Question

Arg(z-1/z+1)=pi/4 then find the radius of the locus

Answer 1

See the attachment for detail solution
Hope it will help you

Answer 2

Answer:

Hence, the Radius of Locus is √2

Step-by-step explanation:

We are given:

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

where z is a complex number written as:

$z=x+iy$

where x,y belongs to real numbers.

Now, we know that:

$Arg[\dfrac{z-1}{z+1}]=Arg[z-1]-Arg[z+1]$

Now,

$Arg[z-1]=Arg[x+iy-1]=Arg[(x-1)+iy]$

and

$Arg[z+1]=Arg[x+iy+1]=Arg[(x+1)+iy]$

Also,

$Arg[(x-1)+iy]=\arctan (\dfrac{y}{x-1})$

and,

$Arg[(x+1)+iy]=\arctan (\dfrac{y}{x+1})$

Hence,

$Arg[z-1]-Arg[z+1]=\arctan (\dfrac{y}{x-1})-\arctan (\dfrac{y}{x+1})$

Now,using the identity:

$\arctan A-\arctan B=\arctan (\dfrac{A-B}{1+AB})$

$\arctan (\dfrac{y}{x-1})-\arctan (\dfrac{y}{x+1})=\arctan (\dfrac{\dfrac{y}{x-1}-\dfrac{y}{x+1}}{1+\dfrac{y}{x-1}\dfrac{y}{x+1}})$

on solving we obtain:

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

Hence, we get:

$\dfrac{2y}{x^2-1+y^2}=\tan (\dfrac{\pi}{4})\\\\\\\dfrac{2y}{x^2-1+y^2}=1\\\\2y=x^2-1+y^2\\\\x^2+y^2-2y=1\\\\x^2+(y-1)^2=2\\\\x^2+(y-1)^2=(\sqrt{2})^2$

Hence, the Radius of Locus is √2