Write $\tau=x+iy\in\mathfrak h$. For
$$
\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma(1),
$$
the imaginary part of the <Möbius transformation> is
$$
\operatorname{Im}(\gamma\tau)=\frac{y}{|c\tau+d|^2}.
$$
Among the primitive integer pairs $(c,d)$, choose one minimizing the nonzero quantity $|c\tau+d|$. Such a minimum exists because only finitely many <lattice point>[lattice points] lie in a bounded region. Complete $(c,d)$ to a matrix $\gamma\in SL_2(\mathbb Z)$. Then $\gamma\tau$ has maximal imaginary part in its <modular group> orbit.

Applying an integral translation does not change that imaginary part, so arrange
$$
-\frac12\leq\operatorname{Re}(\gamma\tau)\leq\frac12.
$$
If $\operatorname{Im}(\gamma\tau)<\sqrt3/2$, then $|\gamma\tau|<1$. The modular inversion $S:z\mapsto-1/z$ would give
$$
\operatorname{Im}(S\gamma\tau)
=\frac{\operatorname{Im}(\gamma\tau)}{|\gamma\tau|^2}
>\operatorname{Im}(\gamma\tau),
$$
contradicting maximality. Hence every orbit meets the stated region. This is the <reduction to the standard modular region> argument.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 1 / a / Solution

Given $\tau\in\mathfrak h$, choose a primitive pair $(c,d)$ minimizing $|c\tau+d|$ and apply a modular matrix with bottom row $(c,d)$. This maximizes the imaginary part within the orbit. Translation puts the real part in $[-1/2,1/2]$, and inversion then shows that the imaginary part is at least $\sqrt3/2$.


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