Question

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

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

Answer 1

The correct option is C

$-\frac{1}{2}$

Explanation for the correct option:

The given complex number $\frac{2+\iota }{a\iota -1}$.

$$\begin{gathered} \frac{2+\iota }{a\iota -1}=\frac{\left(2+\iota \right)\left(-a\iota -1\right)}{\left(a\iota -1\right)\left(-a\iota -1\right)} \\ \Rightarrow \frac{2+\iota }{a\iota -1}=\frac{-2a\iota -2-a{\iota }^2-\iota }{{\left(-1\right)}^2-{\left(a\iota \right)}^2} \\ \Rightarrow \frac{2+\iota }{a\iota -1}=\frac{-2a\iota -2-a{\iota }^2-\iota }{1-{\iota }^2{a}^2} \\ \Rightarrow \frac{2+\iota }{a\iota -1}=\frac{a-2-\iota -2a\iota }{a^2+1}\left[\because {\iota }^2=-1\right] \\ \Rightarrow \frac{2+\iota }{a\iota -1}=\frac{a-2}{a^2+1}+\iota \left(-\frac{2a+1}{a^2+1}\right) \end{gathered}$$

As the imaginary part of $\frac{2+\iota }{a\iota -1}$ is $0$, thus $\left(-\frac{2a+1}{a^2+1}\right)=0$.

$$\begin{gathered} \Rightarrow 2a+1=0 \\ \Rightarrow a=-\frac{1}{2} \end{gathered}$$

Hence, option (C) is the correct answer.