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Math Help - Singularities and residues

  1. #1
    lpd
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    Singularities and residues

    Find the isolated singularities (if any) of the following function. For each isolated singularity, describe its nature; that is is it removeable or a pole (and of what order) or essential? In each case, calculate the residue of the function at the singularity.

    a)  \pi cot(\pi z) - 1/z
    I got a removable singularity at z=0.
    but how do I exactly work out the residue for this?

    b) z^{-\frac{1}{2}}
    I got none. because you cannot have a complex number which is multiplied to a power of a half. I'm not quite sure about this one...

    c) sin(z)sin(\frac{1}{z})
    I got a removable singularity at z=0. Not sure about this one either.... These are some hard functions that I got...

    Please help me out!!
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  2. #2
    lpd
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    for a), should I expand it to Lauren series... and then see what coefficients I get?
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  3. #3
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    Quote Originally Posted by lpd View Post
    Find the isolated singularities (if any) of the following function. For each isolated singularity, describe its nature; that is is it removeable or a pole (and of what order) or essential? In each case, calculate the residue of the function at the singularity.

    a)  \pi cot(\pi z) - 1/z
    I got a removable singularity at z=0.
    but how do I exactly work out the residue for this?

    b) z^{-\frac{1}{2}}
    I got none. because you cannot have a complex number which is multiplied to a power of a half. I'm not quite sure about this one...

    c) sin(z)sin(\frac{1}{z})
    I got a removable singularity at z=0. Not sure about this one either.... These are some hard functions that I got...

    Please help me out!!
    a) The residue at a removable singularity is zero. Your job is to think about why ....

    b) The function is f(z) = \frac{1}{\sqrt{z}} ....

    c) I suggest you try getting the first few terms of the series around z = 0. Is there a 1/z term ....?
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  4. #4
    lpd
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    Quote Originally Posted by mr fantastic View Post

    c) I suggest you try getting the first few terms of the series around z = 0. Is there a 1/z term ....?
    theres no 1/z term when i computed the lauren series. does that imply that res[sin(z)sin(1/z),0]=0?
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  5. #5
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    Quote Originally Posted by lpd View Post
    theres no 1/z term when i computed the lauren series. does that imply that res[sin(z)sin(1/z),0]=0?
    Yes.
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