[proofplan] We integrate the logarithmic derivative of $f$ around a truncated standard fundamental domain for $SL_2(\mathbb{Z})$. [The argument principle](/theorems/356) converts this contour integral into a weighted count of zeros, where the weights come from the local angles of the quotient at ordinary points, at $i$, and at $\rho$. The vertical sides cancel by $T$-periodicity, the horizontal side at height tending to infinity contributes $-\operatorname{ord}_\infty(f)$ through the $q$-coordinate, and the unit-circle boundary paired by $S$ contributes $k/12$ from the automorphy factor. Combining these contributions gives the formula. [/proofplan]

[step:Set up the logarithmic derivative on a truncated fundamental domain]Let \begin{align*} \mathcal{F} := \{z \in \mathbb{H}: |z| \geq 1,\ -1/2 \leq \operatorname{Re}(z) \leq 1/2\} \end{align*} be the closed standard fundamental domain. For $Y > 1$, define its truncation \begin{align*} \mathcal{F}_Y := \{z \in \mathcal{F}: \operatorname{Im}(z) \leq Y\}. \end{align*} Let \begin{align*} L: \mathbb{H} \setminus f^{-1}(0) &\to \mathbb{C} \\ z &\mapsto \frac{f'(z)}{f(z)} \end{align*} be the logarithmic derivative of $f$. Choose $Y>1$ so that $f$ has no zero on the horizontal segment $\{x+iY:-1/2\leq x\leq 1/2\}$. Since $f$ is holomorphic and not identically zero, the zero set of $f$ is discrete in $\mathbb{H}$; hence the compact set $\mathcal{F}_Y$ meets only finitely many zeros of $f$. For each interior zero remove a sufficiently small Euclidean disc. For each zero on the boundary of $\mathcal{F}_Y$, remove the Euclidean sector cut out by $\mathcal{F}_Y$ near that point. These neighbourhoods are chosen disjoint and compatible with the side pairings: if two non-elliptic boundary points are paired by $T$ or $S$, their sectors have the same radius after applying that pairing; at $i$, the deleted sector is symmetric under $S$; and at the two corners $\rho$ and $\rho+1$, the two deleted corner sectors have the same radius after translation by $T$. Denote by $\Omega_{Y,\varepsilon}$ the resulting piecewise smooth domain, oriented positively, where $\varepsilon>0$ records these common paired radii. The function $L$ is meromorphic on a neighbourhood of $\mathcal{F}_Y$ and has only simple poles at zeros of $f$, with residue equal to the corresponding zero order. By construction $f$ has no zero on $\overline{\Omega_{Y,\varepsilon}}$, so $L$ is holomorphic on a neighbourhood of $\overline{\Omega_{Y,\varepsilon}}$. Therefore the [Cauchy integral theorem](/page/Cauchy%20Integral%20Theorem) gives \begin{align*} \frac{1}{2\pi i}\int_{\partial \Omega_{Y,\varepsilon}} L(z)\,dz=0. \end{align*} We decompose $\partial\Omega_{Y,\varepsilon}$ into the outer boundary of $\mathcal{F}_Y$ with the small indenting arcs inserted, and the clockwise boundary components around the removed zeros. If $a$ is a zero of order $m$ and the removed sector has angle $\theta$, then near $a$ we have $L(z)=m/(z-a)+H_a(z)$ for a [holomorphic function](/page/Holomorphic%20Function) $H_a$ near $a$, and the clockwise small-sector contribution tends to $-m\theta/(2\pi)$. Hence the outer boundary integral, after letting $\varepsilon\to0$, equals the sum of the corresponding local weighted zero orders.[/step]

