Why is the degree of a linear holomorphism $f:X\rightarrow X$ such that $[z] \mapsto [cz]$ from the complex torus $X=\mathbb{C}/\Lambda$ to itself equal to |c|^2? (here we assume that $\Lambda=\mathbb{Z}+\tau \mathbb{Z}$ with $\operatorname{Im}(\tau)>0$, $c\Lambda \subseteq \Lambda$, and $[z]$ designs the equivalence class of $z$ in $X$).
This is mentioned on this answer, but I haven’t been able to find the reason by myself.
I’ve tried to compute it by using $\deg(f)=\sum_{p\in f^{-1}({z})}\operatorname{mult}_p(f)$ where $f([z])=[cz]$ and hence $f^{-1}([z])=[c^{-1}z]$, but I’m stuck at trying to find $\operatorname{mult}_p(f)$ for the points in $f^{-1}([z])$ here.
Edit: I already understand why $\operatorname{mult}_p(f)=1$ for all $p\in f^{-1}([z])$ but I don’t see why should $|f^{-1}([z])|=|c|^2$.
First, note that $f$ is a local biholomorphism. Therefore, its degree is the cardinality of $f^{-1}([0])$.
Now, $f^{-1}([0])=c^{-1}\Lambda/\Lambda \cong \Lambda/c\Lambda$.
Let $\chi(x)=x^2+ax+d$ be the characteristic polynomial of the endomorphism $c$ acting on $\Lambda$: then clearly $\Lambda/c\Lambda$ has cardinality $d$, so we need to show that $d=|c|^2$.
This result is clear if $c \in \mathbb{Z}$, so let’s assume that it’s not the case. Since $\chi(c)=0$, $c$ is an algebraic integer contained in a quadratic extension $K$ of $\mathbb{Q}$.
Let $(u,v)$ be a basis of $\Lambda$, then it’s easy to see that $\mathbb{Q}(v/u)=\mathbb{Q}(c)$, hence that $c \notin \mathbb{R}$.
Thus, $d$ is the product of $c$ and its only conjugate over $\mathbb{Q}$ – but its conjugates include $\overline{c}$, so that $d=c\overline{c}=|c|^2$.