Zeros of the Weierstrass $P$ function in special cases

Consider the function \begin{eqnarray*} \wp(z)=\frac{1}{z^2}+\sum_{a\in \Lambda^*}\left( \frac{1}{(z-a)^2}-\frac{1}{a^2}\right) \end{eqnarray*}

I know that finding the zeros of $\wp'(z)$ is quite easy and they're just the points of order 2 on $\mathbb{C}/\Lambda$ .

I also know that finding the zeros of $\wp(z)$ is in general not an easy task, but I am interested in the case of the lattices $\mathbb{Z}\bigoplus i\mathbb{Z}$ and $\mathbb{Z}\bigoplus \omega\mathbb{Z}$ where $\omega = e^{i\frac{2\pi}{3}}$ . I know that in the case of $\mathbb{Z}\bigoplus i\mathbb{Z}$ the zero of $\wp(z)$ in the fundamental domain should be one of the zeros of $\wp'(z)$ , using the symmetry of the lattice, but I don't quite get where it comes from.

$\wp_\tau$ is the unique $\Bbb{Z}+\tau \Bbb{Z}$ periodic meromorphic function whose poles are at $\Bbb{Z}+\tau \Bbb{Z}$ and with $\wp_\tau(z)=z^{-2}+O(z)$ as $z\to 0$

In particular $$\wp_\tau(-z)=\wp_\tau(z),\qquad \wp_i(iz)=-\wp_i(z),\qquad\wp_\omega(\omega z)=\omega \wp_\omega(z)$$

$\wp_\tau$ has two poles per fundamental domain so it must have two zeros.

$$\wp_i(\frac{1+i}{2})=\wp_i(\frac{i-1}{2})=-\wp_i(\frac{1+i}{2})=0$$ This is a double zero because $\wp_i(\frac{1+i}{2}+z)=\wp_i(-\frac{1+i}{2}-z)=\wp_i(\frac{1+i}{2}-z)$. By the previous point it has no other zeros.

$$\wp_\omega(\frac{\omega}{\omega-1})=\omega \wp_\omega(\frac{1}{\omega-1})= \omega \wp_\omega(\frac{\omega}{\omega-1})=0$$ This is a simple zero because $\wp_\omega(-\frac{\omega}{\omega-1})=0$ is another zero.