We say that$ \omega_1$and $\omega_2$ generate the periods of a doubly periodic function if the periods of the function are precisely the complex numbers of the form $ m\omega_1+ n\omega_2 $ where m and n are integers. Show that if$ \omega_1$and$ \omega_2$ generate the periods of a doubly periodic function $f(z)$, and if $\alpha_1$ and $ \alpha_2$are complex numbers, then$\alpha_1 $and $\alpha_2$ generate the periods of $f (z) $if and only if there is a 2 x 2 matrix $\mathrm A$with integer entries and with determinant ±1 such that $A(\omega_1 ,\omega_2) = ( \alpha_1, \alpha_2 ) $.
I think $ \omega_1 $ and $\omega_2$ have form $ m_1\alpha_1 + n_1 \alpha_2 $，$ m_1$and $ n_1$ are integer .
$\Leftarrow$: I want to prove $ \omega_1$and $\omega_2$ have form $ m_1\alpha_1+ n_1\alpha_2 $,i suppose $A= \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end {pmatrix}, \omega_1 =\begin{pmatrix} x_1\\ y_1 \\ \end {pmatrix} $and so on. But i can't find this relation.
$\Rightarrow$: i haven't clue.
Suppose $\omega_1,\omega_2$ generate the periods, and $\alpha_1,\alpha_2$ do as well.
Since $\omega_1,\omega_2$ generate periods, that means any period may be written as an integer combination of them, so in particular $\alpha_1=a_{11}\omega_1+a_{12}\omega_2$ and $\alpha_2=a_{21}\omega_1+a_{22}\omega_2$ for integers $a_{ij}$. That is,
$$ \begin{pmatrix} \alpha_1 \\ \alpha_2 \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} \omega_1 \\ \omega_2 \end{pmatrix} $$
Or $\vec{\alpha}=A\vec{\omega}$ for an integer matrix $A$.
How do we know $A$ must have determinant $\pm1$? Well, the same argument works to show $\vec{\omega}=B\vec{\alpha}$ for some integer matrix $B$ ... can you finish from there?