1

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.

Croc2Alpha
  • 3,927
  • If $z=e^{2\pi i \alpha}$ with $\alpha$ rational, will converge to zero. – mjw Aug 24 '20 at 18:11
  • If $|z|>1,$ diverges. – mjw Aug 24 '20 at 18:12
  • For $\alpha=p/q$ rational, with $p$ and $q$ integer, $n \alpha$ is an integer so $z^n=1.$ Also, $1-z^n = (1-z)(1+z+z^2+ \cdots + z^{n-1})$, so as $n\to\infty$, the series (a term of the product) diverges. – mjw Aug 24 '20 at 18:17
  • 2
    Note that for infinite products, the terminology is usually that the product diverges to $0$. – Aryaman Maithani Aug 24 '20 at 18:18
  • Okay. Not disagreeing. Why is that? (diverges to 0) – mjw Aug 24 '20 at 18:20
  • @mjw: See the comments on this answer. TL;DR seems to be that stating it this way seems to be compatible with most theorems (analogues of those for summations) without having to treat edge cases. – Aryaman Maithani Aug 24 '20 at 18:30
  • kroner: I can't add anything more concrete than what @mjw did. However, I'll refer this which says that ${z^n : n \in \Bbb N}$ is dense in the unit circle $S^1$ when $z = e^{i\pi\alpha}$ with $\alpha$ irrational. However, concluding anything from this seems difficult since we see that $1 - z^n$ can get arbitrarily close to both $0$ and $2$. Since $1 - z^n \not\to 1$, what we can conclude is that the product cannot converge. (But it could diverge to $0$(!)) – Aryaman Maithani Aug 24 '20 at 18:37
  • @kroner: I won't comment on case $|z| \neq 1$ since I didn't really understand the currently given answer. You are correct for the rational argument (though you probably mean rational argument of $\pi$). However, for the irrational, I just said that it could diverge to $0$. I did not prove that. (I don't even know whether to believe that.) All I concluded was that it cannot converge. So, it may either diverge to $0$, to $\infty$, or genuinely not have any limit (some oscillatory behaviour). – Aryaman Maithani Aug 24 '20 at 18:48
  • 1
    @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. – zhw. Aug 24 '20 at 20:09
  • 1
    @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.) – Aryaman Maithani Aug 24 '20 at 20:26
  • 1
    @AryamanMaithani Thanks for your comment. – zhw. Aug 25 '20 at 00:04

2 Answers2

2

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.

mjw
  • 8,647
  • 1
  • 8
  • 23
  • 1
    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 $|z|<1.$ – mjw Aug 24 '20 at 20:23
  • 1
    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. – mjw Aug 24 '20 at 20:29
  • @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. – Aryaman Maithani Aug 25 '20 at 06:29
  • 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.) – mjw Aug 25 '20 at 11:23
1

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.

Angina Seng
  • 158,341