Is Every Point on the Unit Circle an Accumulation Point of Zeros for f(z)?

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Homework Help Overview

The discussion revolves around the function f(z) defined as a Weierstrass Product and the inquiry into whether every point on the unit circle is an accumulation point of its zeros. The participants explore the nature of the zeros of f(z) and their relationship to points on the unit circle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the form of the zeros of f(z) and their proximity to points on the unit circle. There is a question regarding the terminology used, specifically the distinction between 'phase' and 'argument'.

Discussion Status

The discussion is ongoing, with some participants providing insights into the nature of the zeros and their relation to the unit circle. There is a recognition of the mathematical context, but no consensus has been reached on the proof or implications of the original question.

Contextual Notes

Participants note the relevance of the Weierstrass Product in the context of the problem, and there is an acknowledgment of the mathematical background of some contributors, which may influence their perspectives.

Dustinsfl
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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.
 
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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 ...
 
By phase, you mean argument?
 
yes, I'm an electrical engineer :)
 
sunjin09 said:
yes, I'm an electrical engineer :)

Ok thanks. That problem was relatively easy.
 

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