Laurent series and Weierstrass zeta function

Question

Let $\zeta(z)$ be the Weierstrass $\zeta $ function of the lattice $\Omega = \mathbb{Z}\omega+\mathbb{Z}\omega' $

(a) Use the definition of $\zeta(z) $ to write the Laurent series for $\zeta$ near $z=0$ and express it in terms of the Eisenstein sums $s_{n}= s_{n}(\omega, \omega')$

(b) Use (a) to write the Laurent series of $\wp(z)$.

I know how to write down the Laurent series for normal complex valued functions, but how does it work for something like the $\zeta(z)$ function? I'm also completely puzzled by how to do part(b).

Answer 1

I will assume that you defined the Weierstrass zeta function as $$\zeta(z)=\frac{1}{z}+\sum_{\tau \in \Omega\setminus\{0\}}\Big(\frac{1}{z-\tau}+ \frac{1}{\tau}+ \frac{z}{\tau^2}\Big).$$ Additionally, for $n\geq 3$, let $s_n(\omega,\omega')= \sum_{\tau \in \Omega\setminus\{0\}}\tau^{-n}$ denote the Eisenstein series of weight $n$.

1. To prove part a), rewrite $$\zeta(z)=\frac{1}{z}+\sum_{\tau \in \Omega\setminus\{0\}}\frac{1}{\tau^2}\Big(\frac{\tau^2}{z-\tau}+ \tau+ z\Big)= \frac{1}{z}+\sum_{\tau \in \Omega\setminus\{0\}}\frac{z^2}{\tau^2}\Big(\frac{1}{z-\tau}\Big).$$ For all $0<\vert z\vert <r$ with $r=\operatorname{min}\{\vert \tau \vert \colon \tau \in \Omega \text{ and } \tau \neq 0\}$ one thus has $$\zeta(z)= \frac{1}{z}-\sum_{\tau \in \Omega\setminus\{0\}}\frac{z^2}{\tau^3}\Big(\frac{1}{1-\frac{z}{\tau}}\Big)=\frac{1}{z}-\sum_{\tau \in \Omega\setminus\{0\}}\Big(\frac{z^2}{\tau^3}\sum_{k=0}^{\infty}\big(\frac{z}{\tau}\big)^k\Big)= \frac{1}{z}-\sum_{\tau \in \Omega\setminus\{0\}}\sum_{k=0}^{\infty}\frac{z^{k+2}}{\tau^{k+3}}.$$ Since the double series converges absolutely, we can exchange the order of summation $$\zeta(z)= \frac{1}{z}-\sum_{k=0}^{\infty}\sum_{\tau \in \Omega\setminus\{0\}}\frac{z^{k+2}}{\tau^{k+3}}= \frac{1}{z}-\sum_{k=0}^{\infty}s_{k+3}(\omega,\omega')\cdot z^{k+2}.$$ Finally, since every Eisenstein series of odd weight vanishes, the Laurent series of $\zeta(z)$ around zero can be written as $$\zeta(z)= \frac{1}{z}-\sum_{k=1}^{\infty}s_{2k+2}(\omega,\omega')\cdot z^{2k+1}.$$

2. For part b), let me assume that you defined the Weierstrass $\wp$-function as $$\wp(z)=\frac{1}{z^2}+\sum_{\tau\in\Omega\setminus\{0\}}\Big(\frac{1}{(z-\tau)^2}-\frac{1}{\tau^2}\Big).$$ Termwise differentiation then shows $-\zeta'(z)=\wp(z)$ for all $z\in \mathbb{C}$. Since any Laurent series (as well as its termwise differentiated series) converges uniformly on its annulus of convergence, termwise differentiating the Laurent series of $\zeta(z)$ around zero yields the Laurent series of $\zeta'(z)=-\wp(z)$ around zero. In your case, termwise differentiation (and multiplication by $-1$) shows that the series $$\frac{1}{z}+\sum_{k=1}^{\infty}(2k+1)\cdot s_{2k+2}(\omega,\omega')\cdot z^{2k}$$ is the Laurent expansion of $\wp(z)$ around zero.