If the roots of $z^{3} + i z^{2} + 2 i$ = 0 represent the vertices of a triangle in the argand plane, then its area is

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Solution

The correct option is A
2

$z^{3} + i z^{2} + 2 i = 0$ $\Rightarrow$ $(z - i) (z^{2} + 2 i z - 2)$ =0

z = i, $\frac{- 2 i \pm \sqrt{4 i^{2} + 8}}{2}$

$\Rightarrow$ z = i, 1-i, -1-i

∴ area of triangle with vertices (0,1), (1, -1), (-1, -1)

= $\frac{1}{2}$ $∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 1 1 & - 1 & 1 - 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣$ = 2

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Similar questions

z^{3} + i z^{2} + 2 i = 0

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