[guided]We use the same truncated domain and logarithmic derivative as in the exact proof. The closed standard fundamental domain is \begin{align*} \mathcal{F} := \{z \in \mathbb{H}: |z| \geq 1,\ -1/2 \leq \operatorname{Re}(z) \leq 1/2\}, \end{align*} and, for $Y>1$, its truncation is \begin{align*} \mathcal{F}_Y := \{z \in \mathcal{F}: \operatorname{Im}(z) \leq Y\}. \end{align*} The function whose contour integral we use is the logarithmic derivative \begin{align*} L: \mathbb{H} \setminus f^{-1}(0) &\to \mathbb{C} \\ z &\mapsto \frac{f'(z)}{f(z)}. \end{align*} Choose $Y>1$ so that \begin{align*} \{x+iY:-1/2\leq x\leq 1/2\} \end{align*} contains no zero of $f$. Since $f$ is holomorphic and not identically zero, the zero set of $f$ is discrete, and therefore $\mathcal{F}_Y$ contains only finitely many zeros. It is holomorphic away from the zeros of $f$, and near a zero $a \in \mathbb{H}$ of order $m$ we may write \begin{align*} f(z)=(z-a)^m g(z), \end{align*} where $g$ is holomorphic and $g(a)\neq 0$. Hence \begin{align*} L(z)=\frac{m}{z-a}+\frac{g'(z)}{g(z)}. \end{align*} Thus the residue of $L$ at $a$ is exactly $\operatorname{ord}_a(f)$. We do not integrate directly over $\mathcal{F}_Y$ if zeros lie on the boundary, because then $L$ is singular on the path. Instead we remove small neighbourhoods of all zeros touched by the contour. If a zero is in the interior, the removed neighbourhood is a small disc. If a zero lies on a side or corner of the fundamental domain, the removed neighbourhood is the sector cut out by the fundamental domain near that point. The boundary of the punctured region is then a legitimate contour for $L$. As in the exact proof, choose the sectors compatibly with side pairings: non-elliptic paired boundary points have paired radii under $T$ or $S$, the deleted sector at $i$ is symmetric under $S$, and the two deleted corner sectors at $\rho$ and $\rho+1$ have paired radii under $T$. Denote the resulting punctured domain by \begin{align*} \Omega_{Y,\varepsilon}. \end{align*} Then $L$ is holomorphic on a neighbourhood of $\overline{\Omega_{Y,\varepsilon}}$, so the [Cauchy integral theorem](/page/Cauchy%20Integral%20Theorem) gives \begin{align*} \frac{1}{2\pi i}\int_{\partial \Omega_{Y,\varepsilon}} L(z)\,dz=0. \end{align*} Equivalently, this is the local form of the [argument principle](/page/Argument%20Principle) applied to the logarithmic derivative. The boundary components around removed zeros are traversed clockwise, so their integrals appear with negative sign. When $\varepsilon \to 0$, a small circular arc of angle $\theta$ around a zero of order $m$ contributes \begin{align*} -\frac{1}{2\pi i}\int_{\text{arc}} \frac{m}{z-a}\,dz = -\frac{m\theta}{2\pi}. \end{align*} Moving these terms to the other side gives the expected local weight $\theta/(2\pi)$ times the zero order. Equivalently, after decomposing \begin{align*} \partial\Omega_{Y,\varepsilon} \end{align*} into the outer boundary of $\mathcal{F}_Y$ and the clockwise boundaries of the removed sectors, and then letting $\varepsilon\to0$, the outer boundary integral equals the weighted sum of local orders \begin{align*} \sum_a \frac{\theta_a}{2\pi}\operatorname{ord}_a(f), \end{align*} where $\theta_a$ is the angle of $\mathcal{F}_Y$ at the representative $a$. These angles are $2\pi$ at ordinary interior representatives, $\pi$ at $i$, and $2\pi/3$ at the elliptic corner represented by $\rho$.[/guided]

