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

[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.

 

[BvU]

I have also tried combining both fractions into one fraction over z^2-1, but am not getting anywhere

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

 

[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+...

 

[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 $y = z-1$

 

[FactChecker]

##
\displaystyle {1\over z^2 - 1} = - {1\over z^2 \left ( 1 - {1\over z^2}\right ) } ##

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

 

[BvU]

Oops