Results 1 to 5 of 5

Math Help - laurent series

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    142

    laurent series

    Can someone show me how to find the Laurent series of 1/z(z-4) expanded in terms of z-1? I can figure out the expansion for the domain 1 < abs(z) < 3, but I'm having trouble with the other two domains (i.e. 0 < abs(z) < 1 and abs(z) > 3.

    Thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Aug 2008
    Posts
    120
    The has singularities at z=0 and z=4 so you will be able to find two series. One for 0<|z|<4 and for |z|>4. Where did you get your values?
    Last edited by whipflip15; November 11th 2008 at 04:46 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2006
    Posts
    142
    It's expanded in terms of z-1, no?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2008
    Posts
    903
    The objective is to get the expressions in terms of a geometric series of the form \frac{1}{1-f(z)} where |f(z)|<1 in the domain of interest. First write the function as f(z)=-\frac{1}{4z}+\frac{1}{4(z-4)}. For the domain |z-1|<1 start by adding and subtracting one in the denominator:

    f(z)=-\frac{1}{4}\left(\frac{1}{1+z-1}\right)+\frac{1}{4}\left(\frac{1}{-3+z-1}\right)

    =-\frac{1}{4}\left(\frac{1}{1-(-(z-1))}\right)-\frac{1}{12}\left(\frac{1}{1-\frac{z-1}{3}}\right)

    =-1/4\sum_{n=0}^{\infty}(-1)^n(z-1)^n-1/12\sum_{n=0}^{\infty}\frac{(z-1)^n}{3^n};\quad |z-1|<1

    Now in the case of |z-1|>3, you need to have geometric terms which are less than one for |z-1|>3. I'll do the first term:


    -\frac{1}{4}\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{z-1+1}\right)=-\frac{1}{4}\left(\frac{1}{(z-1)(1+\frac{1}{z-1})}\right)

    =-\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{(z-1)^{n+1}}

    Do the second one to get: \frac{1}{4}\sum_{n=0}^{\infty}\frac{3^n}{(z-1)^{n+1}}

    Thus:

    f(z)=-\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{(z-1)^{n+1}}+\frac{1}{4}\sum_{n=0}^{\infty}\frac{3^n}  {(z-1)^{n+1}},\quad |z-1|>3
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2008
    Posts
    120
    The question he asks for |z|>3 not |z-1|>3. That is all i was saying.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Laurent Series
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: September 17th 2011, 01:46 AM
  2. Laurent Series/ Laurent Series Expansion
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 5th 2010, 09:41 PM
  3. Laurent Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 27th 2010, 03:34 PM
  4. Laurent series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 13th 2009, 01:10 PM
  5. Laurent Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 17th 2007, 10:43 AM

Search Tags


/mathhelpforum @mathhelpforum