I just tried to calculate the zeros of the Eisenstein series of weight 4, which should be only at $\rho=e^{i\pi/3}$ and not at $i$. I don't know where the mistake in the following is. Who can help?
We have $(i+1)^4=-4$ and thus \begin{align} G_4(i) &= \sum_{\substack{(m,n)\in\mathbb Z^2\\(m,n)\neq(0,0)}} \frac{1}{(m+in)^4} = \sum_{\substack{(m,n)\in\mathbb Z^2\\(m,n)\neq(0,0)}} \frac{-4}{((i+1)\cdot(m+in))^4}\\ &= \sum_{\substack{(m,n)\in\mathbb Z^2\\(m,n)\neq(0,0)}} \frac{-4}{((m-n)+i(m+n))^4} = \sum_{\substack{(k,l)\in\mathbb Z^2\\(k,l)\neq(0,0)}}\frac{-4}{(k+il)^4}= -4 \cdot G_4(i) \end{align} Thus we get $G_4(i)=0$ and $E_4(i)=0$ (which is wrong, as said above).
Your equation $$ \sum_{\substack{(m,n)\neq(0,0)}} \frac{-4}{((m-n)+i(m+n))^4} = \sum_{\substack{(k,l)\in\mathbb Z^2\\(k,l)\neq(0,0)}}\frac{-4}{(k+il)^4} $$ is wrong. For example, on the right you have a term with $k=1,l=0$. But there is no term on the left corresponding to it. On the left, in $(m-n)+i(m+n)$, the real and imaginary parts have the same parity.