Question

The radius of the locus by the point represented by $z$, when $arg\frac{z-1}{z+1}=\frac{\pi }{4}$, is

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

Answer 1

Answer: (A)

$\sqrt{2}$

Arg$\left[\frac{z-1}{z+1}\right]=\frac{\pi }{4}$

$\Rightarrow$Arg$[z-1\times \frac{1}{z+1}]=\frac{\pi }{4}$

$\Rightarrow$Arg$\left[z-1\right]\times$Arg$\left[\frac{1}{z+1}\right]=\frac{\pi }{4}$

$\Rightarrow$ Arg$[z-1]-$Arg$[z+1]=\frac{\pi }{4}$

Let $z=x+iy$

$\therefore$Arg$[(x-1)+iy]-$Arg$[(x+1)+iy]=\frac{\pi }{4}$

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

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

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

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

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

This represents a circle with center at $(0,1)$ and radius $=\sqrt{0+1-(-1)}=\sqrt{2}$