Convergence of infinite product of complex numbers

Question

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.

Answer 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.

Answer 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.