If complex numbers -3- iy x 2 and x 2- y -4 i are conjugates of each other- where x - y - R- then x - y can beA- -1-4B- 1-4C- -1-4D- 1-4

Given : ( 3+iyx^2) and (x^2+y+4i) are conjugates of each other, then ( 3+iyx^2)=(x^2+y+4i) ⇒ ( 3+iyx^2)=x^2+y 4i ⇒ x^2+y = 3, x^2y = 4 Solving both, we get ( ...

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Byju's Answer

Standard XII

Mathematics

Conjugate of a Complex Number

If complex nu...

Question

If complex numbers $(- 3 + i y x^{2})$ and $(x^{2} + y + 4 i)$ are conjugates of each other, where $x, y \in R,$ then $(x, y)$ can be

(- 1, - 4)

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(1, - 4)

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(- 1, 4)

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(1, 4)

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Solution

The correct options are
A $(- 1, - 4)$
B $(1, - 4)$
Given : $(- 3 + i y x^{2})$ and $(x^{2} + y + 4 i)$ are conjugates of each other, then
$(- 3 + i y x^{2}) = ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ (x^{2} + y + 4 i) \Rightarrow (- 3 + i y x^{2}) = x^{2} + y - 4 i \Rightarrow x^{2} + y = - 3, x^{2} y = - 4$
Solving both, we get
$(- 3 - y) y = - 4 \Rightarrow y^{2} + 3 y - 4 = 0 \Rightarrow (y + 4) (y - 1) = 0 \Rightarrow y = - 4, 1 \Rightarrow x^{2} = 1, - 4$
As $x^{2} \geq 0$ , so
$x^{2} = 1, y = - 4 ∴ (x, y) = (1, - 4), (- 1, - 4)$

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If complex numbers (−3+iyx2) and (x2+y+4i) are conjugates of each other, where x,y∈R, then (x,y) can be

If complex numbers $(- 3 + i y x^{2})$ and $(x^{2} + y + 4 i)$ are conjugates of each other, where $x, y \in R,$ then $(x, y)$ can be