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Accumulation point unit disc

  1. Apr 24, 2012 #1
    Let [itex]f(z) = \prod\limits_{n = 1}^{\infty}(1 - nz^n) [/itex]

    Prove that each point on the unit circle is an accumulation point of zeros of [itex]f [/itex]

    So we have that [itex]z = \sqrt[n]{1/n} [/itex]. Now where do I go from here?

    Probably should note that this is a Weierstrass Product.
     
    Last edited: Apr 24, 2012
  2. jcsd
  3. Apr 24, 2012 #2
    The set of all zeros in of f(z) is [itex]\{\sqrt[n]{1/n} e^{i2\pi\frac{k}{n}}|n,k\in Z_+\}[/itex], now for any [itex]z=e^{i\phi}[/itex] on unit circle, there exit n, k such that [itex]\sqrt[n]{1/n} e^{i2\pi\frac{k}{n}}[/itex] is close enough to [itex]z=e^{i\phi}[/itex] in both amplitude and phase ...
     
  4. Apr 24, 2012 #3
    By phase, you mean argument?
     
  5. Apr 24, 2012 #4
    yes, I'm an electrical engineer :)
     
  6. Apr 24, 2012 #5
    Ok thanks. That problem was relatively easy.
     
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