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Math Help - Residue problem

  1. #1
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    Residue problem

    How do I find the residue of 1/(z^2 sinz) ? It seems to me that it has a 3rd order singularity at z=0, but I cant find the residue. Thanks for any help rendered.
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    Quote Originally Posted by boredaxel View Post
    How do I find the residue of 1/(z^2 sinz) ? It seems to me that it has a 3rd order singularity at z=0, but I cant find the residue. Thanks for any help rendered.
    It has a pole of order 3 at z = 0. And there is a standard formula (it will be in your textbook or class notes) for finding the residue at a pole ....

    Alternatively, you could find and use the first few terms of the Laurent series of \frac{1}{\sin z} (from which it immediately follows that the residue is equal to 1/6).
    Last edited by mr fantastic; April 13th 2010 at 12:18 AM. Reason: Fixed a careless error
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    I tried using the formula but I got this equation eventually (attached along this reply). I could not eliminate the sinz at the denominatior, causing my ans to tend to inf as z tends to 0. Have I used the formula wrongly or is there another formula to get the residue?
    Attached Thumbnails Attached Thumbnails Residue problem-eqn.jpg  
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  4. #4
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    Quote Originally Posted by boredaxel View Post
    I tried using the formula but I got this equation eventually (attached along this reply). I could not eliminate the sinz at the denominatior, causing my ans to tend to inf as z tends to 0. Have I used the formula wrongly or is there another formula to get the residue?
    Following up from my earlier reply:

    Note that \frac{1}{\sin z} has a pole of order 1 at z = 0 since \lim_{z \to 0} \frac{z}{\sin z} is finite (and is equal to 1). Furthermore, \text{Res}\left( \frac{1}{\sin z}, z = 0\right) = 1.

    Therefore \frac{1}{\sin z} = \frac{1}{z} + a + bz + cz^2 + ....

    Therefore \text{Res}\left( \frac{1}{z^2 \sin z}, z = 0\right) = b.

    To get the value of b, substitute \sin z = z - \frac{z^3}{6} + \frac{z^5}{120} - .... into \frac{1}{\sin z} = \frac{1}{z} + a + bz + cz^2 + .... and re-arrange:

    1 = \left(z - \frac{z^3}{6} + \frac{z^5}{120} - .... \right) \left(\frac{1}{z} + a + bz + cz^2 + ....\right)

    Expand and equate coefficients:

    Coefficient of z: a = 0.

    Coefficient of z^2: 0 = b - \frac{1}{6} \Rightarrow b = \frac{1}{6}.
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    Quote Originally Posted by chiph588@ View Post
    What can't Wolfram Alpha do?
    What can't you do? - Wolfram|Alpha
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