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