Question:
If \(\frac{(3-i)^2}{2+i} = A + iB\), where \(A\) and \(B\) are real numbers, then \(A\) and \(B\) are equal to
If \(\frac{(3-i)^2}{2+i} = A + iB\), where \(A\) and \(B\) are real numbers, then \(A\) and \(B\) are equal to
Show Hint
Always rationalize complex denominators using conjugate.
Updated On: Apr 16, 2026
- \(A=-4, B=2\)
- \(A=2, B=-4\)
- \(A=2, B=4\)
- None of these
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The Correct Option is B
Solution and Explanation
Step 1: Expand numerator.
\[ (3-i)^2 = 9 -6i + i^2 = 8 -6i \]
Step 2: Rationalize denominator.
\[ \frac{8-6i}{2+i} \cdot \frac{2-i}{2-i} = \frac{(8-6i)(2-i)}{5} \]
Step 3: Multiply.
\[ = \frac{16 -8i -12i +6i^2}{5} = \frac{16 -20i -6}{5} = \frac{10 -20i}{5} = 2 -4i \] \[ A=2,\quad B=-4 \]
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