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Math Help - Complex analysis - finding the order of poles

  1. #1
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    Complex analysis - finding the order of poles

    I am stuck on the following question:

    Find the poles of z/(1-exp(z)) and determine their orders.

    I have the poles as being all z*2*Pi*i*k with k an integer other than 0 as I am not sure what happens at 0.

    How do I find the order though? It seems to me that writing the series expansion out does not help much.

    Thanks for any and all help (my guess is that the order is 1)
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  2. #2
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    Quote Originally Posted by artvandalay11 View Post
    I am stuck on the following question:

    Find the poles of z/(1-exp(z)) and determine their orders.

    I have the poles as being all z*2*Pi*i*k with k an integer other than 0 as I am not sure what happens at 0.

    How do I find the order though? It seems to me that writing the series expansion out does not help much.

    Thanks for any and all help (my guess is that the order is 1)

    The pole is of order zero it is a removable singularity at 0

    Remember that

    e^{z}=1+z+\frac{z^2}{2}+\frac{z^3}{6}+....

    So

    1-e^{z}=-(z+\frac{z^2}{2}+\frac{z^3}{6}+....)

    Now we have

    \frac{z}{1-e^z}=\frac{-z}{(z+\frac{z^2}{2}+\frac{z^3}{6}+....)}=

    \frac{-1}{(1+\frac{z}{2}+\frac{z^2}{6}+....)}

    Now by long division we get

    \frac{-1}{(1+\frac{z}{2}+\frac{z^2}{6}+....)}=-1+\frac{z}{2}-\frac{1}{12}z^2+\frac{1}{720}z^4-...
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  3. #3
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    thank you for confirming my suspicions... I know I said I thought the pole had order 1, but I also have exactly what you wrote on my paper, I just was not sure if it was removable because calculating the limit as a test for removable singularities seemed to be difficult... again thanks a lot
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    further complex pole thinking

    I was thinking about this some more, and the function should have a pole at z=2\pi\imath since \lim_{z \to 2\pi\imath}|f(z)|=\infty (also at any multiple of z=2\pi\imath)

    The series expansion is correct, but the function still tends to infinity. To show that z=2\pi\imath is a pole we have to be able to show \lim_{z \to 2\pi\imath}(z-2\pi\imath)f(z)=0

    I can't seem to be able to do that
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  5. #5
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    Quote Originally Posted by artvandalay11 View Post
    I was thinking about this some more, and the function should have a pole at z=2\pi\imath since \lim_{z \to 2\pi\imath}|f(z)|=\infty (also at any multiple of z=2\pi\imath)

    The series expansion is correct, but the function still tends to infinity. To show that z=2\pi\imath is a pole we have to be able to show \lim_{z \to 2\pi\imath}(z-2\pi\imath)f(z)=0

    I can't seem to be able to do that
    Write, \frac{z}{1-e^z} = \frac{f(z)}{g(z)}. If g'(z_0) \not = 0 then the residue of the pole at z_0 is \frac{f(z_0)}{g'(z_0)} = -\frac{z_0}{e^{z_0}}.
    Thus, we see that the residues are -2\pi i k for k\in \mathbb{Z}.
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  6. #6
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    Alright, but I need to find the order of the poles
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  7. #7
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    Quote Originally Posted by artvandalay11 View Post
    Alright, but I need to find the order of the poles
    That theorem only works if the order is one!
    (For k=0 we get residue 0 so there is no pole).
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