Q47 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics
Let $z=x+i y$ and a point $P$ represent $z$ in the Argand plane. If the real part of $\frac{z-1}{z+i}$ is 1 , then a point that lies on the locus of $P$ is
- A$(2016,2017)$
- B$(-2016,2017)$
- C$(-2016,-2017)$
- Correct answerD$(2016,-2017)$
Solution
We have, $\frac{z-1}{z+i}$
$$
\begin{aligned}
& =\frac{x+i y-1}{x+i y+i}=\frac{(x-1)+i y}{x+(y+1) i} \\
& =\frac{(x-1)+i y}{x+(y+1) i} \times \frac{x-(y+1) i}{x-(y+1) i} \\
& =\frac{x(x-1)+i x y-(x-1)(y+1) i+y(y+1)}{x^2+(y+1)^2} \\
& =\frac{x(x-1)+y(y+1)}{x^2+(y+1)^2}+\frac{[x y-(x-1)(y+1)] i}{x^2+(y+1)^2}
\end{aligned}
$$
Since,
$$
\operatorname{Re}\left(\frac{z-1}{z+i}\right)=1
$$
$$
\begin{aligned}
& \therefore \quad x(x-1)+y(y+1)=x^2+(y+1)^2 \\
& \Rightarrow \quad x^2-x+y^2+y=x^2+y^2+2 y+1 \\
& \Rightarrow \quad-x+y=2 y+1 \\
& \Rightarrow \quad x+y+1=0 \\
&
\end{aligned}
$$
$\therefore(2016,-2017)$ lies on
$$
x+y+1=0
$$
Hence, the correct answer is option: 'D'.