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

1. ## [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 $\displaystyle x^2 - y^2 = a^2$ on the disk $\displaystyle |w| < 1$ so that the focus corresponds to $\displaystyle w = 0$ and the vertex to $\displaystyle w = -1$.

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

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

Ahlfors also discusses, on p. 90, the use of level curves on the $\displaystyle u,v$ components of the map $\displaystyle f(w) = w^2$, 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.

2. ## 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.

3. People who offer their help in this website do so for free. You should remove the money offering, in my opinion.

4. ## Solution

If you write $\displaystyle z^2 = x^2 - y^2 + i 2xy$ then for any point on the hyperbola, $\displaystyle x^2 + y^2 = a^2$ so for such $\displaystyle z$ on the hyperbola $\displaystyle z^2 = a^2 + i 2xy$ 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.