9. If $x=\frac{1+\sqrt{5} \mathrm{i}}{2}$, then the value of $2 x^{3}+(x+1)^{2}$ is $\qquad$

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9. $\sqrt{5} \mathrm{i}-\frac{5}{2}$.
$\frac{1+\sqrt{5}}{2}$ is a root of the equation $x^{2}-x+\frac{3}{2}=0$, then
$$
\begin{array}{l}
2 x^{3}+(x+1)^{2}=2 x^{3}+x^{2}+2 x+1 \\
=(2 x+3)\left(x^{2}-x+\frac{3}{2}\right)+2 x-\frac{7}{2}=\sqrt{5} \text { i }-\frac{5}{2}
\end{array}
$$