# Constructing Möbius transformation

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.

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.
• 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? – RedLantern Nov 5 '18 at 18:01
• 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). – Maxim Nov 5 '18 at 19:07