If the point p represents the comple $x$ number $z=x+i y$ in the argand plane and if $\frac{z+i}{z-1}$ is a…
If the point p represents the comple $x$ number $z=x+i y$ in the argand plane and if $\frac{z+i}{z-1}$ is a purely imaginary number then the locus of $p$ is
$x^2+y^2+x-y=0$ and $(x, y) \neq(1,0)$
$x^2+y^2-x+y=0$ and $(x, y) \neq(1,0)$
$x^2+y^2-x+y=0$ and $(x, y)=(1,0)$
$x^2+y^2+x+y=0$
Solution
$z=x+i y, p=(x, y)$
$\frac{z+i}{z-1}=\frac{x+i y+i}{x+i y-1}=\frac{x+i(y+1)}{(x-1)+i y} \times \frac{(x-1)-i y}{(x-1)-i y}$
$=\frac{x(x-1)+i(x-1)(y+1)-i x y+y(y+1)}{(x-1)^2+y^2}$
$=\frac{x^2+y^2-x+y}{(x-1)^2+y^2}+\frac{i(x-y+1)}{(x-1)^2+y^2}$
Since $\frac{z+i}{z-1}$ is purely imaginary number
$\therefore \operatorname{Re}\left(\frac{z+i}{z-1}\right)=0 \Rightarrow \frac{x^2+y^2-x+y}{(x-1)^2+y^2}=0$
$\Rightarrow x^2+y^2-x+y=0$ and $(x, y) \neq(1,0)$