Möbius transformation/biholomophic funtion

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? :)

$$T(z)$$ 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.

• Don't I have to show that the image of T is in $\mathbb{D}$? – SvenMath Jan 2 at 15:17
• 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 2 at 23:09
• – user667 Jan 2 at 23:36
• Thank you for your answer and the link. How would you apply this principle? – SvenMath Jan 3 at 0:00
• see my edit for details – user667 Jan 3 at 0:10

I've never heard of that theorem. The natural way of doing this concists in finding the inverse of $$T$$, which is$$z\mapsto\frac{z+z_0}{1+\overline{z_0}z}.$$

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