## accumulation point unit disc

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?

## accumulation point unit disc

yes, I'm an electrical engineer :)

 Quote by sunjin09 yes, I'm an electrical engineer :)
Ok thanks. That problem was relatively easy.