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- poles vs. Zeros, etc.

  1. Oct 19, 2009 #1
    I am having a hard time understanding the difference between poles and zeros, and simple poles versus removable poles. For instance, consider [tex]f(z)=\frac{z^2}{sin(z)} [/tex]. we can expand sine into a power series and pull out a z, so doesn't that remove the singularity at z=0? Also, I don't see why n*pi would not also be removable since it doesn't seem to be a problem in the series expansion (but according to my graded homework, 0 is a zero and n*pi is a simple pole)... Can someone help me out here?
     
  2. jcsd
  3. Oct 19, 2009 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That's how you remove the singularity. But this operation produces a new (partial) function that is not f. (The difference being that this function is defined at 0 whereas f is not)
     
  4. Oct 19, 2009 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    But 0 is not a zero, it is a removable singularity! :/
     
  5. Oct 20, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ?? What is your point? Hurkyl's point was that if f(z) has a "removable singularity" at [itex]z_0[/itex], yes, you can "remove" it but then you get a different function, g(z). g(z)= f(z) for all z except [itex]z_0[/itex]. He never said anything about being a zero.
     
  6. Oct 20, 2009 #5

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    My comment was in response to
    HallofIvy.
     
  7. Oct 26, 2009 #6
    To try and sum up:
    1) Cancel z top and bottom to show that the bottom term -> 1 as z -> 0. So that would remove the singularity and make the function analytic at zero.
    1a) Because the bottom can't go to zero, the function must -> 0 when z -> 0. So there is a zero of the function at z = 0.
    2) But if you don't cancel the z and stick with the original function, the sin(z) will vanish every time z -> n*pi and the function will go through the roof. So there are simple poles when z = n*pi.

    hope this helps.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Complex analysis- poles vs. Zeros, etc.
  1. Complex zeros (Replies: 13)

Loading...