= Solution
Use the normalized <Eisenstein series> $E_4=1+240q+O(q^2)$ and $E_6=1-504q+O(q^2)$, and put
$$
\Delta=\frac{E_4^3-E_6^2}{1728}=q+O(q^2),\qquad j=\frac{E_4^3}{\Delta}.
$$
The <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 $SL_2(\mathbb Z)$.
Set $\rho=e^{2\pi i/3}$. The <stabilizer> of $\rho$ forces $E_4(\rho)=0$, since the weight-four <automorphy factor> for $\begin{pmatrix}0&-1\\1&1\end{pmatrix}$ is $(\rho+1)^4\ne1$. The <stabilizer> of $i$ forces $E_6(i)=0$, since $i^6=-1$. The <valence formula for the modular group> gives
$$
v_\rho(E_4)=1,\qquad v_i(E_6)=1,
$$
with 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 form
$$
h_A=E_4^3-A\Delta
$$
is 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 $\Delta(\rho)\ne0$ 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>. Thus
$$
\boxed{SL_2(\mathbb Z)\backslash\mathbb H\xrightarrow{\;j\;}\mathbb C\text{ is a bijection}.}
$$
This proves <Klein j-invariant classifies complex lattice homothety> using the weighted zeros, including the two exceptional stabilizers.
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