Find the real value of x and y for which of the following equation is satisfied
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$

x = 1, y = 3

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x = - 3, y = 1

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x = 3, y = - 1

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x = 1, y = - 3

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Solution

The correct option is C $x = 3, y = - 1$
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i \Rightarrow (1 + i) (3 - i) x - 2 i (3 - i) + (3 + i) (2 - 2 i) y + i (3 + i) = 10 i \Rightarrow 4 x + 2 i x - 6 i - 2 + 9 y - 7 i y + 3 i - 1 = 10 i$
$\Rightarrow 4 x + 9 y - 3 = 0$ and $2 x - 7 y - 3 = 10$
$\Rightarrow x = 3$ and $y = - 1$

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\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i

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If $\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$ , then the real values of x and y are given by :

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