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If a point $P$ denotes a complex number $z=x+i y$ in the argand plane and if $\frac{z+1}{z+i}$ is a purely real number, then the locus of $P$ is
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Verified Answer
The correct answer is:
$x+y+1=0,(x, y) \neq(0,-1)$
Since
$$
\begin{aligned}
& \frac{z+1}{z+i}=\frac{x+i y+1}{x+i y+i}=\frac{(x+1)}{x+i(y+1)} \\
& \times \frac{x-i(y+1)}{x-i(y+1)}
\end{aligned}
$$
If $\frac{z+1}{z+i}$ is a purely real number, $z \neq-i$
$\begin{aligned} & \text { Then, } \quad I_m\left(\frac{z+1}{z+i}\right)=0, z=-i \\ & \Rightarrow \quad x y-(x+1)(y+1)=0, \quad(x, y) \neq(0,-1) \\ & \Rightarrow \quad x y-x y-x-y-1=0 \\ & \Rightarrow \quad x+y+1=0, \quad(x, y) \neq(0,-1)\end{aligned}$
$$
\begin{aligned}
& \frac{z+1}{z+i}=\frac{x+i y+1}{x+i y+i}=\frac{(x+1)}{x+i(y+1)} \\
& \times \frac{x-i(y+1)}{x-i(y+1)}
\end{aligned}
$$
If $\frac{z+1}{z+i}$ is a purely real number, $z \neq-i$
$\begin{aligned} & \text { Then, } \quad I_m\left(\frac{z+1}{z+i}\right)=0, z=-i \\ & \Rightarrow \quad x y-(x+1)(y+1)=0, \quad(x, y) \neq(0,-1) \\ & \Rightarrow \quad x y-x y-x-y-1=0 \\ & \Rightarrow \quad x+y+1=0, \quad(x, y) \neq(0,-1)\end{aligned}$
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