What Is the Locus of z If |(z+1)/(z-1)|=2?

[ps3stephen]

1. Homework Statement .

if z =x+yi.
Determine the equation of the locus z in terms of x and y such that |(z+1)/(z-1)|=2

3. The Attempt at a Solution .

|(z+1)/(z-1)|=2

(z+1)/(z-1)=2*2

(x+yi+1)/(x+yi-1)=4

[(x+1)+yi]/[(x-1)+yi]=4

[(x+1)+yi]/[(x-1)+yi]×[(x-1)-yi]/[(x-1)-yi]=4

[x*2-1-yi(x+1)+yi(x-1)-y*2i*2]/[(x-1)*2-yi(x-1)+yi(x-1)-y*2i*2]=4

(x*2-1-xyi+xyi-yi-yi-y*2i*2)/(x*2-2x+1-xyi+xyi+yi-yi-y*2i*2)=4

(x*2-1-2yi-y*2i*2)/(x*2-2x+1-y*2i*2)=4

[x*2-1-2yi-y*2(-1)]/[x*2-2x+1-y*2(-1)]=4

(x*2+y*2-1-2yi)/(x*2+y*2-2x+1)=4

(x*2+y*2-1-2yi)=4(x*2+y*2-2x+1)

x*2+y*2-1-2yi=4x*2+4y*2-8x+4

3x*2+3y*2-8x+2yi+5=0

 

[BvU]

Hello Stephen,

modulus of a complex number is defined differently: | x+iy |² = (x+iy)(x-iy) = x²+ y²

 

[PeroK]

Wouldn't it be easier to start by multiplying by |z-1|?