• Support PF! Buy your school textbooks, materials and every day products Here!

Residues of ((log(z))^2)/(1+z^2)^2

  • Thread starter NT123
  • Start date
  • #1
28
0

Homework Statement

This is actually on Wikipedia, but it doesn't show how you actually calculate the residues.

I want to calculate the residues for ((log(z))^2)/(1+z^2)^2. Wikipedia claims the sum of the residues is (-(pi/4) + (i*(pi)^2)/16 - (pi/4) - (i*(pi)^2)/16) = -pi/2


Homework Equations



http://en.wikipedia.org/wiki/Methods_of_contour_integration

The Attempt at a Solution

The singularities are at i and -i. Multiplying first by (z-i)^2 and taking the limit z ---> i, I get (log(i)^2)/4 = (log(e^(i*(pi/2))^2)/4 = -(pi^2)/16.

Then, multiplying by (z+i)^2 and taking the limit z ---> -i I get (log(e^-i*(pi/2))^2)/4 = -(pi^2)/16 again. Any help would be very much appreciated.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
The poles are double poles. Look at http://en.wikipedia.org/wiki/Residue_(complex_analysis [Broken]) under "Limit formula for higher order poles". There's a derivative involved here as well as a limit.
 
Last edited by a moderator:

Related Threads on Residues of ((log(z))^2)/(1+z^2)^2

Replies
2
Views
2K
  • Last Post
Replies
1
Views
3K
Replies
2
Views
530
  • Last Post
Replies
1
Views
723
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
6K
Replies
2
Views
5K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
Top