Question:
Let $x$ and $y$ be real numbers. If $(3+i)x + y + (1-i)y + 3i - 4 = (2x+1)i + (x-y+2)i$, where $i=\sqrt{-1}$, then the pair $(x,y)$ is equal to:
Let $x$ and $y$ be real numbers. If $(3+i)x + y + (1-i)y + 3i - 4 = (2x+1)i + (x-y+2)i$, where $i=\sqrt{-1}$, then the pair $(x,y)$ is equal to:
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Equate real and imaginary parts separately while solving complex equations.
Updated On: Apr 24, 2026
- $(1,2)$
- $(0,2)$
- $(0,-2)$
- $(3,2)$
- $(-1,-2)$
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The Correct Option is B
Solution and Explanation
Concept:
• For two complex numbers to be equal, real and imaginary parts must be equal separately
Step 1: Simplify LHS
\[ (3+i)x + y + (1-i)y + 3i - 4 \] \[ = 3x + ix + y + y - iy + 3i - 4 \] \[ = (3x + 2y - 4) + i(x - y + 3) \]
Step 2: Simplify RHS
\[ (2x+1)i + (x-y+2)i = (3x - y + 3)i \]
Step 3: Compare real parts
\[ 3x + 2y - 4 = 0 \quad ...(1) \]
Step 4: Compare imaginary parts
\[ x - y + 3 = 3x - y + 3 \] \[ \Rightarrow x = 3x \Rightarrow 2x = 0 \Rightarrow x = 0 \]
Step 5: Substitute in (1)
\[ 3(0) + 2y - 4 = 0 \Rightarrow 2y = 4 \Rightarrow y = 2 \] Final Conclusion:
\[ (x,y) = (0,2) \]
• For two complex numbers to be equal, real and imaginary parts must be equal separately
Step 1: Simplify LHS
\[ (3+i)x + y + (1-i)y + 3i - 4 \] \[ = 3x + ix + y + y - iy + 3i - 4 \] \[ = (3x + 2y - 4) + i(x - y + 3) \]
Step 2: Simplify RHS
\[ (2x+1)i + (x-y+2)i = (3x - y + 3)i \]
Step 3: Compare real parts
\[ 3x + 2y - 4 = 0 \quad ...(1) \]
Step 4: Compare imaginary parts
\[ x - y + 3 = 3x - y + 3 \] \[ \Rightarrow x = 3x \Rightarrow 2x = 0 \Rightarrow x = 0 \]
Step 5: Substitute in (1)
\[ 3(0) + 2y - 4 = 0 \Rightarrow 2y = 4 \Rightarrow y = 2 \] Final Conclusion:
\[ (x,y) = (0,2) \]
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