Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 3 / c / Solution

The two cusps of ${\Gamma }_1(2)$ are infinity and zero. At infinity,soThus $f$ is holomorphic at infinity and has a simple zero there.

Use the scaling matrixat the cusp zero. Since $j(-1/\tau )=j(\tau )$,The width of zero is two, so its local parameter is $q_0=e^{\pi i\tau }$. As $\mathrm{Im}\tau \to \infty$,and thereforeIt has a simple pole at zero and is not holomorphic there. This uses the width of a cusp to express the two expansions in their correct local parameters.