The values of x and y for which the numbers 3 | vedclass

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(A) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $z_1 = \overline{z_2}$.Given $z_1 = 3 + i(x^2y)$ and $z_2 = (x^2 + y) + 4i

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$( - 2, - 1)$ or $(2, - 1)$

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(A) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $z_1 = \overline{z_2}$.Given $z_1 = 3 + i(x^2y)$ and $z_2 = (x^2 + y) + 4i$.The conjugate of $z_2$ is $\overline{z_2} = (x^2 + y) - 4i$.Equating $z_1 = \overline{z_2}$,we get $3 + i(x^2y) = (x^2 + y) - 4i$.Comparing real and imaginary parts:$x^2 + y = 3$  (Equation $1$)$x^2y = - 4$  (Equation $2$)From Equation $1$,$x^2 = 3 - y$. Substituting this into Equation $2$:$(3 - y)y = - 4 \implies 3y - y^2 = - 4 \implies y^2 - 3y - 4 = 0$.Factoring the quadratic: $(y - 4)(y + 1) = 0$,so $y = 4$ or $y = - 1$.If $y = 4$,$x^2 = 3 - 4 = - 1$,which gives no real values for $x$.If $y = - 1$,$x^2 = 3 - (- 1) = 4$,so $x = \pm 2$.Thus,the possible values are $(x, y) = (2, - 1)$ or $( - 2, - 1)$.

The values of $x$ and $y$ for which the numbers $3 + i{x^2}y$ and ${x^2} + y + 4i$ are conjugate complex can be

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