(C) Given: $\frac{3+2i}{1+i} = \frac{1}{2}(x+iy)$ $\therefore x+iy = \frac{2(3+2i)}{1+i} \times \frac{1-i}{1-i}$ $= \frac{2(3 - 3i + 2i - 2i^2)}{1 - i

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(C) Given: $\frac{3+2i}{1+i} = \frac{1}{2}(x+iy)$ $\therefore x+iy = \frac{2(3+2i)}{1+i} \times \frac{1-i}{1-i}$ $= \frac{2(3 - 3i + 2i - 2i^2)}{1 - i^2}$ $= \frac{2(3 - i + 2)}{1 + 1} = \frac{2(5 - i)}{2} = 5 - i$ Comparing real and imaginary parts,we get $x = 5$ and $y = -1$. Therefore,$x - y = 5 - (-1) = 5 + 1 = 6$.

Class 11 Mathematics 4-1.Complex numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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If $\frac{3+2i}{1+i} = \frac{1}{2}(x+iy)$,then $x-y =$

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$4$
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$3$
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$6$
D
$5$

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