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

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(C) Given that $\overline{(x + iy)(1 - 2i)} = 1 + i$. Using the property $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$,we have $\overline

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

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(C) Given that $\overline{(x + iy)(1 - 2i)} = 1 + i$. Using the property $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$,we have $\overline{(x + iy)} \cdot \overline{(1 - 2i)} = 1 + i$. This implies $(x - iy)(1 + 2i) = 1 + i$. Therefore,$x - iy = \frac{1 + i}{1 + 2i}$. Taking the conjugate on both sides,we get $\overline{x - iy} = \overline{\left(\frac{1 + i}{1 + 2i}\right)}$. Thus,$x + iy = \frac{1 - i}{1 - 2i}$.

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

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