Riemann surfaces with Riemann Roch theorem, linear fiber over an elliptic curve Let $g:\mathbb{C}\times \mathbb{C^*}\rightarrow \mathbb{C}\times\mathbb{C^*}$ defined by $g(z,w)=(w^n z,\alpha z)$ where $0<|\alpha|<1$. Let $G$ be the cyclic group spanned by $g$ and $A$ the group spanned by $w\mapsto \alpha w$. I have to show,by considering a quotient, that we obtain a linear fiber over an elliptic curve (i.e. a torus).
I have no idea how to begin with that question. 
Thank you for any help. 
 A: $\newcommand{\Cpx}{\mathbf{C}}$Here's a fairly detailed sketch of the underlying framework, together with hints for applying the machinery to your situation.

Generalities: Let $X$ be a connected holomorphic manifold, $\phi:X \to X$ a biholomorphism, and $A$ the cyclic group generated by $\phi$. To get a manifold quotient, we'll assume $A$ acts properly discontinuously, i.e., each point $w$ in $X$ has a neighborhood $U$ whose images under $A$ are pairwise disjoint.
The set of $A$-orbits is the quotient space $Y = X/A$. There is a natural map $\pi:X \to Y$ defined by $\pi(w) = [w]$, the orbit of $w$ under $A$. If $U$ is a neighborhood of $w$ whose images under $A$ are disjoint, then the image of $U$ in $Y$ is a neighborhood of $[w]$ in the quotient topology, and it's straightforward to check that $Y$ is a holomorphic manifold and $\pi$ is a covering map with deck transformation group $A$.
A fundamental domain for the action of $A$ is a region $D$ in $X$ such that $\phi(D) \cap D = \emptyset$ and the images of the closure $\overline{D}$ under $A$ cover $X$. We may think of $Y$ as obtained from $\overline{D}$ by "gluing" boundary points identified by $\phi$.
Next, let $p:L \to X$ be a holomorphic line bundle over $X$, and assume we are given a biholomorphism $g:L \to L$ of the total space that maps the fibre $L_{w}$ over $w$ by a linear transformation (i.e., via multiplication by a non-zero scalar) to $L_{\phi(w)}$. Let $G$ denote the group of biholomorphisms of $L$ generated by $g$. In this situation, there is an induced line bundle over $Y$ with total space $L/G$. (If $U$ is a trivializing neighborhood for $L$ whose images under $A$ are pairwise disjoint, then the image of $U$ in $Y$ is a trivializing neighborhood for $L/G$, since the quotient map restricted to $U$ is a biholomorphism. In writing up a solution, you'll probably want/need to flesh out the details.)

Particulars: Fix a complex number $\alpha$ with $0 < |\alpha| < 1$ and an integer $n$. Let $X = \Cpx^{*}$, $\phi(w) = \alpha w$, and $A$ the infinite cyclic group of biholomorphisms generated by $\phi$. (The "$(\mathbf{Z}, +)$-action" in my comment merely referred to an action by an infinite cyclic group.)
We first seek a fundamental domain for the action of $A$.
One natural approach is to observe that circles centered at $0$ map to circles centered at $0$ under $\phi$ (why?). We might therefore seek a fundamental domain $D$ consisting of a union of circles. (Suggestion: Look at images of the unit circle under $\phi$, and try to find a region $D$ bounded by such circles so that the images of $\overline{D}$ cover all of $\Cpx^{*}$. Then investigate how $\phi$ identifies boundary points of $\overline{D}$, with the aim of convincing yourself the quotient space is a torus.)
Now let $L = \Cpx \times \Cpx^{*}$ be the trivial line bundle over $X$ via $p(z, w) = w$.
The mapping $g(z, w) = (w^{n}z, \alpha w)$ is a biholomorphism of the total space of $L$ that covers $\phi$ and is linear on fibres. (All three claims require a bit of justification, but should be easy.) For the general reasons outlined above, the quotient of $\Cpx \times \Cpx^{*}$ by the infinite cyclic group $G$ generated by $g$ is a holomorphic line bundle over the torus $\Cpx^{*}/A$.
Incidentally, identifying $(z, w)$ with $(w^{n}z, \alpha w)$ "twists" the quotient line bundle: The fibre coordinate $z$ over the unit circle $\{|w| = 1\}$ in the quotient $\Cpx^{*}/A$ gets identified with the fibre coordinate $w^{n}z$ over the (same) circle $\{|w| = |\alpha|\}$. Roughly, the fibre coordinate of $L/G$ with respect to one trivialization "rotates" $n$ times with respect to a second trivialization.
