Find residues from different Laurent series

Question

The link below shows the working steps of finding two different Laurent series with different valid annulus region.

Finding the Laurent series of $f(z)=1/((z-1)(z-2))$

Is it possible to find residues from two different Laurent series?

Answer 1

If you want to find the residue of $f:\mathbb{C}\to\mathbb{C}$ at $z_0$ by looking at the coefficients of some expansion, you need the Laurent series of $f$ in a punctured disc $D(z_0,r)\setminus\{z_0\}$.

For example, if you consider the function $f(z)=\frac{1}{z}$, its residue at $z=1$ is $0$, not $1$.

Answer 2

I do not know if this will answer your question, but consider this if your question is about "different Laurent series for the same function".

Let $f(z)=\sum \limits_{n=-\infty}^{-1}{a_{n}}{z^n}+\sum \limits_{n=0}^{\infty}a_nz^n=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}+\sum \limits_{n=0}^{\infty}a_nz^n$

Define $g(z)$ as

\begin{align} g(z)&=\sum \limits_{n=0}^{\infty}a_nz^n, \forall z, |z|\leqslant R. \\ &=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}, \forall z, |z|>R \end{align}

Clearly $g(z)$ is bounded and analytic on both $|z|\leqslant R$ and $|z|>R$. So is bounded on entire domain.

Since $\sum \limits_{n=-\infty}^{-1}a_nz^n=\sum \limits_{n=1}^{\infty}\dfrac{a_{-n}}{z^n}$ for $|z|=R$, $g(z)$ is analytic on entire domain through analytic continuation. Thus $g(z)$ is bounded entire function, and $g(z)=c$, or $a_n=0$ for all $n$ except $n=0$.

Then if $f(z)\equiv 0$, $a_n=0,$ for all $n$, which means Laurent Series is unique.