Set of Isometries in Complex Plane under Composition forms Group

Theorem
Let $S$ be the set of all complex functions $f: \C \to \C$ which preserve distance when embedded in the complex plane.

That is:
 * $\size {\map f a - \map f b} = \size {a - b}$

Let $\struct {S, \circ}$ be the algebraic structure formed from $S$ and the composition operation $\circ$.

Then $\struct {S, \circ}$ is a group.

Proof
From Complex Plane is Metric Space, $\C$ can be treated as a metric space.

Hence it is seen that a complex function $f: \C \to \C$ which preserves distance is in fact an isometry on $\C$.

Taking the group axioms in turn:

Let $f$ and $g$ be isometries on $\C$.

We have:

Thus $g \circ f$ is an isometry, and so $\struct {S, \circ}$ is closed.

By Composition of Mappings is Associative, $\circ$ is an associative operation.

The identity mapping is the identity element of $\struct {S, \circ}$.

By definition, an isometry is a bijection.

For $f \in S$, we have that $f^{-1}$ is also an isometry.

Thus every element of $f$ has an inverse $f^{-1}$.

All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.