设 $z_1 、 z_2 \in \mathbf{C}$ 且 $z_1=\mathrm{i} \cdot \overline{z_2}$, 满足 $\left|z_1-1\right|=1$, 则 $\left|z_1-z_2\right|$ 的取值范围为
【答案】 $[0,2+\sqrt{2}]$
【解析】
设 $z_1-1=\cos \theta+\sin \theta \mathrm{i}, \therefore z_1=\cos \theta+1+\sin \theta \mathrm{i}=\mathrm{i} \cdot \overline{z_2}$, 即
$$
\begin{aligned}
& z_2=\sin \theta+(\cos \theta+1) \mathrm{i}, \quad \therefore z_1-z_2=(\cos \theta-\sin \theta+1)+(\sin \theta-\cos \theta-1) \mathrm{i} \\
& \left|z_1-z_2\right|^2=(\cos \theta-\sin \theta+1)^2+(\sin \theta-\cos \theta-1)^2=2(\sin \theta-\cos \theta-1)^2=2\left[\sqrt{2} \sin \left(\theta-\frac{\pi}{4}\right)-1\right]^2 \\
& \therefore\left|z_1-z_2\right|=\left|2 \sin \left(\theta-\frac{\pi}{4}\right)-\sqrt{2}\right| \in[0,2+\sqrt{2}]
\end{aligned}
$$