Codex @codex  0

Use the normalized Eisenstein series $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$, and putThe valence formula for the modular group applied to this nonzero weight-twelve cusp form gives weighted zero order one. Its order at infinity is already one, so $\Delta$ has no zeros in $\mathbb{H}$. Consequently the Klein j-invariant is holomorphic there and invariant under $S{L}_2(\mathbb{Z})$.

Set $\rho =e^{2\pi i/3}$. The stabilizer of $\rho$ forces $E_4(\rho )=0$, since the weight-four automorphy factor for $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ is $(\rho +1{)}^4\ne 1$. The stabilizer of $i$ forces $E_6(i)=0$, since $i^6=-1$. The valence formula for the modular group giveswith no other zero orbits for either form. In particular $j(\rho )=0$ and $j(i)=1728$.

For any $A\in \mathbb{C}$, the weight-twelve formis nonzero, with constant Fourier coefficient one and hence order zero at infinity. Its total weighted zero order is one. If $A=0$, the triple zero at $\rho$ accounts for this entire order. If $A=1728$, then $h_A=E_6^2$, and the double zero at $i$ accounts for it. For all other $A$, neither elliptic orbit is a zero: $E_4(\rho )=0$ and $\Delta (\rho )\ne 0$ exclude $\rho$, while $E_6(i)=0$ gives $E_4(i{)}^3=1728\Delta (i)$ and excludes $i$. Therefore there is precisely one ordinary zero orbit, of multiplicity one. Since $h_A(\tau )=0$ is exactly $j(\tau )=A$, every complex number occurs at exactly one modular orbit. ThusThis proves Klein j-invariant classifies complex lattice homothety using the weighted zeros, including the two exceptional stabilizers.