Question

Show that the complex number z, satisfying the condition $arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$ lies on a circle.

Answer 1

Given: $arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$
Let $z=x+iy$

Now,
$\frac{z–1}{z+1}=\frac{x+iy-1}{x+iy+1}$
$\Rightarrow \frac{z–1}{z+1}=\frac{(x-1)+iy}{(x+1)+iy}$
$\Rightarrow \frac{z–1}{z+1}=\frac{(x-1)+iy}{(x+1)+iy}\times \frac{(x+1)-iy}{(x+1)-iy}$
$\Rightarrow \frac{z–1}{z+1}$
$=\frac{(x-1)(x+1)+iy\left[{}_{(x-1)}^{x+1-}\right]-(iy{)}^2}{(x+1{)}^2-(iy{)}^2}$
$=\frac{x^2-1+y^2+i\left(2y\right)}{(x+1{)}^2+y^2}[\because i^2=-1]$
We know,
$arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-}1\frac{\frac{2y}{(x+1{)}^2+y^2}}{\frac{x^2+y^2-1}{(x+1{)}^2+y^2}}=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-}1\frac{2y}{x^2+y^2-1}=\frac{\pi }{4}$
$\Rightarrow \frac{2y}{x^2+y^2-1}=1$
$\Rightarrow 2y=x^2+y^2-1$
$\therefore x^2+y^2-2y-1=0$

Hence, the complex number $z$ lies on the curve $x^2+y^2-2y-1=0$ which is a circle.