Independence of coordinate charts in the definition of the order of a pole of a meromorphic 1-form on a Riemann surface

I am currently reading Forster. Let $X$ be a Riemann suface, and $Y$ an open subset of $X$. Assume we have a meromorphic 1-form $\omega \in \Omega(Y \setminus \{a\})$ with a pole at $a \in Y$. One defines the order of $\omega$ at $a$ as the order of $f$ at $a$, where $\left. \omega \right|_{Y \cap U} = f \, \mathrm{d}z$ in some chart $(U,z)$ around $a$.

How exactly can I prove that this definition is independent on the choice of chart? My strategy was to try to show that different charts give rise to functions that differ by a non-zero constant, so that the respective power series have the same order. However, having done very little differential geometry, I'm not very confident working with the technicalities involved.

Thanks.

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Let $z : U_1 \subset X \to V_1 \subset \mathbb{C}$, $z': U_2 \subset X \to V_2 \subset \mathbb{C}$ be two coordinate systems involved, and $\phi :z(U_1 \cap U_2) \to z'(U_2 \cap U_1)$ sends the coordinate system of $z'$ to that of $z$. Here $\phi$ is a biholomorphic map, i.e. $\frac{d\phi}{dz'}$ is nonvanishing. Then you can check that $$fdz = (f\circ \phi) \frac{d\phi}{dz'}dz'$$ So the coefficient is off by $\frac{d\phi}{dz'}$, a nonvanishing holomorphic function.
I feel really dense for not getting this, but I can't wrap my head around some parts of your answer. I understand how $\frac{d\phi}{dz'}$ makes sense, as $\phi=z\circ z'^{-1}:z'(U \cap U')\rightarrow z(U \cap U')$ is a biholomorphic map of a complex variable, but how does $\frac{dz}{dz'} = \frac{d\phi}{dz'}$? In fact I'm not really sure what $\frac{dz}{dz'}$ actually represents. – Alex Provost Jan 25 '13 at 23:00
@Silencer, sorry for my unclear notation. $\frac{dz}{dz'}$ is by definition $\frac{d\phi}{dz'}$ here. I have edited the answer to replace all of them by $\frac{d\phi}{dz'}$. – user27126 Jan 26 '13 at 2:45