Weierstrass elliptic function in Laurent series form Could anyone help me to figure out how 
        $$  f_0(z) = \wp (\log z; i \pi, \log \rho)              $$
where $\wp$ denotes the Weierstrass elliptic function and $i \pi$, $\log \rho$ are its half-periods.
$$    f_0(z) = \sum_{n=1}^{\infty} \frac{n \rho^{2n}z^{-n}}{1- \rho^{2n}} + \sum_{n=1}^{\infty} \frac{n z^{n}}{1- \rho^{2n}}.     $$
This was obtained in page 214 in Z. Nehari and B. Schwarz, On the coefficients of univalent Laurent series, Proc. Amer. Math. Soc, 1954,212-217.
I got to know that this is a Fourier series in $z$ and a Laurent series in $e^{2\pi i z}$
$$ f_0(z) = \sum_{n=-\infty}^{\infty} a_n e^{2\pi i z} $$
then
$$  \wp\left(\frac{\log z}{2 \pi i}\right) = \sum_{n=-\infty}^{\infty} a_n z^n,                  $$
how to find the Fourier coefficients using 
$$ \sum_{m=-\infty}^{\infty} \frac{1}{(z+m)^2} = \frac{d}{dz} \frac{1}{1-e^{2\pi iz}} = \frac{2 \pi i e^{2\pi iz} }{(1  -e^{2\pi iz})^2} = \sum_{k\geq 0} (2 \pi i k)e^{2\pi iz}. $$
I have also found that 
$$ \wp(z;\tau) = \frac{1}{z^2} + \sum_{(n,m)\neq (0,0)}\left( \frac{1}{(z-m\tau+n)^2} - \frac{1}{(m\tau+n)^2}  \right) $$
and 
$$  \frac{1}{(2 \pi i)^2} \wp(z;\tau)= \frac{1}{12} + \sum_{m=-\infty}^{\infty} \frac{q^m_{\tau}q_z}{(1 - q^m_{\tau}q_z)^2}  - 2 \sum_{n=1}^{\infty}\frac{n q^n_{\tau}}{1-q^n_{\tau}},                 $$
where $q_z = e^{2 \pi i z}$, in Elliptic function book by Serge Lang. Sice $\tau = \frac{\omega_2}{\omega_1} = \frac{\log \rho}{i \pi}$ then 
$$ \frac{1}{(2 \pi i)^2} \wp(z;\tau)= \frac{1}{12} + \sum_{m=-\infty}^{\infty} \frac{z \rho^{2m}}{(1 - z \rho^{2m})^2}  - 2 \sum_{n=1}^{\infty}\frac{n \rho^{2n}}{1- \rho^{2n}}.    $$ 
 A: For $\Im(\tau) > 0$ $$\wp(z,\tau) \overset{def}= \frac{1}{z^2} + \sum_{(n,m)\neq (0,0)}\left( \frac{1}{(z+m\tau+n)^2} - \frac{1}{(m\tau+n)^2}  \right)$$
On the other hand
 $$\sum_{m=-\infty}^{\infty} \frac{1}{(z+m)^2} = \frac{d}{dz} \frac{1}{1-e^{2\pi iz}} $$ is valid for every non-integer $z$ (because the LHS minus the RHS is a 1-periodic bounded entire function thus constant thus equal to its value at $+i\infty$ where it vanishes)
So 
$$\wp(z,\tau) =\sum_m  \frac{d}{dz} \frac{1}{1-e^{2\pi i(z+m\tau)}}+A, \qquad A = \int_0^1 \wp(\tau/2+z,\tau)dz$$
For $\Im(z) \in (0,\Im(\tau))$, expanding $\frac{1}{1-e^{2\pi i(z+m\tau)}}=\frac{-e^{-2\pi i(z+m\tau)}}{1-e^{-2\pi i(z+m\tau)}}$ in geometric series depending on the sign of $\Im(z+m\tau)$ we obtain
$$\wp(z,\tau) = A+\sum_{m \ge 0} \sum_{k=0}^\infty (2i\pi mk)e^{2\pi ik(z+m\tau)} -  \sum_{m <0} \sum_{k=1}^\infty (2i\pi mk)e^{-2\pi ik(z+m\tau)}$$
$$ = A+\sum_{k=1}^\infty e^{2\pi ik z} \sum_{m \ge 0}(2i\pi mk)e^{2\pi ik m\tau}-\sum_{k=1}^\infty e^{-2\pi ik z} \sum_{m \ge 0}(2i\pi mk)e^{-2\pi ik m\tau}$$ which is the Fourier series valid for $\Im(z) \in (0,\Im(\tau))$.
Letting $q=e^{2i \pi z},\rho = e^{2i \pi \tau}$ you get your Laurent series for $$\wp(\frac{\log q}{2i \pi},\frac{\log \rho}{2i\pi})$$ valid for $|\rho|<1,|q| \in (|\rho|,1)$
Not sure how to evaluate $A$ 
