16. Given a complex number $z$ satisfies $\arg (z+a+a \mathrm{i})=\frac{\pi}{4}$, and $\arg (z-a-a \mathrm{i})=\frac{5 \pi}{4}\left(a \in \mathbf{R}^{+}\right)$, find the range of $\arg (z-b+b \mathrm{i})(a>b>0)$.
16. The complex number that simultaneously satisfies the equations $\arg (z+a+a \mathrm{i})=\frac{\pi}{4}, \arg (z-a-a \mathrm{i})=\frac{5 \pi}{4}$ represents a point $P$ on the line segment $A B$, where $A(a, a), B(-a,-a) . \arg (z-b+b \mathrm{i})$ can be viewed as the angle from the positive $x$-axis to the ray $C P$, where $C(b,-b)$. The slopes of $A C$ and $B C$ are $k_{A C}=\frac{a+b}{a-b}, k_{B C}=\frac{a-b}{a+b}$. Therefore, the range of $\arg (z-b+b \mathrm{i})$ is $\left[\arctan \frac{a+b}{a-b}, \arctan \frac{a-b}{a+b}+\pi\right]$.