# Convergence of infinite product of complex numbers

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.

• Note that for infinite products, the terminology is usually that the product diverges to $0$. – Aryaman Maithani Aug 24 '20 at 18:18
• @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. – kroner Aug 24 '20 at 18:53
• @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
• @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
• @AryamanMaithani Thanks for your comment. – zhw. Aug 25 '20 at 0:04

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.

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

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.

• Thanks, I think you mean converges to a nonzero number? – kroner Aug 24 '20 at 20:21
• 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
• 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 6:29
• 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