Solutions To Laurent Series Problems

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

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

b) Find a Laurent series for    0 <  ÷ z +  1 ÷  < 2   

Solutions:    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)  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?

2)  The function: f(z) =  -1/ (z – 1) (z – 2)

 

a) Rewrite it using the partial fraction form

b) Identify the two singular points.

Solutions:

a)Rewrite as partial fraction: 

f(z) = 1/ (z - 1) – 1/( z – 2)

b)     Singular points are z = 1 and z = 2 since by inserting the respective values in the partial fraction form one will obtain infinities.

3) 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    

Solutions:

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

= 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  +  …

Then:

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

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