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Let $\alpha$ be real, $0 \leq \alpha < 1$. Let $U_{\alpha}$ be the open set obtained from the unit disc by deleting the segment $[\alpha, 1)$

Find an isomorphism of $U_{\alpha}$ with the unit disc from which the segment $[0, 1]$ has been deleted.

I am wondering if the following approach is right:

Proof ?: Note that $f(z) = \frac{az+b}{cz+d}$ maps circles to circles in $\mathbb{C}^*$ hence take $f(z) = \frac{\alpha-z}{\alpha -1}$ this mobious transform maps $\alpha \to 0$ and $1 \to 1$ and the image of $f$ should be a proper circle because this is just rotating and scaling and shifting. Moreover it is clear that $[\alpha,1]$ maps to $[0,1]$ because conformal implies angle preserving.

Am I missing anything ?

Samael Manasseh
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  • $f$ maps circles to circles, but not necessarily the unit circle to the unit circle. – Martin R May 04 '24 at 14:28
  • @MartinR Right I see, but I guess since f is defined on the unit disc (with a slit) by the maximum modulus principle $|f(z)|<1$ (as $f(1) = 1$ ) so it still maps to the unit circle no? – Samael Manasseh May 04 '24 at 15:10
  • $f(z) = \frac{\alpha-z}{\alpha -1}$ is a linear function, that cannot be the correct solution. – Note that the conformal automorphism of the unit disk are well-known, see for example https://math.stackexchange.com/q/209308/42969. – Martin R May 04 '24 at 16:51

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Take an automorphism of the unit disk which maps $[\alpha, 1]$ to $[0,1].$

By @Martin R's link it has to be a Möbius transformation of a certain type.

Check that $f(z)=\frac {z-\alpha}{1-\alpha z}$ takes $\alpha, 1,\infty $ to $0,1,-\frac1\alpha,$ respectively.

suckling pig
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