Originally Posted by

**bkarpuz** This shows that product cannot tend to $\displaystyle 0$ (i.e., $\displaystyle 0+\mathrm{i}0$).

On the other hand, we have

$\displaystyle \bigg|\prod_{i=1}^{\infty}z_{i}\bigg|\leq\prod_{i= 1}^{\infty}(|x_{i}|+|1-x_{i}|)=1$

showing that the product can not also diverge (in the sense of complex modulus).

Then, the complex sequence $\displaystyle w_{n}:=\prod_{i=1}^{n}z_{i}$ lies in the annulus$\displaystyle \mathcal{A}:\ \ell\leq|z|\leq1$.

Due to Bolzano-Weierstrass theorem (considered separetely for the real and imaginary parts),

there exists a subsequence $\displaystyle w_{n_{k}}$ of $\displaystyle w_{n}$ which converges to some $\displaystyle w\in\mathcal{A}$.

Recall that $\displaystyle \lim_{n\to\infty}z_{n}=\mathrm{i}$ and $\displaystyle w_{n_{k}+p}=w_{n_{k}}\prod_{i=n_{k}+1}^{n_{k}+p}z_ {i}$,

which shows that $\displaystyle \lim_{k\to\infty}w_{n_{k}+p}=\mathrm{i}^{p}w$ for $\displaystyle p=0,1,\cdots$.

Therefore, the product accumulates to at least (also at most) four different points $\displaystyle \pm w,\pm\mathrm{i}w$ since $\displaystyle w\neq0$. $\displaystyle \rule{0.2cm}{0.2cm}$

But I just wanted to know if it is possible to find (by hand) what $\displaystyle w$ is.