= Jump of Laurent coefficients across poles
{title2=$b_n-a_n=\sum_\zeta\operatorname{Res}_{z=\zeta}\frac{f(z)}{(z-a)^{n+1}}$}
Suppose a <meromorphic function> is <holomorphic> on two concentric <annuli> about $a$, with only finitely many <poles> between chosen inner and outer coefficient circles. Let $a_n,b_n$ denote the corresponding <Laurent coefficients>. Subtract their <contour integral> formulas and apply the <residue theorem> to the intervening annular region. It gives
$$
b_n-a_n=\sum_\zeta\operatorname{Res}_{z=\zeta}\frac{f(z)}{(z-a)^{n+1}}.
$$
The sum is over the <poles> crossed, including their multiplicities through the <residue>. This identity applies to every integer index. Thus changing the <annulus> can change even the coefficients of nonnegative powers; it is not merely a change in the <principal part>.
For example, $f(z)=1/\sin z$ has <simple poles> at $\pm\pi$, each with <residue> $-1$. Moving a coefficient circle centered at zero from radius below $\pi$ to radius between $\pi$ and $2\pi$ changes its $n$th coefficient by $-\pi^{-n-1}-(-\pi)^{-n-1}$. Even indices remain unchanged, while odd indices acquire a nonzero correction.
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