1. (2000 Hebei Province Competition Question) If the complex number $z$ satisfies $|z|=1, A(-1,0), B$ $(0,-1)$ are two points on the complex plane, then the function $f(z)=|(z+1)(\bar{z}-\mathrm{i})|$ takes its maximum value when, the figure formed by points $Z, A, B$ on the complex plane is ( ).
A. Equilateral Triangle
B. Isosceles Triangle
C. Right Triangle
D. Isosceles Right Triangle
1. Choose B. Reason:

Let $z=\cos \theta+i \sin \theta$, then
$$
\begin{aligned}
(z+1)(\bar{z}-\mathrm{i}) & =(1+\cos \theta+\mathrm{i} \sin \theta)[\cos \theta-(1+\sin \theta) \mathrm{i}] \\
& =\cos \theta(1+\cos \theta)+\sin \theta(1+\sin \theta)+\mathrm{i}[\sin \theta \cdot \cos \theta-(1+\cos \theta)(1+\sin \theta)] \\
& =(\sin \theta+\cos \theta+1)-\mathrm{i}(1+\sin \theta+\cos \theta) .
\end{aligned}
$$

So
$$
\begin{aligned}
f(z) & =\sqrt{(1+\sin \theta+\cos \theta)^{2}+(\sin \theta+\cos \theta+1)^{2}} \\
& =\sqrt{2\left[1+\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)\right]^{2}}
\end{aligned}
$$

When $\theta=2 k \pi+\frac{\pi}{4}(k \in \mathbf{Z})$, $f(z)$ is maximized, with the maximum value being $2+\sqrt{2}$. At this time, $z=\cos \frac{\pi}{4}+$ $\operatorname{isin} \frac{\pi}{4}$. The positions of points $A$, $B$, and $Z$ are as shown in the figure, clearly, $\triangle A B Z$ is an isosceles triangle, but it is neither equilateral nor right-angled.