If complex numbers (x - 2y) + i(3x - y) and ( | vedclass

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

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$\sqrt{2}$

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(B) Two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are conjugates if $z_1 = \overline{z_2}$.Given $(x - 2y) + i(3x - y) = \overline{(2x - y) + i(x - y + 6)}$.This implies $(x - 2y) + i(3x - y) = (2x - y) - i(x - y + 6)$.Comparing the real and imaginary parts:Real part: $x - 2y = 2x - y \Rightarrow x + y = 0$.Imaginary part: $3x - y = -(x - y + 6)$ $\Rightarrow 3x - y = -x + y - 6$ $\Rightarrow 4x - 2y = -6$ $\Rightarrow 2x - y = -3$.Adding the two equations: $(x + y) + (2x - y) = 0 - 3$ $\Rightarrow 3x = -3$ $\Rightarrow x = -1$.Substituting $x = -1$ into $x + y = 0$,we get $y = 1$.Thus,$|x + iy| = |-1 + i| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2}$.

If complex numbers $(x - 2y) + i(3x - y)$ and $(2x - y) + i(x - y + 6)$ are conjugates of each other,then $|x + iy|$ is $(x, y \in \mathbb{R})$.

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