Show that a space of holomorphic sections of a line bundle is isomorphic to the space of meromorphic functions on a Riemann surface. Let $\Sigma$ be a Riemann surface, $x \in \Sigma$ and let $z: U \rightarrow D \subset \mathbb{C}$ be a local coordinate system centered at $x$. For every $k \in \mathbb{Z}$, a holomorphic line bundle $L_{kx}$ on $\Sigma$ is defined by gluing the trivial vector bundles $(\Sigma \backslash \{x\}) \times \mathbb{C} \rightarrow \Sigma$ and $U \times \mathbb{C} \rightarrow U$ via the holomorphic transition function 
$$(U \backslash \{x\}) \times \mathbb{C} \rightarrow (U \backslash \{x\}) \times \mathbb{C}, \ (p, v) \mapsto (p, z(p)^k v)$$
Can someone give me some hints how to prove the following claims ? 

The space of global holomorphic sections of $L_{kx}$ is isomorphic to the space of meromorphic functions on $\Sigma$, holomorphic outside of $x$ and has a pole of order at most $k$ if $k \geq 0$ and a zero of order $k$ if $k < 0$. 

Thanks for your help. 
 A: $\newcommand{\C}{\mathbb C}\newcommand{\CP}{\mathbb C\mathbb P}\newcommand{\sm}{\setminus}\newcommand{\ord}{\operatorname{ord}}$A holomorphic section of $L_{kx}$ is given by a pair of functions $s_0:\Sigma\setminus\{x\}\to\C$ and $s_1:U\to\C$ that are compatible with the transition function. Here, being compatible with the transition function means that, for $p\in U\sm\{x\}$, $z(p)^ks_0(p)=s_1(p)$.
So, given a section $(s_0,s_1)$, we can define a meromorphic function $s:\Sigma\to\C$ given by
$$s(p)=\begin{cases}s_0(p)&\text{if }p\neq x\\s_1(p)z(p)^{-k}&\text{if }p=x\end{cases}$$
which is visibly holomorphic outside of $x$. $z(x)=0$ and $s_1$ is holomorphic at $x$ (possibly vanishing there). If $k\ge0$, then $s$ has a pole or order $\le k$ at $x$, and if $k<0$, then $s$ has a zero of order $\ge|k|$ at $x$. In either case, we can say that $\ord_x(s)\ge-k$.
Conversely, given a meromorphic function $s:\Sigma\to\C$ which is holomorphic on $\Sigma\sm\{x\}$, but with $\ord_x(s)\ge-k$, we can construct a section $(s_0,s_1)$ of $L_{kx}$ by setting $s_0(p)=s(p)$ and $s_1(p)=s(p)z(p)^k$. Note that 
$$\ord_{s_1}(x)=\ord_s(x)+k\ord_z(x)=\ord_s(x)+k\ge0,$$
so $s_1$ is holomorphic, and $(s_0,s_1)$ really gives a holomorphic section.
These two constructions are clearly inverse to one another (and linear), so we have are desired isomorphism.
