# Laurent expansions with only one factor in the denominator of the function

1. Jul 28, 2009

### caramello

Hi,

I know how to do the Laurent expansions for a function that has two or more factors in the denominator (for eg: 1/(z-1)(z-2) ). But what if that there is only 1 factor in the denominator, like (z-1)/(z+1)?

I know how to do the computations for both of the cases, however, the thing that I'm confused with is that of the domains of each case:

for the first case (the example): I know that the domains will be |z|<1, 1<|z|<2, and |z|>2

However, what's for the second case then? This is because when I try to find the singularities, it will result in |z|<-1 --> and this is not possible because absolute values won't be less than 0, i.e absolute values can't be negative.
So then will the domain be |z|<1? or |z|>1? because the computational part to find the expansions for |z|<1 will be different from |z|>1, am I right?

Thank you! :)

2. Jul 28, 2009

### HallsofIvy

Laurent expansion about what point? The Laurent expansion of (z-1)/(z+1) about any point other than z= -1 is just its Taylor series about that point. The Laurent series about z= -1 is trivial! (z-1)/(z+1)= (z+1- 2)/(z+1)= (z+1)/(z+1)- 2/(z+1)= 1- 2/(z+1). That is the Laurent series.