An Elliptic Curve has 3 real roots if and only if $\frac{\omega_2}{\omega_1}$ is purely imaginary

Question

Define $\Lambda=\{n\omega_1+m\omega_2 \mid n,m\in \mathbb{Z}\}$ for $\omega_1, \omega_2\in \mathbb{C}$ and $\wp(z)$ is the corresponding Weierstrass elliptic function.

I want to show that $e_1=\wp(\frac{\omega_1}{2}), e_2 = \wp(\frac{\omega_1 + \omega_2}{2}), e_3=\wp(\frac{\omega_2}{2})$ are real if and only if $\omega_1$ is real and $\omega_2$ is purely imaginary.

WLOG, I assumed $\omega_1=1$ and $\omega_2 = i\tau$ for some $\tau\in\mathbb{R}$. Then I deduce that for every $z\in\mathbb{C}$, $\wp(\overline{z})=\overline{\wp(z)}$. By, evaluatin $\wp$ at the half-ratios, I proved that $e_1, e_2, e_3 \in \mathbb{R}$. Is my idea correct?

For proving the other direction I wanted to use the following identities: $$\omega_1 = \int_{\infty}^{e_1}{\frac{1}{\sqrt{(y-e_1)(y-e_2)(y-e_3)}}dy}$$ and $$\omega_2 = \int_{e_1}^{e_2}{\frac{1}{\sqrt{(y-e_1)(y-e_2)(y-e_3)}}dy}$$ but I cannot drive the result from them since I do not know which of $e_i$s is bigger.

Any help is appreciated.

Answer 1

The lattice $\Lambda$ is determined by its elliptic curve (ie. differential equation of $\wp_\Lambda$) $$y^2 = 4x^3-g_2(\Lambda)x-g_3(\Lambda) = 4(x-\wp_\Lambda(\omega_1))(x-\wp_\Lambda(\omega_2))(x-\wp_\Lambda(\frac{\omega_1+\omega_2}2))$$ If $\wp_\Lambda(\omega_1),\wp_\Lambda(\omega_2),\wp_\Lambda(\frac{\omega_1+\omega_2}2)$ are real then the elliptic curve of $\overline{\Lambda}$ is the same, so $\Lambda=\overline{\Lambda}$.

(so either $\Lambda=u\Bbb{Z}+iv\Bbb{Z}$ or $\Lambda=u\Bbb{Z}+(\frac{u}2+iv)\Bbb{Z}$ for some $u,v\in \Bbb{R}$)

Answer 2

I think that for the other directon OP's approach is good. For the other direction, WLOG, we may assume that $e_3<e_2<e_1$. Then, by direct calculation, we can find that $$\omega_1 = \int_{\infty}^{e_1}{\frac{1}{\sqrt{(y-e_1)(y-e_2)(y-e_3)}}dy}\in\Bbb R$$ and $$\omega_2 = i\int_{-\infty}^{e_3}{\frac{1}{\sqrt{e_1-y)(e_2-y)(e_3-y)}}dy}\in i\Bbb R$$