How can you write a Möbius transformation that maps the outside of a circle to the inside of the cirle?

Question

I am supposed to map the outside of the unit disk in the first quadrant onto the unit disk. I don't understand how you map the outside of a disc ( or part disc) to the inside.

Answer 1

A Möbius transformation takes the form $$f(z)=\frac{az+b}{cz+d}$$ where we require $ad-bc\ne 0$. So ask yourself: what is the obvious candidate for $f(\infty)$? And for $f(0)$? These two conditions force $f$ to take a very simple form, which depends only on $a/c$. And you can further tie down $f$ with the condition that $|z|=1\implies |f(z)|=1$, so $|a/c|=1$.

If $a=c=1$, you are mapping the outside of the unit disk in the first quadrant to the inside of the unit disk in the fourth quadrant. So you have to multiply by $i$ (which is also a Möbius transformation!) to get everything back into the first quadrant.

Answer 2

That's "inversion in a circle":http://geometer.org/mathcircles/inversion.pdf

Given a circle or radius "R", if the distance from the center of the circle to point "p" is $r_1$ then p is "inverted" into point p' such that p and p' lie on the same line through the center of the circle at distance $r_2$ so that $r_1r_2= R^2$. Every point outside the circle, $r_1> R$, is mapped to a point such that $r_1< R$, inside the circle. Every point on the circle, $r_1= R$ is mapped to itself, and every point inside the circle is mapped to a point outside the circle (the center of the circle is mapped to "the point at infinity").