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If the roots of \(z^3+i z^2+2 i=0\) represent the vertices of a \(\triangle \mathrm{ABC}\) in the Argand plane, then the area of the triangle is
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\(\begin{aligned}
& \mathrm{z}^3+\mathrm{i} \mathrm{z}^2+2 \mathrm{i}=(\mathrm{z}-\mathrm{i})\left(\mathrm{z}^2+2 \mathrm{iz}-2\right)=0 \\
& \Rightarrow \mathrm{z}=\mathrm{i}, 1-\mathrm{i},-1-\mathrm{i} \\
& \text {Let } \mathrm{A}=(0,1), \mathrm{B}=(1,-1), \mathrm{C}=(-1,-1), \\
& \therefore \triangle \mathrm{ABC}=\frac{1}{2}\left\|\begin{array}{ccc}
0 & 1 & 1 \\
1 & -1 & 1 \\
-1 & -1 & 1
\end{array}\right\| \\
& =\frac{1}{2}|-2-2|=2
\end{aligned}\)
& \mathrm{z}^3+\mathrm{i} \mathrm{z}^2+2 \mathrm{i}=(\mathrm{z}-\mathrm{i})\left(\mathrm{z}^2+2 \mathrm{iz}-2\right)=0 \\
& \Rightarrow \mathrm{z}=\mathrm{i}, 1-\mathrm{i},-1-\mathrm{i} \\
& \text {Let } \mathrm{A}=(0,1), \mathrm{B}=(1,-1), \mathrm{C}=(-1,-1), \\
& \therefore \triangle \mathrm{ABC}=\frac{1}{2}\left\|\begin{array}{ccc}
0 & 1 & 1 \\
1 & -1 & 1 \\
-1 & -1 & 1
\end{array}\right\| \\
& =\frac{1}{2}|-2-2|=2
\end{aligned}\)
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