1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laurent expansion for a complex function

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Expand [itex]f(z)=\frac{1}{z-4}[/itex] in a laurent series valid for (a) [itex]|z|<4[/itex] and (b) [itex]|z|>4[/itex]


    2. Relevant equations
    The formula for laurent expansion...
    [itex]\sum_{n=-∞}^{+∞}a_{n}(z-z_{0})^{n}[/itex]
    where
    [itex]a_{n}=\frac{1}{2\pi i}\oint_c \frac{f(z)}{(z-z_{0})^{n+1}}dz[/itex]


    3. The attempt at a solution
    umm...I have no idea..help please T_T
     
  2. jcsd
  3. Nov 25, 2013 #2
    f(z) = 1/(z-4) = (1/z)##\frac{1}{1 - 4/z}##. The second factor of that product is the sum of a geometric series. So expand that factor and multiply by 1/z. That gives you a Laurent expansion -- it is convergent in a circle out to the first pole which is where?

    The |z| > 4 part is easier because the function is analytic for those values. How does one expand an analytic function?

    It is always possible to get the values of the Laurent coefficients by evaluating the appropriate integral, which I have found mostly to be a pain. But often there is an easier way (as per above) .
     
  4. Nov 25, 2013 #3
    I'm sorry i really don't get what you did with the [itex](1/z)\frac{1}{1-4/z}[/itex] thing...but as far as the situation is for |z|>4, is it just a taylor expansion around [itex]z_{0}=4[/itex] since it is analytic after the singularity at z=4?
     
  5. Nov 25, 2013 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, it's not a Taylor expansion about ##z_0 = 4##. You're being asked to find the Laurent series about the point ##z_0 = 0##. There's a singularity at ##z=4##, which is a distance 4 from ##z_0=0##, so the complex plane is divided into two regions: ##\lvert z-z_0 \rvert < 4## and ##\lvert z-z_0 \rvert >4##. You'll get different series for the two regions.

    This is a really basic problem. It's almost certainly covered in your textbook. If you truly have no idea how to do it, it wouldn't hurt for you to go back and reread your textbook and notes so you'll have some idea about how to tackle it.
     
  6. Nov 25, 2013 #5
    Do you know what a geometric series is and what its sum is? That is at the bottom of this approach.

    Sorry I said some things wrong. The function is analytic everywhere for |z| < 4, and the geometric series should come out to the Taylor's series in that area. While the function is also analytic for |z|> 4 , if you expand it about z = 0 you will get a Laurent series with an exponent for the 1/z term, because of the singularity at z = 4. In both cases you use a geometric sum -- which one depends on whether |z| is > or < 4.

    As Zela says, this is standard material, so do check your textbook or notes.
     
    Last edited: Nov 25, 2013
  7. Nov 25, 2013 #6
    I think i get it now...thanks a bunch guys! Yea its my first shot at this stuff...but i realize that you kinda have to convert them to binomial expansions like (1+z)^n where |z| is determined by the region you are looking at. Thanks again guys!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Laurent expansion for a complex function
  1. Laurent Expansions (Replies: 2)

  2. Laurent Expansion (Replies: 5)

  3. Laurent series expansion (Replies: 13)

Loading...