Question

Let $z=x+iy$ .be a complex number such that arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}\cdot$ Then $x^2+y^2=$

• A. $2$
• B. $\frac{1}{3}$
• C. $1$
• D. $\frac{2}{3}$

Answer 1

Answer: (C)

Given, $z=x+iy$ and
arg$(\frac{z-1}{z+1})=\frac{\pi }{2}...(i)$
$\therefore \frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}\times \frac{(x+1)-iy}{(x+1)-iy}$
$=\frac{(x^2-1)+y^2+i2y}{(x+1{)}^2+y^2}$
Now,
$arg\left(\frac{z-1}{z+1}\right)=arg\left(\frac{\left(x^2+y^2-1\right)}{{\left(x+1\right)}^2+y^2}+i\left(\frac{2y}{{\left(x+1\right)}^2+y^2}\right)\right)$
$\Rightarrow \frac{\pi }{2}=ta{n}^{-1}\left(\frac{2y}{x^2+y^2-1}\right)$
$\Rightarrow tan\frac{\pi }{2}=\frac{2y}{x^2+y^2-1}$ [Using $(i)$]
$\Rightarrow \frac{1}{0}=\frac{2y}{x^2+y^2-1}$
$\Rightarrow x^2+y^2=1$