Suppose $f(z)$ is analytic on $0<|z-z_0|<R$. And we find a Laurent's series for $f(z)$ on annulus $r<|z-z_0|<R$ where $r$ may not be $0$. Then it is said that $a_{-1}$ of such Laurent's series may not be residue unless $r=0$ (Residue is defined as $Res(f,z_0)=\frac{1}{2 \pi i}\int_\gamma f(z)dz$ for any enclosed curve $\gamma$ on $0<|z-z_0|<R$). I find this hard to understand. In particular, fix an enclosed curve $\gamma'$ contained in $r<|z-z_0|<R$ where the Laurent's series applies. Substitute $f$ with this Laurent series into $Res(f,z_0)=\frac{1}{2 \pi i}\int_{\gamma'} f(z)dz$. Then integral of all except the $a_{-1}/(z-z_0)$ terms should evaluate to $0$ as we have finished substitution and are just evaluating the integral with Cauchy formula (and thus are no longer concerned by where Laurent's series apply). In the end since $Res(f,z_0)=\frac{1}{2 \pi i}\int_\gamma f(z)dz$ takes same value for all $\gamma$ in $0<|z-z_0|<R$, our result based on $\gamma'$ applies in general. Where is the mistake in this proof?
Note 1. I have checked explicit expression for $a_{-1}$ in a Laurent expansion where $r>0$. I think $a_{-1}$ should be residue. The statement in Note 2 may be false. I hope the author or somebody with expertise in complex analysis could confirm.
Note 2. The original statement I was referring to can be found in Simon's answer in this post: Calculate residue at essential singularity:
"In fact the residue of $f(z)$ at an isolated singularity $z_0$ of $f$ is defined as the coefficient of the $(z-z_0)^{-1}$ term in the Laurent Series expansion of $f(z)$ in an annulus of the form $0 < |z-z_0|<R$ for some $R > 0$ or $R = \infty$.
If you have another Laurent Series for $f(z)$ which is valid in an annulus $r < |z-z_0|< R$ where $r > 0$, then it might differ from the first Laurent Series, and in particular the coefficient of $(z-z_0)^{-1}$ might be different, and hence not equal to the residue of $f(z)$ at $z_0$."
