If -3+ix^2y and x^2+y+4i are complex conjugat | vedclass

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(B) Given that $-3+ix^2y$ and $x^2+y+4i$ are complex conjugates.Therefore,$-3-ix^2y = x^2+y+4i$.Comparing the real and imaginary parts on both sides:$

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$\pm 1$

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(B) Given that $-3+ix^2y$ and $x^2+y+4i$ are complex conjugates.Therefore,$-3-ix^2y = x^2+y+4i$.Comparing the real and imaginary parts on both sides:$x^2+y = -3$  $(i)$$-x^2y = 4$  $(ii)$From $(ii)$,$y = -\frac{4}{x^2}$.Substituting $y$ in $(i)$:$x^2 - \frac{4}{x^2} = -3$Let $x^2 = t$,then $t - \frac{4}{t} = -3 \implies t^2 + 3t - 4 = 0$.$(t+4)(t-1) = 0$.Since $t = x^2$ must be positive,$t = 1$.$x^2 = 1 \implies x = \pm 1$.

If $-3+ix^2y$ and $x^2+y+4i$ are complex conjugates,then $x=$

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