Weierstrass $\wp$-function: $(\partial_z \wp(z,\omega))^2$ Let $\vartheta(z,\omega)$ be the Riemann theta function. For $j \in \mathbb{Z}$ let $c_j$ be the coefficient of $z^{j}$ in the Laurent expansion of $\partial_z \log \vartheta \left(z + \frac{1 + \omega}{2}, \omega \right)$ at $z$ = 0. The Weierstrass $\wp$ function is
$$
\wp(z, \omega) = - \partial_z^{2} \log  \vartheta \left(z +  \frac{1 + \omega}{2}, \omega \right) + c_1.
$$
I have successfully shown, that $\wp$ is $\omega$- and $1$-periodical and a few other properties. Im stuck at the last property I have to show. 
Let $ e_1 = \wp \left(\frac{1}{2}, \omega \right), \; e_2 = \wp \left(\frac{\omega}{2}, \omega \right)$ and $e_3 = \wp \left(\frac{1+\omega}{2}, \omega \right)$. Then 
$$
(\partial_z \wp(z,\omega))^{2} = 4(\wp(z,\omega) -e_1)(\wp(z,\omega) - e_2)(\wp(z,\omega) -e_3)
$$ 
Any hints on how I could start the proof are very much appreciated!
 A: 
I have successfully shown, that $\wp$ is $\omega$- and $1$-periodical and a few other properties.

If among these other properties are (considering $\wp$ as a function of $z$ only here)


*

*$\wp$ is even,

*$\wp$ has order $2$ (that is, takes each value in $\widehat{\mathbb{C}}$ exactly twice in a fundamental parallelogram, counted with multiplicity; equivalently, it has only one pole, of order $2$, or two simple poles),


you are more or less done.
Consider


*

*the zeros of $\partial_z\wp$ (there are $3$),

*the order of the pole in $0$, and

*the leading coefficient in the Laurent expansion of both sides.



Let $f(z) = \wp'(z)^2 - 4(\wp(z)-e_1)(\wp(z)-e_2)(\wp(z)-e_3)$. Then $f$ is an elliptic function that can have poles only in the lattice points. As the Laurent expansion of $\wp(z)$ starts with $z^{-2} + \dotsc$, then $\wp'(z) = -2z^{-3} + \dotsc$, so the Laurent expansion of $f$ begins $$(-2z^{-3})^2 - 4(z^{-2})^3 + \dotsc = 0z^{-6}+\dotsc,$$ hence the order of the pole of $f$ is $< 6$, therefore $f$ attains the value $\infty$ with multiplicity less than $6$ in a fundamental parallelogram.
On the other hand $f$ has three zeros of order $\geqslant 2$ in the fundamental parallelogram, so it attains the value $0$ with multiplicity at least $6$.
But a non-constant elliptic function has equally many poles and zeros (counted with multiplicity) in a fundamental parallelogram (see below), hence the above shows $f$ is constant (it has more zeros than poles), and since $f$ attains the value $0$, we have $f \equiv 0$.

Let $\Omega = \langle \omega_1, \omega_2\rangle$, where $\operatorname{Im} \dfrac{\omega_2}{\omega_1} > 0$, a lattice, and $g$ a non-constant elliptic function for the lattice $\Omega$. Let $a \in \mathbb{C}$ such that $g$ has neither zeros nor poles on the boundary of the parallelogram $P_a = \{ a + \alpha \omega_1 + \beta \omega_2 : \alpha,\beta \in [0,1]\}$. The logarithmic derivative of $g$ is also an elliptic function for the lattice $\Omega$, and has neither zeros nor poles on the boundary of $P_a$. Let $\zeta_1,\, \dotsc,\, \zeta_k$ be the distinct zeros of $g$ in $P_a$, with multiplicities $\alpha_1,\, \dotsc,\, \alpha_k$, and let $\pi_1,\, \dotsc,\, \pi_m$ the poles of $g$ in $P_a$, with respective multiplicities $\beta_1,\, \dotsc,\, \beta_m$. Then $g'/g$ has simple poles in the $\zeta_\kappa$ and $\pi_\mu$, and is holomorphic everywhere else in a neighbourhood of $\overline{P}_a$. In the $\zeta_\kappa$ resp. $\pi_\mu$, writing $g(z) = (z-w)^r\cdot h(z)$ with $h$ holomorphic and nonzero in a neighbourhood of $w$ reveals that the residue of $g'/g$ in $\zeta_\kappa$ is $\alpha_\kappa$, and the residue in $\pi_\mu$ is $-\beta_\mu$. Hence, by the residue theorem,
$$\frac{1}{2\pi i} \int_{\partial P_a} \frac{g'(z)}{g(z)}\,dz = \sum_{\kappa = 1}^k \alpha_\kappa - \sum_{\mu = 1}^m \beta_\mu.$$
On the other hand, grouping the pairs of parallel sides, we compute
$$\begin{align}
\int_{\partial P_a} \frac{g'(z)}{g(z)}\, dz
&= \int_a^{a+\omega_1} \frac{g'(z)}{g(z)} - \frac{g'(z+\omega_2)}{g(z+\omega_2)}\,dz + \int_{a}^{a+\omega_2} \frac{g'(z+\omega_1)}{g(z+\omega_1)} - \frac{g'(z)}{g(z)}\,dz\\
&= 0 + 0\\
&= 0
\end{align}$$
by the periodicity. Hence
$$\sum_{\kappa = 1}^k \alpha_\kappa = \sum_{\mu = 1}^m \beta_\mu$$
a non-constant elliptic function has equally many zeros as poles (considering $g - c$ shows that all values in the Riemann sphere are attained the same number of times) counting multiplicities.
