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Math Help - singularity proof

  1. #1
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    singularity proof

    Unfortunately I'm not very good with singularities, so any help with this proof would be very much apprectiated. Kudos in advance!!!!!

    S (a,r) denotes a positively orientated circular contour of radius r centered at a
    D'(a,r) denotes a disk such that {z: 0< |z-a|< r}

    Let f be an analytic function on the domain D'(0,2) and suppose that for all n=0,1,2...


    S(0,1)(z^n)f(z)dz = 0

    Show that f has a removable singularity at z=0


    (The s(0.1) is meant to be below the integral sign)
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  2. #2
    Member mfetch22's Avatar
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    Quote Originally Posted by foolsgold View Post
    Unfortunately I'm not very good with singularities, so any help with this proof would be very much apprectiated. Kudos in advance!!!!!

    S (a,r) denotes a positively orientated circular contour of radius r centered at a
    D'(a,r) denotes a disk such that {z: 0< |z-a|< r}

    Let f be an analytic function on the domain D'(0,2) and suppose that for all n=0,1,2...


    S(0,1)(z^n)f(z)dz = 0

    Show that f has a removable singularity at z=0


    (The s(0.1) is meant to be below the integral sign)
    I have no clue how to solve this, but I thought I'd rewrite that last equation for you with the pretty 'latex' program so that other posters that know how to solve it can more easily read it. So, I think you were saying something like this for that last equation:

    [A] \;\; \int_{S(0,1)}^{z^n}f(z)dz = 0

    or did you mean...

    [B] \;\; \int_{S(0,1)}(z^n)f(z)dz = 0

    Or did you mean something else entirely? I'm just trying to ensure "equational clarity" for the posters who can actually prove this problem for you. I hope it helps, just maybe...
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  3. #3
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    hey thanks a lot. it's the second one!
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  4. #4
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    I'm gonna be ommiting the curve over which we are integrating. Now expand f(z) in a Laurent series, notice the unifrom convergence and use that z^m has a primitive in D'(0,2) iff m\neq -1 to conclude that, for n\geq 0:

    0= \int z^nf(z)dz = \sum_{-\infty }^{\infty } a_k\int z^{n+k}dz = a_{-1-n} 2\pi i

    (where a_k are the coefficients of the Laurent series) and so a_{-1-n}=0 for all n\geq 0 which is saying a_k=0 for all k<0. So the Laurent series is a Taylor one, and so f is analytic in the disc D(0,2).
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