The Möbius transformation

$$ w=f(z)=\frac{az+b}{cz+d}, $$

with $a,b,c,d$ complex constants is uniquely determined by three complex parameters.

This is clear because we can divide $f(z)$ by, say $1/b$ and then obviously we will have only three parameters to play with. What I don't understand is the content of the followinf theorem in Gamelin's Complex Analysis, p64:

Theorem. Given any three distinct points $z_0,z_1,z_2$ in the extended complex plane, and given any three distinct values $w_0,w_1,w_2$ in the extended complex plane, there is a unique fractional linear transformation $w=w(z)$ such that $w(z_0)=w_0$, $w(z_1)=w_1$, $w(z_2)=w_2$.

First question: why do I need to fix three points in the complex plane? I know that somehow they must be the three free parameters of the Möbius transformation, but how they are related?

The next questions are related to the proof.

Basically it says that if I have an $f(z)$ s.t. $(z_0,z_1,z_2)\rightarrow(0,1,\infty)$, and $g(w)\rightarrow(w_0,w_1,w_2)$, then $g^{-1}\circ f$, will do the job $w(z_0)=w_0$, $w(z_1)=w_1$, $w(z_2)=w_2$. Now he defines

$$ f(z)=\frac{z-z_0}{z-z_2}\frac{z_1-z_2}{z_1-z_0}, $$

as the "good" $f$ that do the job in $g^{-1}\circ f$ if none of $z_j$ is $\infty$. $g^{-1}\circ f$My problem is that he never defines $g$, and the question is why is it not needed to prove the theorem?


1 Answer 1


Three $z$'s and three $w$'s are needed to assure existence and uniqueness. If you try this with four $z$'s and four $w$'s, there may not be a solution; try this with two and you'll get more than one solution.

As for the meaning of the $g$, it looks like you should define $g$ similarly as $f$ was defined, but it should map $(w_0, w_1, w_2)$ to $(0,1,\infty)$. You should convince yourself that the composition $g^{-1}\circ f$ will map $(z_0, z_1, z_2)$ to $(w_0, w_1, w_2)$ as desired.

It is straightforward to write down the formula for $g^{-1}$. Compose this with $f$ and you should be able to read off the parameters of the final transformation.


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