16. First, we have $P(x)=Q(x) R(x)$ for $Q(x)=x^{m}-|a|^{m} e^{40}$ and $R(x)=$ $x^{m}-|a|^{m} e^{-i \theta}$, where $e^{i \varphi}$ means of course $\cos \varphi+i \sin \varphi$. It remains to factor both $Q$ and $R$. Suppose that $Q(x)=\left(x-q_{1}\right) \cdots\left(x-q_{m}\right)$ and $R(x)=\left(x-r_{1}\right) \cdots\left(x-r_{m}\right)$.
Considering $Q(x)$, we see that $\left|q_{k}^{m}\right|=|a|^{m}$ and also $\left|q_{k}\right|=|a|$ for $k=$ $1, \ldots, m$. Thus we may put $q_{k}=|a| e^{i \beta_{k}}$ and obtain by de Moivre's formula $q_{k}^{m}=|a|^{m} e^{i m \beta_{k}}$. It follows that $m \beta_{k}=\theta+2 j \pi$ for some $j \in Z$, and we have exactly $m$ possibilities for $\beta_{k}$ modulo $2 \pi: \beta_{k}=\frac{\theta+2(k-1) \pi}{m}$ for $k=1,2, \ldots, m$.
Thus $q_{k}=|a| e^{i \beta_{k}}$; analogously we obtain for $R(x)$ that $r_{k}=|a| e^{-i \beta_{k}}$. Consequently,
$$x^{m}-|a|^{m} e^{i \theta}=\prod_{k=1}^{m}\left(x-|a| e^{i \beta_{k}}\right) \quad \text { and } \quad x^{m}-|a|^{m} e^{-i \theta}=\prod_{k=1}^{m}\left(x-|a| e^{-i \beta_{k}}\right) \text {. }$$

Finally, grouping the k.th factors of both polynomials, we get
$$\begin{aligned} P(x) & =\prod_{k=1}^{m}\left(x-|a| e^{i \beta_{k}}\right)\left(x-|a| e^{-i \beta_{k}}\right)=\prod_{k=1}^{m}\left(x^{2}-2|a| x \cos \beta_{k}+a^{2}\right) \\ & =\prod_{k=1}^{m}\left(x^{2}-2|a| x \cos \frac{\theta+2(k-1) \pi}{m}+a^{2}\right) . \end{aligned}$$