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could any one help me to prove the following? Let $f$ be meromorphic at $p$, whose Laurent series in local coordinate is $\sum_{n} c_n(z-z_0)^n$. The order of $f$ at $p$ denoted by $ord_p(f)$, is the minimum exponent actually appearing(with non zero coefficient) in the Laurent series expansion.$$ord_p(f)=min\{n:c_n\neq 0\}$$

Let $f$ be meromorphic at $p$. then $f$ is holomorphic at $p$ iff $ord_p(f)\ge 0$. In this case iff $f(p)=0$ iff $ord_p(f)>0$. $f$ has a pole at $p$ iff $ord_p(f)<0$. $f$ has neither $0$ nor a pole at $p$ iff $ord_p(f)=0$.

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These are rather straightforward results which can found in most books on Complex Analysis.

I'm going to write $m(f)$ to be $\mathrm{ord}_p(f)$.

1) $f$ is holomorphic at $p$ if and only if $m(f) \ge 0$.

This is more or less the definition of being holomorphic. If $f$ is holomorphic at $p$ then $f$ can be expanded into it's Taylor Series centered at $p$ which shows that the negative coefficients are vanishing. Conversely, if $f$ has a Laurent series expansion with vanishing negative terms, then the series is a simple power series from which we deduce that $f$ is holomorphic.

2) $f(p)=0$ if and only if $m(f) > 0$

If $m(f) < 0$ then the series isn't even convergent at $p$ (Either $f\rightarrow\infty$ in the case of a pole or $f$ experiences wild fluctuations in the case of an essential singularity (Picard/Casorati-Weirstrass Theorems)). If $m(f)=0$ then the series has a non-vanishing constant term $c_0$ from which it follows that $f(p) = c_0 \neq 0$. Finally if $m(f) > 0$ then we may write $$f(z) = (z-p)^{m(f)}g(z)$$ where $g$ is some holomorphic function (in fact we can say that $g(p)\neq 0$). It follows then that $f(p)=0$.

3) This is not quite correct as the order needs to be finite for the singularity to be classified as a pole. If $m(f) = -\infty$ then we conventionally say that $p$ is an essential singularity. So let $-\infty<m(f)<0$.

By definition we say that $f$ has an order $k$ pole at $p$ if and only if we can write $f$ as $$f(z) = (z-p)^{-k}g(z)$$ where $g(z)$ is holomorphic. We require $k$ to be the smallest integer satisfying this condition. Equivalently we can say that $f$ has a pole of order $k$ when $|m(f)| = k$. To see this note that since $g$ is holomorphic, we can expand it into a series centered at $p$. Then the extra $(z-p)^{-k}$ term produces a Laurent series with order $m(f) = -k$.

4) $f$ has neither a zero nor a pole if and only if $m(f) = 0$.

This is essentially a special case of the above parts reworded.

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