Question

The perimeter of the locus represented by arg $\left( \frac{z + i}{z-i}\right) = \frac{\pi}{4}$ is equal to

• A. $4 \pi $
• B. $2 \pi \sqrt{2}$
• C. $2 \pi \sqrt{3}$
• D. $\frac{2 \pi}{\sqrt{3}}$

Answer 1

The Correct Option is B

We have, $\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}$
Let $Z = x + i y$
$\therefore \frac{ z + i }{ z - i } =\frac{ x + i y+ i }{ x + i y- i }$
$=\frac{ x +( y + l ) i }{ x +( y - l ) i }$
$=\frac{( x +( y + l ) i )( x -( y - l ) i )}{( x +(y-1) i )( x -( y - l ) i )}$
$=\frac{ x ^{2}+ y ^{2}-1+(- x y+ x + x y + x ) i }{ x ^{2}+( y -1)^{2}}$
$=\frac{ x ^{2}+ y ^{2}- l +2 xi }{ x ^{2}+( y - l )^{2}}$
$x^{2}+(y-1)^{2}$
$\arg \left(\frac{x^{2}+y^{2}-1+2 x i}{x^{2}+(y-1)^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{x^{2}+y^{2}-1}\right)$
$\therefore \tan ^{-1}\left(\frac{2 x}{x^{2}+y^{2}-1}\right)=\frac{\pi}{4}$
$\Rightarrow 2 x=x^{2}+y^{2}-1$
$\Rightarrow (x-1)^{2}+y^{2}=(\sqrt{2})^{2}$
$\therefore$ Locus of $\arg \left(\frac{z+i}{z-i}\right)=\frac{\pi}{4}$ is circle,
whose centre $(1,0)$ and radius $=\sqrt{2}$
Perimetre of circle $=2 \pi r =2 \pi(\sqrt{2}) $
$=2 \pi \sqrt{2}$