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## Homework Statement

Find the Laurent expansion for

[tex]\frac{1}{z^2-1}[/tex]

in the annulus [tex]1 < |z-2| < 3[/tex]

## The Attempt at a Solution

I've gotten to the last parts but getting stuck there.

First I expanded the denominator and did a partial fraction decomposition and arrived at

[tex]\frac{1}{2}(\frac{1}{z-1} - \frac{1}{z+2})[/tex]

Then I got a series for the two terms using geometric series (disregarding the 1/2 factor for now)

[tex]\frac{1}{z-1} = \frac{1}{z-2+1} = \frac{1}{1+\frac{1}{z-2}} = \sum_{0}^{\infty}(-1)^j(z-2)^{-j}[/tex]

similarly for the second term where I get

[tex]\frac{1}{z+1} = \frac{1}{z-2+3} = \frac{1}{3} \frac{1}{1+\frac{z-2}{3}} = \sum_{0}^{\infty}(-1)^j\frac{1}{3^{j+1}}(z-2)^j[/tex]

The problem I have is I can't see how I can put them into a single sum (with the Laurent coefficient different for n<0 and n≤0. It would have been convenient if the j = 0 term in the "principal part" of the Laurent series was zero, but it's 1. Other than that 1 I have the exact answer I should have.