Find the laurents series expantion for the function f(z)=z²/(z-1)(z+2)² in the region 1 < |z -1 | < 3
Answers
Most problems like this can be done with two tools only: partial fractions, and the result 1/(1−w)=1+w+w2+⋯ for |w|<1. So first split your function f into 1/(z−2)−1/(z−1). I will show you how to cope with one of these factors, 1/(z−2). Writing this as −1211−z/2 is tempting but no good: the "1/(1−w)" expansion will converge only for |z/2|<1, not on the regions you are supposed to care about. So you take out a factor of z−1 instead:
1/(z−2)=z−111−2/z
The final term can be expanded with the 1/(1−w) series, valid for |2/z|<1 that is |z|>2. So that does give you a Laurent series valid in the right region (once you multiply bu z−1. These methods can be used to solve all of your problems.
Answer:
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