# Finding all Möbius transformations that fix $0$ and $1$.

I need to find all Möbius transformations that fix $$0$$ and $$1$$.

I'd like to know if my proof is correct:

I used the fact that for any three points $$z_1, z_2, z_3 \in \Bbb C$$ there is a unique Möbius transformation such that $$f(z_1)=1, f(z_2)=0$$ and $$f(z_3)=\infty$$, namely $$f(z)=\frac{(z_1-z_3)(z-z_2)}{(z_1-z_2)(z-z_3)}$$ So, for any $$z_0 \in \Bbb C \backslash \{0,1\}$$, we can define a Möbius transformation that fix $$0$$ and $$1$$, and such that $$f(z_0)=\infty$$, namely

$$f(z)=\frac{(1-z_0)(z-0)}{(1-0)(z-z_0)}=\frac{(1-z_0)z}{z-z_0}.$$

Conversely, if $$f$$ is a Möbius transformation that fix $$0$$ and $$1$$, let $$z_0=f^{-1}(\infty)$$. Since a Möbius transformation has at most two fixed points, $$z_0 \neq \infty$$, so, by uniqueness, $$f(z)=\frac{(1-z_0)z}{z-z_0}.$$ In conclusion, all the Möbius transformations that fix $$0$$ and $$1$$ are of the form $$f(z)=\frac{(1-z_0)z}{z-z_0}.$$ for any $$z_0 \in \Bbb C \backslash \{0,1\}$$.

• Looks fine to me! Aug 31 '20 at 5:32

Your proof is almost correct. You forgot to consider the case that $$z_0 = f^{-1}(\infty) = \infty$$ (in which case $$f$$ is the identity mapping).
Another (but not necessarily better) way is to observe that the Möbius transformation $$S(z) = z/(z-1)$$ maps $$0, 1$$ to $$0, \infty$$. Therefore $$f$$ has fixed points $$0$$ and $$1$$ if and only if the “conjugate” $$S \circ f \circ S^{-1}$$ has fixed points $$0$$ and $$\infty$$, and that are exactly the rotations. Therefore $$\frac{f(z)}{f(z)-1} = \lambda \frac{z}{z-1}$$ for some $$\lambda \in \Bbb C$$, $$\lambda \ne 0$$.