If $Z \neq \pm 1$ is a complex number and $\operatorname{Arg}\left(\frac{Z-1}{Z+1}\right)=\frac{\pi}{4}$,…

If $Z \neq \pm 1$ is a complex number and $\operatorname{Arg}\left(\frac{Z-1}{Z+1}\right)=\frac{\pi}{4}$, then the locus of $Z$ in the Arg and plane is
  1. $x^2+y^2-2 y-1=0$
  2. $x^2+y^2+2 y-1=0$
  3. $x^2+y^2-2 x+1=0$
  4. $x^2+y^2+2 x+1=0$

Solution

$\begin{aligned} & \text { Let } z=x+i y \\ & \begin{aligned} \frac{z-1}{z+1} & =\frac{x+i y-1}{x+i y+1} \\ & =\frac{(x-1)+i y}{(x+1)+i y} \times \frac{(x+1)-i y}{(x+1)-i y} \\ & =\frac{x^2+y^2-1+2 i y}{(x+1)^2+y^2} \\ \arg \left(\frac{z-1}{z+1}\right) & =\tan ^{-1}\left(\frac{2 y}{x^2+y^2-1}\right) \\ & =\pi / 4 \\ \Rightarrow \quad & \frac{2 y}{x^2+y^2-1}=\tan \pi / 4=1 \\ \Rightarrow \quad x^2+y^2-1=2 y & x^2-2 y-1=0 .\end{aligned}\end{aligned}$

Asked in: AP EAMCET 2018 (24 Apr Shift 1)

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