9. If $|z|=2, u=\left|z^{2}-z+1\right|$, then the minimum value of $u$ is $\qquad$ (here $z \in \mathbf{C}$ ).
9. $\frac{3}{2} \sqrt{3} \quad u=|z|\left|z+\frac{1}{z}-1\right|$, let $z=x+y \mathrm{i}, x, y \in \mathbf{R}, u=\sqrt{\left(2 x-\frac{5}{2}\right)^{2}+\frac{27}{4}},|x| \leqslant 2$, so when $x=\frac{5}{4}$, $u$ has the minimum value $\frac{3}{2} \sqrt{3}$.