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.
 A: 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 $f=g$.
A: 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.)
