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In general my approach to construct a Möbius transformation $\varphi$ between two simply connected domains $G_1$ and $G_2$ is to take 3 points on each boundary and map them onto each other. The cross-ratio then gives me the desired $\varphi$.

Question: Is there a trick when taking those points? So far I always chose them so that the equation yielded by the cross ratio is the most simple. But often I end up with a map $\varphi: G_1\to\mathbb C\backslash G_2$.

For example, let $G_1=B_2(-1)$ be the ball of radius 2 centered at $-1$ and $G_2=\{z\in\mathbb C|\Im(z)>0\}$ the upper half-plane. If I choose $$z_1=-3\mapsto w_1=0\\z_2=1\mapsto w_2=-1\\z_3=-1+2i\mapsto w_3=1$$ I get the Möbius transformation $\varphi(z)=\frac{(i-1)z-3+3i}{(3+i)z+(1-5i)}$ which (if I didn't miscalculate) maps $G_1$ to $\mathbb C\backslash G_2$. But if I switch up the points a little it works, for example $$z_1=-3\mapsto w_1=-1\\z_2=1\mapsto w_2=0\\z_3-1+2i\mapsto w_3=1$$

Since getting such a Möbius transform is always tedious to do on paper I would like to have a method where I don't have to try out wich assignment between the points yields the correct transformation.

EDIT: Just in case it was unclear: The cross-ratio equation is given by $$\frac{z-z_1}{z-z_3}\cdot\frac{z_2-z_3}{z_2-z_1}=\frac{w-w_1}{w-w_3}\cdot\frac{w_2-w_3}{w_2-w_1}.$$ Solving this for $w=\varphi(z)$ gets me the transformation.

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There is a typo in the last formula ($w_2 - w_3$ occurs twice), but your $\varphi(z)$ correctly maps $z_i$ to $w_i$. However, conformal mappings (of the first kind) preserve the orientation. Traversing the points $1, -3, -1 + 2 i$ keeps the disk $G_1$ on the right and traversing the points $-1, 0, 1$ keeps the lower half-plane on the right, therefore $G_2$ in your first example is the lower half-plane.

Once you've derived the formula for Mobius transformations mapping the unit disk to the upper half-plane, you can take such a transformation $\phi$ and construct the composition of a linear transform mapping $G_1$ to the unit disk with $\phi$.

You can also exploit the properties of the Mobius transformation. Choose $w(-3) = 0$ and $w(1) = \infty$, then the transformation is $k (z + 3)/(z - 1)$. Then $-1$ is mapped to $-k$ and $\infty$ is mapped to $k$, therefore taking any $k = i a, \,a < 0$ gives a valid answer.

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  • $\begingroup$ Thanks for your answer! I corrected the typo. Also I didn't know these transformations preserve the orientation. Now it's clear to me how I have to assign the points. Thanks for that piece of info! Also your $\phi$ is just the inverse Cayley transformation, right? And the linear transformation should be $h(z)=\frac{1}{2}z+1$. In that case your approach on this is far faster. Is this correct? $\endgroup$ – RedLantern Nov 5 '18 at 18:01
  • $\begingroup$ Correct, the Cayley transform is not the only possibility (I've clarified the answer a bit), but since the goal is to construct one transformation, not to find the general form, you can take the inverse of the Cayley transform. A linear function is also not unique, because of rotations, but $z/2 + 1$ won't work; $(z + 1)/2$ will (first translating the center of $G_1$ to the origin and then scaling). $\endgroup$ – Maxim Nov 5 '18 at 19:07

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