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

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Let f(z) = \prod\limits_{n = 1}^{\infty}(1 - nz^n)

Prove that each point on the unit circle is an accumulation point of zeros of f

So we have that z = \sqrt[n]{1/n}. 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 \{\sqrt[n]{1/n} e^{i2\pi\frac{k}{n}}|n,k\in Z_+\}, now for any z=e^{i\phi} on unit circle, there exit n, k such that \sqrt[n]{1/n} e^{i2\pi\frac{k}{n}} is close enough to z=e^{i\phi} 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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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