Question

If z = x + iy is a variable complex number such that arg $\frac{z-1}{z+1} = \frac{\pi}{4}$ then :

• A. $x^2 - y^2 - 2x = 1$
• B. $x^2 + y^2 - 2x = 1$
• C. $x^2 + y^2 - 2y = 1$
• D. $x^2 + y^2 + 2x = 1$

Answer 1

Answer: (C)

Given $z = x + iy$ is a variable complex number
where arg $\frac{z-1}{z+1} = \frac{\pi}{4}$
Now, consider
$\frac{z-1}{z+1} = \frac{x+iy-1}{x+iy+1} \times \frac{x-iy+1}{x-iy+1}$ (By rationalizing)
$= \frac{\left[x+\left(iy-1\right)\right]\left[x-\left(iy-1\right)\right]}{\left(x+1+iy\right)\left(x+1-iy\right)} = \frac{x^{2}-\left(iy-1\right)^{2}}{\left(x+1\right)^{2}-i^{2}y^{2}}$
$ \frac{z-1}{z+1} = \frac{x^{2}+y^{2}-1+2iy}{x^{2}+y^{2}+2x+1}$
=$\frac{x^{2}+y^{2}-1+2iy}{x^{2}+y^{2}+2x+1} +\frac{2yi}{x^{2}+y^{2}+2x+1}$
$\therefore \quad arg\left(\frac{z-1}{z+1}\right) = tan^{-1}\left[\frac{\frac{2y}{x^{2}+y^{2}+2x+1}}{\frac{x^{2}+y^{2}-1}{x^{2}+y^{2}+2x+1}}\right]$
$= tan^{-1}\left(\frac{2y}{x^{2}+y^{2}-1}\right)$
But given arg $\left(\frac{z-1}{z+1}\right) = \frac{\pi }{4}$
$\therefore tan^{-1} \frac{2y}{x^{2}+y^{2}-1} =\frac{\pi }{4} \Rightarrow \frac{2y}{x^{2}+y^{2}-1} = tan \frac{\pi }{4}$
$\Rightarrow \quad\frac{2y}{x^{2}+y^{2}-1} = 1\quad\left[\because tan \frac{\pi }{4}=1 \right]$
$\Rightarrow \quad x^{2} + y^{2} -1 = 2y\quad\Rightarrow \quad x^{2} + y^{2} - 2y = 1$