[step:Compute the contribution from the cusp]Let $n := \operatorname{ord}_\infty(f)$. Since $f$ is holomorphic at the cusp, it has a Fourier expansion \begin{align*} f(z)=\sum_{r=n}^{\infty} a_r q^r, \qquad q=e^{2\pi i z}, \qquad a_n \neq 0. \end{align*} Define \begin{align*} F: \{q \in \mathbb{C}:0<|q|<1\} &\to \mathbb{C} \\ q &\mapsto f(z)\quad \text{where }q=e^{2\pi i z}. \end{align*} Then \begin{align*} F(q)=q^n h(q) \end{align*} for a holomorphic function $h$ near $0$ satisfying $h(0)=a_n\neq 0$. Since $h(0)\neq0$ and $h$ is continuous at $0$, choose $\delta>0$ such that $h(q)\neq0$ whenever $|q|<\delta$. Choose $Y_0>1$ such that $e^{-2\pi Y_0}<\delta$. Then for every $Y\geq Y_0$, the region $\{z\in\mathcal{F}:\operatorname{Im}(z)>Y\}$ contains no zeros of $f$ in $\mathbb{H}$; the only contribution from that end is the cusp order $n=\operatorname{ord}_\infty(f)$. Since \begin{align*} \frac{dq}{q}=2\pi i\,dz, \end{align*} we have \begin{align*} L(z)\,dz = \frac{F'(q)}{F(q)}\,dq. \end{align*} Along the top edge of $\mathcal{F}_Y$, oriented from $1/2+iY$ to $-1/2+iY$, the variable $q=e^{2\pi iz}$ traverses the circle $|q|=e^{-2\pi Y}$ once clockwise. Therefore \begin{align*} \lim_{Y\to\infty} \frac{1}{2\pi i} \int_{1/2+iY}^{-1/2+iY} L(z)\,dz = -n = -\operatorname{ord}_\infty(f). \end{align*}[/step]

[guided]The cusp is handled by changing from the coordinate $z$ to the local coordinate \begin{align*} q=e^{2\pi i z}. \end{align*} Since $f$ is a modular form, it is holomorphic at the cusp, so its Fourier expansion has the form \begin{align*} f(z)=\sum_{r=n}^{\infty} a_r q^r, \qquad a_n\neq 0, \end{align*} where $n=\operatorname{ord}_\infty(f)$. Equivalently, in the punctured $q$-disc we define \begin{align*} F: \{q \in \mathbb{C}:0<|q|<1\} &\to \mathbb{C} \\ q &\mapsto f(z)\quad \text{where }q=e^{2\pi i z}, \end{align*} and write \begin{align*} F(q)=q^n h(q), \end{align*} where $h$ is holomorphic near $0$ and $h(0)\neq 0$. Because $h(0)\neq0$ and $h$ is continuous at $0$, there is a number $\delta>0$ such that $h(q)\neq0$ for $|q|<\delta$. Choose $Y_0>1$ with $e^{-2\pi Y_0}<\delta$. If $\operatorname{Im}(z)>Y_0$, then $|q|=e^{-2\pi\operatorname{Im}(z)}<\delta$, so $F(q)=q^n h(q)$ is nonzero for $q\neq0$. Thus no zeros of $f$ in $\mathbb{H}$ lie sufficiently high in the cusp; the missing point $q=0$ records exactly the cusp order $n=\operatorname{ord}_\infty(f)$. Differentiating $q=e^{2\pi i z}$ gives \begin{align*} dq=2\pi i q\,dz. \end{align*} Thus \begin{align*} L(z)\,dz = \frac{f'(z)}{f(z)}\,dz = \frac{F'(q)}{F(q)}\,dq. \end{align*} The top horizontal side of the truncated fundamental domain is oriented from $1/2+iY$ to $-1/2+iY$. Under $q=e^{2\pi iz}$, this path is the circle $|q|=e^{-2\pi Y}$ traversed clockwise. Hence \begin{align*} \frac{1}{2\pi i} \int_{1/2+iY}^{-1/2+iY} L(z)\,dz = \frac{1}{2\pi i} \int_{|q|=e^{-2\pi Y}\ \mathrm{clockwise}} \frac{F'(q)}{F(q)}\,dq. \end{align*} Since \begin{align*} \frac{F'(q)}{F(q)} = \frac{n}{q}+\frac{h'(q)}{h(q)} \end{align*} and $h'/h$ is holomorphic near $0$, the integral of $h'/h$ over the small circle tends to $0$, while the clockwise integral of $n/q$ is $-2\pi i n$. Therefore the limiting contribution of the top side is \begin{align*} -\operatorname{ord}_\infty(f). \end{align*}[/guided]

[step:Cancel the punctured vertical sides by translation invariance]Let $T \in SL_2(\mathbb{Z})$ be the matrix acting by $Tz=z+1$. Since $f$ has weight $k$ for $SL_2(\mathbb{Z})$ and $T$ has automorphy factor $1$, we have \begin{align*} f(z+1)=f(z). \end{align*} Differentiating gives $f'(z+1)=f'(z)$, hence \begin{align*} L(z+1)=L(z). \end{align*} Let $V_{L,\varepsilon}$ and $V_{R,\varepsilon}$ denote the portions of the left and right vertical sides that remain in $\partial\Omega_{Y,\varepsilon}$ after deleting the paired boundary sectors. By the compatible choice of sectors, $T(V_{L,\varepsilon})=V_{R,\varepsilon}$. The left side is oriented downward and the right side upward, so the substitution $w=Tz=z+1$ gives \begin{align*} \int_{V_{R,\varepsilon}}L(w)\,dw+\int_{V_{L,\varepsilon}}L(z)\,dz=0. \end{align*} The missing pieces are precisely the small indentation arcs around zeros on the paired vertical boundary; their limits are not part of the side cancellation and are recorded by the local sector weights below.[/step]

