I've been working through Hilary Priestley's Book Complex Analysis (fantastic read) and have reached her discussion of the Laurent Expansion for holomorphic functions. Considering the function
$f(z)=\frac{1}{\sin(z)}$
I'm trying to establish how the Laurent series changes when we define the expansion on different annuli. For example on the punctured disc D'(0,\pi)=\{z:0
$\frac{1}{\sin(z)}=\frac{1}{z} \big(1-\frac{z^2}{3!}+\frac{z^4}{5!}+O(z^6) \big)^{-1}$
which after some manipulation we can express as
$\frac{1}{\sin(z)}=\frac{1}{z} \big(1+\frac{z^2}{6}+\frac{7z^4}{360}+O(z^6) \big)$
I now look to consider the Laurent expansion of $f$ on $D(\pi,2\pi)$ in the form
$\sum^{\infty}_{n=-\infty}d_nz^n$
If we express the expansion on D'(o,\pi) in the form
$\frac{c_{-1}}{z}+\sum^{\infty}_{n=0}c_nz^n$ as we have just demonstrated is possible, I want to express the coefficient $d_n$ in terms of $c_n$. It seems the best approach to this would be to consider expressions for $c_n$ and $d_n$ as contour integrals around circles with radii satisfying $0
If anyone can offer assistance with this approach, I would be very grateful. Regards as always, MM.