= Solution
For a <complex lattice>, put $g_2=60G_4$ and $g_3=140G_6$. The <complex lattice> form of the <Klein j-invariant> is
$$
j(\Lambda)=1728\frac{g_2^3}{g_2^3-27g_3^2}.
$$
The 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 $\Lambda'=\alpha\Lambda$ for some $\alpha\ne0$. Scaling the absolutely convergent <lattice Eisenstein sums> gives
$$
G_4(\alpha\Lambda)=\alpha^{-4}G_4(\Lambda),\qquad
G_6(\alpha\Lambda)=\alpha^{-6}G_6(\Lambda).
$$
If both sums are nonzero, $\alpha^4=\alpha^6=1$, so $\alpha^2=1$ and $\alpha\Lambda=\Lambda$.
If $G_6=0$, then $G_4\ne0$ 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 $G_6\ne0$ 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 case
$$
\boxed{\Lambda'=\Lambda.}
$$
The 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.
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