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Math Help - [SOLVED] Cpx Analysis: Conformally Map Component of Hyperbola to Disk

  1. #1
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    [SOLVED] Cpx Analysis: Conformally Map Component of Hyperbola to Disk

    This problem is from Ahlfors; Chapter 3, Section 5, Page 97, Problem 5 .
    Statement: Map the inside of the right-hand branch of the hyperbola x^2 - y^2 = a^2 on the disk |w| < 1 so that the focus corresponds to w = 0 and the vertex to w = -1.

    Recall that we can describe this hyperbola with center 0, focus c \in \mathbb{R}, and a real constant k > 0 as |z+c| - |z-c| = 2k from the basic definition of a hyperbola. For k < |c|, we can derive H(z) = \{ z \in \mathbb{C} \mid \frac{x^2}{k^2} -  \frac{y^2}{c^2 - k^2}  = 1 \} where z = x + iy. Let k^2 = \frac{c^2}{2} then H(z) =\{ z \mid \frac{x^2}{\frac{c^2}{2}} -  \frac{y^2}{\frac{c^2}{2}} \}. Set c^2 = 2a^2 where a is specified in the hypothesis. Then H(z) =   \{ z \mid \left( \frac{x}{a} \right)^2 -  \left( \frac{y}{a} \right)^2  = 1 \} which has center 0, vertex a, and focus c = \sqrt{2} a.

    Call the open region to be mapped into the disk \Omega . My first (failed) attempt was to try and find a point c' symmetric to the focus and outside \Omega so that I could map c \mapsto 0, c' \mapsto \infty, and a \mapsto -1. However, no such symmetric point exists that I could identify.

    Ahlfors also discusses, on p. 90, the use of level curves on the u,v components of the map f(w) = w^2, but I don't see how I can use that either.

    Ideas and / or a solution are both welcome at this point. I appreciate your help.
    Last edited by huram2215; May 11th 2010 at 08:39 AM. Reason: clarifications
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  2. #2
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    Reward for Solution

    I decided to pull the reward idea back. It didn't quite feel right and Defunct was quick to provide sound advice. So I still need help on this - it's an open problem for me.
    Last edited by huram2215; May 12th 2010 at 06:32 AM. Reason: Advice from an experienced user.
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  3. #3
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    People who offer their help in this website do so for free. You should remove the money offering, in my opinion.
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  4. #4
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    Solution

    If you write z^2 = x^2 - y^2 + i 2xy then for any point on the hyperbola, x^2 + y^2 = a^2 so for such z on the hyperbola z^2 = a^2 + i 2xy which is a vertical line. So the interior of the right branch of the hyperbola maps to a region to the right of this line. The natural maps take this to the unit circle.
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