Induced isomorphism on elliptic curves Let $\Lambda_1$ and $\Lambda_2$ be lattices over $\mathbb{C}$ and $F(z)$ an entire function such that for all $\omega_1 \in \Lambda_1$ there exists an $\omega_2 \in \Lambda_2$ such that $F(z+\omega_1)=F(z)+\omega_2$ for all $z \in \mathbb{C}$.
(i) Show that $F(z)$ is of the form $F(z)=Az+B$ for some $A,B \in \mathbb{C}$, and that $A\Lambda_1 \subset \Lambda_2$.
In this case, $F$ induces a map between elliptic curves $\underline{F}:\mathbb{C}/\Lambda_1 \rightarrow \mathbb{C}/\Lambda_2$ which is an isomorphism if $A\Lambda_1=\Lambda_2$.
(ii) Suppose that $\Lambda_1=<1,\tau_1>$, $\Lambda_2=<1,\tau_2>$. Show that $\underline{F}$ is an isomorphism iff $\tau_1=\frac{a\tau_2+b}{c\tau_2+d}$ where $ad-bc=\pm1$.
I have already shown (i) to be true. For (ii), it is clear that since $A1,A\tau_1 \in \Lambda_2$, there exist integers a,b,c,d such that 
\begin{align*}
A\tau_1&=a\tau_2+b\\
A&=c\tau_2+d
\end{align*}
Which gives $\tau_1=\frac{A\tau_1}{A}=\frac{a\tau_2+b}{c\tau_2+d}$, but I'm not sure how to proceed from here.
 A: Suppose $F$ is an isomorphism. From $(i)$ we have that $A\Lambda_1=\Lambda_2.$
So, you got $$A\tau_1=a\tau_2+b$$ $$A=c\tau_2+d,$$ for some $a,b,c,d \in\mathbb{Z}.$ 
That can also be written differently, as 
$$
A\begin{bmatrix}
\tau_1 \\ 1
\end{bmatrix} =
\begin{bmatrix}
a&b \\ c&d
\end{bmatrix}
\begin{bmatrix}
\tau_2 \\ 1
\end{bmatrix}
$$
All we have to do now is to see that $ad-bc=\pm1.$
But also, the following is true: $ \Lambda_1=\frac{1}{A}\Lambda_2, $ so similarly we conclude $$\frac{1}{A}\tau_2=a'\tau_1+b'$$ $$\frac{1}{A}=c'\tau_1+d',$$ for some $a',b',c',d' \in \mathbb{Z},$ i.e. 
$$
\dfrac{1}{A}\begin{bmatrix}
\tau_2 \\ 1
\end{bmatrix} =
\begin{bmatrix}
a'&b' \\ c'&d'
\end{bmatrix}
\begin{bmatrix}
\tau_1 \\ 1
\end{bmatrix}.
$$
By combining the two matrix equations that we have, we get
$$
\begin{bmatrix}
\tau_1 \\ 1
\end{bmatrix} =
\begin{bmatrix}
a&b \\ c&d
\end{bmatrix}
\begin{bmatrix}
a'&b' \\ c'&d'
\end{bmatrix}
\begin{bmatrix}
\tau_1 \\ 1
\end{bmatrix}.
$$
Since, $\{1,\tau_1\}$ is be a basis for $\mathbb{C}$ over $\mathbb{R}$ (otherwise it wouldn't define a lattice), it's obvious that $$\begin{bmatrix}
a&b \\ c&d
\end{bmatrix}
\begin{bmatrix}
a'&b' \\ c'&d'
\end{bmatrix}=I.$$
Now $$ \det\begin{bmatrix}
a&b \\ c&d
\end{bmatrix}
\begin{bmatrix}
a'&b' \\ c'&d'
\end{bmatrix}=\det I=1. $$ Since all matrix entries are integers, the determinants are also integer numbers and we conclude $$\det\begin{bmatrix}
a&b \\ c&d
\end{bmatrix}=\pm 1,$$ so $ad-bc=\pm1.$
Conversely, suppose that $\dfrac{\tau_1}{1}=\dfrac{a\tau_2+b}{c\tau_2+d},$ where $ad-bc=\pm1.$
Equivalently, we have $$A\tau_1=a\tau_2+b,$$ $$A=c\tau_2+d,$$ for some $A\in\mathbb{C}$ (because the numerators and the denominators only have to be proportional, not equal). Now, obviously, $A\Lambda_1\subseteq\Lambda_2.$
Written differently, the above two equalities become$$
A\begin{bmatrix}
\tau_1 \\ 1
\end{bmatrix} =
\begin{bmatrix}
a&b \\ c&d
\end{bmatrix}
\begin{bmatrix}
\tau_2 \\ 1
\end{bmatrix},
$$ i.e.
$$
A
\begin{bmatrix}
d&-b \\ -c&a
\end{bmatrix}
\begin{bmatrix}
\tau_1 \\ 1
\end{bmatrix} =
\begin{bmatrix}
\tau_2 \\ 1
\end{bmatrix}.$$
Similarly, we get $\Lambda_2\subseteq A\Lambda_1.$
