Given nonzero complex constants $\omega_1,\omega_2$, with nonreal ratio, we define $$\wp(z;\omega_1,\omega_2)=\frac{1}{z^2}+ \sum_\omega \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2} $$ where the sum is taken over all nonzero linear combinations $\omega=n_1 \omega_1+n_2 \omega_2$ with integer coefficients.
It is known that $\wp$ is of order 2, which means that for any $c \in \hat{\mathbb C}$ the equation $\wp(z)=c$ has two non-congruent solutions (two points are called congruent if their difference is linear combination of $\omega_1,\omega_2$ with integer coefficients).
In addition, it is known that $\wp$ is even, and that the poles on the "lattice" $ \omega_1 \mathbb Z+\omega_2 \mathbb Z$ are all of order 2.
My question is: Could there be a point $z_0$ such that $\wp'(z_0)=0$? This would imply that the value $\wp(z_0)$ is taken twice at $z_0$.
I tried using the evenness of the function to show that there isn't such point. However starting with $z_0=\frac{1}{2} \omega_1+\frac{1}{2} \omega_2$ this approach fails.
Thanks.
In fact, since $\wp'$ is of order $3$, it has three zeros. Since $\wp'$ is odd and periodic, the zeros are
$$\frac{\omega_1}{2},\, \frac{\omega_2}{2},\, \frac{\omega_1+\omega_2}{2}.$$
By the oddness of $\wp'$, we have
$$\wp'\left(-\frac{\omega_1}{2} \right) = -\wp'\left(\frac{\omega_1}{2} \right).$$
On the other hand,
$$\wp'\left(\frac{\omega_1}{2} \right) = \wp'\left(-\frac{\omega_1}{2} +\omega_1\right) = \wp'\left(-\frac{\omega_1}{2} \right)$$
by the periodicity. Together, that implies
$$\wp'\left(\frac{\omega_1}{2} \right) = 0.$$
The argument for the other two points is analogous. Since $\wp'$ has only three zeros (in a fundamental mesh), there are no others.