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Question

If the roots of $$\displaystyle\ z^{3}+iz^{2}+2i=0$$ represents the vertices of a $$\displaystyle\ \Delta ABC$$ in the Argand plane then the area of the triangle is 


A
 372
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B
 374
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C
 2
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D
none of these
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Solution

The correct option is D $$\displaystyle\ 2$$
One obvious root is $$i$$
Then 
$$(z-i)(z^{2}+2iz-2)=0$$
$$z=i$$
And 
$$z^{2}+2iz-2=0$$
$$(z+(i+1))(z+(i-1))=0$$
$$z=-i-1$$ and $$z=-i+1$$
Hence the triangle is formed by the vertices 
$$A(0,1),B(-1,-1),C(1,-1)$$
Hence this is an isosceles triangle. with $$AB=AC$$
Considering $$BC$$ as the base.
The vertical height will be given by the distance between A(0,1) and the midpoint of BC
Hence $$H=2$$
And Base$$=BC=2$$
Hence area 
$$=\dfrac{B\times H}{2}$$
$$=2$$.

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