Understanding what is the order of a meromorphic function I am reading a book on Complex Curves by Miranda. The book defined meromorphic function:
A function on an open set W of a Riemann surface X is said to be meromorphic at $p \in W$ if there exists a chart $\phi: U \rightarrow V$ where $p \in U$ such that $f \circ \phi^{-1}$ is either holomorphic or has removable singularity at $\phi(p)$ or a pole at $\phi(p)$.
Then we defined the order of a meromorphic function f at p to be the minimal power of the Laurent series ie $ord_p(f)=\min\{n: c_n\}$ for the $\sum c_n(z-z_0)^n$.
So it was noted that this is independent of the chart. I am confused on what this order is. For example if we work with the Riemann Sphere and we have a $ord_p(f)=3$ what does this mean for f?
I guess from what I have read, the $ord_p(f)$ is the order of the pole if it negative value. Is this what I should understand from the order function?
 A: Question
"For example if we work with the Riemann Sphere and we have a $ord_p(f)=3$ what does this mean for f?"
Answer
$\bullet$ If $p\in \mathbb C$ it means that $f(z)=(z-p)^3\frac {u(z)}{v(z)}$ 
where $u(z),v(z)\in \mathbb C[z]$ are polynomials satisfying $u(p)\neq 0,v(p)\neq0$.
For example  if $p=2$ we might have $f(z)=(z-2)^3\frac {-z^8}{z^4-15}$.
Beware however that the order of $g(z)=(z-2)^3\frac {-z^8}{z^4-16}$ is $ord_2(g)=2$, not $3$ .
$\bullet$$\bullet$  If $p=\infty$ it means that $f(z)=\frac {u(z)}{v(z)}$ 
where $u(z),v(z)\in \mathbb C[z]$ are polynomials satisfying $u\neq 0,v\neq0$ and $\operatorname {degree} v(z)= \operatorname {degree} u(z)+3$.
For example $f(z)=\frac {-iz^2-3z+17}{(1+i)z^5-2z^4+\sqrt {\pi}}$.    
Remark
Note the pleasant fact that in both $\bullet$  and $\bullet$$\bullet$  fractions are not required to be in lowest terms, i.e. $u,v$ are not required to be relatively prime.
For example $ord_2 (z-2)^3 \frac {(z^4+5)^8}{(z^4+5)^4}=3$         and $ord_\infty (\frac {z^{12}}{z^{15}})=3$
