Construction of the Weierstrass $\wp$ function

Question

In the construction of the Weierstrass $\wp$-function, why do we have to add a leading term $\frac{1}{z^2}$ instead of letting $\wp(z)=\sum_{\omega\neq0}\left\{\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right\}$? I am guessing this is to take care of the origin too.

Answer 1

It is not obvious that $$\wp(z) = \frac{1}{z^2}+\sum_{(n,m) \in \mathbb{Z}^2 \setminus (0,0)} \frac{1}{(z-nw_1-mw_2)^2}-\frac{1}{(nw_1-mw_2)^2}\tag{1}$$ is doubly periodic.

What is obvious is that $\frac{1}{(z-nw_1-mw_2)^2}-\frac{1}{(nw_1-mw_2)^2} = \mathcal{O}(\frac{1}{|nw_1+mw_2|^3})$ so that $(1)$ converges uniformly on any compact $\subset \mathbb{C} \setminus \Lambda$,

thus $\wp(z)$ is meromorphic and $$\wp'(z) = \sum_{(n,m) \in \mathbb{Z}^2} \frac{-2}{(z-nw_1-mw_2)^3}$$ which is obviously $w_1,w_2$ periodic and meromorphic.

Finally, say that for any $z \in \mathbb{C} \setminus \Lambda$ : $$\wp(z+w_1)-\wp(z) =\int_z^{z+w_1} \wp'(s)ds $$ $$= \sum_{(n,m) \in \mathbb{Z}^2} \frac{1}{(z+w_1-nw_1-mw_2)^2}-\frac{1}{(z-nw_1-mw_2)^2} = 0$$ and hence $\wp(z)$ is $w_1$ periodic. Do the same with $w_2$, and you get that $\wp(z)$ is a meromorphic function that is $w_1,w_2$ periodic with one double pole at each $z =nw_1+mw_2$.