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Sequence of complex numbers

  1. Mar 12, 2012 #1
    Let <zn> be a sequence complex numbers for which Im(zn) is bounded below.
    Prove <e^(i*zn)> has a convergent subsequence.

    My question on this is what possible help could the boundedness of the Im(zn) to this proof and what theorem might be of help?
     
  2. jcsd
  3. Mar 12, 2012 #2

    micromass

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    Do you know anything about compactness or Bolzano-Weierstrass??
     
  4. Mar 12, 2012 #3

    mathwonk

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    do you know how e^z behaves geometrically?
     
  5. Mar 13, 2012 #4
    Yes, I do know the B-W theorem. I looked at it but it applies to bounded sequences. This sequence is only bounded below and only on the Im part. I suppose since you asked you see a way it is still applicable.

    It was my understanding e^z forms a circle in the complex. Is this what you mean?
     
    Last edited: Mar 13, 2012
  6. Mar 13, 2012 #5

    micromass

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    Try to show that the sequence is bounded. Try to show that there is a constant C such that

    [tex]|e^{iz_n}|\leq C[/tex]
     
  7. Mar 13, 2012 #6
    Edit: The latex was not loading but now I see what you mean. I apologize if this sounds ignorant but I did not think e^(i*zn) was bounded. If you are saying I should show it is then I assume it can be.
    When I think of this sequence I must be seeing it all wrong. could you give any explanation of how it behaves. If it helps I see nothing but a circle when I think of it. What attribute am I missing?
     
    Last edited: Mar 13, 2012
  8. Mar 13, 2012 #7

    mathwonk

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    circles are bounded, right?
     
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