# Complex Analysis: Laurent Expansion for $\frac{1}{\sin(z)}$ on different annuli.

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 $D'(0,\pi)=\{z:0<z<\pi \}$, we can express the Laurent expansion by:

$$\frac{1}{\sin(z)}=\frac{1}{z} \big(1-\frac{z^2}{3!}+\frac{z^4}{5!}+O(z^6) \big)^{-1}$$

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 $0<R_1<\pi<R_2<2\pi$ to allow for this, and also to find explicit expressions for $d_n$ for $n \leq{-1}$.

If anyone can offer assistance with this approach, I would be very grateful. Regards as always, MM.

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This is also exercise 8 in Chapter 7 of the book ( revised edition) – user3203476 Jul 5 '15 at 7:22

I do not exactly understand how you want to calculate $d_n$, but here is how I would do it:
Define the function $$g(z) = \frac{1}{z} - \frac{1}{z-\pi} - \frac{1}{z+\pi} = \frac{\pi^2+z^2}{z (\pi^2 -z^2)}.$$ Then $h(z) = f(z) - g(z)$ is holomorphic on $D'(0,2\pi)$ and thus can be expanded in a Taylor series. The coefficients are given by $$h_n= \frac{1}{n!}\frac{d^n h}{d z^n}(0),$$ i.e., $h_0=0$, $h_1 = (1/6-2/\pi^2)$, ...
The Laurent series of $g(z)$ can be obtained easily and reads $$g(z)= \frac1z + \frac1\pi \sum_{n=1}^\infty \left(-\frac\pi{z}\right)^n - \frac1\pi \sum_{n=1}^\infty \left(\frac\pi{z}\right)^n.$$