NTA Abhyas JEE Main2020MathematicsComplex NumberPractice
The real values of x and y satisfying the equation 1 + i x - 2 i 3 + i + 2 - 3 i y + i 3 - i = i are
Options
- Ax = - 1 , y = 3
- Bx = 3 , y = - 1
- Cx = 0 , y = 1
- Dx = 1 , y = 0
Correct answer
B. x = 3 , y = - 1
Step-by-step solution
Given, 1 + i x - 2 i 3 + i + 2 - 3 i y + i 3 - i = i ⇒ ( 1 + i ) ( 3 − i ) x − 2 i ( 3 − i ) + ( 2 − 3 i ) ( 3 + i ) y + i ( 3 + i ) ( 9 + 1 ) = i ⇒ 4 + 2 i x + 9 - 7 i y - 3 i - 3 = 10 i ⇒ ( 4 x + 9 y − 3 ) + i ( 2 x − 7 y - 3 ) = 10 i Equating the real and imaginary parts, we get 2 x - 7 y = 13 and 4 x + 9 y = 3 , hence x = 3 and y = - 1