## Tuesday, December 31, 2013

### 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 2nd and 3rd 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 2nd, 3rd 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