Consider the Hopf curve $X=(\mathbb{C} -\{0\})/\mathbb{Z}$, where $k \in \mathbb{Z}$ acts by $z \mapsto \lambda^kz$, for $\lambda \in (0,\infty)$. Show that $X$ is isomorphic to an elliptic curve $\mathbb{C}/\Gamma$ and determine $\Gamma$ explicitly.
I'm working on this problem and cannot find any tools to solve it. What they are asking me to show is that $(\mathbb{C} -\{0\})/\mathbb{Z} \cong \mathbb{C}/\Gamma$, but I really don't know anything about $\Gamma$ other than that it is a lattice.
Is there some theorem related to these actions that will provide useful here? I did try to find something about them, but the resource did not have any results related to these.
I shall first claim that, the lattice $\Lambda$ is generated by $\log\lambda$ and $2\pi \sqrt{-1}$.
Recall that, the logarithm function $\log$ is only well-defined on the Riemann surface $\Sigma_{\log}$, which is an infinite cover of $\mathbb{C}^*$: $$\pi:\Sigma_{\log}\longrightarrow\mathbb{C}^*$$ Next, denoted by $[z]_h$ the point on the Hopf curve $\mathbb{C}^*/\mathbb{Z}$, and $[z]_{\Lambda}$ the point on the torus $\mathbb{C}/\Lambda$, where $$\Lambda=\mathbb{Z}\langle\log\lambda\rangle\oplus\mathbb{Z}\left\langle2\pi\sqrt{-1}\right\rangle$$ define the map to be $$\begin{aligned}\phi:\mathbb{C}^*/\mathbb{Z}&\longrightarrow\mathbb{C}/\Lambda\\ [z]_h&\mapsto\left[\log\left(\pi^{-1}(z)\right)\right]_{\Lambda}\end{aligned}$$
Then I claim this $\phi$ is a well-defined biholomorphism.
Indeed, the choice of preimage $\pi^{-1}(z)\in\Sigma_{\log}$ will not impact on the value of $\phi$, since for two different preimages $w_1,w_2\in\Sigma_{\log}$, we have $$\log w_1-\log w_2=2m\pi\sqrt{-1}$$ for some $m\in\mathbb{Z}$, by definition, they define the same equivalent class in $\mathbb{C}/\Lambda$.
For two different $z,z'\in[z]_h$, there exists some $k\in\mathbb{Z}$ such that $z=\lambda^kz'$, hence we have $$\phi([z]_h)=\left[k\log\lambda+\log \pi^{-1}(z')\right]_{\Lambda}=\phi([z']_h)$$
thus it is well-defined.
It is not hard to check $\phi$ is surjective and injective.