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Contour integration

  1. Nov 30, 2007 #1

    quasar987

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    [SOLVED] contour integration

    1. The problem statement, all variables and given/known data
    I'm really rusty with this. I need to calculate

    [tex]2\pi i Res\left(\frac{1}{1+z^4},e^{i\pi/4}\right)[/tex]



    3. The attempt at a solution

    Well,

    [tex]2\pi iRes\left(\frac{1}{1+z^4},e^{i\pi/4}\right)=\int_{C_{\rho}}\frac{1}{1+z^4}dz[/tex]

    where [tex]C_{\rho}[/tex] is a little circle of radius rho centered on [tex]e^{i\pi/4}[/tex], on which there are no singularities. Fine, so let's take [tex]\rho=1/\sqrt{2}[/tex].

    Now I need to parametrize C_rho. Take

    [tex]\gamma(t)=e^{i\pi/4}+\frac{e^{i2\pi t}}{\sqrt{2}} \ , \ \ \ \ 0\leq t < 1[/tex]

    We have

    [tex]\frac{d\gamma}{dt}=i\sqrt{2}\pi e^{i2\pi t}[/tex]

    So that

    [tex]\int_{C_{\rho}}\frac{1}{1+z^4}dz=\int_0^1\frac{i\sqrt{2}\pi e^{i2\pi t}}{1+(e^{i\pi/4}+\frac{1}{\sqrt{2}}e^{i2\pi t})^4}dt[/tex]

    Now what?? :grumpy:

    I believe the answer is supposed to be [tex]-e^{i\pi/4}/4[/tex]
     
    Last edited: Nov 30, 2007
  2. jcsd
  3. Dec 1, 2007 #2
    Why don't you just calculate the residue directly?
     
  4. Dec 1, 2007 #3

    quasar987

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    Refresh my memory please? :smile:
     
  5. Dec 1, 2007 #4
    The singularity at [tex]e^{\frac{{i\pi }}{4}} [/tex] is a simple pole and the numerator of the integrand is non zero and holomorphic at the singularity. So you can calculate the residue by using the formula

    [tex]
    {\mathop{\rm Re}\nolimits} s\left( {\frac{1}{{z^{^4 } + 1}}} \right) = \frac{{1_{z = z_0 } }}{{\frac{d}{{dx}}\left( {z^4 + 1} \right)}}_{z = z_0 } ,z_0 = e^{\frac{{i\pi }}{4}}
    [/tex]

    In the calculation you can simplify the algebra a little by using [tex]z_0 ^4 = - 1[/tex].

    The result should coincide with the answer that you expected to get as indicated in your original post.
     
    Last edited: Dec 1, 2007
  6. Dec 1, 2007 #5

    quasar987

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    Nice, thank you so much.
     
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