I would like to research behavior of $\prod_{k=0}^{\infty}(1+q^k), q \in \mathbb C$.
It is clear that for $q \in \mathbb R, |q|<1, \prod_{k=0}^{\infty}(1+q^k)$ converges because $\sum_{k=0}^{\infty}\log(1+q^k) \le \sum_{k=0}^{\infty}q^k$ converges.
It is also clear that for $q \ge 1, \prod_{k=0}^{\infty}(1+q^k) \ge \prod_{k=0}^{\infty}2 = \infty$ and for $q < -1, \prod_{k=0}^{\infty}(1+q^k)$ diverges, while for $q=-1, \prod_{k=0}^{\infty}(1+q^k) = 0$.
I am not sure how to approach this problem for $q \in \mathbb C$.
Edit
After some thinking I have realized that for $q \in \mathbb C, |q| > 1, \prod_{k=0}^{\infty}(1+q^k)$ diverges, because for $k_0$ large enough $|(1+q^k)| \ge |q|^k-1 \ge2, \forall k \ge k_0$ and thus $$|\prod_{k=0}^{n}(1+q^k)|= \prod_{k=0}^{n}|(1+q^k)| \ge A \prod_{k=k_0}^n2$$ for some $A > 0$. Thus $|\prod_{k=0}^{\infty}(1+q^k)|=\infty$ which makes impossible $\prod_{k=0}^{\infty}(1+q^k)$ to converge.
So what is left is to understand behavior for $|q| \le 1$.