= Solution
Let $\Lambda=\mathbb Z\omega_1+\mathbb Z\omega_2$ be a <complex lattice>. Orient its basis so that $\operatorname{Im}(\omega_1/\omega_2)>0$. Multiplication by $\omega_2^{-1}$ gives the normalized <complex lattice> $\mathbb Z\tau+\mathbb Z$ with $\tau=\omega_1/\omega_2\in\mathbb H$. Thus every <homothety of complex lattices> class has a normalized representative.
If $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL_2(\mathbb Z)$, then
$$
\mathbb Z(\gamma\tau)+\mathbb Z
=(c\tau+d)^{-1}\bigl(\mathbb Z(a\tau+b)+\mathbb Z(c\tau+d)\bigr)
=(c\tau+d)^{-1}(\mathbb Z\tau+\mathbb Z).
$$
The last equality follows because the two integer <vectors> are an integral basis: their <determinant> is one. Hence points in the same modular <orbit> give <homothetic complex lattices>.
Conversely, suppose $\Lambda_{\tau'}=\alpha\Lambda_\tau$ for $\alpha\ne0$. There is an integer basis-change <matrix> of <determinant> $\pm1$ such that
$$
\tau'=\alpha(a\tau+b),\qquad 1=\alpha(c\tau+d).
$$
Both normalized bases have positive <orientation>, and multiplication by $\alpha$ preserves real <orientation> because its real <determinant> is $|\alpha|^2>0$. The basis-change <determinant> is therefore $+1$. Dividing the two equalities gives $\tau'=(a\tau+b)/(c\tau+d)$. Thus \b[the modular orbits and <homothety of complex lattices> classes correspond bijectively].
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