The quotient of $\mathbb{C}$ under the cyclic group is the punctured plane?

Question

For $G$ is a group generated by a translation, that is, $$G=\langle f \rangle$$ where $f(z)=z+a$ and $a\in \mathbb{C}$.

Then consider the quotient

$$X= \mathbb{C}/G.$$

According to the uniformization theorem of Riemann surfaces, it should be biholomorphic to the punctured plane $\mathbb{C}^{\ast}$ (because the universal covering is $\mathbb{C}$ and the deck transformation group is G).

But how to prove this quotient space is biholomorphic to the punctured plane?

Answer 1

Hint: manipulate the exponential mapping $\exp:\mathbb{C}/2\pi i\mathbb{Z}\to\mathbb{C}^{\times}$.

Answer 2

Consider the map $\exp: \mathbb C\to\mathbb C^\ast$ given by $z\mapsto \exp(\frac{2\pi i z}{a})$. This is a covering map with deck transformation group isomorphic to $G=\langle z+a \rangle_\mathbb Z$. By the general holomorphic covering space theory, the base space $\mathbb C^\ast$ is biholomorphic to the total space modulo the deck transformation group, which is $\mathbb C/ G$.