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Math Help - [SOLVED] complex analysis help...

  1. #1
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    [SOLVED] complex analysis help...

    Hello, thank you for reading.
    f(z) is a holomorfic (<=>analytic) function in a disk D (D = {z: |z-z0|<R}, for some R) (notice that it's an open disk), and f is not identically zero.
    I need to prove/disprove the next two claims.
    My trouble is that I can in no way understand the difference between them! So what I'd actually like is an explenation of the difference of the two claims...
    Thank you very much for reading/responding.
    Here are the claims:
    1) in every sub-Disk that is contained in D (not sure how to say that in English, but "totally contained", meaning it doesn't equal D) there is a finite number of "zeroes" of f(z). (z is zero <=> f(z)=0).
    2) every circle that is contained in D (circle: D'={z: |z-z1|=R for some R), circles a finite number of zeroes of f(z).

    Simply cannot see the difference between the claims. D is open - that's the problem...

    Thanks in advance!
    Tomer.
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  2. #2
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    Quote Originally Posted by aurora View Post
    Hello, thank you for reading.
    f(z) is a holomorfic (<=>analytic) function in a disk D (D = {z: |z-z0|<R}, for some R) (notice that it's an open disk), and f is not identically zero.
    I need to prove/disprove the next two claims.
    My trouble is that I can in no way understand the difference between them! So what I'd actually like is an explanation of the difference of the two claims...
    Thank you very much for reading/responding.
    Here are the claims:
    1) in every sub-Disk that is contained in D (not sure how to say that in English, but "totally contained", meaning it doesn't equal D) there is a finite number of "zeroes" of f(z). (z is zero <=> f(z)=0).
    2) every circle that is contained in D (circle: D'={z: |z-z1|=R for some R), circles a finite number of zeroes of f(z).

    Simply cannot see the difference between the claims. D is open - that's the problem...
    To deal with the language question first, the English expression is "properly contained in".

    The difference between (1) and (2) is this. For (1), you are looking at an open disk that is properly contained in the open unit disk, but could go right up to the boundary of it. For example, it could be the open disk of radius 1/2 centred at the point z=1/2. But the circle in (2), together with its interior, forms a closed disk. It has to stay strictly away from the boundary of the unit disk. More precisely, it must be contained within the disk (centred at the origin) of radius 1δ for some δ>0.
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  3. #3
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    Opalg, thank you so much, you are of course right....
    in the example you mentioned, there's that point, (1,0), of whom every neighborhood of it containes both points outside the main disk and inside the small disk. In the second claim I would be able to create such a "seperating" neighborehood.

    I got it! (not the solution quite yet, but at least I know how to attack it...)
    THANK YOU!!!
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  4. #4
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    Let D_0 be a subdisk of D.
    Now consider \bar D_0.
    Assume that are infinitely many zeros in \bar D_0.
    Let \{z_n\} be a sequence of distinct points in \bar D_0.
    By Bolzano-Weierstrass theorem there is a convergent subsequence with limit in \bar D_0.
    Argue for a contradition.
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