Question

If the imaginary part of $\frac{2z+1}{\iota z+1}$ is $\left(-2\right)$ then the locus of the point representing $z$ in the complex plane is

• A. circle
• B. straight line
• C. parabola
• D. ellipse

Answer 1

The correct option is B

straight line

Explanation for the correct option:

Let us assume that $z=x+\iota y$.

Thus, $\frac{2z+1}{\iota z+1}=\frac{2\left(x+\iota y\right)+1}{\iota \left(x+\iota y\right)+1}$.

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

Thus, the imaginary part of $\frac{2z+1}{\iota z+1}$ is $\left(\frac{2y-2y^2-2x^2-x}{x^2+y^2-2y+1}\right)=-2$.

$$\begin{gathered} \Rightarrow 2y-2y^2-2x^2-x=-2x^2-2y^2+4y-2 \\ \Rightarrow x+4y-2y=2 \\ \Rightarrow x+2y=2 \end{gathered}$$

Therefore, the locus of the point representing $z$ in the complex plane is a straight line.

Hence, option (B) is the correct answer.