1. (16 points) Given a complex number $z$ satisfying $|z|=1$. Find
$$
u=\left|z^{3}-3 z+2\right|
$$

the maximum value.
Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
From $|z|=1$, we know $x^{2}+y^{2}=1$, and $|x| \leqslant 1$.
Then $u=\left|z^{3}-3 z+2\right|=\left|z^{3}-z-2(z-1)\right|$
\[
\begin{array}{l}
=\left|(z-1)\left(z^{2}+z-2\right)\right| \\
=\left|(z-1)^{2}(z+2)\right| \\
=\left(\sqrt{(x-1)^{2}+y^{2}}\right)^{2} \sqrt{(x+2)^{2}+y^{2}} \\
=\sqrt{2-2 x} \sqrt{2-2 x} \sqrt{4 x+5} \\
\leqslant \sqrt{\left(\frac{(2-2 x)+(2-2 x)+(4 x+5)}{3}\right)^{3}} \\
=3 \sqrt{3} .
\end{array}
\]

When and only when $2-2 x=4 x+5$, i.e., $x=-\frac{1}{2}$,
the equality holds.
Therefore, when $z=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$, $u$ achieves its maximum value $3 \sqrt{3}$.