Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Residue questions

  1. Dec 9, 2009 #1
    Sorry I don't have equation editor, for some reason every time I install it on Microsoft Word it never appears...

    1. The problem statement, all variables and given/known data
    Calculate the residue at each isolated singularity in the complex plane


    2. Relevant equations
    #1 Simple pole at z0 then,
    Res[f(z), z0] = lim (z - z0)(f(z)) as z goes to z0.
    #2 Double pole at z0 then,
    Res[f(z), z0] = lim d/dz [(z - z0)^2*f(z)] as z goes to z0.
    #3 If f(z) and g(z) are analytic at z0, and if g(z) has a simple zero at z0 then
    Res[f(z)/g(z), z0] = f(z0)/g'(z0)
    #4 If g(z) is analytic and has a simple zero at z0, then
    Res[1/g(z), z0] = 1/g'(z0).

    3. The attempt at a solution

    The problem occurs when z = 0 so looking at

    Res[e^(1/z), 0], Using #1, #3, #4 don't help the problem. So using #2

    lim as z goes to z0 [d/dz z^2 * e^(1/z)], there's still a problem... I'm completely lost at this point.

    1. The problem statement, all variables and given/known data
    Evaluate the following integral, using the residue theorem
    Integral |z| = 1 (sin(z)/z^2)dz

    2. Relevant equations
    See Above

    3. The attempt at a solution
    How would I start this? z = 0 gives a problem, would I take the integral first and then evaluate?
  2. jcsd
  3. Dec 9, 2009 #2
    z=0 isn't a pole for this function.

    write out the laurent series. it has all negative powers. thus z=0 is an essential isolated singularity.

    knowing the laurent series, how do you find the residue?
  4. Dec 9, 2009 #3
    New Edit: I agree with Latentcorpse (he replied whe I was still typing the message)

    You know that exp(z) is an analytical function everywhere on the complex plane. This means that you can replace z by 1/z in the series expansion without any problem. It will yield a Laurent series that converges everywhere (except at z = 0). You can then read-off the residue at z = 0.

    The Laurent expansion contains arbitrarily large negative powers of z, we then say that the function has an "essential singularity" at z = 0.

    You can still compute the residue at this essential singularity like in case of ordinary poles, by considering the mapping z ---> 1/z

    The residue is 1/(2 pi i) times a contour integral that encircles the singularity in an anti-clockwise way. If then perform a change of variables z -->1/z, the integral changes to:

    1/(2 pi i) exp(z) (-1/z^2) dz

    which is now traversed clockwise. Changing it to anticlockwise will yield a minus sign. This means that the residue at zero of exp(1/z) is the same as the residue at zero of the function


    This function has an ordinary pole (a "double pole") at zero.
  5. Dec 9, 2009 #4
    Thank you, appreciate the comments... I'm completely lost in Complex Analysis (First Grad Level class, never took it as an undergrad)
  6. Dec 10, 2009 #5
    read off the [itex]c_{-1}[/itex] coefficient i.e. the coefficient of the [itex]z^{-1}[/itex] term in the laurent series for [itex]\frac{e^z}{z^2}[/itex] and you'll get the residue is 1. hopefully.
  7. Dec 10, 2009 #6
    Yeah, I figured it out thanks to you guys, much appreciated, I have another question up on the page, was wondering if you could maybe help with that one? If not it's okay, Thanks again!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook