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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.