If Z = x + iy is a complex number and sqrt{x^ | vedclass

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(B) Given,$\sqrt{x^2 - 2x + 8} + (x + 4)i = y(2 + i)$.Equating real and imaginary parts:$\sqrt{x^2 - 2x + 8} = 2y$  $(1)$$x + 4 = y$  $(2)$Substitute

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$-2 + 2i$

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(B) Given,$\sqrt{x^2 - 2x + 8} + (x + 4)i = y(2 + i)$.Equating real and imaginary parts:$\sqrt{x^2 - 2x + 8} = 2y$  $(1)$$x + 4 = y$  $(2)$Substitute $y = x + 4$ into equation $(1)$:$\sqrt{x^2 - 2x + 8} = 2(x + 4)$Squaring both sides:$x^2 - 2x + 8 = 4(x^2 + 8x + 16)$$x^2 - 2x + 8 = 4x^2 + 32x + 64$$3x^2 + 34x + 56 = 0$$(3x + 28)(x + 2) = 0$So,$x = -2$ or $x = -\frac{28}{3}$.If $x = -2$,then $y = -2 + 4 = 2$. Thus,$Z = -2 + 2i$.If $x = -\frac{28}{3}$,then $y = -\frac{28}{3} + 4 = -\frac{16}{3}$. Thus,$Z = -\frac{28}{3} - \frac{16}{3}i$.Comparing with the given options,$Z = -2 + 2i$ is the correct choice.

If $Z = x + iy$ is a complex number and $\sqrt{x^2 - 2x + 8} + (x + 4)i = y(2 + i)$,then $Z$ is equal to

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