Question

If $z=x-iy$ such that $|z+1|=|z-1|$ and amp $\frac{z-1}{z+1}=\frac{\pi }{4}$ 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

The correct option is B $x=0,y=\sqrt{2}+1$

Given $z=x-iy$

$|z+1|=|z-1|$

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

$\Rightarrow 2x\left(2\right)=0$

$\Rightarrow 4x=0$

$\therefore x=0$

$\frac{x-1+iy}{x+1+iy}=\frac{\pi }{4}$

Now $x=0$, we get

$\Rightarrow \frac{-1+iy}{1+iy}$

$\Rightarrow \frac{1-y^2}{1+y^2}+\frac{-2iy}{1+y^2}$

Argument is $\frac{\pi }{4}$

$\therefore \frac{-2y}{1-y^2}=1$

$\Rightarrow 1-y^2=-2y$

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

$\Rightarrow (y-1{)}^2=2$

$\Rightarrow y-1=\sqrt{2}$

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

Hence, option 'B' is correct.