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Question: Find the real value of x and y for which the following equation is satisfied \(\dfrac{{\left( {1 + i...

Find the real value of x and y for which the following equation is satisfied (1+i)x−2i3+i+(2−3i)y+i3−i=i\dfrac{{\left( {1 + i} \right)x - 2i}}{{3 + i}} + \dfrac{{\left( {2 - 3i} \right)y + i}}{{3 - i}} = i
A. x=1, y=3
B. x=-3, y=1
C. x=3, y=-1
D. x=1, y=-3

Explanation

Solution

Hint: Here the complex number equation is given, we have to simplify the equation to find the value of x and y and then compare with the general equation of complex number.

Complete step-by-step answer:
Given, (1+i)x−2i3+i+(2−3i)y+i3−i=i\dfrac{{\left( {1 + i} \right)x - 2i}}{{3 + i}} + \dfrac{{\left( {2 - 3i} \right)y + i}}{{3 - i}} = i
Taking LCM of the above equation, we get
⇒(3−i)(1+i)x−2i(3−i)+(3+i)(2−3i)y+i(3+i)(3+i)(3−i)=i\Rightarrow \dfrac{{\left( {3 - i} \right)\left( {1 + i} \right)x - 2i\left( {3 - i} \right) + \left( {3 + i} \right)\left( {2 - 3i} \right)y + i\left( {3 + i} \right)}}{{\left( {3 + i} \right)\left( {3 - i} \right)}} = i
⇒(3−i+3i−i2)x−6i+2i2+(6+2i−9i−3i2)y+3i+i2(3+i)(3−i)=i\Rightarrow \dfrac{{\left( {3 - i + 3i - {i^2}} \right)x - 6i + 2{i^2} + \left( {6 + 2i - 9i - 3{i^2}} \right)y + 3i + {i^2}}}{{\left( {3 + i} \right)\left( {3 - i} \right)}} = i
As we know that, i2=−1{i^2} = - 1 and (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
⇒(3−i+3i+1)x−6i−2+(6+2i−9i+3)y+3i−1(3+i)(3−i)=i\Rightarrow \dfrac{{\left( {3 - i + 3i + 1} \right)x - 6i - 2 + \left( {6 + 2i - 9i + 3} \right)y + 3i - 1}}{{\left( {3 + i} \right)\left( {3 - i} \right)}} = i
⇒(4+2i)x−3i−3+(9−7i)y32−i2=i⇒4x+2ix−3i−3+9y−7iy9+1=i\Rightarrow \dfrac{{\left( {4 + 2i} \right)x - 3i - 3 + \left( {9 - 7i} \right)y}}{{{3^2} - {i^2}}} = i \Rightarrow \dfrac{{4x + 2ix - 3i - 3 + 9y - 7iy}}{{9 + 1}} = i
⇒4x+9y−3+(2x−3−7y)i10=i⇒4x+9y−310+(2x−3−7y)i10=i\Rightarrow \dfrac{{4x + 9y - 3 + \left( {2x - 3 - 7y} \right)i}}{{10}} = i \Rightarrow \dfrac{{4x + 9y - 3}}{{10}} + \dfrac{{\left( {2x - 3 - 7y} \right)i}}{{10}} = i ……. (1)
Now comparing the real part of the complex number on LHS with the real part of the complex number on RHS and the imaginary number on the RHS of equation (1), we can say
4x+9y−310=0\dfrac{{4x + 9y - 3}}{{10}} = 0 And (2x−3−7y)10=1\dfrac{{\left( {2x - 3 - 7y} \right)}}{{10}} = 1
⇒4x+9y−3=0\Rightarrow 4x + 9y - 3 = 0 ……. (2)
And 2x−3−7y=102x - 3 - 7y = 10 ……. (3)
Now multiply equation (3) by 2 and subtract this from equation (2), we get
⇒4x+9y−3−2(2x−3−7y)=0−(2×10)\Rightarrow 4x + 9y - 3 - 2\left( {2x - 3 - 7y} \right) = 0 - \left( {2 \times 10} \right)
⇒4x+9y−3−4x+6+14y=−20\Rightarrow 4x + 9y - 3 - 4x + 6 + 14y = - 20
‘4x’ will cancel out from LHS and whole equation will be in terms of y,
⇒23y+3=−20⇒23y=−23⇒y=−1\Rightarrow 23y + 3 = - 20 \Rightarrow 23y = -23 \Rightarrow y = - 1
Substituting the value of y in equation (3) and x=3x=3,
Then the correct answer is Option C.

Note: In these types of problems where some equation in terms of complex numbers is given, solve it by simply comparing the complex numbers finally obtained on the LHS and the RHS of the equation.