For a lattice $\Lambda = [\lambda_1, \lambda_2] \subset \mathbb C$, the Weierstrass $\wp$-function defined as \begin{equation} \wp(z) = \frac{1}{z^2} + \sum_{\lambda \in \Lambda \setminus \{0\}} \left( \frac{1}{(z - \lambda)^2} - \frac{1}{\lambda^2} \right) \end{equation}
and the theta function $\sigma$ given by \begin{equation} \sigma(z) = z \prod_{\lambda \in \Lambda \setminus \{0\}} \left( 1 - \frac{z}{\lambda} \right) e^{z/\lambda + \frac{1}{2} z^2/\lambda^2}, \end{equation}
I want to show that, for $a \notin \Lambda$, \begin{equation} \wp(z) - \wp(a) = - \frac{\sigma(z - a)\sigma(z + a)}{\sigma(a)^2 \sigma(z)^2}, \end{equation}
c.f. exercise 2.12 in http://pub.math.leidenuniv.nl/~luijkrmvan/elliptic/2011/ec.pdf.
I have shown both sides share zeros and poles (counting multiplicities), and now I want to show they are proportional by a factor 1 to prove the identity. To do so I consider the expression near the pole at $z = 0$. Then \begin{equation} \wp(z) - \wp(a) \sim \frac{1}{z^2} \quad \text{and} \quad - \frac{\sigma(z - a)\sigma(z + a)}{\sigma(a)^2 \sigma(z)^2} \sim \frac{1}{z^2} \prod_{\lambda \in \Lambda \setminus \{0\}} \frac{\lambda^2 - a^2}{\lambda^2(\lambda - a)^2}. \end{equation}
However, I cannot find a way to deduce \begin{equation} \prod_{\lambda \in \Lambda \setminus \{0\}} \frac{\lambda^2 - a^2}{\lambda^2(\lambda - a)^2} = -1, \end{equation}
and I am also not sure whether I have made a mistake. Any help would be greatly appreciated!
Define $f(z):=\wp(z)-\wp(a)$. Clearly $f$ is an elliptic function with order 2. Looking at poles and zeros actually shows that $$ f(z)=C\frac{\sigma(z-a)\sigma(z+a)}{\sigma(z)^2} $$ for a constant $C$. We now need to compute such constant. One the one hand, since $\wp$ has a pole of order $2$ at $0$, we find $$ \lim_{z\to 0}z^2f(z)=\lim_{z \to 0}z^2 \wp(z)-\lim_{z \to 0}z^2 \wp(a)=\lim_{z \to 0}z^2 \wp(z)-0=1 $$ On the other hand, using the definition of $\sigma$ we get $\lim_{z \to 0}\frac{\sigma(z)}{z}=1$ and therefore $$ \lim_{z\to 0}z^2f(z)=C \lim_{z \to 0}\frac{\sigma(z-a)\sigma(z+a)}{\frac{\sigma(z)^2}{z^2}}=-C\sigma(a)^2 $$ Thus, $$ C=\frac{-1}{\sigma(a)^2} $$ and the desired result now follows.