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Let $f$ be a modular function on the upper half-plane, $\rho = e^{2\pi i/3}$ and $v_\rho(f)$ the order of $f$ at $\rho$, i.e. the integer $n$ such that $f/(z-\rho)^n$ is holomorphic and non-zero at $\rho$. Then by the residue theorem we have $\frac1{2i\pi} \oint_{\mathcal{C}_r} \frac{f^\prime}{f}dz = -v_\rho(f)$, where $\mathcal{C}_r$ is a circle around $\rho$ with radius $r$ and $r$ sufficiently small.

Let $B_r$ and $B^\prime_r$ be the points of intersection of $\mathcal{C}_r$ and the fundamental domain of $PSL_2(\mathbb{Z})$. How can we derive that $\lim\limits_{r\to 0} \int_B^{B^\prime} \frac{f^\prime}{f}dz = -\frac{1}{6}v_\rho(f)$. This is used in "A Course in Arithmetic" by Serre, p. 86.

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1 Answer 1

It's certainly not true in general: try it e.g. with $f(z) = z+1$, $C$ the circle with centre 0 and radius 2. There must be some symmetry that you're not telling us about.

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OK, in those cases where it is true, what can be deduced about $f$? –  Michael Hardy Aug 24 '11 at 2:07
If I'm not mistaken, we must have $f(z) = c z^n \exp(h(z^6))$ for some constant $c$, integer $n$ and meromorphic $h$. –  Robert Israel Aug 24 '11 at 7:23

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