Question

If the imaginary part of $\frac{2+i}{ai-1}$ is zero, where $a$ is a real number, then the value of $a$ is equal to

• A. $\frac{1}{2}$
• B. $2$
• C. $-\frac{1}{2}$
• D. $-2$
• E. $\frac{3}{2}$

Answer 1

Answer: (C)

Given, let $z=\frac{2+i}{a i-1}$ and $Im (z)=0, a \in R \,\,\,\,\,\,\dots(i)$
$z=\frac{(2+i)(a i+1)}{(a i-1)(a i+1)} $
$z=\frac{2 a i+a i^{2}+2+i}{a^{2} \dot{2}^{2}-1} $
$z=\frac{(2-a)+i(2 a+1)}{\left(-a^{2}-1\right)}$
$\left(\because i^{2}=-1\right)$
$z=\frac{(a-2)}{\left(a^{2}+1\right)}-i \frac{(2 a+1)}{\left(a^{2}+1\right)}$
From Eq. (i), $-\frac{(2 a+1)}{\left(a^{2}+1\right)}=0$
$\Rightarrow \, 2 a+1=0$
$\Rightarrow \, a=-\frac{1}{2}$