Q65 / 160 · TS EAMCET 2017 · Mathematics
If the imaginary part of $\frac{2 z+1}{i z+1}$ is -2 , then the locus of the point representing $z$ in the complex plane is
- Aa circle
- Correct answerBa parabola
- Ca straight line
- Dan ellipse
Solution
Let $z=x+i y$
$$
\begin{aligned}
\therefore \frac{2 z+1}{i z+1} & =\frac{2(x+i y)+1}{i(x+i y)+1}=\frac{2 x+1+i y}{(-y+1)+i x} \\
& =\frac{[(2 x+1)+i y][(-y+1)-i x]}{[(-y+1)+i x][(-y+1)-i x]}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{(2 x+1)(-y+1)-(2 x+1) x i+y(-y+1) j+x y}{(-y+1)^2+x^2} \\
& =\frac{(2 x+1)(-y+1)+x y+\left(-y^2+y-2 x^2-x\right) j}{(-y+1)^2+x^2}
\end{aligned}
$$
It is given that imaginary part of $\frac{2 z+1}{i z+1}=-2$
$$
\begin{aligned}
& \therefore \frac{-y^2+y-2 x^2-x}{(-y+1)^2+x^2}=-2 \\
& \Rightarrow-y^2+y-2 x^2-x=-2(-y+1)^2-2 x^2 \\
& \Rightarrow-y^2+y-x=-2 y^2-2+4 y \\
& \Rightarrow y^2-3 y-x+2=0
\end{aligned}
$$
which represent the equations of parabola.
Hence, the correct answer is option: 'B'.