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Residue Theorem

  1. Jun 4, 2006 #1

    AKG

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    In the complex plane, let C be the circle |z| = 2 with positive (counterclockwise) orientation. Show that:

    [tex]\int _C \frac{dz}{(z-1)(z+3)^2} = \frac{\pi i}{8}[/tex]


    This isn't homework, it was a problem in one of the practice GREs. It looks like a straightforward application of the residue theorem, the only problem is that I never understood the second half of complex analysis? The only pole of the integrand contained in the interior of C is 1, so I have to find the residue of the integrand at 1? What's a residue, and how do I find it? Mathworld gives a definition in terms of the Laurent series, but I'm sure there's a simpler way when it comes to basic rational functions like these. And once I have the residue, what do I do, multiply by [itex]2\pi i[/itex]? Thanks.
     
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  3. Jun 4, 2006 #2

    benorin

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    [tex]\frac{1}{(z-1)(z+3)^2} = \frac{1}{16(z-1)}-\frac{1}{16(z+3)}-\frac{1}{4(z-3)^2} [/tex]

    EDIT: Thanks, Tide, for chatching my typo.

    by the method of partial fractions, and this also gives the residue at z=1 is 1/16, hence

    [tex]\int _C \frac{dz}{(z-1)(z+3)^2} = 2\pi i\left( \frac{1}{16}\right)=\frac{\pi i}{8}[/tex]
     
    Last edited: Jun 4, 2006
  4. Jun 4, 2006 #3

    TD

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    Or, without partial fractions: if z = a is a pole of order 1 of f(z), then:

    [tex]
    {\mathop{\rm Res}\nolimits} \left( {f,a} \right) = \mathop {\lim }\limits_{z \to a} \left( {\left( {z - a} \right)f\left( z \right)} \right)
    [/tex]

    So:

    [tex]
    \mathop {\lim }\limits_{z \to 1} \frac{{z - 1}}{{\left( {z - 1} \right)\left( {z + 3} \right)^2 }} = \frac{1}{{\left( {1 + 3} \right)^2 }} = \frac{1}{{16}}
    [/tex]
     
  5. Jun 4, 2006 #4

    Tide

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    benorin,

    Your last term in the partial fraction expansion should have z+3 instead of z-1.
     
  6. Jun 4, 2006 #5

    AKG

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    Something doesn't look right with your partial fraction decomposition. (z-1) occurs to the first power (in the denominator) and (z+3) occurs to the second. On the right side of your equation, it appears you would get (z-1) occuring to the second and (z+3) to the first. Also, how does this give you a residue? Assuming what you have on the right side is correct, do you know somehow that in expanding [itex]\frac{1}{16(z+3)}[/itex] about 1 (i.e. in a Laurent series in powers of (z-1)) that the coefficient of 1/(z-1) will be 0?

    Oh wait, is it because if

    [tex]\frac{1}{16(z+3)} = \sum a_n(z-1)^n[/tex]

    the the coefficient of 1/(z-1) is zero because:

    [tex]\frac{z-1}{16(z+3)} = \sum a_n(z-1)^{n+1}[/tex]

    so we can find a-1 by plugging in z=1 on the left which gives zero. So if this is right, I get how you got the residue to be 1/16. But I have one more sort of stupid question: how did you do the partial fraction decomposition? I would have done:

    [tex]\frac{1}{(z-1)(z+3)^2} = \frac{Az + B}{z-1} + \frac{Cz + D}{z+3} + \frac{Ez^2 + Fz + G}{(z+3)^2}[/tex]

    and then tried to solve for A, B, C, D, E, F, and G. Is there a quicker way?
     
    Last edited: Jun 4, 2006
  7. Jun 4, 2006 #6

    AKG

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    Thanks. Can this be easily generalized to poles of higher order?
     
  8. Jun 4, 2006 #7

    TD

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    It can be generalized, but it won't be as easy as this. For higher order poles, derivatives will be involved.

    If z = a is a pole of order n of f(z), then:

    [tex]
    {\mathop{\rm Res}\nolimits} \left( {f,a} \right) = \frac{1}{{\left( {n - 1} \right)!}}\mathop {\lim }\limits_{z \to a} \left( {\frac{{d^{n - 1} \left( {z - a} \right)^n f\left( z \right)}}{{dz^{n - 1} }}} \right)
    [/tex]

    Of course, for n = 1, we find the previous formula again.
     
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