Im: (z+i)/(1-z) = 1
|z|=1
What i did so far:
im: (a+bi+i)(1-a+bi)/(1-a-bi)(1-a+bi)=1
im: (a-b-(a^2)-(b^2)-ai+bi+i)/((a^2)+(b^2)-2a+1)=1
-a+b+1=(a^2)+(b^2)-2a+1
(a^2)+(b^2)-a-b=0
From here on forward i'm not sure how to proceed because i don't know how to implement the |z|=1 in the equation.What i did is this but i'm fairly certain it's not correct:
(a-(1/2))^2+((b-(1/2))^2 - 1/2 = 0
(a-(1/2))^2+((b-(1/2))^2 = (sqr.root 2/2)^2
My assumption for the solution is: all points on radius of (sqr.root 2/2)^2 with center at (1/2,1/2)
Would appreciate your help. Thank you
here is my solution, which avoids "a's and b's".
Given:
.
the second equation tells us .
thus: .
now:
(since z* = 1/z)
the fraction we are subtracting is a bit ugly, so let's simplify it, first:
oh look, we have a common denominator, now. so:
now 1/i = -i. so we can replace 1/(2i) with -i/2. so we have:
remember, this whole mess is the imaginary part of (z+i)/(1-z), and is equal to 1. thus:
. this equation is linear in z, so we most certainly can solve it (over C). put all the z terms on one side, and the non-z terms on the other:
. this easily simplifies to:
.
but....we're not done, yet. let's simplify the fraction on the right:
, so:
.
but:
.
therefore:
.
verify that this indeed is a solution.
(with all due respect for Plato, his method leads to one "false" solution, which must be eliminated: z = 1. this is due to "squaring introducing extraneous solutions").
I think i figured it out, but i would appreciate some feedback regarding correctness.
(b-a+1)/((1-a)^2 + b^2) = 1 for all (a,b) =/= (1,0) (denominator can't be 0)
b-a+1=(1-a)^2 + b^2
a^2 + b^2 - a - b = 0
|z|=1 = sqr.root (a^2 + b^2 )
1=sqr.root (a^2 + b^2 ) /^2
1=a^2 + b^2
a^2 + b^2 - a - b = 0
1 - a - b = 0
b = 1 - a
b = 1 - sqr.root(1 - b^2)
sqr.root(1 - b^2)= 1 - b /^2
1 - b^2 = 1 -2b + b^2
2b^2 - 2b = 0
b^2 - b = 0
b(b - 1) = 0, b can't be 0 so the only solution is
b = 1
1 - a - b = 0
1 - a - 1 = 0
a = 0
Solution:
z=i
Could this be correct?
Edit: Ignore this post. I haven't seen Deveno's explanation until after i posted. Thank you Deveno.
your solution is fine: you correctly identified the "extraneous solution" by realizing that b = 0 implies z = 1, which would make (z+i)/(1-z) undefined (and 1 certainly ISN'T undefined).
the point is, we only have ONE possible z that works. i prefer to show such things by working with a single variable (z alone) rather than two (a and b), to avoid such difficulties (although i will admit, the algebra can get hairy this way).