Prove that any isometry is of one of the forms $$z \mapsto az+b, \ \ \ \ \ z \mapsto a\bar z +b$$ where $|a|=1$
How should one prove this for all possible isometries?
My textbook, Algebra and Geometry, gives a proof that I cannot understand.
Suppose $F$ is any isometry, and let $$F_1(z)={{F(z)-F(0)}\over {F(1)-F(0)}}$$ Then $|F(1)-F(0)|=1$ so that $F_1$ is an isometry with $F_1(0)=0$ and $F_1(1)=1$. This implies that $F_1(i)$ is $i$ or $-i$, and we deduce that either $F_1(z)=z$ or $F_1(z)=\bar z$ for all $z$
My problem with the above proof is, how could one deduce that $|F(1)-F(0)|=1$ and how can one deduce, then, that $F_1(i)=i \ \text{or} -1$??
Well, $F$ is an isometry - what this means, is that the distance between $a$ and $b$ is the same as the distance between $F(a)$ and $F(b)$.
So now you want to calculate $\vert F(0)-F(1)\vert$. I claim this expression is equal to the distance between two numbers. What are these numbers? How does that help us? (HINT: what does $\vert a-b\vert$ represent, in general?)