Question

If $z=x+iy$ is a vatiable complex number such that $arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$, then

• A. $x^2-y^2-2x-1=0$
• B. $x^2+y^2-2x-1=0$
• C. $x^2+y^2-2y-1=0$
• D. $x^2+y^2+2x-1=0$

Answer 1

Answer: (C)

$x^2+y^2-2y-1=0$

We have,

$\Rightarrow z=x+y\Rightarrow \left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}\Rightarrow \left(\frac{\left[\left(x-\right)+iy\right]\left[\left(x+1\right)-iy\right]}{\left({\left(x+1\right)}^2+y\right)}\right)=\frac{\pi }{4}\Rightarrow \left(\frac{\left(x^2-1\right)+i\left(-yx+y+xy+y\right)+y^2}{{\left(x+1\right)}^2+y^2}\right)=\frac{\pi }{4}\Rightarrow \left(\frac{\left(x^2+y^2-1\right)+i2y}{{\left(x+1\right)}^2+y^2}\right)=\frac{\pi }{4}\Rightarrow \left(\frac{\left(x^2+y^2-1\right)+i2y}{{\left(x+1\right)}^2+y}\right)=\frac{\pi }{4}\Rightarrow \frac{\frac{27}{{\left(x+1\right)}^2+y^2}}{\frac{x^2+y^2-1}{{\left(x+1\right)}^2+y^2}}=1\Rightarrow \frac{2y}{\left[{\left(x+1\right)}^2+y^2\right]}\times \frac{\left[{\left(x+1\right)}^2+y^2\right]}{x^2+y^2-1}=1\Rightarrow x^2+y^2-1=2y$

$Hence,weget$,

$\Rightarrow x^2+y^2-2y-1=0$

Hence, this is the answer.