9. (16 points) Let
$$
\begin{array}{l} 
P(z)=z^{4}-(6 \mathrm{i}+6) z^{3}+24 \mathrm{iz}^{2}- \\
(18 \mathrm{i}-18) z-13 .
\end{array}
$$

Find the area of the convex quadrilateral formed by the four points in the complex plane corresponding to the four roots of $P(z)=0$.
II. 9. Notice that, $P(1)=0$.
Then $P(z)=(z-1)\left(z^{3}-(6 \mathrm{i}+6) z^{2}+24 \mathrm{iz}^{2}+\right.$ $(18 \mathrm{i}-5) z+13)$.

Let $Q(z)=z^{3}-(6 \mathrm{i}+5) z^{2}+(18 \mathrm{i}-5) z+13$.
Then $Q(\mathrm{i})=0$.
Hence $Q(z)=(z-\mathrm{i})\left(z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}\right)$.
Using the quadratic formula for $z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}=0$ yields $z=3+2 \mathrm{i}$ or $2+3 \mathrm{i}$.

Thus, the four roots of $P(z)=0$ are $1, \mathrm{i}, 3+2 \mathrm{i}, 2+3 \mathrm{i}$, which form a rectangle with a length of $2 \sqrt{2}$ and a width of $\sqrt{2}$, giving an area of $\frac{1}{2} \sqrt{2} \times 2 \sqrt{2}=2$.