Results 1 to 3 of 3

Thread: simple pole

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    41

    simple pole

    Hi, can someone please explain to me very slowly why there is a simple pole at z = +-i for the function:

     f(z) = \frac{cos(z)}{1+z^2} + 4z

    I dont get it sighs.. please help me
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,713
    Thanks
    1882
    Quote Originally Posted by Dgphru View Post
    Hi, can someone please explain to me very slowly why there is a simple pole at z = +-i for the function:

     f(z) = \frac{cos(z)}{1+z^2} + 4z

    I dont get it sighs.. please help me
    Poles occur where the denominator is 0 + 0i.


    So if 1 + z^2 = 0

    z^2 = -1

    z = \pm i.


    So there are simple poles at z = i and z = -i.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,448
    Thanks
    2531
    More generally, simple poles (poles of order 1) occur where the Laurent series for a function has a " (z- a)^{-1}" term but no other negative exponents. (The order of a pole is the "highest" negative exponent in the Laurent series expansion.)

    Here, the function is \frac{cos(z)}{z^2+ 1}+ 4z= \frac{cos(z)}{(z- i)(z+ i)}+ 4z.

    cos(z) is an analytic function for all z so its "Laurent" series is actually its Taylor series about that point and has no negative exponents. \frac{1}{z+ i} is analytic everywhere except at z= -i, so it has a Taylor series expansion about z= i and therefore so does the product \frac{cos(z)}{z+i}. Multiplying that Taylor series by \frac{1}{z- i} introduces a term with (z- i)^{-1}. Finally, just adding "4z" to the series does not change that. This function has a pole of order 1 at z= i.

    For z= -i, just swap "i" and "-i".
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. pole
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 19th 2010, 03:28 AM
  2. Pole
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: Mar 31st 2010, 01:45 PM
  3. Problem with pole
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Feb 23rd 2010, 05:40 AM
  4. pole
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 16th 2009, 12:19 PM
  5. length of pole
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Aug 20th 2008, 05:08 PM

Search Tags


/mathhelpforum @mathhelpforum