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Math Help - Trouble calculating residues

  1. #1
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    Trouble calculating residues

    Hello,

    I'ld like to calculate the reside of f(z):=\frac{z^{n-1}}{\sin^n(z)} at z_0=0.

    Using the definition I get \text{res}_{0}f = \frac{1}{2\pi i} \int_\kappa \frac{z^{n-1}}{\sin^n(z)} dz, but I cant compute this integral.

    My next idea was to develop the function into a Laurent-Series and use the coefficent a_{-1} but thats basically the same calculation.

    I would like to get some help on how to solve this and similar problems.

    Thank you!
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  2. #2
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    Quote Originally Posted by vsywod View Post
    Hello,

    I'ld like to calculate the reside of f(z):=\frac{z^{n-1}}{\sin^n(z)} at z_0=0.

    Using the definition I get \text{res}_{0}f = \frac{1}{2\pi i} \int_\kappa \frac{z^{n-1}}{\sin^n(z)} dz, but I cant compute this integral.

    My next idea was to develop the function into a Laurent-Series and use the coefficent a_{-1} but thats basically the same calculation.

    I would like to get some help on how to solve this and similar problems.

    Thank you!
    It's a simple pole right? lim_{z\to 0} zf(z)=1 and that's the residue although you can use the formula (k+1)\frac{g^{(k)}}{h^{(k+1)}}. The integral formula should only be used by those who love Complex Analysis and want dessert.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Remembering the 'infinite product'...

    \sin z = z \cdot \prod_{k=1}^{\infty} (1-\frac{z^{2}}{k^{2} \pi^{2}}) (1)

    ... we obtain...

    \sin^{n} z = z^{n} \cdot \prod_{k=1}^{\infty} (1-\frac{z^{2}}{k^{2} \pi^{2}})^{n} (2)

    ... so that is...

     f(z)= \frac{1}{z\cdot \prod_{k=1}^{\infty} (1-\frac{z^{2}}{k^{2} \pi^{2}})^{n}} (3)

    ... and finally ...

    Res_{z=0} f(z) = \lim_{z \rightarrow 0} z\cdot f(z) = 1 (4)

    Kind regards

    \chi  \sigma
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  4. #4
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    Quote Originally Posted by shawsend View Post
    It's a simple pole right?
    Well, actually I'm not sure it's a simple pole. I guess it is, but how can I show it?
    I'ld have to give a holomorphic function g thats doesn't vanish in 0 and which satisfies
    f(z) = z^{-1} \cdot g(z).
    This would mean g(z) = \frac{z^n}{\sin^n(z)} which is clearly holomorphic outside of 0. But why is \lim_{z\to 0} g(z) = 1?
    I think this is a stupid question, but my head is really spinning at the moment...
    Thank you!
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  5. #5
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    Quote Originally Posted by vsywod View Post
    Well, actually I'm not sure it's a simple pole. I guess it is, but how can I show it?
    I'ld have to give a holomorphic function g thats doesn't vanish in 0 and which satisfies
    f(z) = z^{-1} \cdot g(z).
    This would mean g(z) = \frac{z^n}{\sin^n(z)} which is clearly holomorphic outside of 0. But why is \lim_{z\to 0} g(z) = 1?
    I think this is a stupid question, but my head is really spinning at the moment...
    Thank you!
    . . . you makin' me come to Mars ain't ya. You know how I hate Mars . . . where's my Calculus text? Let's see, the limit of a composite is the composite at the limit so w^n is continuous at 0 and thus \lim_{z\to 0}\frac{z^n}{\sin(z)^n}=\left(\lim_{z\to 0} \frac{z}{sin(z)}\right)^n=1. I don't know. Is that ok?
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  6. #6
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    Yes, I got it myself now, thank you
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