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Hyperbolic Circle <=> Euclidean Circle.

  1. Jul 18, 2008 #1

    MathematicalPhysicist

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    I have this question which is rather simple, basically reiterating a general theorem.

    Show that S={z in H||z-i|=3/5} is a hyperbolic circle S={w in H| p(w,w0)=r}
    for r>0 and find sinh(r/2) and w0.

    Now to show that it's hyperbolic is the easy task, I just want to see if I got my calculations correct for sh(r/2) and w0.

    Now it's best to move to the unit disk model of poincare by: f(z)=(z-i)/(z+i) because this is an isometry it keeps the same lengths here.
    (D={z in C| |z|<1} so f(S)={u in D||f(z)-f(i)|=|(u-0)|=3/5} so it's a unit eulidean circle around zero, now we have the following relationship between this radius and the hyperbolic radius r=log((1+3/5)/(1-3/5))=2log(2) and sh(r/2)=3/4 and to get back w0 we need to use the fact that f(w0)=(w0-i)/(w0+i)=0 thus w0=i.
    Is this approach valid?

    thanks in advance.
    btw, there's a great book from jim anderson from southhampton university on hyperbolic geometry.
    In case someone wants a recommendation (for those who want to learn the subject via models and not by the axiomatic method).
     
  2. jcsd
  3. Jul 21, 2008 #2

    MathematicalPhysicist

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