Suppose $f$ is analytic on the annulus $\{z : 1/2 < |z|<2\}$ except for the simple pole at $1$. Suppose that the residue of $f$ at $z=1$ is $1$. Let $\sum a_n z^n$ and $\sum b_n z^n$ be the Laurent expansion of $f$ on the annuli $\{ z: 1/2<|z|<1\}$ and $\{z: 1<|z|<2\}$ respectively. Compute $b_n-a_n$ for every integer $n$.
Here is my thought: I know $$ a_n = \frac{1}{2\pi i}\int_{|z|=3/4}\frac{f(z)}{z^{n+1}}dz $$ $$ b_n = \frac{1}{2\pi i}\int_{|z|=3/2}\frac{f(z)}{z^{n+1}}dz $$ Hence $b_n - a_n$ should be the residue of $\dfrac{f(z)}{z^{n+1}}$ at $z=1$, which is $1$. Hence $b_n - a_n=1$ for all $n$.
Is this argument correct?
The computation is correct, with the justification for residue as follows:
Since $f(z)$ has a simple pole at $z=1$, $\dfrac{1}{z^{n+1}}$ is holomorphic around $z=1$, using the limit formula $Res(\frac{f(z)}{z^{n+1}})=\lim_{z \to 1} \dfrac{f(z)(z-1)}{z^{n+1}}=\dfrac{\lim_{z \to 1} f(z)(z-1)}{\lim_{z \to 1} z^{n+1}} = \dfrac{Res(f,1)}{1^{n+1}} = 1$.