I need to prove that, for r ∈ (0, 1) and z in the unit disk:
|(z - r) / (1 - rz)| ≤ |z| + r
Even the smallest hint will be greatly appreciated! Thanks guys.
Hi MacstersUndead,
I've tried that for hours on end, but can never make it work. For if Re(z) > 0 then |1 - rz| < 1, and |z / (1 - rz)| > |z|.
(|z| + r) seems to be an extremely tight bound on the LHS; every time I try to use a basic inequality, it blows up in my face. :/
Suppose we use the notation ϕr(z) = (z-r) / (1-rz). Then it is a fact from complex analysis that:arctanh|ϕr(z)| ≤ arctanh|z| + arctanh|r|(This holds because Mobius transformations are isometries of the disk under the Poincaré metric.) Do you have any idea how to go from that inequality to the one desired in my original post?|ϕr(z)| ≤ |z| + r
I'm sorry, in that aspect I cannot help you. I'm not at a high level of mathematics yet; I'm in Analysis I.
However, this looks a lot like...
what do we know about inverse hyperbolic tan?
is it a function such that f(a+b) = f(a) + f(b)?
In that case tanh-1|z| + tanh-1r = tanh-1(|z| + r)
so then
tanh-1|ϕr(z)| ≤ tanh-1(|z| + r)
so it follows that
|ϕr(z)| ≤ |z| + r
Well, if we can show that:[1] a ≤ b implies tanh(a) ≤ tanh(b)for a, b ∈ (0, 1) then we can just take tanh of both sides of the inequality
[2] tanh(a + b) ≤ tanh(a) + tanh(b)arctanh|ϕr(z)| ≤ arctanh|z| + arctanh|r|and we'll be done. Clearly #1 holds (a graph here: Hyperbolic Tangent -- from Wolfram MathWorld). But I'm not sure about #2. Hmm...maybe I'll try to talk to a teacher. Meh. This problem is breaking my will to live.