Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hyperbolic Geometry Question.

  1. Jul 12, 2008 #1
    Let b>a>0 real numbers, and z a point in H={z in CU{infinity}| Im(z)>0}
    let I be in H, a euclidean interval which connects az with bz. find the point w in I such that it divides I into two intervals with the same hyperbolic length.

    Now what I did is as follows:
    let f be a mobius transformation such that: f(az)=|z|(ia) and f(bz)=|z|(ib)
    it keeps the length of I, which is: (b-a)|z| (a euclidean length), now because f is an isometry in H, the f(w) divides the interval on {z=Im(z)>0} to two parts.
    now the hyperbolic length should be calculated by using the hyperbolic riemanian metric, after some manipulations I get that: (f(w))^2=(ab)|z|^2 f(w)=sqrt(ab)|z|.
    now we know that f(w)=(a'w+b')/(c'w+d') for a'd'-b'c'=1 all in CU{inifnity}.

    now from f(az)=|z|(ia) and f(bz)=|z|(ib) we have two equations with four variables: (a',b',c',d') so we can choose two of them arbitrarily and from the last equations get the other two (we can also use the fact on a'd'-b'c'=1), now if I choose: d'=1,c'=0 then i get that a'=1 and b'=|z|(ia)-az=|z|(ib)-bz
    from here I get that: w=sqrt(ab)|z|-|z|(ia)+az
    not sure if this is correct cause, from the next to last equation i get that: z=i|z| which means that z=Im(z).

    any hints?
  2. jcsd
  3. Jul 12, 2008 #2
    I haven't sorted out the rest of your ideas here. But I do see an error in the very beginning: f does not map I to the interval connecting a|z|i to b|z|i. In fact, it maps I to a portion of a circle that connects a|z|i to b|z|i, since I is not a geodesic but vertical line segments are.

    You have the right initial idea, though, in using the isometries of H rather first using the hyperbolic distance formula (which actually you don't even need to solve this). Think of it this way: can you find an isometry f so that f(az) is an element of I and f(f(az))=bz? If so, that the answer should become obvious.
  4. Jul 12, 2008 #3
    I'm a slow learner, anyway: if I choose such an isometry that
    f(az)=w (the w in question), and f(w)=bz
    now we find that (well I'm quite dumb, cause I think I figured it out), because
    the f is an ismotery it keeps the same distances, so:

    Now to find f.
    not sure, how exactly.
  5. Jul 13, 2008 #4
    Well, f is an isometry of H, i.e., f is a Mobius transformation. So, your task is simply to find its coefficients -- just as you were originally trying to do. Be careful, though, it's not just the *distance* between the points that matters here, it's the fact that all of the points need to lie in or on the boundary of I, i.e., f(az) needs to be an element of I.

    The genius of Polya's problem-solving techniques might help you here: Have you actually drawn a picture of the situation yet? Unconsciously, you have already thought about one case (where the original complex point z is purely imaginary). What would the picture look like for other possible z's? There is a very important property of I (and the entire Euclidean line that it lives on) that you may be overlooking and might help you find f.

    One thing I've learned from my mathematical elders repeatedly is that, when one is stuck on a problem, one should work out a specific case of the problem to see if anything enlightening gets teased out.
  6. Jul 13, 2008 #5
    Well no I havent tried to draw, and I'm not that good anyway in drawing.

    You said that:
    "it's not just the *distance* between the points that matters here, it's the fact that all of the points need to lie in or on the boundary of I, i.e., f(az) needs to be an element of I."

    well obviously by our definition that: f(az)=w and f(w)=bz we know that f must be on I.

    but how to find the coefficients, what I have is:
    now I have four varibales a',b',c',d' (a'd'-b'c'=1).
    w isn't given.
    what is given is z,a,b
    from f(az)=w I may know what is w if I know a',b',c',d'.
    Do I need to use the equation:
    p(f(az),f(w))=p(az,w) also?
    Don't know.
  7. Jul 13, 2008 #6
    Well, presumably, you can graph to points that are real multiples of each other on the complex plane and connect them with a line and see where that line intersects the x-axis...

    Being "not that good" at drawing is a poor excuse. Get into the habit of drawing out mathematical situations -- and you'll find that you will become "that good" at it. Unless of course there is a physical reason why you can't draw -- in that case, you need to practice visualizing the situation.

    As for your algebraic attempts, you are leaving out the key assumption about w, that is, that w lies in I.
  8. Jul 13, 2008 #7
    Ok Iv'e looked again at my notebook.
    and it states that for these kind of transforamtion thing, we need to find another two points which the function gives zero and infinity, i.e the points which instersect the real line, as you pointed
    (ax0,ay0), b(x0,y0) (z=x0+iy0)
    so the equation of the line is: y=(b-a)(y0/x0)(x-ax0)+ay0
    from here I can find that: y=0 when x=-ay0(x0/y0(b-a))+ax0
    which is one point where f(z)=infinity on this line.
    which means: (a'x+b')/(c'x+d')=infinity
    i.e c'x+d'=0 from here d'=-c'x.
    I have a'd'-b'c'=1.
    I think I also to find the point of intersection with y, i.e x=0, such that:
    f(z)=0, from there I could get a connection between: a' and b'.

    am I in anywwhere close to the answer?
  9. Jul 13, 2008 #8
    you mean I need to use the fact that the line is invariant under f?
  10. Jul 14, 2008 #9

    I can't help noticing from your comments that you have yet to draw a picture. This is geometry -- visualizing the problem will solve it very quickly.

