Prove that for $k \geq 4$, $E_k(z) = \displaystyle\frac{1}{2} \displaystyle\sum_{m, n \in \mathbb{Z}\\ gcd(m,n) = 1} (mz + n)^{-k}$.
The usual definition of $E_k(Z)$ is
$E_k(Z) = \displaystyle\frac{1}{2\zeta(k)}G_k(z)$ where
$G_k(z) = \displaystyle\sum_{m, n}'(mz+n)^{-k}$, the primed summation means that the summation is taken over pairs of $m, n$ not both zero.
I have done this in the following way:
$G_k(z) = \displaystyle\sum_{m, n}'(mz+n)^{-k}$
$ = \displaystyle \sum_{n \in \mathbb{Z}} n^{-k} + z^{-k} \displaystyle \sum_{m \in \mathbb{Z}} m^{-k} +\displaystyle \sum_{m, n \in \mathbb{Z}\setminus\{0\}} (mz+n)^{-k}$
$=2\zeta(k) + 2z^{-k}\zeta(k) + \displaystyle \sum_{m, n \in \mathbb{Z}\setminus\{0\}} (mz+n)^{-k}$
Thus $ E_k(z) = \displaystyle\frac{G_k(z)}{2\zeta(k)} = 1 + z^{-k} + \displaystyle\frac{1}{2\zeta(k)} \sum_{m, n \in \mathbb{Z}\setminus\{0\}} (mz+n)^{-k}$.
Now I do not understand how to proceed.
Helps are appreciated. Thanks.