I have trouble understanding why the proof of the absolute and uniform convergence of Laurent series isn't just an easy corollary of the absolute and uniform convergence of power series. In my teacher's notes the proof is long and done by calculating some superior limits.
More specifically the result I mean is this (I use $0$ as base point to simplify):
Let $f$ be an holomorphic function in the annulus $r<|z|<R$. We have already shown in class that there exist some $a_n\in \mathbb{C}$ such that $$f(z)=\sum_{n=-\infty}^{+\infty}a_nz^n,\ \forall\ r<|z|<R,$$ where $$a_n=\frac{1}{2\pi i}\int_{|w|=\rho}\frac{f(w)}{w^{n+1}}dw,$$ for every $r<\rho<R$. Now, I need to prove that the convergence of the above series is absolute in the annulus $r<|z|<R$ and uniform in the compact sets within the annulus.
Here's my attempt:
The regular part is just a power series, so by the standard result on power series, it converges absolutely at least in the disk $|z|<R$ and uniformly in the compact sets within the disk.
On the other hand, the singular part is $$\sum_{n=1}\frac{a_{-n}}{z^n}.$$ Since the singular part converges for $r<|z|<R$, the following series $$\sum_{n=1}a_{-n}\xi^n,$$ converges when $1/R<|\xi|<1/r$. So again, by applying the power series result, the series above converge absolutely in the disk $|\xi|<1/r$ and uniformly in its compact sets. Therefore, the singular part converges absolutely when $|z|>r$ and in the compact sets within the region $|z|>r$.
Is my approach correct?