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This is a question about problem 3 from Ahlfors section 3.1, which states

Prove that the most general transformation which leaves the origin fixed and preserves all distances is either a rotation or a rotation followed by reflexion in the real axis.

Now it has been answered here before where they show that if $f$ is indeed a fractional linear transformation then $f(z)=az$ where $|a|=1$. But when it's not, they somehow conclude the only two options are $f(z)=az$ or $a\bar{z}$(here the conjugate is what you get after reflecting in the real axis). Firstly, Ahlfors doesn't even define what "general transformations" are, but assuming they are perhaps continuous/holomorphic maps $\mathbb{C}\to \Bbb{C}$. I don't see why it has to be only one of these two kinds in such a scenario. So I guess my question is :

If $f:\mathbb{C}\to \Bbb{C}$ a continuous map preserving the origin and preserving lengths, then $f(z)$ or $f(\bar{z})$ is a linear fractional transformation. Then since for linear fractional transformations we know the answer from the linked answer, the full claim follows.

Is this a true claim? Is there any way to prove it? I think as stated the problem doesn't even allow $f$ to have any kind of structure, linear or holomorphic etc. So I don't see why we should even consider fractional transformations to begin with. I feel like this shouldn't be as difficult of a problem if only Ahlfors defined the terminology clearly and this is making me lose my mind. Any help would be appreciated.

Chanel Rose
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    I don't think one even needs to assume continuity to derive the result. $f(1)$ has to be on the unit circle; after a rotation, we may assume $f(1)=1$. Then for any $z$, $f(z)$ must be on the circle around the origin of radius $|z|$, and on the circle around $1$ of radius $|z-1|$, and those two circles intersect only at $z$ and $\bar z$. – Gerry Myerson Feb 08 '23 at 00:49
  • Is $f$ a linear map? what kind of transformation is $f$? But I understand your proof, you're right. 4 – Chanel Rose Feb 08 '23 at 00:57
  • My argument makes no assumptions about $f$, certainly no assumption that $f$ is linear, but it proves that $f(z)$ is $az$ or $a\bar z$ for some complex $a$ with $|a|=1$. – Gerry Myerson Feb 08 '23 at 00:59
  • Wow I see, that's a very strong result then, thank you very much. You indeed proved it for the most general transformation possible. If you want to write it as an answer I can accept it, this is amazing. I still don't know what Ahlfors meant by general transformation, but I guess that is irrelevant now. – Chanel Rose Feb 08 '23 at 01:11

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