## 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?