Complex Locus 2 (1 Viewer)

[qwe]

OK heres another question I'm having trouble with:

"Given that z cannot =-2 and w= (z-2)/(z+2) (srry I don't know how to make a vinculum) and modulus(w) =1. Find the locus of z."

This is turning out to be an extreme algebra bash, so my question is there a better way besides letting w=a+bi and keep bashing?

(Note: the answer is (x^2 +y^2 -4)^2 +16y^2)

 

[Lukybear]

Which book is this from?

 

[study-freak]

lol, untouchablecuz did it nicely

 

[Lukybear]

yea wth... if study freak cant solve it then no one can...

 

[untouchablecuz]

now off to the gym

 

[ninetypercent]

mod w = l z-2l/lz +2l

mod w = 1

lz-2l / lz+2l = 1
lz - 2l - lz +2l = 0
let z = a + bi

la - 2 + bil - l a + 2 + bil = 0
root (a-2)^2 + b^2 - root (a+2)^2 + b^2 = 0
a^2 - 4a + 4 + b^2 - a^2 + 4a + 4 + b^2 + 2root [(a-2)^2 + b^2][(a+2)^2 + b^2] = 0
4[(a-2)^2 + b^2][(a+2)^2 + b^2] = 0
(a-2)^2(a+2)^2 + b^2(a+2)^2 + b^2(a-2)^2 + b^4 = 0
(a^2 - 4)^2 + b^2[(a+2)^2 + (a-2)^2] + b^4 = 0
(a^2 - 4)^2 + b^2(a^2 + 4a + 4 + a^2 - 4a + 4) + b^4 = 0
(a^2 - 4)^2 + b^2(2a^2 + 8) + b^4 = 0
a^4 - 8a^2 + 16 + 2a^2b^2 +8b^2 + b^4 = 0
(a^2 + b^2 - 4)^2 + 8a^2 = 0

ive done something wrong

 

[Lukybear]

mod w = l z-2l/lz +2l

mod w = 1

lz-2l / lz+2l = 1
lz - 2l - lz +2l = 0
let z = a + bi

What did you do there?

 

[ninetypercent]

lz-2l / lz+2l = 1
lz-2l = lz+2l
lz - 2l - lz +2l = 0
let z = a + bi

and i dont get the correct answer...

 

[GUSSSSSSSSSSSSS]

mod w = l z-2l/lz +2l

mod w = 1

lz-2l / lz+2l = 1
lz - 2l - lz +2l = 0
let z = a + bi

la - 2 + bil - l a + 2 + bil = 0
root (a-2)^2 + b^2 - root (a+2)^2 + b^2 = 0
a^2 - 4a + 4 + b^2 - a^2 + 4a + 4 + b^2 + 2root [(a-2)^2 + b^2][(a+2)^2 + b^2] = 0
4[(a-2)^2 + b^2][(a+2)^2 + b^2] = 0
(a-2)^2(a+2)^2 + b^2(a+2)^2 + b^2(a-2)^2 + b^4 = 0
(a^2 - 4)^2 + b^2[(a+2)^2 + (a-2)^2] + b^4 = 0
(a^2 - 4)^2 + b^2(a^2 + 4a + 4 + a^2 - 4a + 4) + b^4 = 0
(a^2 - 4)^2 + b^2(2a^2 + 8) + b^4 = 0
a^4 - 8a^2 + 16 + 2a^2b^2 +8b^2 + b^4 = 0
(a^2 + b^2 - 4)^2 + 8a^2 = 0

ive done something wrong

in the first bolded line, why not just take the -la+2+bil to the other side then square both sides?? wud simplify the algebra quite a bit

u probably made a mistake somewhere else in the algebra but im sry i cbb checking the whole thing =S lol