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)
$=\frac{\left[x+\left(iy-1\right)\right]\left[x-\left(iy-1\right)\right]}{\left(x+1+iy\right)\left(x+1-iy\right)}=\frac{x^2-{\left(iy-1\right)}^2}{{\left(x+1\right)}^2-i^2{y}^2}$
$\frac{z-1}{z+1}=\frac{x^2+y^2-1+2iy}{x^2+y^2+2x+1}$
=$\frac{x^2+y^2-1+2iy}{x^2+y^2+2x+1}+\frac{2yi}{x^2+y^2+2x+1}$
$\therefore arg\left(\frac{z-1}{z+1}\right)=ta{n}^{-1}\left[\frac{\frac{2y}{x^2+y^2+2x+1}}{\frac{x^2+y^2-1}{x^2+y^2+2x+1}}\right]$
$=ta{n}^{-1}\left(\frac{2y}{x^2+y^2-1}\right)$
But given arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$
$\therefore ta{n}^{-1}\frac{2y}{x^2+y^2-1}=\frac{\pi }{4}\Rightarrow \frac{2y}{x^2+y^2-1}=tan\frac{\pi }{4}$
$\Rightarrow \frac{2y}{x^2+y^2-1}=1\left[\because tan\frac{\pi }{4}=1\right]$
$\Rightarrow x^2+y^2-1=2y\Rightarrow x^2+y^2-2y=1$