[guided]The vertical-side cancellation must be done on the actual contour, not on the undeleted sides. Let $V_{L,\varepsilon}$ be the union of the remaining pieces of the line $\operatorname{Re}(z)=-1/2$ in $\partial\Omega_{Y,\varepsilon}$, and let $V_{R,\varepsilon}$ be the corresponding union on $\operatorname{Re}(z)=1/2$. The sectors were chosen compatibly, so translation by one sends the left remaining pieces exactly onto the right remaining pieces: \begin{align*} T(V_{L,\varepsilon})=V_{R,\varepsilon}. \end{align*} The modular transformation law for $T$ has automorphy factor $1$, hence \begin{align*} f(z+1)=f(z). \end{align*} Differentiating gives \begin{align*} f'(z+1)=f'(z), \end{align*} and therefore \begin{align*} L(z+1)=L(z). \end{align*} Now the orientations matter. The right vertical side is traversed upward, while the left vertical side is traversed downward. Under the substitution $w=z+1$, the integral over $V_{L,\varepsilon}$ becomes the negative of the integral over $V_{R,\varepsilon}$. Thus \begin{align*} \int_{V_{R,\varepsilon}}L(w)\,dw+\int_{V_{L,\varepsilon}}L(z)\,dz=0. \end{align*} This proves cancellation for the non-singular portions of the actual punctured contour. If a zero lies on a vertical side, the small indentation arcs around the paired representatives are deliberately excluded from $V_{L,\varepsilon}$ and $V_{R,\varepsilon}$; their limits are counted separately as two half-angle contributions in the local-weight step.[/guided]

[step:Pair the unit-circle arcs and extract the automorphy contribution]Let $S \in SL_2(\mathbb{Z})$ be the matrix acting by $Sz=-1/z$. Split the lower boundary arc of $\mathcal{F}$ into \begin{align*} A_L &:= \{e^{i\theta}: 2\pi/3 \geq \theta \geq \pi/2\},\\ A_R &:= \{e^{i\theta}: \pi/2 \geq \theta \geq \pi/3\}, \end{align*} both oriented as part of the positively oriented boundary of $\mathcal{F}_Y$. The map $S$ sends $A_R$ onto $A_L$ with the opposite orientation. The modular transformation law for $S$ is \begin{align*} f(Sz)=z^k f(z). \end{align*} Differentiating this identity on $A_R$ gives \begin{align*} L(Sz)S'(z)=\frac{k}{z}+L(z), \qquad S'(z)=\frac{1}{z^2}. \end{align*} Hence \begin{align*} L(Sz)\,d(Sz)=\left(\frac{k}{z}+L(z)\right)\,dz. \end{align*} Let $A_{L,\varepsilon}$ and $A_{R,\varepsilon}$ denote the portions of $A_L$ and $A_R$ that remain in $\partial\Omega_{Y,\varepsilon}$ after the deleted sectors have been removed. The compatible choice of deleted sectors gives $S(A_{R,\varepsilon})=A_{L,\varepsilon}$ with opposite orientation, except that the indentation arcs themselves are omitted from these two sets and will be counted locally. Using the opposite orientation of $S(A_{R,\varepsilon})=A_{L,\varepsilon}$, we obtain \begin{align*} \int_{A_{L,\varepsilon}} L(w)\,dw+\int_{A_{R,\varepsilon}}L(z)\,dz &= -\int_{A_{R,\varepsilon}}L(Sz)\,d(Sz)+\int_{A_{R,\varepsilon}}L(z)\,dz\\ &= -k\int_{A_{R,\varepsilon}}\frac{dz}{z}. \end{align*} The omitted subarcs have total length tending to $0$ as $\varepsilon\to0$, and the function $z\mapsto 1/z$ is bounded on the unit circle, so \begin{align*} \lim_{\varepsilon\to0}\int_{A_{R,\varepsilon}}\frac{dz}{z} = \int_{A_R}\frac{dz}{z}. