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

Question

Aicharya · Accepted Answer

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} 	imes \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}ight)=1 $ $ \begin{aligned} & 	herefore \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} $ $	herefore(2016,-2017)$ lies on $ x+y+1=0 $

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 …

Solution