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

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.

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 $$u,v\in \Bbb{R}$$)
I think that for the other directon OP's approach is good. For the other direction, WLOG, we may assume that $$e_3. Then, by direct calculation, we can find that $$\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$$