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

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