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In my studies of complex analysis, I have encountered this question:

We are asked to find the complex numbers $ z $ for which the infinite countable product converges $$\prod_{n=1}^{\infty} (1-z^n)$$ to a nonzero number.

I know what it means for a product to converge (its sequence of partial products converges to nonzero number) but I cannot find any numbers for which this converges, perhaps the ratio test? Though when I try to apply it it doesn't seem to work. I thought to split to cases when $|z|>1,|z|<1,|z|=1$ but again intractable. I need to find all complex numbers for which the product converges and to show that is indeed everything. Thanks to all helpers. ******EDIT: fixed it to converge to nonzero so complex analysts won't disagree with me on terminology.

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    $\begingroup$ Note that for infinite products, the terminology is usually that the product diverges to $0$. $\endgroup$ – Aryaman Maithani Aug 24 '20 at 18:18
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    $\begingroup$ @AryamanMaithani: yes for rationals multiples of $\pi$ we have divergence to zero but for irrational all we can conclude is that it cannot converge? Thank you very much. $\endgroup$ – kroner Aug 24 '20 at 18:53
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    $\begingroup$ @AryamanMaithani I think you overestimate the usage of the terminology "diverges to $0.$" It seems to me to be a weird old school thing. Rudin, for example, doesn't mention it in RCA. $\endgroup$ – zhw. Aug 24 '20 at 20:09
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    $\begingroup$ @zhw: possibly. I wouldn't defend my stance much. However, it does have the nice benefit here that $a_n \not\to 1$ lets us conclude that $\prod a_n$ doesn't converge. (Analogous to the summation case with $0$.) If we counted $0$ as convergence, it would have to be addressed separately. (But yes, I agree that it's not difficult to add that; just personal preference.) $\endgroup$ – Aryaman Maithani Aug 24 '20 at 20:26
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    $\begingroup$ @AryamanMaithani Thanks for your comment. $\endgroup$ – zhw. Aug 25 '20 at 0:04
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It converges as an infinite product if $|z|<1$. It is zero when $z$ is a root of unity, but complex analysts would claim it diverges then too. It is certainly divergent for all other $z$. For $|z|<1$ the (principal) logarithms of $1-z^n$ are asymptotic to $-z^n$ so the product converges.

For $|z|>1$ the terms do not converge to $1$, while for $|z|=1$ things are much more delicate.

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  • $\begingroup$ Thank you, so how can I handle $|z|=1$? Can you please show me? $\endgroup$ – kroner Aug 24 '20 at 18:15
  • $\begingroup$ can you please clarify why for $|z|=1$ we have divergence? $\endgroup$ – kroner Aug 24 '20 at 18:28
  • $\begingroup$ Fixed the terminology in the question so for $|z|<1$ it converges to nonzero? $\endgroup$ – kroner Aug 24 '20 at 18:34
  • $\begingroup$ so for $|z|<1$ it diverges to zero or converges to nonzero? $\endgroup$ – kroner Aug 24 '20 at 18:58
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Just to discuss $|z|<1$:

$$\displaystyle Q = \prod_{n=1}^\infty(1-z^n)$$

$$\log Q = \sum_{n=1}^\infty \log (1-z^n)$$

The $n$th term of the series is $a_n=\log(1-z^n)$ and $\lim \sup |a_n|^{1/n} =|z|$, so if $|z|<1$, the series (and thus the product) converges.

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  • $\begingroup$ Thanks, I think you mean converges to a nonzero number? $\endgroup$ – kroner Aug 24 '20 at 20:21
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    $\begingroup$ Well, yes. To converge or diverge (see heated discussion above for vocabulary!) to zero, one of the terms needs to go to zero, or $z^n\to 1$, which is impossible for $|z|<1.$ $\endgroup$ – mjw Aug 24 '20 at 20:23
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    $\begingroup$ Looking again at the other answer, it already has been stated that $\log (1-z^n) \sim -z^n$, so that really covered it already. $\endgroup$ – mjw Aug 24 '20 at 20:29
  • $\begingroup$ @mjw: I am not sure if I agree with "one of the terms needs to go to zero". Consider the product $\prod a_n$ where each $a_n$ is $1/2$. The infinite product diverges to $0$ but the terms are all constant. $\endgroup$ – Aryaman Maithani Aug 25 '20 at 6:29
  • $\begingroup$ Yes, you are right. Thank you. Let's rephrase that. Here $\forall n, |b_n| >0$ and $b_n\to 1$, where $b_n$ is a term in the product. (Writing $b_n$ where $a_n=\log b_n$ and $a_n$ is defined above to be a term in the sum.) $\endgroup$ – mjw Aug 25 '20 at 11:23

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