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Complex conjugation equality problem

  1. Apr 12, 2006 #1
    :rolleyes: let be the equality where "*" is complex conjugation....z=a+ib but z*=a-ib with a and b real, then:

    [tex] \int_{0}^{c}dxf(x)g(x)=\int_{0}^{c}dxf(x)[g(x)]* [/tex]

    with c a real number then my question is if the equality above implies necessarily that:

    [tex] g(x)=[g(x)]* [/tex] so g is real where:

    -f(x)>0 or 0 on the interval (0,c) and f is a real-valued function.
    -c is an arbitrary and positive real number.
  2. jcsd
  3. Apr 12, 2006 #2


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    Let [tex]g(x)=u(x)+iv(x)[/tex] so that [tex][g(x)]*=u(x)-iv(x)[/tex]. The equality then becomes

    [tex] \int_{0}^{c}f(x)(u(x)+iv(x)) dx=\int_{0}^{c}f(x)(u(x)-iv(x)) dx \\ \Rightarrow \int_{0}^{c}f(x)u(x) dx+i\int_{0}^{c}f(x)v(x) dx = \int_{0}^{c}f(x)u(x) dx-i\int_{0}^{c}f(x)v(x) dx \Rightarrow i\int_{0}^{c}f(x)v(x) dx = -i\int_{0}^{c}f(x)v(x) dx \Rightarrow 2i\int_{0}^{c}f(x)v(x) dx=0[/tex]

    so we require that [tex]\int_{0}^{c}f(x)v(x) dx=0[/tex] for said equality to hold, in terms of g(x) this is [tex]\int_{0}^{c}f(x)\Im [g(x)] dx=0[/tex]

    which could hold for any number of different functions f and g.
  4. Apr 12, 2006 #3

    matt grime

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    As stated it is exceptionally trivial to find counter examples and benorin was being far to rigorous in proving something here. g=0 for all but some finite number of points where it is i will do. And you wonder why people don't trust your non-rigorous 'proofs'?
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