Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why is every conformal bijection between disks actually a linear fractional transformation?

I thought I could justify this claim with the following idea.

Suppose $f$ is a conformal bijection from a disk $A$ to a disk $B$. Let $z_0\in A$ be arbitrary. Now there is a LFT $g$ from the unit disk to $A$ mapping 0 to $z_0$. Also, there is a LFT $h$ from the unit disk to $B$ mapping $0$ to $f(z_0)$. So altogether, $F=h^{-1}\circ f\circ g$ is a bijection on the unit disk fixing $0$, so by Schwarz' lemma, $|F(z)|\leq |z|$. Since $F^{-1}$ shares the same property, we have $|F^{-1}(F(z))=|z|\leq |F(z)|$, so $|F(z)|=|z|$, so by Schwarz' lemma, $F(z)=cz$ for some $c$. So $F$ is a LFT, and thus $f$ is as well.

Is this valid? I was surprised to conclude $|F(z)|=|z|$ for all $z$, I wasn't expecting to find $F$ to be an isometry. Thanks all.

share|cite|improve this question
If you're proposing $F$ is a rotation, it shouldn't surprise you that $|F(x)|=|z|$. I believe the proof in books is similar to what you've written. – ShawnD Mar 15 '12 at 3:17
The conformity of a linear fractional transformation is dependent on that it preserves circles. – mathon Jul 14 '12 at 8:17

1 Answer 1

Yes this is valid. ${}{}{}{}{}{}{}{}{}{}{}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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