automorphism of complex torus I am reading Miranda's book on algebraic curves and Riemann surfaces. In the section where the automorphisms of tori are discussed, it is shown that any of such a map $F: \mathbb{C}/\Lambda \rightarrow \mathbb{C}/\Lambda$ is induced by a linear complex map $G(z) = az + b$ satisfying $a\Lambda = \Lambda$.
So far it is clear for me the proof of this fact; however, then it is mentioned that $|a| = 1$ (also clear) and that $a$ is a root of unity. This final statement is unclear for me, why does $a$ have to be a root of unity?
 A: Let me first review the proof that $|a|=1$.
Since $\Lambda$ is discrete, the set of distances $\{d(u,v) \mid u,v \in \Lambda\}$ is a discrete set of real numbers. The map $g(z) = az+b$ multiplies each distance by the same amount, equal to $|a|$, or more precisely
$$d(g(u),g(v)) = |a| d(u,v), \quad u,v \in \Lambda
$$
It follows that if $u,v \in \Lambda$ attain the minimum positive value of $d(u,v)$ then
$$d(u,v) = d(g(u),g(v) = |a| \, d(u,v)
$$
and therefore $|a|=1$.
Next, we may suppose that $b=0$ and that $G(z)=az$, because $G(0)= b$, and the function $z \mapsto z-b$ covers an automorphism of $\mathbb C / \Lambda$, and so we can replace the function $z \mapsto G(z)=az+b$ by the function $z \mapsto G(z)-b=(az+b)-b = az$ which also covers an automorphism of $\mathbb C / \Lambda$.
Let $\Delta$ denote the minimum positive value of $|z|$ for $z \in \Lambda$, and let $\Lambda_\Delta$ denote the finite set of all $z \in \Lambda$ such that $|z|=\Delta$. If $|z|=\Delta$ then $|G(z)|=|a| \, |z| = \Delta$, and so $G$ permutes the finite set $\Lambda_\Delta$. Since this is a finite set, some power $G^k$ restricts to the identity map on $\Lambda_\Delta$. So for any $z \in \Lambda_\Delta$ we have
$$a^k z = G^k(z) = z
$$
and hence $a^k=1$.
A: Mosher proof is very clever.
I just wanted to expand on the proof given by Miranda of Proposition (1.12) at the end of page 63. What he shows is that $a$ satisfies an (monic integral) algebraic equation of degree $2$, and he says that

This forces $||\gamma|| = 1$, and in fact $\gamma$ must be a root of unity$\ldots$
  The only roots of unity which satisfy quadratic equations are the $4^{\text{th}}$ and $6^{\text{th}}$ roots of unity.

The first statement is justified by Mosher, the second one (if you include the $3^{\text{th}}$ root of unity case) by the fact that the $n$-th cyclotomic polynomial has degree $\varphi(n)$.
There is also an algebraic result due to Kronecker which says

Let $\lambda \neq 0$ be a root of a monic polynomial $f(z)$ with integer coefficient. If all the roots of $f(z)$ are in the unit disk $|z| \leq 1$ then $|\lambda| = 1$ (because it is a root of unity).

Finally, you can give a uglier proof as follows.

We start as Miranda:
Let $\lambda \in \Lambda^{\ast} = \Lambda \setminus \{0\}$ be an element of minimum norm. Then by linearity $a \lambda$ is an element of minimum norm of $a \Lambda^{\ast} = \Lambda^{\ast}$, hence $| a \lambda | = |\lambda|$ and $|a| = 1$ is unimodular.
If $a = 1, -1$ then the automorphism is just the identity or the opposite map. Assume that it is not one of them, i.e. that $a$ it is not real.
Then it is easy to see that
$$ \Lambda = \langle \lambda, a \lambda \rangle = \lambda \, \langle 1, a\rangle$$
and since $a^2 \lambda \in L$ we have that $a^2 \lambda = m a \lambda + n \lambda$,  hence $a^2 - m a - n = 0$.
This, with $a \neq \overline{a}$, means that $a$ is root of the polynomial
$$z^2 -  mz - n = (z - a)(z - \overline{a})$$
and by Vieta's formulas
$$
\begin{align}
n &= - a \overline{a} = - 1 \\
|m| &= |a + \overline{a}| \leq |a| + |\overline{a}| = 2.
\end{align}
$$
Now you can try by hand the cases $m = -2, \ldots, 2$. You should find that


*

*for $m = \pm 2$ you have $a = \pm 1$,

*for $m = 0$ you have $a = \pm i$, fourth root of unity,

*for $m = 1$ you have that $a$ is a sixth root of unity,

*for $m = -1$ you have that $a$ is a third root of unity and $1 + a$ is a sixth root of unity.

