isometry of the complex plane 
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 $F_1(i)=i \ \text{or} -1$??
 A: 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?)
A: So for each $z,w\in \mathbb{C}$ we have:
$$ |f(z)-f(w)| = |z-w|$$

Make a substitution $ g(z)= f(z)-f(0)$, so for each $z,w\in \mathbb{C}$ we have:
$$ |g(z)-g(w)| = |z-w|$$
and $g(0)=0$. Since $g$ is injective we have $g(1)\ne 0$ so we can make $$h(z)=g(z)/g(1)$$ Now we see $|g(1)|=1$ and for each $z,w\in \mathbb{C}$ we have:
$$ |h(z)-h(w)| = |z-w|$$
with $h(1)=1$ and $h(0)=0$. Say $w=0$ we get for each $z\in \mathbb{C}$:
$$ |h(z)| = |z|$$
so $0$ is on perpendicular bisector of a segment beetwen $\alpha = z$ and $\beta = h(z)$. If we put $w=1$ we get $$ |h(z)-1| = |z-1|$$
so $1$ is also on perpendicular bisector of a segment beetwen $\alpha = z$
and $\beta = h(z)$. So the real axis is is on perpendicular bisector of a segment beetwen
$\alpha \beta$ which means $h(z)=z$ or $h(z)=\overline{z}$
for each $z\in \mathbb{C}$
Say exist $z,w\in
\mathbb{C}\setminus \mathbb{R}$, such that $ h(z)=z$ and
$h(w)=\overline{w}$. Then
$$ |z-\overline{w}| = |z-w|$$
so $z$ is on a perpendicular bisector of a segment beetwen $w$ and
$\overline{w}$ which is real axis so $z\in
\mathbb{R}$. A contradiction! So $h(z)\equiv z$ or $h(z)\equiv
\overline{z}$, which means that only solution are 
$$ f(z) = a z+b$$
and
$$ f(z) = a\;\overline{z}+b$$
where $a$ and $b$ are given complex numbers and $|a|=1$.
