# Finding a hyperbolic isometry that fixes the point $x = 2$ and $x = 17$

I know that a Möbius transformation is hyperbolic if the trace is $> 2$ which is $a + d$. But I'm not sure of the next steps involved to arrive at the answer.

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–  Cameron Buie Apr 21 '13 at 23:35

Hint: Consider $$f(z)=\frac{az+b}{cz+d}$$ with $ad-bc=1.$ If $c\ne 0$, then this transformation has at least one fixed point, and all fixed points will be roots of the quadratic equation $$cz^2-(a-d)z-b=0.$$
The quadratic equation you want will be of the form $$0=c(z-2)(z-17)=cz^2-(19c)+34c$$ Thus, we need $b=-34c,$ $a-d=19c$, $a+d>2$, $ad-bc=1$, and $c\ne 0$. Can you go from there?
A fixed point of $f$ is precisely some $z$ such that $f(z)=z$. As for your choices of $a,b,c,d$, that's pretty much the idea. Of course, once we choose such values, you'll want to confirm that the resulting function $f$ actually does what we want it to do. –  Cameron Buie Apr 21 '13 at 22:38
So when you say to check whether the resulting function $f$ does what it is supposed to do, does that mean put the values of a,b,c,d in the original equation and make sure that they equal to 2 or 17 ? –  devcoder Apr 21 '13 at 22:43
I mean you need to check that $f(z)$ is actually a hyperbolic isometry, and that $f(2)=2$ and $f(17)=17$. –  Cameron Buie Apr 21 '13 at 22:54
Checking for hyperbolic isometry, isn't that just checking $a+d > 2$ and $ad-bc \ne 0$, or is there something more involved than that ? –  devcoder Apr 21 '13 at 23:12