Solution of Laurent Series Challenge Problem:

Let f(z) = 7 z ²  + 9 z  - 18 /  z ³ -  9z

 

Find Laurent series for the convergence regions:

a)  0  < ÷ z ÷   <    3  and

b)    ÷ z  ÷      >   3    

 

We approach using partial fraction decomposition:



7z ²  + 9 z  - 18 /  z ³ -  9z = z (z  + 9) + 18/ [z {z + 3) (z – 3) ]

 

= A/ z  +      B/ (z + 3) + C / (z – 3)

7 z ²  + 9 z  - 18  =  A(z + 3) (z – 3) + Bz (z – 3) + Cz (z + 3)

 

For z = 0: - 9A = -18 so that A = 2

 

For z = -3:  18B = 7(-3)² + 9(-3) – 18 = 

 

63 – 27 – 18 = 18   so that B = 1

For z = 3:  18 C =  7(3)² + 9(3) – 18 =  63 + 27 – 18 = 72

 

So that C = 4

Then:

 

7 z ²  + 9 z  - 18 / [z {z + 3) (z – 3) ]

 

 

= 2/ z  + 1 / (z + 3) + 4/ (z -3)

 

 

 

Rewrite as:

 

2/ z +  1/3 (1 / 1 + z/3) – 4/3 (1 / 1 – z/3)

 

For term 1:   ÷ z ÷    <   1

 

For term 2 : ÷ z ÷    <   3
 

For term 3:    ÷ z ÷    <   3

 

Expand 2^nd and 3^rd terms and expand using 1 / (1 – z) and substituting:

 

2/ z + 1/3 (1 – z/ 3  +   z ²  / 3²  +  …) -  4/3 (1 + z/ 3 +   z ²  / 3²  +  …)

 

Combining Terms:

 

2/z – 1 – 5z/ 3²  +  3 z ²  / 3³  +  …)

 

 

Which series can be represented:

 

2/ z +  [ å^¥ _{n = 0}    (-1) ⁿ – 4   / 3 ^{n + 1}  ] z ⁴     

For: 0  < ÷ z ÷   <    3 

 

 

Now, rewrite the original partial-fraction f(z) in the form:

 

2/ z + 1/ z (1 / 1 + 3/z) + 4/z ( 1 / 1 – 3/z)

 

Expand 2^nd, 3^rd etc terms using 1/ 1 – z:

 

Þ  2/ z +  1/z (1 – 3/z  + 3²  / z ²  +   ….)  +

......+   4/z (1 + 3/z + 3²  / z ²  +  …)

 

= 2/z  + 1/z  - 3/ z ²  +  3²  / z ³  +  +…….

+ 4/z +  12/ z ²  +  36  / z ³   + …..

 

= 2/z +  [5/z + 9/ z ²  +    45  / z ³  +  ………]

 

Which can be represented in the form:

 

2/ z +   å^¥ _{n = 0}    3 ⁿ  (4 +  (-1) ⁿ  ) /  z ^{n + 1}   

 

For:  ÷ z  ÷      >   3