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Equal sums give equal Klein j-invariants, hence the complex lattices are homothetic. If both sums are nonzero, the scaling factor obeys $a^4=a^6=1$ and is $\pm 1$. If only $G_6$ vanishes, the lattice is homothetic to the square lattice and the possible fourth roots of unity already preserve it. If only $G_4$ vanishes, it is homothetic to the hexagonal lattice and the possible sixth roots of unity preserve it. The two sums cannot vanish together because the elliptic discriminant is nonzero.

Two complex lattices are homothetic when multiplication by one nonzero complex number takes one onto the other. Changes between oriented integral bases have determinant one, so normalized period ratios classify these classes by $S{L}_2(\mathbb{Z})\backslash \mathbb{H}$. The lattice Eisenstein sums scale as $G_k(a\Lambda )=a^{-k}{G}_k(\Lambda )$.

Let $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$ be a complex lattice. Orient its basis so that $\mathrm{Im}({\omega }_1/{\omega }_2)>0$. 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 $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in S{L}_2(\mathbb{Z})$, thenThe 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 ${\Lambda }_{{\tau }^{\prime }}=\alpha {\Lambda }_{\tau }$ for $\alpha \ne 0$. There is an integer basis-change matrix of determinant $\pm 1$ such thatBoth 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 ${\tau }^{\prime }=(a\tau +b)/(c\tau +d)$. Thus the modular orbits and homothety of complex lattices classes correspond bijectively.

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.