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Residue Problem.

  1. Jul 7, 2008 #1
    hi all, my first post; had a minor headache with this problem lol.

    PROBLEM 1:
    Finding Residue:
    -----------------
    find Res(g,0) for [itex] g(z) = z^{-2}coshz[/itex]

    My Attempt/Solution:
    -----------------
    I know [itex] coshz = 1 + \frac{x^2}{2!} + \frac{z^4}{4!} .... [/itex]

    so now [itex] z^{-2}coshz = z^{-2} (1 + \frac{z^2}{2!} + \frac{z^4}{4!} ... ) = \frac{1}{z^2} + \frac{z^2}{2!z^{2}} + \frac{z^4}{4!z^{2}} ... = \frac{1}{z^2} + \frac{1}{2!} + \frac{z^2}{4!}[/itex]

    we know the residue is the coefficient of the -1th term (or the coefficient of [itex]z^{-1}[/itex]) but there is no -1th term as you can see. So does that mean the residue is 0, or am I missing something?


    PROBLEM 2:
    Finding Integral:
    ----------------------------
    Evaluate by integrating around a suitable closed contour:
    [itex]\int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4}[/itex]

    My Attempt/Solution:
    -----------------
    First consider the intergral,
    [itex] I_R = \int_{\gamma^R} \frac{e^{3iz}}{(z+2i)(z-2i)}dz[/itex] where [itex] \gamma^R = \gamma^R_1 + \gamma^R_2 [/itex] and [itex]\gamma^R_1 = \{|z| = R , Im(z) > 0\} [/itex] and [itex]\gamma^R_2 = \{z = Re(z) , -R\leq z \leq R\}[/itex]

    Now, let [itex]g(z) = \frac{e^{3iz}}{(z+2i)(z-2i)}[/itex]. The integrand has 2 simple poles, one at 2i and one at -2i. Only 2i is inside the contour so:

    [itex]Res(g,2i) = \lim_{z \to 2i}\frac{(z-2i)e^{3iz}}{(z+2i)(z-2i)} = \lim_{z \to 2i}\frac{e^{3iz}}{(z+2i)} = \frac{e^{-6}}{4i}[/itex]

    By Cauchy's Residue Theorem, we have:
    [itex] I_R = (2\pi i)\frac{e^{-6}}{4i} = \frac{\pi e^{-6}}{2}[/itex]

    To show that [itex]\int_{\gamma^R_1}g(z)dz \to 0, R \to \infty[/itex], we apply Jordan's Lemma : [itex]M(R) \leq \frac{1}{R^2-4}[/itex].

    So now this means, [itex] \int_{-\infty}^{\infty} \frac{cos3x}{x^2 + 4} = \frac{\pi e^{-6}}{2}[/itex]

    Please could you verify whether I have worked these out correctly. Thanks a great deal, highly appreciated :).
     
    Last edited: Jul 7, 2008
  2. jcsd
  3. Jul 7, 2008 #2
    Both look good to me.
     
  4. Jul 8, 2008 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In problem 1, the fact that " no -1th term" because its coefficient is 0.
     
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