I have to show, that the Möbius transformation $$ T(z) = \frac{z-z_0}{1-\bar{z_0}z}$$ is an biholomorphic function on $ \mathbb{D}$. $ \mathbb{D}:=\{ z \in \mathbb{C}: |z|<1 \}$ and $z_0 \in \mathbb{D}$. I know the following theorem: Is $ \mathbb{D}$ convex, T an holomorphic funtion and $Re T'(z)>0 $ in $ \mathbb{D}$. Then T is biholomorphic. So I have to calculate $Re T'(z) $ - $$ T(x+iy) = \frac{x+iy-(a+ib)}{1-\bar{(a+ib)}(x+iy)} $$ with $ z=x+iy, \ z_0=a+ib$ Is this the right way? :)
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$T(z)$ is the quotient of two holomorphic functions, so it is holomorphic on $\mathbb{D}$. The inverse, $T^{-1}(z)=\frac{z+z_0}{1+\bar{z_0}z}$, is also holomorphic on $\mathbb{D}$ for the same reason. Thus, T(z) is biholomorphic on $\mathbb{D}$.
To show that the image of T is $\mathbb{D}$ using maximum modulus principle: If $|z|=1$, then $z=e^{i\theta}$. Then we have, $$T(z)=\frac{e^{i\theta}-z_0}{1-\bar{z_0}e^{i\theta}}$$ From which, $$T(z)=\frac{e^{i\theta}-z_0}{e^{i\theta}(e^{-i\theta}-\bar{z_0})}$$ Let $\alpha=e^{i\theta}-z_0$. Then, $$T(z)=\frac{\alpha}{e^{i\theta}\bar{\alpha}}=e^{-i\theta}\frac{\alpha}{\bar{\alpha}}$$. and we conclude that $|T(z)|=|e^{-i\theta}||\frac{\alpha}{\bar{\alpha}}|=1$. By maximum modulus principle, for $z\in \mathbb{D}$ we must have $|T(z)|<1$ as desired.
user667
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Don't I have to show that the image of T is in $ \mathbb{D}$? – Sven Jan 02 '19 at 15:17
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If you are asking if T is an automorphism of $\mathbb{D}$ then yes. I would recommend using the maximum modulus principle to show this. – user667 Jan 02 '19 at 23:09
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See also: https://math.stackexchange.com/questions/506058/show-that-left-frac-alpha-beta1-bar-alpha-beta-right-1-when – user667 Jan 02 '19 at 23:36
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Thank you for your answer and the link. How would you apply this principle? – Sven Jan 03 '19 at 00:00
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see my edit for details – user667 Jan 03 '19 at 00:10
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Thank you:). Can i say that T is conformal, because T is a Möbius transformation? – Sven Jan 03 '19 at 09:40
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What definition of conformal mapping are you using? – user667 Jan 03 '19 at 13:31
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Hello, f is an holomorphic function and $ f'(z_0) \ne 0$, then f is conformal. – Sven Jan 03 '19 at 14:41
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Ok, in that case you can prove that if $f$ is injective then $f'(z) \neq 0$. Then $T(z)$ will satisfy your definition. – user667 Jan 03 '19 at 17:20
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Would you show this by definition, also $ T(z_1)=T(z_2) \rightarrow z_1=z_2 $ – Sven Jan 03 '19 at 18:19
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No, it is more involved. See: https://math.stackexchange.com/questions/398637/why-do-injective-holomorphic-functions-have-nonzero-derivative – user667 Jan 03 '19 at 18:26
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When i prove this theorem, i have to show anyway that f is injektive? – Sven Jan 03 '19 at 18:35
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$T$ is a bijection from $\mathbb{D}$ to $\mathbb{D}$ ... by definition it is injective. – user667 Jan 03 '19 at 20:41
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Ok sorry. I forgot that. Thank you. I think it's clear now. – Sven Jan 04 '19 at 12:37
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I've never heard of that theorem. The natural way of doing this concists in finding the inverse of $T$, which is$$z\mapsto\frac{z+z_0}{1+\overline{z_0}z}.$$
José Carlos Santos
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Because it is holomorphic and its inverse is holomorphic too. – José Carlos Santos Jan 02 '19 at 15:01
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I see. T is also bijectiv. Then i can say, that T is biholomorphic?. Why is T biholomorphic in $\mathbb{D} $? – Sven Jan 02 '19 at 15:10
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You'll have to prove that $\lvert z\rvert<1\implies\bigl\lvert f(z)\bigr\rvert<1$. – José Carlos Santos Jan 02 '19 at 15:18
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I would argue like that: $ |z-z_0|<1 $ , $ $|\bar{z_0}z|<1 \rightarrow |1-\bar{z_0}z| <1 $. But how can i argue that the qoutient is less then 1. – Sven Jan 02 '19 at 15:28