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$\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\C}{\mathbb{C}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\s}{\sigma}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\F}{\mathbb{F}}$

Remmert chapter 0 exercise 1 page 16

Given $$T(z) = az+b\overline{z}$$ for $a,b \in \C $

For C-linearity , it must hold that: $$T(y+x)=a(x+y)+b(\overline{y+x}) = T(y)+T(x); T(yx)=a(yx)+b(\overline{yx})= y(ax+b\overline{x})=yT(x)$$

and we can see that both of these conditions does only hold for $a,b = 0$

For angle preservation the condition is: $$|w||z|<T(z),T(w)>=|T(z)||T(w)|<w,z> ; w,z \in \C$$

and so this is also true only for $a,b = 0$

When is T(z) bijective : we need show injectivity and surjectivity, so $$injectvitiy: T(z)=az+b\bar{z}=ay+b\bar{y}=T(y) \Rightarrow z=y$$ $$surjectivity: T:D\rightarrow M, z\mapsto az+b\overline{z}$$ then for all x in M there must be at least one y in D which connect to it.

Injectivity doesn't seem to make problem, but for surjectivity we can try find $ T^{-1}$!

We put: $$w=az+b\overline{z}, \bar{w}=\overline{az}+\bar{b}z$$

we get: $$z= \frac{\bar{w}-\bar{a}(z)}{\bar{b}} = \frac{b\bar{w}-\bar{a}w+\bar{a}az}{\bar{b}b} = \frac{b\bar{w}-\bar{a}w}{(1-\bar{a}a)b\bar{b}}$$

so we can say that for $a\bar{a}\ne 1 , b \ne 0 $ $T^{-1}$ exist and T(z) is bijective.

When is $|T(z)|= |az+b\bar{z}|=|z|$ for all $z\in \C$?

If we put $Im(z)=0$ then $z=\bar{z}$ and it follows that: $|z(a+b)| =|z|$ so we can put $|a+b|=1$ so we have $|z||a+b|=|z| \ \ \ \forall z \in \C$

How can we show that this is the only cases surely?

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