Question

If $z=x+iy$ such that $|z+1|=|z-1|$ and $arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$, then

• A. $x=\sqrt{2}+1,y=0$
• B. $x=0,y=\sqrt{2}+1$
• C. $x=0,y=\sqrt{2}-1$
• D. $x=\sqrt{2}-1,y=0$

Answer 1

Answer: (B)

$x=0,y=\sqrt{2}+1$

$|z+1|=|z-1|\Rightarrow {(x+1)}^2+y^2={(x-1)}^2+y^2$

arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$

$\Rightarrow$ arg $\left(\frac{[(x-1)+iy][(x+1)-iy]}{{(x+1)}^2+y^2}\right)=\frac{\pi }{4}$

$\Rightarrow$ arg $\left(\frac{(x^2+y^2-1)+2iy}{{(x+1)}^2+y^2}\right)=\frac{\pi }{4}$

$\Rightarrow y>0,x^2+y^2-1>0,\frac{2y}{x^2+y^2-1}=1\Rightarrow x^2+y^2-2y-1>0,y>0,x^2+y^2-1>0$

$\therefore$ arg $\left(\frac{z-1}{z+1}\right)=\frac{\pi }{4}$

is the are $ABC$ of the circle $x^2+y^2-2y-1=0.$

Solving with $x=0,$ we get $y=\frac{2\pm \sqrt{8}}{2}=1\pm \sqrt{2},y>0$

$\therefore y=1+\sqrt{2}$