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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.

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  • $\begingroup$ Don't I have to show that the image of T is in $ \mathbb{D}$? $\endgroup$ – SvenMath Jan 2 at 15:17
  • $\begingroup$ 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. $\endgroup$ – user667 Jan 2 at 23:09
  • $\begingroup$ See also: math.stackexchange.com/questions/506058/… $\endgroup$ – user667 Jan 2 at 23:36
  • $\begingroup$ Thank you for your answer and the link. How would you apply this principle? $\endgroup$ – SvenMath Jan 3 at 0:00
  • $\begingroup$ see my edit for details $\endgroup$ – user667 Jan 3 at 0:10
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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$$z\mapsto\frac{z+z_0}{1+\overline{z_0}z}.$$

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  • $\begingroup$ Thank you. But how do i know now that T is biholomorphic? $\endgroup$ – SvenMath Jan 2 at 15:00
  • $\begingroup$ Because it is holomorphic and its inverse is holomorphic too. $\endgroup$ – José Carlos Santos Jan 2 at 15:01
  • $\begingroup$ I see. T is also bijectiv. Then i can say, that T is biholomorphic?. Why is T biholomorphic in $\mathbb{D} $? $\endgroup$ – SvenMath Jan 2 at 15:10
  • $\begingroup$ You'll have to prove that $\lvert z\rvert<1\implies\bigl\lvert f(z)\bigr\rvert<1$. $\endgroup$ – José Carlos Santos Jan 2 at 15:18
  • $\begingroup$ 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. $\endgroup$ – SvenMath Jan 2 at 15:28

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