Does a complex lattice determine the vectors it is generated by?

Question

A lattice $\Lambda$ in the complex plane is defined as $$\Lambda=\{n\omega_1+m\omega_2:(n,m)\in {\mathbb Z}^2\}$$ where $\omega_1$, $\omega_2$ are complex numbers that are ${\mathbb R}$- linearly independent (this means that $\omega_1\neq 0 \neq \omega_2$ and $\omega_1/\omega_2 \not \in {\mathbb R}$). Is it true that $\omega_1$ and $\omega_2$ are determined by the lattice, up to a negative sign, meaning that if
$$\{n\omega_1+m\omega_2:(n,m)\in {\mathbb Z}^2\}=\{n\omega_1'+m\omega_2':(n,m)\in {\mathbb Z}^2\}$$ then $\{\omega_1',\omega_2'\}=\{\pm \omega_1, \pm \omega_2\}$? I have looked at various sources where they discuss lattices in the complex plane, but I could not find a discussion about this.

Answer 1

Consider $\omega_1=1$ and $\omega_2=i+1$. It is easy to see that this lattice is the same as the lattice generated by $\omega_1'=1$ and $\omega_2'=i$