If 1- i x -2 i -3- i -2-3 i y - i -3- i - i- then the real values of x and y are given by -A- x-3- y-1B- x-3- y-1C- x-3- y-1D- x-1- y-3

The correct option is B x= 3, y= 1 Basic concept (1 + i) x 2i /3 + i + (2 3i)y + i/3 i = i , then 4x + 9y 3 = 0 and 2x 7y 13 = 0 so x = 3 and y ...

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If 1+ i x -2 ...

Question

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 :

x = -3, y= - 1

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

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

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

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