If the conjugate of (x + iy)(1 - 2i) is 1 + i | vedclass

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(B) Given that the conjugate of $(x + iy)(1 - 2i)$ is $1 + i$.Let $z = (x + iy)(1 - 2i)$.Then $\bar{z} = 1 + i$.Taking the conjugate on both sides of

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$x + iy = \frac{1 - i}{1 - 2i}$

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(B) Given that the conjugate of $(x + iy)(1 - 2i)$ is $1 + i$.Let $z = (x + iy)(1 - 2i)$.Then $\bar{z} = 1 + i$.Taking the conjugate on both sides of $z = (x + iy)(1 - 2i)$,we get $\bar{z} = \overline{(x + iy)(1 - 2i)} = \overline{(x + iy)} \cdot \overline{(1 - 2i)}$.Since $\bar{z} = 1 + i$,we have $(x - iy)(1 + 2i) = 1 + i$.Alternatively,we can write $z = \overline{1 + i} = 1 - i$.Thus,$(x + iy)(1 - 2i) = 1 - i$.Dividing both sides by $(1 - 2i)$,we get $x + iy = \frac{1 - i}{1 - 2i}$.This matches Option $(B)$.

If the conjugate of $(x + iy)(1 - 2i)$ is $1 + i$,then

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