(1) $\displaystyle\underset{1\leq j \leq k_n}{\max}|\theta_{nj}|\rightarrow0;$
(2) $\displaystyle\sum_{j=1}^{k_n}|\theta_{nj}|\leq M<\infty$, where $M$ does not depend on $n$;
(3) $\displaystyle\sum_{j=1}^{k_n}\theta_{nj}\rightarrow\theta$, where $\theta$ is a (finite) complex number.
Then we have $$\prod_{j=1}^{k_n}(1+\theta_{nj})\rightarrow e^\theta.$$
$\bullet$ Proof.
By (1), $\exists\,n_0$ such that $\forall\,n\geq n_0$, $|\theta_{nj}|\leq\frac{1}{2}$ for all $j$ then $$(1+\theta_{nj})\neq0$$so that the product $\prod_{j=1}^{k_n}(1+\theta_{nj})$ is zero-free. For this $n\geq n_0$, by Taylor's expansion, $\displaystyle\left(f(x)=f(0)+\frac{f'(0)}{1!}x+\sum_{m=2}^\infty\frac{f^{(m)}(0)}{m!}x^m\right)$ $$\log{(1+\theta_{nj})}=\theta_{nj}+\Lambda|\theta_{nj}|^2,\;|\Lambda|\leq1,$$because $$\begin{array}{rl}\left|\log{(1+\theta_{nj})}-\theta_{nj}\right|&=\left|\sum_{m=2}^\infty\frac{(-1)^{m-1}}{m}\theta_{nj}^m\right|\leq\sum_{m=2}^\infty\frac{\left|\theta_{nj}\right|^m}{m}=\frac{|\theta_{nj}|^2}{2}\sum_{m=2}^\infty\frac{\left|\theta_{nj}\right|^{m-2}}{m/2}\\ \left(\because\,|\theta_{nj}|\leq\frac{1}{2},\,m\geq2\right)& \leq\frac{|\theta_{nj}|^2}{2}\sum_{m=2}^\infty\left(\frac{1}{2}\right)^{m-2}=|\theta_{nj}|^2\leq1.\end{array}$$Hence, $$\sum_{j=1}^{k_n}\log{(1+\theta_{nj})}=\sum_{j=1}^{k_n}\theta_{nj}+\Lambda'\sum_{j=1}^{k_n}|\theta_{nj}|^2,\;|\Lambda'|\leq1$$where $\Lambda'$ might be different to $\Lambda$. By the conditions, we have (1) and (2) implies $$\sum_{j=1}^{k_n}|\theta_{nj}|^2\leq\underset{1\leq j \leq k_n}{\max}|\theta_{nj}|\sum_{j=1}^{k_n}|\theta_{nj}|\leq M\underset{1\leq j \leq k_n}{\max}|\theta_{nj}|\rightarrow0.$$Thus by (3), $$\sum_{j=1}^{k_n}\log{(1+\theta_{nj})}\rightarrow\theta\implies\prod_{j=1}^{k_n}(1+\theta_{nj})\rightarrow e^\theta.$$
$\Box$
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