Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $M$ is a hyperbolic Moebius transformation with fixed points at $(0, 0), (1, 0)$ which, when applied to the complex $(x_0, y_0)$, yields the result $(x_1, y_1)$.

How do I solve for $M$ given $x_0, y_0, x_1$, and $y_1$?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

The transformation we are looking for is $$\frac{w}{w-1}=k~\frac{z}{z-1}$$ with $0<k$. Let $z_0=x_0+y_0~i$ and $z_1=x_1+y_1~i$. We can suposse $z_0\neq z_1$ because if $z_0=z_1$ then the function is the identity. Then $$k=\frac{z_1(z_0-1)}{z_0(z_1-1)}.$$ The inverse of $\displaystyle f(z)=\frac{z}{z-1}$ is itself. Therefore $$w=f(k\cdot f(z))=\frac{k~z}{(k-1)z+1}$$ It remains the question that $k$ must be a positive real number. As $0$ and $1$ are fixed points, then $w$ is a traslation in $\mathbb{H}^2$ along the geodesic with infinity points $0$ and $1$. How the equidistant through $z_0$ to this geodesic is invariant, then $z_1$ is in the same equidistant. Then is necessary for $0<k$ that $z_1$ lives in the circumference through $0$, $1$ and $z_0$.

share|improve this answer

Hint: You can take your matrix representation of the Möbius transformation to be an element of $PSL(2,\mathbb{R})$, or your transformation to be $T(z)= \frac{az+b}{cz+d}$ with $ad-bc=1$. Then with the conditions $T(0)=0, T(1)=1 $ and $ T(z_0)=z_1$ you should be able to solve for $a, b , c$ and $d$.

share|improve this answer
    
P.S. I'm assuming T is an isometry of $\mathbb{H}$, the upper half plane. –  Sean Jun 2 '12 at 18:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.