    Let me repeat my advice thus far:

    1. Sketch the situation. No matter what the initial z is and the real coefficients a and b, the line that connects az and bz has a special property.

    2. Once you figure out that property, it should be relatively easy (without much algebra, in fact) to figure out which transformations preserve that line.

    3. From there, figure out which specific transformation f will preserve the line such that f(az) is in I and f(f(az))=bz.

    4. Your solution will be w=f(az).

    You have now reached the threshold of my help. There is more than enough information to solve the problem, if you work at it diligently. Follow steps 1-4 and you'll solve it.
  11. Jul 14, 2008 #10
    here's my drawing, as I said not that good.
    from there, I have the liberty to choose f, as long as it preserves az in the same line, correct?
    so we want f(az)=(a'(az)+b')/(c'(az)+d')
    I don't get something, we want that f(az)=w, so I don't think I have a free choice here of f.

    Attached Files:

  12. Jul 17, 2008 #11
    OK, I think I figured it up.
    let me know if it's close.

    let f(u)=(u-p)/(u-q)
    where p,q are points of intersection of the hyprebolic line passing through this euclidean interval.

    now we have that: p-q=1 and w=(az-p)/(qz-q) and (w-p)/(w-q)=bz
    we insert the former equation for w in the latter, and find p or q (after using p-q=1), and this way we find both p and q, and we also find w.

    Is this correct?

    thanks in advance.
  13. Jul 18, 2008 #12
    p,q are points in the real line of x which the hyperbolic line intersects.
  14. Jul 18, 2008 #13
    Look at your picture again -- you've left out a crucial aspect of the initial conditions while drawing it.?

    What do we know about a and b? How then are az and bz related geometrically?
  15. Jul 18, 2008 #14
    well, b>a, and that means that also Re{az}<Re{bz} the same goes for Im.
    you mean that p and q are determined by Re{az} and Re{bz}, i.e, p=Re{az} and q=Re{bz}, well I think that this is'nt correct and I have a feeling that my last post (#12) is correct, we set f(u)=(u-p)/(u-q) where p and q are the points of intersection of the hyperbolic line connecting az to bz, with the real line, and from p-q=1
    and: f(w)=bz=((az-p)/(az-q)-p)/((az-p)/(az-q)-q) from these two I can find p,q in terms of a,b,z and thus find w.
    though it does seem to be a bit long here.
  16. Jul 18, 2008 #15
    to be precise the absolute values of Re satisfy the inequality I wrote.
  17. Jul 18, 2008 #16
    b>a>0 ----> a and b are real numbers -----> the two points az and bz are *real* multiples of z and hence real mulitples of each other. Now, go back to your picture. Did you draw az and bz in such a way that they are real multiples of each other?

    Even better, what does the set of all points that are real multiples of z look like? And which isometries preserve this set?
  18. Jul 18, 2008 #17
    the set of points of real multiples of z is a line which passes through the origin.
    an isometry that preserves this line is a reflection, and reflections of mobious transformations are given by f(z)=(a'z*+b')/(c'z*+d'), where * is the conjugate operation.
    (where a',b',c',d' are real).
    I still don't see how it helps me here.
    I think that I need to draw a hyperbolic line that passes through this interval and intersects the real line at two points, other than that, nothing comes to my mind.
  19. Jul 18, 2008 #18
    Um, that's both too many and too few. Not all reflections preserve this line and we can restrict ourselves to orientation-preserving transformations. So, just look at the Moebius transformations z |--> (az+b)/(cz+d). (Obviously, different a, b, & z from the set-up of the problem).

    Now, relax and think about it a bit. Which of these transformations preserve *any* line passing through the origin? In fact, we can be even easier on ourselves: which of these transformations fix both 0 and infinity?
  20. Jul 18, 2008 #19
    well obviously (a/d)z for ad=1 preserves 0 and infinity.
    So now I only need to use the fact that:
    and then I get that: (b/a)=(a'/d')^2 so w=sqrt(b/a)(az)=z*sqrt(ab).
    So just a minute, the line which I want to preserve is it a euclidean line (I mean when I continue the interval through the origin)?
  21. Jul 19, 2008 #20
    I have a similar question (assuming the previous question is solved):

    I need to find z in H such that p(i,z)=p(z,9i)=1/2p(i,9i)
    where p is the hyprebolic distance.

    again we need an isometry here, which i think should be satisfying: f(9i)=9i and f(i)=i, but I need another equation to find the coefficients, we also want that f(z)=iR for some R in (1,9), I know that 1/2p(i,9i)=log(3) so because: p(z,9i)=p(iR,9i)=log(9/R)=log(R)
    then R would be equal 3, and if I knew what is f then I would find z.
    perhaps f should be the same as the previous question i.e in the form: f(w)=(a'/d')w, well I guess it's not cause it will fix also z itself cause then a'/d'=1.
    what equation is missing in this question?

    Thanks in advance.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Hyperbolic Geometry Question.