Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex Analysis Residue Query

  1. May 5, 2012 #1
  2. jcsd
  3. May 5, 2012 #2


    The point [itex]\,\,z=2i\,\,[/itex] is a pole of order 2 of [itex]\,\,\frac{1}{(z^2+4)^2}=\frac{1}{(x-2i)^2(z+2i)^2}\,\,[/itex] . Thus, from the well-known

    formula that stems from Laurent series, the residue is given by
    [tex]\lim_{z\to 2i}\frac{1}{(2-1)!}\frac{d}{dz^2}\left(\frac{(z-2i)^2}{(z^2+4)^2}\right)=\lim_{z\to 2i}-\frac{2}{(z+2i)^3}=-\frac{i}{32}[/tex]

    DonAntonio
     
  4. May 5, 2012 #3
    Very good, thank you. Just have 2 queries

    1) shouldnt ##/dz^2## be just ##/dz##?

    2) Why did you choose pole ##z-2i## instead of ##z+2i## for the limit..?
    Is it to do with ##0 < |z-z_0|<R##

    Thanks
     
  5. May 5, 2012 #4
    ....
     
  6. May 5, 2012 #5
    Ok that makes sense, thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex Analysis Residue Query
  1. Complex analysis (Replies: 1)

Loading...