Prove that for all $d \geq 1$ there exist a torus $X= \mathbb{C} / \Lambda$ and a holomorphic map $X = \mathbb{C} / \Lambda\rightarrow X= \mathbb{C} / \Lambda$ of degree $d.$
Attempt: Let $\Lambda$ denote the lattice of $\mathbb{C}$ generated by $1, \lambda \in \mathbb{C}$ with Im $\lambda \neq 0$ i.e $\Lambda= \mathbb{Z} + \mathbb{Z}\lambda.$ Then $X= \mathbb{C}/ \Lambda$ is a compact Riemann Surface.
Question 1: Is true that every holomorphic map from a torus to itself is linear?
Question 2: (I'm not quite sure if the following map is holomorphic) Consider $f: X= \mathbb{C}/ \Lambda \rightarrow X= \mathbb{C}/ \Lambda$ given by $z+ \Lambda \mapsto z^d + \Lambda.$ Is f holomorphic of degree d?
Q1: A continuous map $X\to X$ can be lifted to a map $\Bbb C\to \Bbb C$ such that $f(z+a)-f(z)\in\Lambda$ whenever $a\in\Lambda$. That's quite a restriction. In fact it follows that for fixed $a$ the element $f(z+a)-f(z)\in\Lambda$ depends continuously on $z$, hence is the same for all $z$. Thus we obtain a map $\phi\colon \Lambda\to\Lambda$. We see that $\phi(a+b)=\phi(a)+\phi(b)$, i.e., $\phi$ is additive. As a consequence, $|\phi(a)|=O(|a|)$ and so $|f(z)|=O(|z|)$. For holomorphic $f$ this implies that $f$is indeed linear.
Q2: Per Q1 this is not holomorphic. Consider linear maps $z\mapsto cz$ with suitable $c\in\Bbb C$ (a constant term does not matter). As seen above we need that $c\Lambda\subseteq \Lambda$. Verifiy that the degree of this map is $|c|^2$. How can we choose $\Lambda$ (i.e., $\lambda$) so that there is a nice $c\in\Bbb C$ with $|c|^2=d$ and $c\Lambda\in \Lambda$?