Möbius transformation from disk to itself defined defined by interior points?

Question

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)?

Answer 1

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, $f(\Bbb D) = \Bbb D$.

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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 $$ \frac{(f(z)-w_1)(1-|w_2|^2)}{(f(z)-w_2)(1-w_1 \overline{w_2})} = \frac{(z-z_1)(1-|z_2|^2)}{(z-z_2)(1-z_1 \overline{z_2})} \, . $$

Answer 2

If my calculations are correct, it is sufficient to specify two points and their images inside the unit disc (which I will call $D$). I will be using the following facts:

1. A corollary of the Schwarz Lemma which says that automorphisms of the unit disc which fix 0 are of the form $f(z)=\lambda z$ with $\vert\lambda\vert=1$.
2. If $M:=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ is invertible, then $\varphi_M:=\frac{az+b}{cz+d}$ defines a Möbius transformation, and if $A,B$ are such matrices, then $\varphi_A\circ\varphi_B=\varphi_{AB}$. In particular, $\varphi_M^{-1}=\varphi_{M^{-1}}$.
3. If $\kappa\neq0$ and $M$ an invertible $2\times2$ matrix, then $M$ and $\kappa M$ define the same Möbius transformation.
4. For $a\in D$, the matrix $M_a:=\begin{pmatrix}1&-a\\\bar a&-1\end{pmatrix}$ defines a Möbius transformation from the unit disc to itself with $\varphi_{M_a}(a)=0$ and $\varphi_{M_a}(0)=a$. As shorthand, I will write $\varphi_a:=\varphi_{M_a}$.

On to the calculations: Let $z_1,z_2,w_1,w_2\in D$ and let $\varphi_M:D\to D$ be a Möbius transformation with $\varphi_M(z_1)=w_1,~\varphi_M(z_2)=w_2$. We want to find the matrix $M$ defining this Möbius transformation. Due to $\varphi_M(z_1)=w_1$ we have

$$\varphi_{w_1}\circ\varphi_M\circ\varphi_{z_1}(0)=0.$$

According to 1. above, since $\varphi_{w_1}\circ\varphi_M\circ\varphi_{z_1}$ is an automorphism of the unit disc which fixes 0, we get

$$\varphi_{w_1}\circ\varphi_M\circ\varphi_{z_1}=\varphi_{\Lambda},\qquad\Lambda:=\begin{pmatrix}\lambda&0\\0&1\end{pmatrix},~\vert\lambda\vert=1.$$

The correspondence between matrices and Möbius transformations (fact 2.) allows us to translate this to

$$M_{w_1}MM_{z_1}=\Lambda,$$

from which we get

$$M=M_{w_1}^{-1}\Lambda M_{z_1}^{-1}.$$

I'll spare you the exact calculations (though you might want to check them to be sure), but this yields

$$M=\frac{1}{(1-\vert w_1\vert^2)(1-\vert z_1\vert^2)}\begin{pmatrix}\lambda-w_1\bar z_1 & w_1-\lambda z_1\\ \lambda\bar w_1-\bar z_1&1-\lambda\bar w_1 z_1\end{pmatrix}.$$

Fact 3. allows us to drop the factor $\frac{1}{(1-\vert w_1\vert^2)(1-\vert z_1\vert^2)}$, so we end up with

$$M=\begin{pmatrix}\lambda-w_1\bar z_1 & w_1-\lambda z_1\\ \lambda\bar w_1-\bar z_1&1-\lambda\bar w_1 z_1\end{pmatrix}.$$

What remains is to determine $\lambda$. We haven't used the fact that $\varphi_M(z_2)=w_2$ yet, so that's what we're going to be using now. We get

$$\varphi_{M}(z_2)=\frac{(\lambda-w_1\bar z_1)z_2~+~w_1-\lambda z_1}{(\lambda\bar w_1-\bar z_1)z_2~+~1-\lambda\bar w_1 z_1}\overset !=w_2.$$ Solving for $\lambda$ yields

$$\lambda=\frac{\bar z_1z_2(w_1-w_2)+1-\bar w_1z_1}{\bar w_1w_2+z_2-z_1}.$$

Keep in mind though, that you need $\vert\lambda\vert=1$. If this condition is not fulfilled, transformation will map the unit disc to some other disc instead.

With this, we have determined $M$, and thus $\varphi_M$. But I won't write out the full expression because it's gross. Though I suspect that it could be written down nicely after some manipulations (Edit: like in the other answer).