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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


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


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
up vote 5 down vote accepted

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 $$g(z)= \frac1z + \frac1\pi \sum_{n=1}^\infty \left(-\frac\pi{z}\right)^n - \frac1\pi \sum_{n=1}^\infty \left(\frac\pi{z}\right)^n.$$

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