## Tuesday, March 9, 2021

### 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 -    u/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 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  +  …

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 / n +  ] z

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    n  (4 +  (-1) )  /  n + 1

For:  ÷ z  ÷   >   3