Question

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

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

Let $z=x+iy$

Given that $|z+1|=|z-1|$

$\Rightarrow |(x+1)+iy|=|(x-1)+iy| $

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

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

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

$\Rightarrow x=0$

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

$\Rightarrow amp\left(\frac{z-1}{z+1}\right)=amp\left(\frac{x+iy-1}{x+iy+1}\right)$

But $x=0$

$\Rightarrow amp\left(\frac{z-1}{z+1}\right)=amp\left(\frac{-1+iy}{1+iy}\right)$

$\Rightarrow amp\left(\frac{-1+iy}{1+iy}\right)=\frac{\pi }{4}$

$\Rightarrow amp(\frac{-1+iy}{1+iy}\times \frac{1-iy}{1-iy})=\frac{\pi }{4}$

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

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

$\Rightarrow {\tan }^{-1}\left(\frac{2y}{y^2-1}\right)=\frac{\pi }{4}$

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

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

On solving we get,

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

Since $\theta$ lies in $Q_1$

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