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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


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    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
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