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Let $\Lambda = \{m \omega_1+n\omega_2; m,n \in \mathbb{Z}\}$ with $\omega_i \in \mathbb{C}$ with $\omega_2/\omega_1 \notin \mathbb{R}$ be a lattice. Define the Weierstrass $\mathscr{P}$ function on the torus $\mathbb{C}/\Lambda$ as $$\mathscr{P}(z) = \frac{1}{z^2}+\sum\limits_{\omega \in \Lambda} (\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2})$$

I am asked to show that if we choose $a \in \mathbb{C}/\Lambda$ with $a \notin \frac{1}{2}\Lambda$ then the elliptic function $$h(z) = (\mathscr{P}(z-a) − \mathscr{P}(z+a))(\mathscr{P}(z)-\mathscr{P}(a))^2-\mathscr{P}'(z)\mathscr{P}'(a)$$

has no poles and is hence constant. To do this I tried to show that $h$ is analytic. I attempted to do this by taking Laurent expansions around $0,a,-a$ and showing that $h$ was analytic in a disc around all those points (as those are the only possible points where $h$ can have a pole) but was not able to get this result. I don't have any other ideas for what to do.

How should I go about doing this problem? Any hints are appreciated!

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This function (with the sum over $\omega\in\Lambda\color{red}{\setminus\{0\}}$) is commonly denoted $\wp(z)$.

You have $\wp(-z)=\wp(z)$, $$\wp(z)=z^{-2}+O(z^2),\quad\wp'(z)=-2z^{-3}+O(z)\qquad(z\to 0)$$ and $\wp'''(z)=12\wp(z)\wp'(z)$ (follows from $\wp''(z)=6\wp^2(z)-g_2/2$ which in turn follows from well-known $\wp'^2(z)=4\wp^3(z)-g_2\wp(z)-g_3$, or, the more elementary way, can just be deduced from Laurent expansions of $\wp$, $\wp'$ and $\wp'''$ — all the negative powers of $z$ except $-24z^{-5}$ vanish).

This is sufficient. With $z\to 0$ you have $$\begin{align}h(z)&=\big(-2\wp'(a)z-\wp'''(a)z^3/3+O(z^5)\big)\big(z^{-2}-\wp(a)+O(z^2)\big)^2+2\wp'(a)z^{-3}+O(z)\\ &=-2\wp'(a)z^{-3}\big(1+2\wp(a)z^2+O(z^4)\big)\big(1-2\wp(a)z^2+O(z^4)\big)+2\wp'(a)z^{-3}+O(z)\\ &=8\wp^2(a)\wp'(a)z+O(z)=O(z),\end{align}$$ and even easier things at $z\to\pm a$ when the double pole of $\wp(z\mp a)$ is compensated by (the) double zero of $\big(\wp(z)-\wp(a)\big)^2$.

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