Solution to First Laurent Series Additional Problem

Let f(z) = exp (-1/z ²) /  z⁵

a) Show the Laurent series can be written: 

å^¥ _{n = 0}    (-1) ⁿ   /   n !  z ^{2n + 5}   

 

We can write:

 

exp (-1/ z ²)  = 1 -   z ^-2   - z ^{- 4}/  2!  - z ^{- 6}/  3!  +  ……..

 

So that:

 

exp (-z ²) /  z⁵     =  1/  z⁵  (1 -   z ^-2   - z ^{- 4}/  2!  - z ^{- 6}/  3!  +  ……..  )

 

=    z ^{– 5} -  z ^{– 7} -  - z ^{- 9}/  2!   - z ^{- 11}/  3!   +  …….

Or:

 

exp (-z ²) /  z⁵   =    z ^{– 5}  (å^¥ _{n = 0}    (-z ^-2   ) ⁿ   /   n !  )

 

 

=   å^¥ _{n = 0}    (-1) ⁿ   /   n !  z ^{2n + 5}   

 

The challenge problem for math mavens remains and I will provide the solution for that on Tuesday! See if you can work it out!