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I Identical zero function in the complex plane

  1. Jun 1, 2017 #1
    Hi! If a holomorphic function ##f:G \to C##, where ##G## is a region in the complex plane is equal to zero for all values ##z## in a disk ##D_{[z_0,r]}##, inside ##G##, is it zero everywhere in the region G? And if this is true, does it mean that if an entire function is zero in a disk, it is zero in the whole complex plane? Thank you!
     
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  3. Jun 1, 2017 #2

    WWGD

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    Yes, it is enough for f to be zero on a dense subset inside of the region G , i.e., a subset containing a limit point of G, in G. This is often called the Identity Theorem. EDIT: A related theorem is that the same holds for an entire function with uncountably-many zeros ( in the plane, of course, not extended-value).
     
    Last edited: Jun 1, 2017
  4. Jun 2, 2017 #3

    WWGD

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    I leave it as a challenge for you to find a Real-valued continuous, even differentiable non-constant function that has uncountably-many zeros.
     
  5. Jun 4, 2017 #4

    mathwonk

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    (as you know), even a countable convergent sequence of zeroes suffices to make a holomorphic function dead zero. the basic point is that an analytic entire function is determined by its power series coefficients, i.e. its derivatives at the center of the expansion, say at z = zero.
     
  6. Jun 4, 2017 #5

    WWGD

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    Yes, I was using Weirstrass' theorem to show that an uncountable subset of the plane must have a limit point in it.
     
  7. Jun 4, 2017 #6

    mathwonk

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    thats why i mentioned the more fundamental result, since, as you say, yours is a corollary.
     
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