We can find a unique Möbius transformation from the unit disk to itself by specifying three points and their images on the unit circle of the $z$ and $w$-plane, respectively, to find the transformation parameters $a$, $b$, $c$, and $d$.
Is there a way of finding maps from the unit disk to itself by specifying points and images within the unit disk instead (i.e., in the unit disk but not on the unit circle)?
Best Answer
A Möbius transformation $f$ which maps the unit disk $\Bbb D$ onto itself is uniquely determined by two distinct points $z_1, z_2$ and their images $w_1, w_2$ in $\Bbb D$. The image points cannot be arbitrarily chosen, however.
As an example, $z_1=w_1=0$ implies that $f$ is a rotation, so that $z_2$ and $w_2$ must have the same modulus.
The general case becomes relatively easy with some geometrical arguments: Preservation of symmetry and preservation of the cross-ratio.
Möbius transformations preserve symmetry with respect to circles, therefore $f(z_1) = w_1$ and $f(z_2) = w_2$ implies that also $$ f(1/\overline{z_1}) = 1/\overline{w_1} \, , \, f(1/\overline{z_2}) = 1/\overline{w_2} \, . $$
That implies the uniqueness: If both $f$ and $g$ have those properties then $g^{-1} \circ f$ is a Möbius transformation with 4 fixed points, and therefore the identity.
Möbius transformations also preserve the cross-ratio, therefore $$ \tag{*} (1/\overline{z_1}, 1/\overline{z_2}; z_1, z_2) = (1/\overline{w_1}, 1/\overline{w_2}; w_1, w_2) $$ so that this is a necessary condition for the existence of $f$.
It is also sufficient: If $z_1 \ne z_2, w_1 \ne w_2 \in \Bbb D$ satisfy $(*)$ then the Möbius transformation by $$ \tag{**} (z, 1/\overline{z_2}; z_1, z_2) = (f(z), 1/\overline{w_2}; w_1, w_2) $$ satisfies $f(z_1) = w_1)$ and $f(z_2) = w_2$.
The function $f$ also satisfies $f(1/\overline{z_1}) = f(1/\overline{w_1})$ and $ f(1/\overline{z_2}) = f(1/\overline{w_2})$ so that two “symmetry pairs” with respect to the unit circle are mapped to pairs which are also symmetric wrt to the unit circle. It follows that the image of the unit circle is again the unit circle, and consequently, $f(\Bbb D) = \Bbb D$.
Summary: A Möbius transformation $f$ which maps the unit disk $\Bbb D$ onto itself is uniquely determined by two distinct points $z_1, z_2$ and their images $w_1, w_2$. For given $z_1 \ne z_2, w_1 \ne w_2 \in \Bbb D$ such a Möbius transformation exists if and only if $(*)$ is satisfied, i.e. if $$ \frac{(1-|z_1|^2)(1-|z_2|^2)}{|1-z_1 \overline{z_2}|^2} = \frac{(1-|w_1|^2)(1-|w_2|^2)}{|1-w_1 \overline{w_2}|^2} \, . $$ If that condition is satisfied then $f$ is given by $(**)$, that is $$ \frac{(f(z)-w_1)(1-|w_2|^2)}{(f(z)-w_2)(1-w_1 \overline{w_2})} = \frac{(z-z_1)(1-|z_2|^2)}{(z-z_2)(1-z_1 \overline{z_2})} \, . $$