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Math Help - Complex Analysis-Analytic Functions

  1. #1
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    Complex Analysis-Analytic Functions

    1. Prove that there is no analytic function inside
    Code:
    z:0<|z|<1
    such as every z in this region satysfies:
    Code:
    f^2(z)=z
    .

    2. Let f be analytic at the annulus:
    Code:
     D=z:0<|z-a|<r
    .
    Prove that if f has an antideriative function then Res(f,a)=0 .

    Hope you'll be able to help me

    Thanks a lot!
    Last edited by WannaBe; June 19th 2010 at 08:50 AM.
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  2. #2
    A Plied Mathematician
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    Regarding #2, what is a "former" function?
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  3. #3
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    iI meant antideriative...sry
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    For #1, suppose that such a function exists, and show that the Cauchy-Riemann equations are not satisfied at z=0.
    Or you can differentiate implicitly with respect to z and show that f'(z) cannot be analytic at 0.
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    For #2, suppose that an antiderivative exists. Show that this implies that the integral along a simple closed contour inside the annulus vanishes... so what if the interior of the contour contains the point z=0?
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  6. #6
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    About 1-
    We can't use Cauchy-Riemann,because f isn't analytic at z=0...
    How can we solve this problem?

    2 is completely understandable...


    Thanks @!
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  7. #7
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by WannaBe View Post
    About 1-
    We can't use Cauchy-Riemann,because f isn't analytic at z=0...
    How can we solve this problem?

    2 is completely understandable...


    Thanks @!
    I'm sorry, I hadn't seen the point 0 was excluded. I hadn't had my coffee just yet!

    One way which I can think of is this : a function analytic inside a domain maps a closed contour to a closed contour. So assume there exists such a function. Now take a simple closed contour around the origin, and show that f does not map it to a closed contour (use the argument principle, and the fact that 2f(z)f'(z)=1 inside the punctured disc).
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  8. #8
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    Thanks...
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