What is the sum $\sum_{(m, n)\in\mathbb{Z}\\(m,n)\neq(0,0)}\frac{1}{(m+in)^\alpha}$ equal to? I had a look at this webpage, which led me to wonder what
$$\sum_{(m, n)\in\mathbb{Z}\\(m,n)\neq(0,0)}\frac{1}{(m+in)^\alpha}$$
would equal to.
My thoughts are as follows:
\begin{align}
\sum_{(m, n)\in\mathbb{Z}\\(m,n)\neq(0,0)}\frac{1}{(m+in)^\alpha}&=\sum_{(m, n)\in\mathbb{Z}\\(m,n)\neq(0,0)}\frac{(m-in)^\alpha}{\left(m^2+n^2\right)^\alpha} \\
&=\sum_{m\in\mathbb{Z}/\{0\}}\frac{1}{m^\alpha}+\sum_{n\in\mathbb{Z}/\{0\}}\frac{1}{(in)^\alpha}+\sum^\infty_{m=-\infty}\sum^\infty_{n=1}\frac{(m-in)^\alpha+(m+in)^\alpha+(-m+in)^\alpha+(-m-in)^\alpha}{\left(m^2+n^2\right)^\alpha} \\
&=\sum_{m=1}^\infty\frac{1}{m^\alpha}+\sum_{m=1}^\infty\frac{1}{(-m)^\alpha}+\sum_{n=1}^\infty\frac{1}{(in)^\alpha}+\sum_{n=1}^\infty\frac{1}{(-in)^\alpha}+\sum^\infty_{m=1}\sum^\infty_{n=1}\frac{\left(1+(-1)^\alpha\right)\left[(m-in)^\alpha+(m+in)^\alpha\right]}{\left(m^2+n^2\right)^\alpha} \\
&=\left(1+i^\alpha+(-1)^\alpha+(-i)^\alpha\right)\zeta(\alpha)+\sum^\infty_{m=1}\sum^\infty_{n=1}\frac{\left(1+(-1)^\alpha\right)\cdot2\sqrt{m^2+n^2}^\alpha\cos\left[\alpha\arctan\left(\frac nm\right)\right]}{\left(m^2+n^2\right)^\alpha} \\
&=\left(1+i^\alpha+(-1)^\alpha+(-i)^\alpha\right)\zeta(\alpha)+2\left(1+(-1)^\alpha\right)\sum^\infty_{m=1}\sum^\infty_{n=1}\frac{\cos\left[\alpha\arctan\left(\frac nm\right)\right]}{\left(m^2+n^2\right)^\frac\alpha2} \\
&=\left(1+i^\alpha+(-1)^\alpha+(-i)^\alpha\right)\zeta(\alpha)+2\left(1+(-1)^\alpha\right)\sum^\infty_{k=1}\sum_{\gcd(m,n)=1}\frac{\cos\left[\alpha\arctan\left(\frac{kn}{km}\right)\right]}{\left(k^2m^2+k^2n^2\right)^\frac\alpha2} \\
&=\left(1+i^\alpha+(-1)^\alpha+(-i)^\alpha\right)\zeta(\alpha)+2\left(1+(-1)^\alpha\right)\sum^\infty_{k=1}\sum_{\gcd(m,n)=1}\frac{\cos\left[\alpha\arctan\left(\frac{n}{m}\right)\right]}{\left(k^2m^2+k^2n^2\right)^\frac\alpha2} \\
&=\left(1+i^\alpha+(-1)^\alpha+(-i)^\alpha\right)\zeta(\alpha)+2\left(1+(-1)^\alpha\right)\sum^\infty_{k=1}\frac{1}{k^\alpha}\sum_{\gcd(m,n)=1}\frac{\cos\left[\alpha\arctan\left(\frac{n}{m}\right)\right]}{\left(m^2+n^2\right)^\frac\alpha2} \\
&=\left(1+i^\alpha+(-1)^\alpha+(-i)^\alpha+2\left(1+(-1)^\alpha\right)\sum_{\gcd(m,n)=1}\frac{\cos\left[\alpha\arctan\left(\frac{n}{m}\right)\right]}{\left(m^2+n^2\right)^\frac\alpha2}\right)\zeta(\alpha) \\
\end{align}
Though I couldn't (and presumably lack the analytical ability to) simply the expression further.
This seems tangentially related to the sum
$$\sum_{(m, n)\in\mathbb{Z}\\(m,n)\neq(0,0)}\frac{1}{(m^2+n^2)^\frac{\alpha}{2}}$$
which is equal to $4\beta\left(\frac{\alpha}{2}\right)\zeta\left(\frac{\alpha}{2}\right)$, though I don't know how this sum can be applied.
 A: (As stated by Peter Wu in the comments, you actually have here Eisenstein series $G_{2k}$ defined at $i$.)
A result by Hurwitz shows that
$$ G_k=\sum\limits_{m,n\ \in\ \mathbb{Z} \backslash [0,0]} \frac{1}{(m+ni)^k} = \frac{(2\varpi)^k}{k!}H_k $$
, where $$\varpi = 2\int\limits_{0}^{1}\frac{1}{\sqrt{1-x^4}}\ dx=\frac{1}{2}B(\frac{1}{4}, \frac{1}{2})=\frac{\Gamma(1/4)^2}{2\sqrt{2\pi}}$$ , and $H_k$ are the Hurwitz numbers, defined as the coefficients of the Laurent series of the Weierstrass $\wp$-function, or the Weierstrass elliptic function, given as
$$\wp(z)=\frac{1}{z^2}+\sum\limits_{\omega\neq0} \left(\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right)$$
, where $\omega = m\varpi+ni\varpi$ and $m, n\in\mathbb{Z}.$
Note that $H_k = 0$ if $k \not\equiv 0 \pmod4$, since $\wp(-z)=\wp(z)$ and $\wp(iz)=-\wp(z)$. It's also well-known that $H_4=\frac{1}{10}$ and $H_8 = \frac{3}{10}$, but one can evaluate the other values of $H_{4k}$ through the coefficients of the Laurent series of Weierstrass' elliptic function.
A few explicit evaluations - we have that $G_4=\frac{1}{15}\varpi^4, G_8=\frac{1}{525}\varpi^8, G_{12} = \frac{2}{53625}\varpi^{12}$
