# Comparing Laurent series coefficients on different annuli

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 residue of $f$ at $1$ is $1$. For $n \neq -1$, why do you expect the residue of $\frac{f(z)}{z^{n+1}}$ at $1$ to be $1$? – Daniel Fischer Aug 9 '17 at 21:28
• You could say $f(z) = \frac 1{z-1}$ and check it. – Doug M Aug 9 '17 at 21:31
• @Doug M For $f(z)=\frac{1}{z-1}$, on the small annulus, $f(z) = -1-z-z^2-\cdots$; on the bigger annulus, $f(z) = z^{-1} + z^{-2}+\cdots$, so it looks like the difference is indeed $1$. Is it correct? – Yuxin Wang Aug 9 '17 at 21:39
• Yes, that is correct. Also in the more general situation in your question. But not in even more general situations. What is it that gives you $$\operatorname{Res}\biggl( \frac{f(z)}{z^{n+1}}; 1\biggr) = 1$$ for all $n\in \mathbb{Z}$ here? – Daniel Fischer Aug 9 '17 at 21:41
• @Daniel Fischer Since $f(z)$ has a simple pole at $z=1$, $\frac{1}{z^{n+1}}$ is holomorphic around $z=1$, using the limit formula $Res(\frac{f(z)}{z^{n+1}})=\lim_{z \to 1} \frac{f(z)(z-1)}{z^{n+1}}=\frac{\lim_{z \to 1} f(z)(z-1)}{\lim_{z \to 1} z^{n+1}} = \frac{Res(f,1)}{1^{n+1}} = 1$. – Yuxin Wang Aug 9 '17 at 21:42

Since $f(z)$ has a simple pole at $z=1$, $\dfrac{1}{z^{n+1}}$ is holomorphic around $z=1$, using the limit formula $Res(\frac{f(z)}{z^{n+1}})=\lim_{z \to 1} \dfrac{f(z)(z-1)}{z^{n+1}}=\dfrac{\lim_{z \to 1} f(z)(z-1)}{\lim_{z \to 1} z^{n+1}} = \dfrac{Res(f,1)}{1^{n+1}} = 1$.