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प्रश्न
Find the real value of x and y, if `((1+i)x-2i)/(3+i) + ((2-3i)y+i)/(3-i) = i, xy ∈ R, i = sqrt-1`
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उत्तर
\[ \frac{\left( 1 + i \right)x - 2i}{3 + i} + \frac{\left( 2 - 3i \right)y + i}{3 - i} = i\]
\[ \Rightarrow \frac{\left( 1 + i \right)\left( 3 - i \right)x - 2i\left( 3 - i \right) + \left( 2 - 3i \right)\left( 3 + i \right)y + i\left( 3 + i \right)}{\left( 3 + i \right)\left( 3 - i \right)} = i\]
\[ \Rightarrow \frac{3x - ix + 3ix - i^2 x - 6i + 2 i^2 + 6y + 2iy - 9iy - 3 i^2 y + 3i + i^2}{9 - i^2} = i\]
\[ \Rightarrow \frac{4x + 2ix - 3i + 9y - 7iy - 3}{10} = i\]
\[ \Rightarrow \left( 4x + 9y - 3 \right) + i\left( 2x - 3 - 7y \right) = 10i\]
\[\text { Comparing both the sides: } \]
\[4x + 9y - 3 = 0\]
\[ \Rightarrow 4x + 9y = 3 . . . (1) \]
\[2x - 3 - 7y = 10\]
\[ \Rightarrow 2x - 7y = 13 . . . (2)\]
\[\text{Multiplying equation (2) by 2:} \]
\[4x - 14y = 26 . . . (3) \]
\[\text { Subtracting equation (3) from (1): } \]
\[ 4x + 9y = 3 \]
\[ 4x - 14y = 26 \]
\[ - + - \]
\[ 23y = - 23\]
\[ \therefore y = - 1\]
\[\text { Substituting the value of y in equation (1) }: \]
\[4x - 9 = 3\]
\[ \Rightarrow 4x = 12\]
\[ \Rightarrow x = 3\]
\[ \therefore x = 3 \text { and y } = - 1\]
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