1. Not finding help here? Sign up for a free 30min 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!

What exactly is a residue - what are its applications[complex analysis]

  1. Apr 27, 2007 #1
    In fundamental of Complex analysis - residues is introduced as an exercise at the end of a chapter and thats it! (or it may resurface in a later chapter),

    My question is that saff and snider looks at it as the numerator of the partial fraction exapansion of a polynomail fraction.

    But in Schaums series we have a nice little function like this:

    Code (Text):

    a = lim   1/(k-1)! . [color=red](d^(k-1) /dz^(k-1)) [/color] {(z-a)^k f(z)}
    where the term in red is the differential operator and the order is determined by k-1

    so whats this used for? which method is right? why choose one method over the other? And what is it beside the sum of all the residues at the singularities = the integral of the function that contains it - i.e. f(z) ?

    sorry if this is a silly question.
    Last edited: Apr 27, 2007
  2. jcsd
  3. Apr 27, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    If you integrate zn on a closed curve around 0 what do you get?
    More generally, if you integrate (z-z0)n on a close curve containing z0 what do you get?

    That's easy to answer. Since (z-z0)n is analytic everywhere except at z0, we get the same for all such curves so we can look at the circle, of radius 1 around z0, taking [itex]z= z_0+ e^{i\theta}[/itex] so that [itex](z-z_0)^n= e^{ni\theta}[/itex] and [itex]dz= ie^{i\theta}d\theta[/itex]:
    [tex]\int (z-z_0)^n dz= \int_0^{2\pi}ie^{(n+1)i\theta}d\theta[/tex]
    As long as n is not equal to -1, that is
    evaluated at [itex]\theta= 0[/itex] and [itex]\theta= 2\pi[/itex]. But [itex]e^{(n+1)i\theta}[/itex] is 0 at both ends so the integral is 0.

    If n= -1, then [itex]e^{(n+1)i\theta}= e^0= 1[/itex] so the integral is just
    [tex]\int_0^{2\pi} d\theta= 2\pi i[/itex].

    Now, suppose f(z) is a function having a "pole of order n" at z= z0. That means it can be written as a power series with powers of z down to -n: a "Laurent series".
    [tex]f(z)= a_{-n}(z-z_0)^{-n}+ \cdot\cdot\cdot+ a_{-1}(z-z_0)^{-1}+ \cdot\cdot\cdot[/tex]
    Integrating that, term by term, on a contour containing z0, every term gives 0 except the (z-z0)-1 term. That term gives [itex]2\pi i a_{-1}[/itex] so the integral is that.

    The "residue" of f(z) at z0 is precisely the coefficient of (z-z0)-1 in a Laurent series expansion of f(z). It can be found using a formula similar to the formula for Taylor's series coefficients- That's the formula Schaum's outline is giving you. If f(z) is analytic inside a countour except for some points at which it has poles, the integral of f around that contour is just [itex]2\pi i[/itex] times the sum of the residues at each of those poles.
  4. Apr 28, 2007 #3
    thanks a tonne halls of ivy
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: What exactly is a residue - what are its applications[complex analysis]