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} + i {2y}}{(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} = tan^{-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$