Could anyone please help me to figure out how $$f_0(z) = \wp(\log z; i\pi, \log \rho)$$ where $\wp$ denotes the Weierstrass elliptic function and $i \pi$ and $\log \rho$ are its half periods. $$ f_0(z) = \displaystyle \sum_{n=1}^{\infty} \frac{n \rho^{2n}z^{-n}}{1-\rho^{2n}} + \sum_{n=1}^{\infty} \frac{n z^{n}}{1-\rho^{2n}}. $$
Can $\log z$ have Laurent expansion at the origin?
Z. Nehari and B. Schwarz, On the coefficients of univalent Laurent series, Proc. Amer. Math. Soc., 5, 1954, 212 - 217 (page 214).
$$\wp_{\tau}(z) = \frac1{z^2} + \sum_{(n,m) \ne (0,0)}\frac1{(z+n\tau+m)^2}-\frac1{(n\tau+m)^2}$$ is $1$ periodic and analytic on $\Im(z) \in (0,|\Im(\tau)|)$ so it has a Fourier series valid for $\Im(z) \in (0,|\Im(\tau)|)$ which is a Laurent series for $\wp_{\tau}(\frac{\log u}{2i\pi})$ valid for $|e^{2i \pi u}| \in (1,e^{|\Im(\tau)|})$
The Fourier coefficients are found from the pole expansion of inverse of trigonometric functions. For $\Im(z) > 0$ $$\sum_{m}\frac{1}{(z+m)^2}= \frac{d}{dz} \frac1{1-e^{2i \pi z}}= \sum_{k \ge 0} (2i\pi k) e^{2 i \pi kz}$$ For $\Im(z) < 0$ $$\sum_{m}\frac{1}{(z+m)^2}= \frac{d}{dz} \frac1{1-e^{2i \pi z}}= \sum_{k \ge 1} (-2i\pi k) e^{-2 i \pi kz}$$