For what real values of x and y are the number $- 3 + i x^{2} y$ and $x^{2} + y + 4 i$ conjugate complex?

x = - 1, y = \pm 4

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x = 1, y = \pm 4

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x = \pm 1, y = 4

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x = \pm 1, y = - 4

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Solution

The correct option is C $x = \pm 1, y = - 4$
Hence applying the given condition
$- x^{2} y = 4$ ...(i)
and
$x^{2} + y = - 3$
Or
$x^{2} = - (3 + y)$ substituting in i, we get
$y^{2} + 3 y = 4$
$y^{2} + 3 y - 4 = 0$
$(y + 4) (y - 1) = 0$
$y = - 4$ and $y = 1$
Hence substituting $y = - 4$ we get
$x^{2} = 1$
$x = \pm 1$

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