I'm not familiar with complex analysis, but I need to catch up a basic fact about holomorphic and meromorphic functions.
Let $\Omega\subseteq\mathbb C$, $f:\Omega\to\mathbb C$ and $z_0\in\Omega$.
Am I correct that
- if $f$ is holomorphic at $z_0$, then there is a $\varepsilon>0$ and $(a_n)_{n\in\mathbb N_0}\subseteq\mathbb C$ with $$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\;\;\;\text{for all }z\in B_\varepsilon(z_0)\tag1;$$
- if $f$ is meromorphic at $z_0$, then there is a $\varepsilon>0$, $n_0\in\mathbb N_0$, $a_{-n_0},\ldots,a_{-1}\in\mathbb C$ and $(a_n)_{n\in\mathbb N_0}\subseteq\mathbb C$ with $$f(z)=\sum_{n=-n_0}^\infty a_n(z-z_0)^n\;\;\;\text{for all }z\in B_\varepsilon(z_0)\tag2.$$
If so, does the converse hold as well? If not, how do we need to correct the claims? In any case, I'd highly appreciate if someone could confirm whether I'm right or wrong and point me to a reference which provides rigorous proofs.
Claim 1. is correct. This is a very famous result of complex analysis, namely that holomorphic functions are analytic. Claim 2. on the other hand is a little strange. If $f$ is holomorphic, no terms of negative exponent is needed in its Laurent series. On the other hand, if it is, the function is said to have a pole at this point, which means it is not holomorphic there. More precisely:
If $f$ is holomorphic on $r < |z-z_0| < R$ and its Laurent series in this domain is given by
$$f(z)= \sum_{n=-m}^{\infty} a_n(z-z_0)^n,$$
with $m>0$, $f$ is said to have a pole of order $m$ at $z_0$.
When it comes to the converse, you are really asking if series are holomorphic, and the answer is a little subtle. Uniformly convergent series are holomorphic, but more specifically series that "look like" Laurent series, are in fact Laurent series for a holomorphic function.