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Q on proof of periods of non-constant meromorphic functions Watch

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    Theorem attached.

    I know the theorem holds for a discrete subgroup of C more generally, C the complex plane, and that the set of periods of a non-constant meromorphic function are a discrete subset.

    I have a question on part of the proof (showing the second type, integral muliples nw)

    On proving the second case \Omega, the set of periods \Omega = nw , \Omega \neq \{0\} is considered and we show that exists w_1 \in \Omega /\{0\} with the least value of |w_1|.

    A step of doing this in my text is, and this is my problem:

    Since \Omega is discrete, there is some \epsilon > 0 st for any disc |z| < \epsilon contains no elements of \Omega / \{0\}, it follows that for any w \in \Omega , the disc |z-w|< \epsilon contains no elements of \Omega / \{w\}

    . I don't understand why/ how this follows from the fact that is a group, why this 'translation' is possible with the same radius \epsilon follows from the fact it is a group?

    No idea where to start, any help much appreciated, but my guess would be that the proof will depend on what the operation of the group is, but this isn't specified in the theorem? or is just assumed to be addition?


    Many thanks
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    (Original post by xfootiecrazeesarax)
    Since \Omega is discrete, there is some \epsilon > 0 st for any disc |z| < \epsilon contains no elements of \Omega / \{0\}, it follows that for any w \in \Omega , the disc |z-w|< \epsilon contains no elements of \Omega / \{w\}
    Is this actually true? It seems easy to construct a pair of periods where it fails e.g. take \Omega=\{ 1, 1+i/4 \} and \epsilon = 1/2.

    But I know very little about periodic complex functions, so maybe I'm confused.
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    (Original post by xfootiecrazeesarax)
    A step of doing this in my text is, and this is my problem:

    Since \Omega is discrete, there is some \epsilon > 0 st for any disc |z| < \epsilon contains no elements of \Omega / \{0\}, it follows that for any w \in \Omega , the disc |z-w|< \epsilon contains no elements of \Omega / \{w\}

    . I don't understand why/ how this follows from the fact that is a group, why this 'translation' is possible with the same radius \epsilon follows from the fact it is a group?
    Suppose it's false, then we can find z with |z-w| < \epsilon and z \in \Omega. But then since \Omega is a group, if z, w \in \Omega then z-w \in \Omega. But this then contradicts the definition of epsilon.[/QUOTE]

    (Original post by atsruser)
    Is this actually true? It seems easy to construct a pair of periods where it fails e.g. take \Omega=\{ 1, 1+i/4 \} and \epsilon = 1/2.
    I'm using this to give an explict example of what's going on. If Omega is as you describe, then f(z) = f(z+1) and f(z) = f(z+1+i/4) for all z. But then f(z+1) = f(z+1+i/4) for all z and so f(z) = f(z+i/4) for all z. So i/4 \in \Omega and so you can't take epsilon = 1/2 as you suggest.
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    (Original post by DFranklin)
    I'm using this to give an explict example of what's going on. If Omega is as you describe, then f(z) = f(z+1) and f(z) = f(z+1+i/4) for all z. But then f(z+1) = f(z+1+i/4) for all z and so f(z) = f(z+i/4) for all z. So i/4 \in \Omega and so you can't take epsilon = 1/2 as you suggest.
    I'm glad there's someone here who knows what they're talking about. Yes, my example fails since it's not a group of periods - I guess it's the generators at best.

    And PRSOM
 
 
 
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