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Math Help - residues

  1. #1
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    residues

    Hey guys.
    How can I calculate the residues of this function (in the pic) in all of its singularity points?
    I'm kind of a newbie in this this residues stuff and I can really use an example.

    Thanks in advance.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by asi123 View Post
    Hey guys.
    How can I calculate the residues of this function (in the pic) in all of its singularity points?
    I'm kind of a newbie in this this residues stuff and I can really use an example.

    Thanks in advance.
    z=0 is a double pole.
    z=1 is a simple pole.

    The general formula for residues in poles of order n is :

    \text{Res}_{z=a} f(z)=\frac{1}{(n-1)!} \cdot \frac{d^{n-1}}{dz^{n-1}} \left\{(z-a)^n f(z)\right\}
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  3. #3
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    Also, be careful of the removable singularities !

    Poles are points where the limit is undefined.
    But removable singularities are points where the limit exists. For example \frac{z^2-3z+2}{(z-2)(z-4)}
    2 annulates the denominator, so you can think that it's a pole. But (z-2) also divides zē-3z+2 (because zē-3z+2=(z-1)(z-2))
    So the limit when z goes to 2 is defined. It's a removable singularity.
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    Quote Originally Posted by Moo View Post
    Also, be careful of the removable singularities !

    Poles are points where the limit is undefined.
    But removable singularities are points where the limit exists. For example \frac{z^2-3z+2}{(z-2)(z-4)}
    2 annulates the denominator, so you can think that it's a pole. But (z-2) also divides zē-3z+2 (because zē-3z+2=(z-1)(z-2))
    So the limit when z goes to 2 is defined. It's a removable singularity.
    But I don't have any removable singularities in my function, right?

    Thanks.
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  5. #5
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    Quote Originally Posted by asi123 View Post
    But I don't have any removable singularities in my function, right?

    Thanks.
    No you don't, because z^2+z-1=\left(z-\frac{-1+\sqrt{5}}{2}\right)\left(z-\frac{-1-\sqrt{5}}{2}\right)

    it was just in case...because you might meet some in the future :P


    There are also essential singularities, which are singularities that are neither poles nor removable singularities.
    http://en.wikipedia.org/wiki/Mathema...mplex_analysis
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  6. #6
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    Ok, this is what I did (in the pic).

    Now I need to sum the residues to get the answer?

    Thanks.
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  7. #7
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    It depends on where you integrate o.O
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