Q73 / 170 · Manipal (MET) 2020 · Mathematics
If $\frac{(3-i)^2}{2+i}=A+i B$, where $A$ and $B$ are real numbers, then $A$ and $B$ are equal to
- A$A=-4, B=2$
- Correct answerB$A=2, B=-4$
- C$A=2, B=4$
- DNone of these
Solution
Given equation is
$\frac{(3-i)^2}{2+i}=A+i B$
$\Rightarrow \quad \frac{9-1-6 i}{2+i}=A+i B$
$\Rightarrow \quad \frac{8-6 i}{2+i}=A+i B$
$\begin{array}{cc}\Rightarrow & \frac{2(4-3 i)}{2+i} \times \frac{2-i}{2-i}=A+i B \\ \Rightarrow & \frac{28-4 i-6 i-3]}{4+1}=A+i B \\ \Rightarrow & \frac{2[5-10 i]}{5}=A+i B \\ \Rightarrow & 2-4 i=A+i B\end{array}$
On equating the real and imginary parts from both sides, we get
$A=2$ and $B=-4$
Hence, the correct answer is option: 'B'.