Find real values of $x$ and $y$ for which the complex numbers $- 3 + i x^{2} y$ and $x^{2} + y + 4 i$ are conjugate of each other.

x = 1, y = 4

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x = 1, y = - 4

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x = - 1, y = - 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 1, y = 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
B $x = 1, y = - 4$
D $x = - 1, y = 4$
Given, $- 3 + i x^{2} y = ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ x^{2} + y + 4 i$
$- 3 + i x^{2} y = x^{2} + y - 4 i$
$- 3 = x^{2} + y$
and $x^{2} y = - 4$
$∴ - 3 = x^{2} - \frac{4}{x^{2}}$ [Putting $y = - 4 / x^{2}$ from (2) in (1)]
$x^{4} + 3 x^{2} - 4 = 0$
$(x^{2} + 4) (x^{2} - 1) = 0$
$x^{2} - 1 = 0$
$x = \pm 1$
From (2), $y = - 4$ , when $x = \pm 1$ .
Hence, $x = 1, y = - 4$ or $x = - 1, y = 4$ .

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