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. Writing $z=e^{i\theta}$ on $A_R$, where $\theta$ decreases from $\pi/2$ to $\pi/3$, gives \begin{align*} \int_{A_R}\frac{dz}{z} = \int_{\pi/2}^{\pi/3} i\,d\mathcal{L}^1(\theta) = -\frac{\pi i}{6}. \end{align*} Therefore \begin{align*} \lim_{\varepsilon\to0} \frac{1}{2\pi i} \left( \int_{A_{L,\varepsilon}} L(w)\,dw+\int_{A_{R,\varepsilon}}L(z)\,dz \right) = \frac{k}{12}. \end{align*} The indentation arcs around zeros on the unit-circle boundary are not included in this paired-arc computation; their limits are exactly the sector contributions recorded in the local-weight step.[/step]

[guided]The only non-cancelling boundary contribution away from the cusp comes from the automorphy factor on the unit-circle arc. We split that arc into the left half \begin{align*} A_L := \{e^{i\theta}: 2\pi/3 \geq \theta \geq \pi/2\} \end{align*} and the right half \begin{align*} A_R := \{e^{i\theta}: \pi/2 \geq \theta \geq \pi/3\}. \end{align*} Both arcs are oriented from left to right along the lower boundary of the fundamental domain. The matrix \begin{align*} S= \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} \end{align*} acts by $Sz=-1/z$. If $z=e^{i\theta}\in A_R$, then \begin{align*} Sz=-e^{-i\theta}=e^{i(\pi-\theta)}\in A_L. \end{align*} As $\theta$ decreases from $\pi/2$ to $\pi/3$, the angle $\pi-\theta$ increases from $\pi/2$ to $2\pi/3$, so $S$ maps $A_R$ onto $A_L$ with the opposite orientation. The modular transformation law for weight $k$ gives \begin{align*} f(Sz)=z^k f(z). \end{align*} We differentiate both sides with respect to $z$. The left side differentiates by the chain rule: \begin{align*} \frac{d}{dz}f(Sz)=f'(Sz)S'(z), \qquad S'(z)=\frac{1}{z^2}. \end{align*} The right side differentiates by the product rule: \begin{align*} \frac{d}{dz}\bigl(z^k f(z)\bigr) = kz^{k-1}f(z)+z^k f'(z). \end{align*} Dividing by $f(Sz)=z^k f(z)$ gives \begin{align*} L(Sz)S'(z)=\frac{k}{z}+L(z). \end{align*} Multiplying by $dz$ yields \begin{align*} L(Sz)\,d(Sz)=\left(\frac{k}{z}+L(z)\right)\,dz. \end{align*} Now use the orientation reversal. Since $S(A_R)=A_L$ with opposite orientation, \begin{align*} \int_{A_L}L(w)\,dw = -\int_{A_R}L(Sz)\,d(Sz). \end{align*} Therefore \begin{align*} \int_{A_L} L(w)\,dw+\int_{A_R}L(z)\,dz &= -\int_{A_R}\left(\frac{k}{z}+L(z)\right)\,dz +\int_{A_R}L(z)\,dz\\ &= -k\int_{A_R}\frac{dz}{z}. \end{align*} Let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. On $A_R$ we write $z=e^{i\theta}$ with $\theta$ decreasing from $\pi/2$ to $\pi/3$. Since $dz/z=i\,d\theta$, we get \begin{align*} \int_{A_R}\frac{dz}{z} = \int_{\pi/2}^{\pi/3} i\,d\mathcal{L}^1(\theta) = -\frac{\pi i}{6}. \end{align*} Thus the paired unit-circle arcs contribute \begin{align*} \frac{1}{2\pi i} \left( \int_{A_L} L(w)\,dw+\int_{A_R}L(z)\,dz \right) = \frac{k}{12}. \end{align*} This is the source of the constant $k/12$ in the valence formula.[/guided]

[step:Record the local weights of zeros in the quotient]Let $a \in \overline{\mathcal{F}}\cap \mathbb{H}$ be a zero of $f$ of order $m=\operatorname{ord}_a(f)$. The local contribution attached to $a$ is the limit of the clockwise indentation arc around $a$ after it is moved to the other side of the identity from [Cauchy's theorem](/page/Cauchy's%20Theorem). If the deleted sector at $a$ has interior angle $\theta_a$, the local contribution is \begin{align*} \operatorname{contrib}(a) = \frac{\theta_a}{2\pi}\operatorname{ord}_a(f). \end{align*} If $a$ lies in the interior of $\mathcal{F}$ and is not equivalent to another boundary point, then the deleted neighbourhood is a full disc, so $\theta_a=2\pi$ and the contribution is $m$. If $a$ lies on a non-elliptic boundary point, then exactly two boundary points $a_L$ and $a_R$ of $\mathcal{F}$ represent the same $SL_2(\mathbb{Z})$-orbit. The compatible deletion used in the side-pairing steps removes one half-disc at each representative, so $\theta_{a_L}=\theta_{a_R}=\pi$. The paired side integrals cancel on the remaining nonsingular sides, and the two indentation arcs add to \begin{align*} \frac{\pi}{2\pi}m+\frac{\pi}{2\pi}m=m. \end{align*} Thus an ordinary non-elliptic orbit contributes $\operatorname{ord}_a(f)$ once. At $i$, the deleted sector is the sector of $\mathcal{F}$ between the two unit-circle arcs. Its interior angle is $\pi$, and the punctured unit-circle arcs have already been paired by $S$ away from this indentation. Hence the local quotient contribution is \begin{align*} \frac{\pi}{2\pi}\operatorname{ord}_i(f) = \frac{1}{2}\operatorname{ord}_i(f). \end{align*} At $\rho$ and $\rho+1$, the two corners of $\mathcal{F}$ represent the same orbit. The compatible deletion removes a corner sector of angle $\pi/3$ at each corner, while the remaining sides are paired by $T$ and $S$. Their total indentation contribution is \begin{align*} 2\cdot \frac{\pi/3}{2\pi}\operatorname{ord}_\rho(f) = \frac{1}{3}\operatorname{ord}_\rho(f). \end{align*}[/step]

[guided]The contour integral records zeros through the indentation arcs. Suppose $a$ is a zero of order $m$ and the deleted sector of $\mathcal{F}$ at $a$ has angle $\theta$. Near $a$ we have \begin{align*} L(z)=\frac{m}{z-a}+H_a(z), \end{align*} where $H_a$ is holomorphic near $a$. The clockwise indentation arc therefore contributes \begin{align*} -\frac{1}{2\pi i}\int_{\text{indentation arc}}L(z)\,dz \longrightarrow -\frac{\theta}{2\pi}m. \end{align*} When this term is moved to the other side of the contour identity, it contributes \begin{align*} \frac{\theta}{2\pi}m. \end{align*} For an interior point of $\mathcal{F}$, the deleted sector is a full disc, so $\theta=2\pi$ and the contribution is \begin{align*} \frac{2\pi}{2\pi}m=m. \end{align*} This accounts for ordinary orbits represented by interior points. For a non-elliptic point on the boundary, the point is paired with exactly one other boundary point by a side-pairing transformation. The paired straight or circular pieces cancel or combine on the remaining punctured contour; only the two indentation arcs are left as local zero contributions. Each representative contributes a sector of angle $\pi$. Since the two boundary representatives describe one orbit in the quotient, the total contribution of that orbit is \begin{align*} \frac{\pi}{2\pi}m+\frac{\pi}{2\pi}m=m. \end{align*} Thus ordinary boundary orbits are still counted with coefficient $1$. At $i$, the point is fixed by the order-two elliptic transformation $S:z\mapsto -1/z$. In the fundamental domain the two circular arcs meet at $i$ with interior angle $\pi$, and the remaining punctured arcs have already been paired by $S$. Therefore the local sector contribution is \begin{align*} \frac{\pi}{2\pi}\operatorname{ord}_i(f) = \frac{1}{2}\operatorname{ord}_i(f). \end{align*} This is why the valence formula counts a zero at $i$ with weight $1/2$. At $\rho=e^{2\pi i/3}$, the orbit also contains $\rho+1=e^{\pi i/3}$ on the other corner of the fundamental domain. Each of these two corners has interior angle $\pi/3$, and the remaining punctured sides are handled by the $T$- and $S$-pairings. The two corner sectors together therefore contribute \begin{align*} 2\cdot \frac{\pi/3}{2\pi}\operatorname{ord}_\rho(f) = \frac{1}{3}\operatorname{ord}_\rho(f). \end{align*} This matches the fact that the image of $\rho$ in the quotient has stabilizer of order $3$.[/guided]

