Question

If $arg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}$ then the locus of $z$ is:

• A. circle with radius $2$
• B. circle with radius $1$
• C. straight line
• D. pair of lines

Answer 1

The correct option is B circle with radius $1$

$z=x+iyarg\left(\frac{z-1}{z+1}\right)=\frac{\pi }{2}arg(z-1)-arg(z+1)=\frac{\pi }{2}{\tan }^{-1}\frac{y}{x-1}-{\tan }^{-1}\frac{y}{x+1}=\frac{\pi }{2}-{\tan }^{-1}\frac{y}{x+1}=\frac{\pi }{2}-{\tan }^{-1}\frac{y}{x-1}-{\tan }^{-1}\frac{y}{x+1}={\cot }^{-1}\frac{y}{x-1}-\frac{y}{x+1}=\tan \left({\cot }^{-1}\frac{y}{x-1}\right)=\frac{x-1}{y}-y^2=x^2-1x^2+y^2=1$

Thus it is a circle with radius $1$.