Question

If the complex numbers \( (2 + i)x + (1 - i)y + 2i - 3 \) and \( x + (-1 + 2i)y + 1 + i \) are equal, then \( (x, y) \) is

• A. \( (1, -2) \)
• B. \( (-1, 2) \)
• C. \( (2, -1) \)
• D. \( (2, -2) \)
• E. \( (2, 1) \)

Hint:

To solve for complex numbers, equate real and imaginary parts separately and solve the system of equations.

Answer 1

Answer: (E)

Step 1: We are given that two complex numbers are equal. So, equate the real and imaginary parts of both sides of the equation. \[ (2 + i)x + (1 - i)y + 2i - 3 = x + (-1 + 2i)y + 1 + i \] Step 2: Simplify both sides: \[ (2x + ix) + (y - iy) + 2i - 3 = x - y + 2iy + 1 + i \] Step 3: Group the real and imaginary terms: \[ (2x + y - 3) + i(x - y + 2) = (x - y + 1) + i(2y + 1) \] Step 4: Equate the real and imaginary parts: 1. \( 2x + y - 3 = x - y + 1 \) 2. \( x - y + 2 = 2y + 1 \) 
Step 5: Solve the system of equations: From equation 1: \[ 2x + y - 3 = x - y + 1 \quad \Rightarrow \quad x + 2y = 4 \quad \cdots (1) \] From equation 2: \[ x - y + 2 = 2y + 1 \quad \Rightarrow \quad x - 3y = -1 \quad \cdots (2) \] Step 6: Solve the system of linear equations. Using equation (1): \[ x = 4 - 2y \] Substitute \( x = 4 - 2y \) in equation (2): \[ (4 - 2y) - 3y = -1 \quad \Rightarrow \quad 4 - 5y = -1 \quad \Rightarrow \quad y = 1 \] Substitute \( y = 1 \) into equation (1): \[ x + 2(1) = 4 \quad \Rightarrow \quad x = 2 \] Thus, \( (x, y) = (2, 1) \).