Getting Residues Complex Analysis

  • #1
Hi everyone!

I still didn't fully understand how to get the residues of a complex function. For example the function [tex]f(z)=\frac{1}{(z^{2}-1)^{2}}[/tex] in the region [tex]0<|z-1|<2[/tex] has a pole of order 2. So the residue of f(z) in 1 should be given by the limit:

[tex]\lim_{z \to 1}(z-1)^{2}f(z)=1/4 [/tex]​

But when I get the Laurent serie:

[tex] \sum_{n=1}^{+\infty} (-1)^{n+1}n\frac{(z-1)^{n-3}}{2^{n+1}}[/tex]​

i don't know how to get the residue directly from it.
 

Answers and Replies

  • #2
351
1
Hi.

The formula for the residue of a function at [itex]z_0[/itex], a pole of order [itex]n[/itex] is:

[tex] res(f;z_0) = \lim_{z \rightarrow z_0} \frac{1}{(n-1)!} \frac{d^{n-1}}{dz^{n-1}} (z-z_0)^n f(z) [/tex]

Here the factorial term is just 1! = 1. So what we need is the derivative of [itex](z-z_0)^2 f(z) [/itex].

So we get:
[tex] \lim_{z\rightarrow z_0} \frac{d}{dz} \frac{1}{(z+1)^2} = \lim_{z\rightarrow z_0} \frac{-(2z+2)}{(z+1)^4} = -\frac{1}{4} [/tex]

Now since you have the Laurent series (I didn't check this), you can pull the residue right off the series. The residue of a function at a point [itex]z_0[/itex] is the coefficient of the term [itex](z-z_0)^{-1}[/itex] in its Laurent series expansion around [itex]z_0[/itex].

The way you have written the expansion means that that term is when n=2.

So the residue is the coefficient of the term when n=2, which is
[tex] \frac {(-1)^{2+1} (2)}{2^{2+1}} = -\frac{1}{4} [/tex]
 
  • #3
Thanks! Your explanation really helped me. Now it makes sense.
 

Related Threads on Getting Residues Complex Analysis

  • Last Post
Replies
9
Views
7K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
4K
Replies
9
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
3K
Top