## Tuesday, December 24, 2013

### Laurent Series Problem Solutions

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

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.

c) If  the function is analytic in the domains:

÷ z ÷    <   1;   1   < ÷ z ÷   <    2   and:   2   < ÷ z ÷   <    ¥

Draw a sketch showing the different domains.

The sketch is shown in the accompanying graphic where:

d1 is for  ÷ z ÷    <   1;

d2 is for :   1   < ÷ z ÷   <    2

d3 is for:  2   < ÷ z ÷   <    ¥

2)Find the Laurent series that represents the function:

f(z) =  z2  sin (1 / z2 )

in the domain:  : 0   < ÷ z ÷   <    ¥

We have (from the hint):

sin(z) = z -  z3 / 3!  +  z5 / 5!  …..  -  (-1) n- 1 z 2n -1/  (2n -1)!  +

Then:

sin (1/z) =  1/ z -    1/ z3 3!   +  1/ z5 5!   -  1/ +  z7 7!  +    ……..

And:

sin (1 / z2 )  =  1/ z2    - 1/ z6 3!    +  1/ z 10  5!    - ……..

whence (on  substituting):

z2  sin (1 / z2 )  =   z2     [1/ z2    - 1/ z6 3!    +  1/ z 10  5!    - ……..

=  1    -   1/ z4 3!    +  1/ z 8  5!     -   1/ z 12  7!      +   …..

Which can be written as 1 followed by a representative series, i.e.:

1  +  å¥ n = 1     (-1) n/  (2n +  1)! z 4n

Check values for the term n = 2:

(-1) 2 =  1  and (2n +  1)!  = (2(2) + 1)! = 5!

z 4n   =  z 4(2)  =  8

So term 2 in the series is: 1/ z 8  5!

The reader can check additional terms as he desires to see the series works!