## edr1c 3 years ago expand the following f(z) in a Laurent series expansion valid for |z-1|>1

1. edr1c

$f(z)=\frac{ z }{ (z-1)(2-z) }$

2. edr1c

oh kay i think i know where i went wrong.

3. edr1c

let u=z-1 so that z=u+1 $\frac{ z }{ (z-1)(2-z) }=\frac{ u+1 }{ u(1-u) }=\frac{ 1 }{ u }+\frac{ 2 }{ 1-u }$$=\frac{ 1 }{ u }+2\sum_{n=0}^{\infty}u ^{n}$$=\frac{ 1 }{ z-1 }+2\sum_{n=0}^{\infty}(z-1) ^{n}$ did i make any error or its not simplified yet? the answer provided is$=\frac{ 1 }{ z-1 }-\sum_{n=0}^{\infty}2(\frac{ 1 }{ z-1 }) ^{n+1}$

4. satellite73

5. satellite73

you get $\frac{1}{z-1}-\frac{2}{z-2}$

6. edr1c

yep

7. satellite73

oops now i got stuck, i am not getting the right thing

8. edr1c

ouh substitution after the partial fraction works.

9. satellite73

second part is $$\frac{2}{1+(1-z)}$$ right?

10. edr1c

yea

11. satellite73

so i am confused as to why this doesn't alternate, but if you got it, then i guess i have something wrong probably in my mind

12. edr1c

for the region of validity, like in this case |z-1|>1, after we let u=z-1, any series we obtain using that substitution will always be in that region? my english aint good enough lol

13. satellite73

yes, i believe so

14. edr1c

oh kay thx alot =)

15. satellite73

oh wait i am confusing myself. it is $$|z-1|>1$$ not less than that is why you need to flip it

16. edr1c

so if i turn the partial fraction into $\frac{ 1 }{ u }-\frac{ 2 }{ u }(\frac{ 1 }{ 1-\frac{ 1 }{ u } })$should be ok right?