Question 108: Given a complex number $z$ satisfying $|z|=1$, then the range of values for $\left|(z-2)(z+1)^{2}\right|$ is $\qquad$

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Question 108, Solution Method One: First, it is clear that $\left|(z-2)(z+1)^{2}\right| \geq 0$, with equality holding when $z=-1$.

On the other hand, establish the complex plane, with $A: 2, B:-1, C: z$, then $\left|(z-2)(z+1)^{2}\right|=a^{2} b_{0}$, and according to Stewart's Theorem:
$$
\begin{array}{l}
1=|O C|^{2}=\frac{|B C|^{2} \cdot|O A|+\left|A C^{2}\right| \cdot|O B|}{|A B|}-|O A| \cdot|O B|=\frac{2 a^{2}+b^{2}}{3}-2 \times 1 \\
\Rightarrow 9=2 a^{2}+b^{2} \geq 3 \sqrt[3]{a^{2} \cdot a^{2} \cdot b^{2}} \\
\Rightarrow a^{2} b \leq 3 \sqrt{3}
\end{array}
$$

Equality holds when $z=\frac{1 \pm \sqrt{3} i}{2}$. Therefore, the range of $\left|(z-2)(z+1)^{2}\right|$ is $[0,3 \sqrt{3}]$.
Solution Method Two: First, it is clear that $\left|(z-2)(z+1)^{2}\right| \geq 0$, with equality holding when $z=-1$.
On the other hand,
$$
\begin{array}{l}
\left|(\mathrm{z}-2)(\mathrm{z}+1)^{2}\right|^{2}=(\mathrm{z}-2)(\overline{\mathrm{z}}-2) \cdot((\mathrm{z}+1)(\overline{\mathrm{z}}+1))^{2} \\
=(5-2(\mathrm{z}+\overline{\mathrm{z}})) \cdot(2+(\mathrm{z}+\overline{\mathrm{z}}))^{2} \\
\leq\left(\frac{(5-2(\mathrm{z}+\bar{z}))+(2+(\mathrm{z}+\mathrm{z}))+(2+(z+\bar{z}))}{3}\right)^{3}=27
\end{array}
$$

Thus, $\left|(z-2)(z+1)^{2}\right| \leq 3 \sqrt{3}$, with equality holding when $z=\frac{1 \pm \sqrt{3} i}{2}$.
Therefore, the range of $\left|(z-2)(z+1)^{2}\right|$ is $[0,3 \sqrt{3}]$.