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Math Help - show why domain D is not simply connected.

  1. #1
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    show why domain D is not simply connected.

    Suppose we are given a domain D containing a simple closed contour C. Suppose that in addition we are given a point Zo with Zo NOT in D and \int_{|z|=1} \frac{1}{z-z_0}dz \neq 0

    Explain why this ensures that D is not simply connected.

    My attempt:
    If C is contained in D, but point Zo is not in D then there are two options:

    a) Zo is outside C and outside D, or
    b) Zo is inside C, but in a region inside C that is not in D ( here i picture D as an annulus of two circles, C within the annulus)

    For a) since Zo is outside C, f(z) = \frac{1}{z-z_0} is analytic on and within C, and by the Cauchy-Goursat theorem \int_C f(z)dz = 0
    But since here \int_{|z|=1} \frac{1}{z-z_0}dz \neq 0 it cannot be that Zo is outside C, hence option b) must b true.

    Therefore it is the case that Zo is inside C, but in a region inside C that is not in D, which guarentees that D is not a simply connected domain.

    I'm not convinced this is correct. I especially wonder if my assumption of only two cases a) and b) are correct.It makes sense to me but i feel i might be overlooking something. Would appreciate any comments!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: show why domain D is not simply connected.

    Quote Originally Posted by punkstart View Post
    Suppose we are given a domain D containing a simple closed contour C. Suppose that in addition we are given a point Zo with Zo NOT in D and \int_{|z|=1} \frac{1}{z-z_0}dz \neq 0
    Did you mean \int_{C} \frac{1}{z-z_0}dz \neq 0 ?
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  3. #3
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    Re: show why domain D is not simply connected.

    Yes here C is the curve |z|=1, not sure why C was given, surely it is not necessary to know that C is |z|=1 to solve this problem ? But for the sake of including all information i retype exactly my problems as they are in the text.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: show why domain D is not simply connected.

    Quote Originally Posted by punkstart View Post
    Yes here C is the curve |z|=1
    Well, the result is valid for any C\subset D and your reasoning is essentially right. However, it is sufficient to say: D simply connected implies \int_{C}\frac{dz}{z-z_0}=0 by the Cauchy-Goursat Theorem, which contradicts the hypothesis \int_{C}\frac{dz}{z-z_0}\neq 0 .
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