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For a complex lattice, put $g_2=60G_4$ and $g_3=140G_6$. The complex lattice form of the Klein j-invariant isThe denominator is nonzero for a complex lattice: after normalization it is a nonzero scalar multiple of $E_4^3-E_6^2$, whose absence of zeros was proved in part (b). The equal lattice Eisenstein sums therefore give equal $j$ values, so parts (a) and (b) give ${\Lambda }^{\prime }=\alpha \Lambda$ for some $\alpha \ne 0$. Scaling the absolutely convergent lattice Eisenstein sums givesIf both sums are nonzero, ${\alpha }^4={\alpha }^6=1$, so ${\alpha }^2=1$ and $\alpha \Lambda =\Lambda$.

If $G_6=0$, then $G_4\ne 0$ and $j=1728$. The complex lattice is homothetic to $\mathbb{Z}i+\mathbb{Z}$, which is preserved by multiplication by $i$. The constraint ${\alpha }^4=1$ allows only powers of $i$, each already preserving the complex lattice. If $G_4=0$, then $G_6\ne 0$ and $j=0$. The complex lattice is homothetic to $\mathbb{Z}\rho +\mathbb{Z}$. Multiplication by $\rho$ preserves this complex lattice because ${\rho }^2=-1-\rho$, and multiplication by $-1$ always preserves it. The constraint ${\alpha }^6=1$ allows exactly the sixth roots generated by these two multiplications. Again $\alpha \Lambda =\Lambda$. Both sums cannot be zero simultaneously because the discriminant is nonzero. Hence in every caseThe exceptional cases are essential in a complex lattice is determined by its fourth and sixth Eisenstein sums; merely deducing homothety of complex lattices would not prove equality.