p57 and on when they start expanding around z0. This is the pole, the roots of the denominator. You can use the integral or algebraic tricks to get the Laurent series.

]]>They do mention expanding around a singularity.

]]>tinyurl. com /bqwts5a

(remove spaces)

and go to Part 6. In there is the exercise I am working on. Unfortunately they have not provided solutions to Parts 5-7, only 1-4.

]]>I just do not see any negative powers in your series. Why is it a Laurent series?

]]>The poles of that are -1 and 3.

Trouble is, when I expand around either one I do not get your principal part.

]]>valid in the annuli;

(a) 0 ≤ |z| < 1,

(b) 1 < |z| < 3,

(c) 3 < |z|.

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I've found a Laurent expansion but I'm not sure what to do about the different annulus ranges.

which forms the geometric series

or

which I think simplifies to

but I don't know how to address the problem of the annulus ranges given for (a), (b) and (c). I think mine is valid for the range in (a), but I don't know what to do about the others.

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