1. Let $a, b \in \mathbf{R}$, the complex number $z=a+4 \mathrm{i}$, and $\frac{z}{z+b}=4 \mathrm{i}$. Then $b=(\quad)$.
(A) -16
(B) 1
(C) 16
(D) 17
$-1 . \mathrm{D}$.
From the problem, we know
$$
\begin{array}{l}
a+4 \mathrm{i}=[(a+b)+4 \mathrm{i}] 4 \mathrm{i}=-16+4(a+b) \mathrm{i} \\
\Rightarrow a=-16,4(a+b)=4 \Rightarrow b=17 .
\end{array}
$$