# Conformal automorphism of unit disk that interchanges two given points

Let $$a$$ and $$b$$ be distinct points in the unit disk $$D$$. Show that there exists a conformal automorphism $$f$$ of $$D$$ that interchanges $$a$$ and $$b$$; that is, $$f(a) = b$$ and $$f(b) = a$$.

Idea: we know that $$g(z)=\frac{\alpha-z}{1-\bar{\alpha}z}$$ interchanges $$0$$ and $$\alpha$$ and by composition we can find out the map $$f(a) = b$$ for any $$a$$ and $$b$$ in the unit disk $$D$$. But how can I get the other way by the same map? Thanks.

Let $$a' = 1/\overline a, \,b' = 1/\overline b$$. Take the Mobius transformation $$f$$ that maps the points $$a, b, a'$$ to $$b, a, b'$$. Since $$f$$ preserves the cross-ratio, we get $$(a, b; a', b') = (b, a; b', f(b')) = (a, b; f(b'), b'),$$ therefore $$f(b') = a'$$. Since $$b$$ and $$b'$$ are symmetric (conjugate) wrt the unit circle $$\mathcal C$$, their images $$a$$ and $$a'$$ are symmetric wrt $$f(\mathcal C)$$. In the same way, $$b$$ and $$b'$$ are symmetric wrt $$f(\mathcal C)$$.
These two pairs of symmetric points uniquely determine the circle, therefore $$f(\mathcal C) = \mathcal C$$. (If $$a, b$$ and the origin $$O$$ are not collinear, then $$f(\mathcal C)$$ has to be a circle with the center at the intersection of $$aa'$$ and $$bb'$$, which is $$O$$, and with the radius $$\sqrt{ |a| \cdot |a'|} = 1$$. If $$a, b, O$$ are collinear, the center is found from a linear equation.)
As you already noticed, for $$\alpha \in \Bbb D$$ the Möbius transformation $$T_\alpha(z) = \frac{\alpha - z}{1- \bar \alpha z}$$ is an automorphism of $$\Bbb D$$ which interchanges the points $$0$$ and $$\alpha$$. This can be used to construct an automorphism interchanging two given points $$a, b \in \Bbb D$$: With $$c = T_a(b) = \frac{a- b}{1- \bar a b}$$ the Möbius transformation $$f = T_a^{-1} \circ T_c \circ T_a$$ has the desired properties: $$\begin{matrix} & T_a & & T_c & & T_a^{-1}\\ a & \to & 0 & \to & c & \to & b\\ b & \to & c & \to & 0 & \to & a \end{matrix}$$ Each $$T_\alpha$$ is its own inverse, so that $$f= T_a \circ T_c \circ T_a$$, and from that one can compute the explicit representation $$f(z) = \frac{a(1-|b|^2) + b(1-|a|^2) - (1-|ab|^2)z}{(1-|ab|^2) - \bigl(\bar a(1-|b|^2) + \bar b(1-|a|^2)\bigr) z} \, ,$$ i.e. $$f = T_\alpha$$ with $$\alpha = \frac{a(1-|b|^2) + b(1-|a|^2)}{1-|ab|^2} \, .$$
Remark: This Möbius transformation is the only automorphism of the unit disk which interchanges $$a$$ and $$b$$, i.e. the solution is unique:
Assume that $$f$$ and $$g$$ are two automorphism of the unit disk which both interchange the distinct points $$a, b \in \Bbb D$$. Then $$T= g^{-1} \circ f$$ is a Möbius transformation which fixes $$a, b$$ and their mirror points with respect to the unit circle, i.e. $$T$$ has four distinct fixed points. It follows that $$T = id$$ and $$f=g$$.