# Complex Analysis. Laurent Series Expansion in region(22C).

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1. Mar 31, 2017

### Kemba Huskie

<Moderator's note: moved from a technical forum, so homework template missing>

Hi. I have solved the others but I am really struggling on 22c. I need it to converge for |z|>2. This is the part I am really struggling with. I am trying to get both fractions into a geometric series with 1/(1-(1/(z/2)) so that it converges for 1/(z/2) <1; which becomes, (z>2). I however cannot manipulate either series into this form that I am looking for. I have tried many times. I have also tried combining both fractions into one fraction over z^2-1, but am not getting anywhere. Any pointers, or tips so that I can get it to converge for |z|>2 by manipulating the fractions (just simply cannot get in right form). Thank you.

Last edited by a moderator: Apr 3, 2017
2. Apr 1, 2017

### BvU

Worries me. You get $\displaystyle {2\over z^2 - 1}$, right ?
Perhaps a simple transform to $y = z-1$ can shed some light ?

3. Apr 1, 2017

### FactChecker

Two approaches:
1) You can do long division of polynomials the same way the you do it with integers.
2) Use 1/(1-r) = 1+r+r2+r3+... and 1/(1+r) = 1-r+r2-r3+...

4. Apr 1, 2017

### BvU

Yet another ? From some searching in PF I really like the suggestion $\displaystyle {1\over z^2 - 1} = - {1\over z^2 \left ( 1 - {1\over z^2}\right ) }$

And please forget about my $y = z-1$

5. Apr 1, 2017

### FactChecker

Ha! Of course! That is the easiest way. (but the leading minus sign on the right is not needed)

6. Apr 1, 2017

Oops