Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 29 / 3 / a / Solution

The infinite product converges locally uniformly in $\mathbb{H}$, because ${\sum }_{n\ge 1}|q{|}^n<\infty$ uniformly on compact subsets. None of its factors vanishes there, and the convergent logarithm of its tail shows that the product is nonzero. Logarithmic differentiation, justified by locally uniform convergence, givesThe middle equality groups the absolutely convergent terms $n{q}^{nr}$ by $m=nr$. Thus $F^{\prime }/F=2\pi i{E}_2$.

Let $H(\tau )={\tau }^{-12}F(-1/\tau )/F(\tau )$. It is nonzero and holomorphic on the simply connected upper half-plane. Using the printed transformation $E_2(-1/\tau )={\tau }^2{E}_2(\tau )+6\tau /(\pi i)$, its logarithmic derivative isSo $H$ is constant. At $\tau =i$ we have $-1/i=i$ and $i^{-12}=1$, giving $H(i)=1$. Therefore $F(-1/\tau )={\tau }^{12}F(\tau )$. Also $F(\tau +1)=F(\tau )$ directly from $q=e^{2\pi i\tau }$. Since these two transformations generate $S{L}_2(\mathbb{Z})$, $F$ has the weight-twelve modular form transformation law.

At the unique modular cusp its expansion is $F=q+O(q^2)$, so it extends holomorphically with zero value. HenceThis is the modular discriminant. The transformed argument contributes $6\tau /(\pi i)$; the converted TeX's numerator $67$ is a transcription error, not the printed relation.