Riemann Roch Meromorphic section on a line bundle. Let $g:\mathbb{C}\times \mathbb{C^*}\rightarrow \mathbb{C}\times\mathbb{C^*}$ defined by $g(z,w)=(w^n z,\alpha w)$ 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 the following result : Any meromorphic map with the condition $\forall w\in \mathbb C^*,f(\alpha w)=w^n f(w)$ defines a meromoprhic section on the bundle $L:=\mathbb C \times\mathbb C^*/G$ (L is a bundle over the elliptic curve $E:=\mathbb C^*/A$).
1) I have to verify that the function $f$ defined by : 
$f(z,w)=z.\prod_{k\in\mathbb{Z}}(\frac{w}{w-\alpha^{k}\zeta})^n,    \forall \zeta$ fixed in $\mathbb C$,$\forall n$
is a section of $L$ over $E$. 
Do i have to check that $f(z,\alpha w)=w^n f(z,w)$? Or use the fact that the points $(z,w)$ and $g(z,w)$ are identified in $L$ ? I have some difficulties here.
2) After, how can i deduce the degree of $L$?
Thank you.
 A: I'm going to skip the nonsensical first question and answer the second.
Consider a nonzero meromorphic global section $f$ (i.e. a meromorphic function $f : \Bbb C^* \to \Bbb P^1(\Bbb C)$ such that $f(\alpha.w) = w^nf(w)$).
We pick a fundamental domain $\Omega$ for $\Bbb C^*/A$, for example the region $|\alpha| < |w| < 1$. Then the degree of $f$ is $\deg(f) = \frac 1 {2i\pi} \int_{\partial  \Omega} \frac {f'(w)}{f (w)}dw$ .
Since the boundary of $\Omega$ is two circles, $2i\pi \deg(f) = \int_{|w|=1} \frac{f'(w)}{f(w)}dw - \int_{|w|=|\alpha|} \frac{f'(w)}{f(w)}dw $, where each circle is oriented in the standard way.
Using a change of variables, $\int_{|w|=|\alpha|} \frac{f'(w)}{f(w)}dw = \int_{|w|=1} \frac{f'(\alpha w)}{f(\alpha w)}d(\alpha w) = \int_{|w|=1} \frac{\alpha f'(\alpha w)}{f(\alpha w)}dw$.
Now, since $f(\alpha w) = w^n f(w)$ we have $\alpha f'(\alpha w) = nw^{n-1}f(w) + w^nf'(w)$.
Hence  $\int_{|w|=|\alpha|} \frac{f'(w)}{f(w)}dw = \int_{|w|=1}( \frac nw + \frac {f'(w)}{f(w)}) dw$.
we get $2i\pi \deg(f) = - \int_{|w|=1} \frac nw dw = -2i\pi n$, and so $\deg(f) = \deg(L) = -n$.
In particular, this tells us that $f$ has to have at least $n$ poles in $\Omega$, and so there is no holomorphic global section of $L$
