Question

If $arg\left(\frac{z-1}{z+1}\right)=\pm \frac{\pi }{4}$ then length of the path traced by the points in the locus is

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

Answer 1

The correct option is A $2\sqrt{2}\pi$

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

or, $arg(-z1)-arg(z+1)=\pm \frac{\pi }{4}$

Let $z=x+iy\Rightarrow arg(x+iy-1)-arg(x+iy+1)=\pm \frac{\pi }{4}$.

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

or, ${\tan }^{-1}\left(\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\frac{y}{x-1}\times \frac{y}{x+1}}\right)=\pm \frac{\pi }{4}$

or, $\frac{2y}{(x-1)(x+1)+y^2}=\pm 1$ $[\because \tan (\pm \frac{\pi }{4})=\pm 1]$

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

$\therefore$ Centre at $(0,1)$ and radius $=\sqrt{2}$

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

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

$\therefore$ Centre at $(0,-1)$ and radius $=\sqrt{2}$

So, radius is same in both case.

$\therefore$ Length of Path traced $=2\pi r=2\pi \sqrt{2}=2\sqrt{2}x$ Ans.