Question

Expand: f(z) = z / (z-1) (2-z) as Laurent series for |z| < 1​

Answer 1

Partial fractions:

`f(z) = \frac z{(z-1)(2-z)} = -\frac1{1-z} + \frac2{2-z}`

For |z| < 1, we have

`\displaystyle \frac1{1-z} = \sum_(n=0)^\infty z^n`

and

`\displaystyle \frac2{2-z} = \frac1{1-\frac z2} = \sum_(n=0)^\infty \left(\frac z2\right)^n`

(The latter series is valid for |z/2| < 1 or |z| < 2, but |z| < 1 is a subset of this region.)

Then

`\boxed{f(z) = \displaystyle \sum_(n=0)^\infty \left(\frac1{2^n} - 1\right) z^n}`

Answer 2

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.