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Analytic function ( zeros)

  1. Apr 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Let f be a nonconstant function. Prove that f has at most countably many zeros

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 11, 2010 #2

    Dick

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    What do you know about analytic functions where the zeros have a accumulation point? Can an uncountable set not have a accumulation point?
     
  4. Apr 12, 2010 #3
    I don't know what you mean by that since zeros of ananlytic function are isolated points and non of them is a limit point.
     
  5. Apr 12, 2010 #4

    HallsofIvy

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    That was exactly Dick's point! If there were an uncountable set of zeros, there would have to be an accumulation point, contradicting the fact that the zeros are isolated.
     
  6. Apr 12, 2010 #5
    You mean that we can prove it by contradicion.

    Let S be the set of zeros of f and suppose that it is uncountable.

    then we can get a sequence of these zeros convergent to a point in z.

    This gives that the zeros are not isolated.

    but, if this is what you mean, how can we be sure that we have a convergent series in S.
     
  7. Apr 12, 2010 #6

    Dick

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    Right. So the trick is to show there is a convergent sequence. Divide the plane up into regions of finite area that cover the whole plane. Like squares of edge size one centered on each point m+n*i, where m and n are integers. Can you show at least one of those squares contains an infinite number of zeros?
     
  8. Apr 12, 2010 #7
    suppose that f doesn't have caountably many zeros.
    then it has infinitely many zeros in one of this partition.
    Infinitely many zeros in a bounded set implies an accumulation point of zeroes, and an accumulation point of zeros for an analytic function implies that that function is zero everywhere.
    contradiction

    So f has at most countably many zeros.
     
  9. Apr 12, 2010 #8

    Dick

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    Ok. You didn't make it terribly clear why one element of the partition contains infinitely many zeros. Do you know why?
     
  10. Apr 13, 2010 #9
    I think this comes from the infinitely many zeros ( uncountable assumption)
     
  11. Apr 13, 2010 #10

    Dick

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    Still unclear. You can have one zero in each square. That's an infinite number of zeros. Sure, it's the uncountable assumption but how does it work?
     
  12. Jun 14, 2010 #11
    http://math.nyu.edu/student_resources/wwiki/index.php/Complex_Variables:_2006_January:_Problem_5" [Broken] using Taylor's theorem.
     
    Last edited by a moderator: May 4, 2017
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