$(x^{4} + 2 x i) - (3 x^{2} + y i) = (3 - 5 i) + (1 + 2 y i)$ . Find x and y

x = 2, y = 3

x = - 2, y = \frac{1}{3} .

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x = 2, y = 3

x = 2, y = \frac{1}{3} .

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x = - 2, y = 3

x = 2, y = \frac{1}{3} .

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

x = - 2, y = 3

x = - 2, y = \frac{1}{3} .

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

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Solution

The correct option is A $x = 2, y = 3$ or $x = - 2, y = \frac{1}{3} .$
Rewriting the given expression, one get
$(x^{4} - {3 x}^{2}) + (2 x - y) i = 4 (2 y - 5) i$
Equation real and imaginary parts one get
$x^{4} - {3 x}^{2} = 4$ and $2 x - y = 2 y - 5$
$\Rightarrow (x^{2} + 1) (x - 2) (x + 2) = 0$ and $2 x = (3 y - 5)$
$∴ x = \pm i$ or $x = 2$ or $x = - 2$
But x being real, so either $x = 2$ or $x = - 2$ consequently $y = 3$ or $y = \frac{1}{3}$

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