[step:Pass to the limit and obtain the formula]Choose $Y\geq Y_0$ as in the cusp analysis, and perturb $Y$ if necessary so that the horizontal segment $\{x+iY:-1/2\leq x\leq 1/2\}$ contains no zero of $f$. The nonvanishing of $h$ for $|q|<e^{-2\pi Y_0}$ shows that increasing $Y$ past $Y_0$ creates no additional zeros in $\mathbb{H}$. Since $\mathcal{F}_{Y_0}$ is compact and $f$ is not identically zero, only finitely many zeros occur in $\mathcal{F}_{Y_0}$. Thus all zero contributions are finite sums of local residue limits, and every nonsingular boundary piece converges after the compatible deletions: paired vertical pieces cancel exactly for every $\varepsilon$, paired unit-circle pieces converge to $k/12$, and indentation arcs converge to the local angle weights. Combining the preceding boundary computations and then letting $\varepsilon\to 0$ and $Y\to\infty$, the normalized outer contour integral gives \begin{align*} \sum_{z \in \mathcal{R}}\operatorname{ord}_z(f) + \frac{1}{2}\operatorname{ord}_i(f) + \frac{1}{3}\operatorname{ord}_\rho(f) = \frac{k}{12} - \operatorname{ord}_\infty(f). \end{align*} Moving the cusp term to the left-hand side gives \begin{align*} \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{z \in \mathcal{R}}\operatorname{ord}_{z}(f) = \frac{k}{12}. \end{align*} This is the desired valence formula.[/step]

[guided]We now assemble the contour computation and justify the limiting process. Choose $Y\geq Y_0$ as in the cusp analysis, and perturb $Y$ if necessary so that the top horizontal segment \begin{align*} \{x+iY:-1/2\leq x\leq 1/2\} \end{align*} contains no zero of $f$. The cusp analysis showed that $f$ has no zeros in the region $\operatorname{Im}(z)>Y_0$ inside the fundamental domain, so increasing $Y$ beyond $Y_0$ does not introduce new finite zeros. There are only finitely many zero contributions to track. Indeed, $\mathcal{F}_{Y_0}$ is compact and the zero set of the nonzero holomorphic function $f$ is discrete in $\mathbb{H}$, so $\mathcal{F}_{Y_0}$ meets only finitely many zeros. Around these finitely many points we chose compatible deleted sectors. For each fixed $\varepsilon$, the function $L$ is holomorphic on a neighbourhood of $\overline{\Omega_{Y,\varepsilon}}$, so the [Cauchy integral theorem](/page/Cauchy%20Integral%20Theorem) gives \begin{align*} \frac{1}{2\pi i}\int_{\partial\Omega_{Y,\varepsilon}}L(z)\,dz=0. \end{align*} Now let $\varepsilon\to0$. The paired vertical pieces cancel exactly for every $\varepsilon$ by $T$-periodicity. The paired unit-circle pieces converge to $k/12$ because the remaining integral is $-k\int dz/z$ over punctured arcs whose omitted lengths tend to zero. Each indentation arc around a zero converges to its sector-angle contribution by the local residue computation for $L(z)=m/(z-a)+H_a(z)$. Finally, the top horizontal edge tends to $-\operatorname{ord}_\infty(f)$ by the $q$-coordinate calculation. Therefore the finite zero contributions satisfy \begin{align*} \sum_{z \in \mathcal{R}}\operatorname{ord}_z(f) + \frac{1}{2}\operatorname{ord}_i(f) + \frac{1}{3}\operatorname{ord}_\rho(f) = \frac{k}{12} - \operatorname{ord}_\infty(f). \end{align*} Moving the cusp contribution to the left-hand side gives \begin{align*} \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{z \in \mathcal{R}}\operatorname{ord}_{z}(f) = \frac{k}{12}. \end{align*} This is exactly the valence formula.[/guided]