears as the negative of the top-segment limit.