Friday, December 13, 2013

Solutions to Laurent Series Problems


Consider the function: f(z) = 1/ (z+ 1) ((z + 3)

 
a) Find a  Laurent series  for:   1  <  ÷ z ÷    < 3


Solution:    We first resolve the function f(z) by partial fractions, so:

 
f(z) =  1/ (z+ 1) ((z + 3) =   ½ [1/ z + 1]  -  ½ [1/ z + 3] 



÷ z ÷     > 1:  

 

 1/ 2 ÷ z +  1 ÷    =  1/ 2z (1 + 1/z)

 

=  1/ 2z [ 1 – 1/z  + 1/ z2 + …….]

 

This is the  principal part of the series.
 
 
Next, consider:  ÷ z ÷     <   3:  
 
1/ 2 (z + 3) =  1/ 6(1 + z/3) =  1/6  - z/ 18 +   z2 /54  +  ……

 
 
This is the analytic part of the series. So we just combine the two parts to get:
 
f(z) =  1/ 2z [ 1 – 1/z  + 1/ z2 + …….] +   1/6  - z/ 18 +   z2 /54  +  ……
 
 
 
b)  Laurent series for 0  <   ÷ z  + 1÷     <   2

 
 
Consider first:   ÷ z  + 1÷       >   0   

We let (z + 1)  = u  then write:

 
1/ (z+ 1) ((z + 3)  =  1/ u (u + 2) – 1/ 2u (1 + u/2)


 
= 1/ 2u (1 – u/2 +  u2/4 -    u3 /8 + ……)

 
Replace u  with z above:
 
 
1 / 2(z + 1)  - ¼    + (z – 1)/ 8  -   (z + 1) 2 / 16 + ……

 
Now take:
 
 ÷   z  + 1÷       <   2   or     ÷ u÷     <   2    (letting z + 1 = u)
 
 
 
Then for the same series above, since ÷ z  + 1÷       <   2    we require:  z  ¹  -1
 
 
WHY?
 
 


 
 
 
 

 

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