When the product $|\prod_{k=1}^n(1-z^k)|$ tends to infinity?

Question

Let $z\in\mathbb{C}$. I would like to know when the product

$$|\prod_{k=1}^n(1-z^k)|$$

tends to infinity?

My attempt : We have:

$$|\prod_{k=1}^n(1-z^k)|= \prod_{k=1}^n |1-z^k|$$

Since we have:

$$|1-z^k|\leq 1 +|z^k|= 1+|z|^k$$

we obtain that: \begin{equation} |\prod_{k=1}^n(1-z^k)|\leq 1+ \prod_{k=1}^n |z|^k \end{equation}

Let $|z|=R$. Then:

\begin{equation} |\prod_{k=1}^n(1-z^k)|\leq 1+ \prod_{k=1}^n R^k=1+R^{n(n+1)/2} \end{equation}

Case 1: If $R<1$ then the product is bounded and cannot be divergent. Case 2: If $R>1$ then the product is finite and I couldn't conclude when it diverges.

Is that correct? My main question is to find when this product tends to infinity, even if we can consider a sequence of $z_{n_k}$ such that the product diverges

Many thank's for helping me.

Answer 1

If $|z| >2$ then $|\prod_{k=1}^{n} (1-z^{k})|>(|z|-1)^{n} \to \infty$.

A stronger result: if $|z| >1$ choose $m$ such that $|z^{m} |>2$. Then $|\prod_{k=m}^{n} (1-z^{k})|> (|z|^{m} -1)^{n-m+1} \to \infty$. Hence $|z| >1$ is good enough.