locus : | (z-2)/(z+2) | <= 1 (1 Viewer)

[Mumma]

locus of z

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

thank you

 

[Trev]

Let z=x+iy:
|(x-2+iy)|/|(x+2+iy)| <= 1
[(x-2)²+y²]/[(x+2)²+y²] <=1
Since the denominator is always positive (both variables are squared) then it is okay to multiply the denominator out with an inequality.
(x-2)²+y² <= (x+2)²+y²
-8x <= 0; x>=0.

I think that's right but I may have done something wrong somewhere, I haven't done these in quite a while.

 

[YBK]

I think you can do this geometrically.

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

then you draw that, and when the 2 sides are equal the result is x = 0 but |z+2| > |z-2| therefore x>= 0

easier like this