Question

expand the following f(z) in a Laurent series expansion valid for |z-1|>1

Answer 1

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

Answer 2

oh kay i think i know where i went wrong.

Answer 3

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\[=\frac{ 1 }{ z-1 }-\sum_{n=0}^{\infty}2(\frac{ 1 }{ z-1 }) ^{n+1}\]

Answer 4

start with partial fractions i think

Answer 5

you get
\[\frac{1}{z-1}-\frac{2}{z-2}\]

Answer 6

yep

Answer 7

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

Answer 8

ouh substitution after the partial fraction works.

Answer 9

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

Answer 10

yea

Answer 11

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

Answer 12

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

Answer 13

yes, i believe so

Answer 14

oh kay thx alot =)

Answer 15

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

Answer 16

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