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Math Help - Residues at simple poles

  1. #1
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    Residues at simple poles

    Hi folks,

    Having real problems with a complex analysis homework. can the residue of a complex function at a simple pole be 0?

    ie for g(z)= sin(z)/z i think the pole is just 0 of order one which is simple, but by calculating the residue i get 0..am i doing something wrong?

    I'm using the formula if c is the simple pole and g(z)= f(z)/h(z) then the residue at c of g(z) = f(c)/h'(c)..is that okay?

    Thanks for any help

    Donsie
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  2. #2
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    Quote Originally Posted by Donsie View Post
    Hi folks,

    Having real problems with a complex analysis homework. can the residue of a complex function at a simple pole be 0?

    ie for g(z)= sin(z)/z i think the pole is just 0 of order one which is simple, but by calculating the residue i get 0..am i doing something wrong?

    I'm using the formula if c is the simple pole and g(z)= f(z)/h(z) then the residue at c of g(z) = f(c)/h'(c)..is that okay?

    Thanks for any help

    Donsie
    There is no pole at all at z= 0 for \frac{sin(z)}{z}, just a "removable discontinuity". Remember, from Calculus I, that \lim_{z\to\infty}\frac{sin(z)}{z}= 1. "Remove" the discontinuity by defining g(0)= 1.

    The Taylor's series for sin(z) is z- \frac{1}{6} z^3+ \frac{1}{5!}z^5+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}z^{2n+1}+ \cdot\cdot\cdot.

    So \frac{sin(z)}{z}= 1- \frac{1}{6}z^2+ \frac{1}{5!}z^4+ \cdot\cdot\cdot+ \frac{(-1)^n}{(2n+1)!}z^{2n}+ \cdot\cdot\cdot, an analytic function. Again, that has the value "1" at z= 0.
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  3. #3
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    ..

    thank you!

    So to make sure i have it right..

    the function f(z)= 1/1+z^3 has a simple pole at z= -1. correct?

    Then to get the residue at this pole can i use the formula res(f,c) where c is the simple pole = g(c)/h'(c) where g(c) is just 1, h'(z)= 3z^2 so h'(c)=3. so the residue is just 1/3? is that also correct?

    Thank you for your patience!!

    Donsie
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  4. #4
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    Quote Originally Posted by Donsie View Post
    thank you!

    So to make sure i have it right..

    the function f(z)= 1/1+z^3 has a simple pole at z= -1. correct?

    Then to get the residue at this pole can i use the formula res(f,c) where c is the simple pole = g(c)/h'(c) where g(c) is just 1, h'(z)= 3z^2 so h'(c)=3. so the residue is just 1/3? is that also correct?

    Thank you for your patience!!

    Donsie
    Yes. But note that z = -1 is not the only simple pole. There are simple poles at values of z such that z^3 + 1 = 0.
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