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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 $z_0 \in \Bbb C \backslash \{0,1\}$.

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  • $\begingroup$ Looks fine to me! $\endgroup$ Aug 31 '20 at 5:32
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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 $\lambda \in \Bbb C$, $\lambda \ne 0$.

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